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https://math.stackexchange.com/questions/1368988/evaluating-lim-n-to-infty-fracn2-sqrt2-2-cos-left-frac360-circn | # Evaluating $\lim_{n \to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$
I was thinking about different ways of finding $\pi$ and stumbled upon what I'm sure is a very old method: dividing a circle of radius $r$ up into $n$ isosceles triangles each with radial side length $r$ and central angle $$\theta=\frac{360^\circ}{n}$$ Use $s$ for the side opposite to $\theta$.
Then we can approximate the circumference as $sn$. By the law of cosines: $$s=\sqrt{2r^2-2r^2 \cos{\theta}}=r\sqrt{2-2\cos{\left(\frac{360^\circ}{n}\right)}}$$
We know $\pi=\text{circumference}/\text{diameter} \approx \frac{sn}{2r}=\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$. This becomes exact at the limit: $$\pi=\lim_{n \to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$$
Now for my question: How would you solve the opposite problem? To make my meaning more clear, above I used the definition of $\pi$ to determine a limit that gives its value. But if I had just been given the limit, what technique(s) could I have used to determine that it evaluates to $\pi$?
• You can approximate the opposite side length to be $r sin \theta$ which is actually the height of the triangle. This gives a weaker limit $\pi = \frac {180 sin \theta }{ \theta}$ , where $\theta$ tends to zero. – N.S.JOHN Dec 12 '16 at 9:03
Notice, $$\lim_{n\to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{2\pi}{n}\right)}$$ $$=\lim_{n\to \infty}\frac{n}{2}\sqrt{2\left(1-\cos\left(\frac{2\pi}{n}\right)\right)}$$ $$=\lim_{n\to \infty}\frac{n}{2}\sqrt{2\left(2\sin^2\left(\frac{\pi}{n}\right)\right)}$$ $$=\lim_{n\to \infty}\frac{2n}{2}\sin\left(\frac{\pi}{n}\right)$$ $$=\lim_{n\to \infty}\frac{\sin\left(\frac{\pi}{n}\right)}{\frac{1}{n}}$$ $$=\pi\lim_{n\to \infty}\frac{\sin\left(\frac{\pi}{n}\right)}{\left(\frac{\pi}{n}\right)}$$ Let $\frac{\pi}{n}=t\implies t\to 0\ as\ n\to \infty$ $$=\pi \lim_{t\to 0}\frac{\sin t}{t}$$ $$=\pi\times 1=\pi$$
• Ah, the third step is the connection I was failing to make. Thanks! – hexaflexagonal Jul 21 '15 at 19:10
Since there were already two solid answers that provided efficient approaches, I thought that it would be instructive to see a different way forward.
Here, we will use the expansion of the cosine as
$$\cos x=1-\frac12 x^2+O(x^4) \tag 1$$
Letting $x=\frac{2\pi}{n}$ in $(1)$ yields
$$2-2\cos \left(\frac{2\pi}{n}\right)=\frac{4\pi^2}{n^2} +O(n^{-4})$$
Finally, we have
\begin{align} \frac{n}{2}\sqrt{2-2\cos \left(\frac{2\pi}{n}\right)}&=\frac n2 \sqrt{\frac{4\pi^2}{n^2} +O(n^{-4})}\\\\ &=\frac n2 \frac{2\pi}{n}\left(1+O(n^{-2})\right)\\\\ &=\pi+O(n^{-3})\to \pi \end{align}
as expected!
• Love it! Yeah, I was definitely interested in seeing several approaches (and almost said as much in the original post). Thanks! – hexaflexagonal Jul 22 '15 at 2:18
• @DivergentQueries Wow!! Thank you. And you're welcome! It was my pleasure. I am still home and recovering from surgery 3 weeks ago. Your comment just made my day. – Mark Viola Jul 22 '15 at 2:20
• nice answer. i have edited a little – Bhaskara-III Sep 21 '16 at 13:16
• @Mats Granvik Please refrain from accepting edits without checking first to see their effects. In this case, you accepted an edit that rendered an equation unreadable and changed language indiscriminantly (e.g., changed "letting" to "substituting," which had no positive impact, added the word "same," which made the sentence awkward). I have reversed out all of the edits and restored the original version. – Mark Viola Sep 21 '16 at 14:39
• @Bhaskara-III Please refrain from making edits that either add nothing or actually degrade the quality of the post. In this case, you made edits that rendered an equation unreadable and changed language indiscriminantly (e.g., removed the "$$" from equation (1) making it unreadable, changed "letting" to "substituting," which had no positive impact, added the word "same," which made the sentence awkward). I have reversed out all of the edits and restored the original version. – Mark Viola Sep 21 '16 at 14:41$$\sqrt{2-2\cos x}=2\left|\sin\frac{x}{2}\right|$$hence everything boils down to:$$ \lim_{x\to 0}\frac{\sin x}{x}=1.$$Just recall double angle trig. identity: \cos 2x=1-2\sin^2x, Starting from,$$\begin{align} \lim_{n\to \infty}\frac{n}{2}\sqrt{2-2\cos\left(2\pi/n\right)}\\=\lim_{n\to \infty}\frac{n}{2}\sqrt{2-2+4\sin^2\left(\pi/n\right)}\\ =\lim_{n\to \infty}\frac{n}{2}\cdot 2\sin\left(\pi/n\right)\\ =\pi\cdot \lim_{n\to \infty}\frac{\sin(\pi/n)}{\pi/n}\\ =\pi\cdot \lim_{t\to 0}\frac{\sin(t)}{t}\\ =\pi\cdot 1\\=\pi \end{align} | 2019-11-21T21:03:05 | {
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https://math.stackexchange.com/questions/2599895/prove-that-the-function-is-surjective-but-not-injective | # prove that the function is surjective but not injective.
$f: \mathbb N \to \mathbb N$ be defined as $f(n) = \begin{cases} \dfrac{n}{2}, & \text{if$n$is even} \\ \dfrac{n+1}{2}, & \text{if$n$is odd} \end{cases}$
How do I prove that the function is surjective but not injective?
Attempt:
It's not injective because $f(1)=f(2)$ but I doubt that it's a valid proof.
I am new to proof writing in functions therefore I am unable to frame the language for surjective proof. I know that for a surjective function range of function = co domain of function.
• You are right! it is true if $f(1)=f(2)$ then it is not injective. – GhD Jan 10 '18 at 17:00
• @GhD Is that the only way to write the proof? – Abcd Jan 10 '18 at 17:02
• No, but it is enough to prove that it is not injective. – GhD Jan 10 '18 at 20:12
For each $n \in \Bbb N,$
$$f (2n)=n$$
so, it is surjective.
• I don't know what can be explained. He literally proves surjection by definition. – user370967 Jan 10 '18 at 17:11
When someone says "all cats are white", if you want to prove that his/her statement is false, all you need to do is to find a cat that is not white. I am telling this because you said you are new to proof writing but in some cases, proofs are really that straight-forward. So your proof is valid for injectivity because all you needed was an example contradicting the assumption of the fuction being injective, and you found one as $f(1) = f(2)$.
And for $f: \mathbb{N} \to \mathbb{N}$ to be surjection, range of $f$ must be $\mathbb{N}$ and even when the function is defined as $f(n) = \frac{n}{2}$ where $n$ is even, this function is surjective because if $n$ is even, then we can write $n = 2k$ where $k \in \mathbb{N}$. Then the function becomes $f(2k) = k$, $k \in \mathbb{N}$. Notice that RHS of this equation is range of $f$ and we have $k \in \mathbb{N}$. So range is $\mathbb{N}$. So $f$ is surjective.
That's basically enough for a proof. If you want to make it more explicit, you can use three sentences:
If function is injective, then $f(x) = f(y)$ means $x = y$. But $f(2) = f(1)$, even though $2 \neq 1$. So the function cannot be injective.
The nice thing about this proof is that each sentence acts like a fact or a deduction. We introduce the definition of injective functions in the first sentence. We introduce our true sentence about the function that we just found out (you can put any calculation here). And in the final sentence, we use those two facts and deduce a property of $f$.
• That's only one half of what needs to be proved wrt the given question statement, and the asker was spot on about why it is not injective. You haven't though, addressed surjectivity. – Namaste Jan 10 '18 at 17:14 | 2019-08-20T21:19:03 | {
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https://math.stackexchange.com/questions/3117260/how-is-rearranging-56-times-100-div-8-into-56-div-8-times-100-allowed-by-the/3117265 | # How is rearranging $56\times 100\div 8$ into $56\div 8\times 100$ allowed by the commutative property?
So according to the commutative property for multiplication:
$$a \times b = b \times a$$
However this does not hold for division
$$a \div b \neq b \div a$$
Why is it that in the following case:
$$56 \times 100 \div 8 = 56 \div 8\times 100$$
It seems like division is breaking the rule. There is something I am misunderstanding here.
Is it because $$a\times b\div c=a\div c\times b$$ is allowed since $$b\div c$$ are not being rearranged so that $$c\div b$$?
If this is the case are you allowed to rearrange values in equations so long as no values have the form $$a \div b = b \div a$$ and $$a - b = b -a$$ ?
• To would-be editors: please, please do not convert the $\div$ operators in this question into the fractional form $\frac ab.$ It changes the question completely. Feb 18, 2019 at 14:46
• Also due to the non commutivity, and non associativity of division, the notation used above is decidably ambiguous. That means there are a number of equally valid ways to interpret expressions ( this is why different calculators can give different results). You can explore the possibilities. In higer level math fractions, parents, etc. are used and there is no ambiguity. Feb 18, 2019 at 15:11
Notice that you have always $$\div 8$$, no matter the order of the other terms. You don't divide by a different number. It might help to think $$\div c$$ as a multiplication with $$d=1/c$$. Then everything would look easier: $$a\times b\div c=a\times b\times d=a\times d\times b=a\div c\times b$$
Hope this makes sense.
$$a\times b ÷ c$$
$$=a\times b\times\dfrac{1}{c}$$
$$=(a\times\dfrac{1}{c})\times b$$
$$=\dfrac{a}{c}\times b$$
$$=a÷c\times b$$
• The first step is not elementary. The literal interpretation of the order of operations in $a\times b \div c$ is $(a\times b) \div c$. what gives us the ability to take the $b$ out of the first operation and put it into $\frac bc$? I think this answer would be better if it went straight from $a\times b \div c$ to $a\times b \times \frac1c$. Feb 18, 2019 at 14:53
Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.
So $$a\div b \times c$$ is being interpreted from left to right as $$(a\div b)\times c=\cfrac {ac}b$$, but done from right to left $$a\div (b\times c)=\cfrac a{bc}$$ and the two results are not the same.
Likewise with $$a\div b \div c$$ we have $$(a\div b)\div c=\cfrac a{bc} \neq a\div (b\div c)=\cfrac {ac}b$$.
So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.
• I don't understand what you are trying to say. How is this any different than with addition and subtraction? If you interpret $a-b+c$ as $a-(b+c)$ then you get a different answer. But you shouldn't do that, so what? See also my answer. Mar 4, 2019 at 13:29
• @MarcvanLeeuwen You are right that it is essentially the same and also that you shouldn't do it - but the fact is that people do get these things confused. If you use BIDMAS (brackets, indices, division, multiplication, addition, subtraction) as a rule, applied without proper procedural knowledge, as I have seen in English Primary Schools, then in your example you would get the second, wrong, answer by doing the addition first and the subtraction afterwards. Mar 4, 2019 at 17:06
Because division is the inverse of multiplication, that is: $$X \div Y =X\cdot \frac 1Y$$
So you have: $$56\cdot 100 \cdot \frac18 =56\cdot\frac18\cdot100$$ Which is obvious.
• While it is true that "division is the inverse of multiplication" the emphasis should be on what follows from it: That division is equivalent to multiplication with the inverse and can be replaced by it. Feb 18, 2019 at 8:14
Swap the first two factors which is definitely allowed: 100 × 56 ÷ 8.
Now put parentheses around part of it: 100 × (56 ÷ 8).
Again, swap: (56 ÷ 8) × 100.
Take off the parentheses: 56 ÷ 8 × 100.
And there you go! transforming one into the other using only multiplicative commutation (plus the fact that one of the factors commutated can be a product, not just a simple value).
• Now explain why putting parentheses around the last two terms in $100\times56\div8$ is OK but putting parentheses around the last two terms in $56\div8\times100$ is not OK. For that matter, why is it OK to swap the first two terms in $100\times56\div8$ but not the last two terms of $56\div8\times100$? Feb 18, 2019 at 14:43
• The short answer to adding the parentheses is, "because it doesn't change the outcome". But you're right, a more rigorous explanation is needed, and I don't have one. To swap the last two terms of 56 ÷ 8 × 100 does require parentheses around them, and if they aren't already there, they can't be arbitrarily added without changing the whole thing. Feb 18, 2019 at 14:51
• A formal reason is that $56\times100\div8$ means $(56\times100)\div8$ by convention, and $(56\times100)\div8=(100\times56)\div8=(100\times56)\times\frac18=100\times(56\times\frac18)=100\times(56\div8).$ But that's making it more complicated than necessary. I think it's simpler if we start with $(56\times100)\div8=(56\times100)\times\frac18$. Feb 18, 2019 at 14:58
The exact same situation occurs with addition and subtraction instead of multiplication and division. Do you find it hard to justify rewriting say $$a+b-c$$ to $$a-c+b$$? And if not, which laws are you using to justify this? I would say associativity, commutativity, and the special equivalence $$x-y=x+(-y)$$ (which can be construed to be the definition of subtraction, although historically and in teaching subtraction is introduced before introducing negative numbers). In detail one has inserting parentheses to make explicit the one implied by convention) \eqalign{ a+b-c &= (a+b)-c \\&= (a+b)+(-c) \\&= a+(b+(-c)) \\& = a+((-c)+b) \\&= (a+(-c))+b \\&= (a-c)+b &= a-c+b} By perfect analogy, let us do that with '$$\times$$' replacing '$$+$$', and '$$\div$$' replacing '$$-$$': \eqalign{ a\times b\div c &= (a\times b)\div c \\&= (a\times b)\times (\div c) \\&= a\times (b\times (\div c)) \\& = a\times ((\div c)\times b) \\&= (a\times (\div c))\times b \\&= (a\div c)\times b &= a\div c\times b} Note how I just invented the unary use of '$$\div$$', where $$\div x$$ of course means the multiplicative inverse of$$~x$$, just as $$-x$$ means its additive inverse. Now why did nobody think of that before? It could have avoided the ugliness of$$~x^{-1}$$, which has to borrow exponentiation and additive inverse to get multiplicative inverse.
The non-zero real numbers form an abelian (commutative) group under multiplication. The notation $$a\div b$$ is shorthand for $$ab^{-1}$$.
So $$ab^{-1}\ne ba^{-1}$$ but $$56\times100 \times8^{-1}=56\times 8^{-1}\times 100$$
• (-1) this answer is almost certainly nearly useless; the set of people that would ask this question will be nearly disjoint from the set of people that understand this answer Feb 18, 2019 at 9:05
• Additionally, it's not even clear he asking over reals, could be over natural numbers in which case there is no group with multiplication as group operation. Feb 18, 2019 at 11:18
With multiplication and division alone (no addition or subtraction) they are associative with respect to one another, BUT division itself is NOT commutative.
So basically you can do the multiplication or division in either order, BUT you must respect which way you interpret the inputs to the left or right of the operator's sign. If you flip this it has the effect of flipping the inputs which is equal precisely when the operator is commutative. As you noted, with division, it is not commutative.
• No, as Mark Bennet explained. Feb 18, 2019 at 11:51
• Mark Bennets answer is incorrect. You CAN read from either right or left, provided you interpret the same operator symbol the same. So input on left of operator goes to numerator, etc. If you don't change convention you can read the combined expression from either left or right. It's indisputable that there are two levels of right and left. This is not addressed. Yet fundamental to the combined notion of associative and commutative. Feb 18, 2019 at 12:07
• Mark Bennet's answer shows that $a\div (b\times c)$ is not generally equal to $(a\div b)\times c$. His arithmetic is completely correct. You seem to be taking a view in which the operators $\times$ and $\div$ are not binary operators but are merely flags of some kind within an $n$-tuple of factors. I don't think that view is widely shared. Feb 18, 2019 at 14:37
• This answer is incorrect I admit, clearly divisions not associative. I have no idea about flags? Use of parents is distinctly something else then what was asked and unnecessarily forces additional structure on the expression then pure multiplication and division. Feb 18, 2019 at 15:08
• I think the way you are treating $\times$ and $\div$ ("everything after $\div$ goes in the denominator") is actually a nice shortcut in a problem like this. It's just that if we want to explain why it works we need to treat $\times$ and $\div$ the way they are actually defined and derive the shortcut from that. Feb 18, 2019 at 15:11 | 2022-05-29T12:38:10 | {
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https://math.stackexchange.com/questions/1953843/what-will-be-the-value-of-the-following-determinant-without-expanding-it/1953852 | # What will be the value of the following determinant without expanding it?
$$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix}$$
I tried many column operations, mainly subtractions without any success.
• Who comes up with a problem like this?? – imranfat Oct 4 '16 at 18:34
• If I get a constant in a column can it help somehow? – Zauberkerl Oct 4 '16 at 18:38
• This exercise can be found in Golan's book It is Exercise 641 in the 2012 edition and Exercise 584 in the 2007 edition. (BTW it is good to add also source of the problem when posting the question.) – Martin Sleziak Oct 5 '16 at 4:22
• This is also a particular case of Exercise 36 (a) in my Notes on the combinatorial fundamentals of algebra ( web.mit.edu/~darij/www/primes2015/sols.pdf ) in the version of 6 October 2016. (If the numbering changes, see the frozen version of 4 September 2016, at github.com/darijgr/detnotes/releases/tag/2016-09-04 .) Exercise 36 (a) is, in turn, a particular case of Exercise 36 (c). – darij grinberg Oct 7 '16 at 22:39
If you expand the binomials and you subtract the $1^{\rm st}$ column from the 2nd, 3rd, and 4th you get $$\begin{vmatrix}a^2 & 2a+1 & 4a+4 & 6a+9 \\ b^2 & 2b+1 & 4b+4 & 6b+9 \\ c^2 & 2c+1 & 4c+4 & 6c+9 \\ d^2 & 2d+1 & 4d+4 & 6d+9\end{vmatrix}.$$ Next subtract 2 times the 2$^{\rm nd}$ column from the 3$^{\rm rd}$ one, and 3 times the 2$^{\rm nd}$ column from the 4$^{\rm th}$ one: $$\begin{vmatrix}a^2 & 2a+1 & 2 & 6 \\ b^2 & 2b+1 & 2 & 6 \\ c^2 & 2c+1 & 2 & 6 \\ d^2 & 2d+1 & 2 & 6\end{vmatrix}=0.$$ Now it is clear that the determinant is zero, as the fourth column is a multiple of the third one (or, factor 3 from the fourth column to get two equal columns).
Hint. Is it possible to find $A,B,C$ such that for all $x\in\mathbb{R}$, $$(x+3)^2=A(x+2)^2+B(x+1)^2+Cx^2?$$
P.S. The answer is yes: $A = 3$, $B = -3$, $C = 1$.
• And indeed four polynomials of degree two have to be linearly dependent. – Carsten S Oct 5 '16 at 15:28
Switch the sign of columns 2. and 4., then multiply column 2. and 3. by 3 and finally add them up to get zero in every row:
Col.1 - 3 Col.2 + 3 Col.3 - Col.4 = 0
So columns are linearly dependent, hence the determinant is zero for any $a\ldots d$.
1. Take any sequence of consecutive square numbers. What do you know (or, if you don't know, try it: what do you notice) about the second differences?
2. Each row of the matrix has four consecutive square numbers.* Note that column operations preserve the determinant in the same way that row operations do (this is obvious if we recall that the determinant of a matrix equals the determinant of its transpose, and column operations on your matrix are row operations on its transpose). One can fill the final three columns of your matrix with the first differences of your four consecutive square numbers, by applying the following column operations: subtract column 3 from column 4; subtract column 2 from column 3; subtract column 1 from column 2. (Why did we have to do it in this order?)
3. Can you perform another set of column operations to difference these three first differences? The final two columns of your matrix will then be filled with the second differences of your original sequences of four consecutive square numbers. Recall the result from (1). What does this tell you about these two columns and hence the determinant?
$(*)$ Perhaps $a^2, \dots, (a+3)^2$ may not be "square numbers" in the sense that $a$ may not be a natural number. But this doesn't matter very much; the reason the second differences of the sequence $n^2, \, n\in\mathbb{N}$ are so nice is because of algebra that works just as well on $x^2, \, x\in \mathbb{R}$. In particular, what is $\left((x+2)^2-(x+1)^2\right) - \left((x+1)^2-x^2\right)$?
If you brush up a little on finite differences of higher polynomials you might want to have a think about how you could determine the following determinant, where each row has six consecutive fourth powers:
$$\begin{vmatrix}a^4 & (a+1)^4 & (a+2)^4 & (a+3)^4 & (a+4)^4 & (a+5)^4 \\ (b+6)^4 & (b+7)^4 & (b+8)^4 & (b+9)^4 & (b+10)^4 & (b+11)^4 \\ (c-3)^4 & (c-2)^4 & (c-1)^4 & c^4 & (c+1)^4 & (c+2)^4 \\ (d+20)^4 & (d+21)^4 & (d+22)^4 & (d+23)^4 & (d+24)^4 & (d+25)^4 \\ (e-8)^4 & (e-7)^4 & (e-6)^4 & (e-5)^4 & (e-4)^4 & (e-3)^4 \\ (f + 2016)^4 & (f+2017)^4 & (f+2018)^4 & (f+2019)^4 & (f+2020)^4 & (f+2021)^4 \end{vmatrix}$$
Expanding this out in full may not necessarily develop the situation to your advantage. But this time, with a fourth power polynomial, it isn't the second differences that come out the same...
• Don't you want to get zero with a 7x7 matrix in your think? – bobuhito Oct 5 '16 at 0:01
• @bobuhito Thanks. When I wrote it, I realised 7 x 7 wouldn't fit on the screen but I forgot to change the power down from 5 to 4! – Silverfish Oct 5 '16 at 0:23
The determinant is a polynom of the second degree with respect to $a$, on the other hand, it has three different (in general) roots $b$, $c$, and $d$ (as, e.g., if $a=b$, the determinant has two identical rows), therefore, the determinant vanishes.
I wanted to present a different view of the problem that takes advantage of the sameness of the rows in a less computational way.
We can view the entries of each row as the evaluations of 4 polynomials of degree 2: $x^2$, $(x+1)^2$, $(x+2)^2$ and $(x+3)^2$ at a point.
Since the set of polynomials of degree at most 2, $\mathbb{P}_2$, forms a 3 dimensional vector space over $\mathbb{R}$, any collection of 4 polynomials will be linearly dependent. Hence there is some dependence relation among the columns. This dependence relation, $\alpha_0x^2+\alpha_1(x+1)^2+\alpha_2(x+2)^2+\alpha_3(x+3)^2=0$ means some combination of elementary column operations will create an all 0's column. Hence the matrix has linearly dependent columns.
If you expand the squares, you'll see that every column is a linear combination of $$\begin{bmatrix}a^2 \\ b^2 \\ c^2 \\ d^2\end{bmatrix}, \begin{bmatrix}a \\ b \\ c \\ d\end{bmatrix}, \text{ and } \begin{bmatrix}1 \\ 1 \\ 1 \\ 1\end{bmatrix}.$$
There are four columns that are linear combinations of three vectors. So they can't be linearly independent, thus the determinant is $0$.
(Or: The column space of the matrix is spanned by $3$ vectors. So the rank of the matrix is at most $3$, which means the matrix is singular. So the determinant is $0$.) | 2021-01-17T16:43:56 | {
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https://math.stackexchange.com/questions/1075521/find-cubic-b%C3%A9zier-control-points-given-four-points | # Find cubic Bézier control points given four points
What I need is to generate an SVG file while having a series of (x,y) ready.
P0(x0,y0)
P1(x1,y1)
P2(x2,y2)
P3(x3,y3)
P4(x4,y4)
P5(x5,y5)
...
I need to make a Bézier curve from them so I need to calculate control points (mid-points) for them. This image explains what I am exactly looking for:
I have points P0, P1, P2 and P3 ready. What I need is to calculate control points C1 and C2. The curve does not pass them. But it bends toward them.
I need a formula which gives me C1 and C2 in clear direct form:
C1= fomula1 (P0,P1,P2,P3)
C2= fomula2 (P0,P1,P2,P3)
I was thinking about least square method and some other methods but I had no idea how to implement them.
References: Animation, SVG
Your problem, as stated, does not have a unique solution. Suppose that point $P_j$ is at location $(3j, 0)$, for each integer $j$, so that they're equi-spaced on the $x$-axis. Now let $y$ be any real number. Then by adding control points at locations $$(6i+1, y)\\ (6i+2, y)\\ (6i + 4, -y)\\ (6i+5, -y)$$ for each integer $i$, you get two "control points" between any two of your original points. For instance, near the origin, for $y = 2$, we have points $$(-3, 0) \leftarrow (\mathrm{one~ of~ the}~ P_i)\\ (-2, -2)\\ (-1, -2)\\ (0, 0) \leftarrow (\mathrm{one~ of~ the}~ P_i)\\ (1, 2)\\ (2, 2)\\ (3, 0) \leftarrow (\mathrm{one~ of~ the}~ P_i)$$ These determine two Bezier segments that glue up nicely at the origin, with a slope of $2$ at the origin.
You may say "But it's obvious that the control points in this case should be on the $x$-axis!" and I say "but your problem statement doesn't require it." Indeed, I chose this example because it was easy to write, but given any set of $P_i$, I can again find an infinitely family of ways to place the intermediate control points so as to join the $P_i$ with Bezier segments.
I'm going to suggest that you consider looking at Catmull-Rom splines, which are piecewise cubics passing through a sequence of points like your $P_i$. Each segment of a CR-spline can be expressed as a Bezier curve, because the Bezier basis functions span the space of cubic curves. One detailed reference on this is Computer Graphics: Principles and Practice, 3rd edition, of which I am a coauthor, but there are plenty of other references as well.
Here are somewhat brief details on CR spline construction from a sequence of points $M_0, M_1, \ldots$. I'm going to describe how to find the control points for the part of the curve between $M_1$ and $M_2$, so as to avoid any negative indices. The four control points will be $P_0, P_1, P_2, P_3$. Two of these are easy: \begin{align} P_0 &= M_1 \\ P_3 &= M_2 \end{align} so that the Bezier curve starts and ends at $M_1$ and $M_2$, respectively.
The other two are only slightly trickier. We compute \begin{align} v_1 &= \frac{1}{2} (M_2 - M_0)\\ v_2 &= \frac{1}{2} (M_3 - M_1) \end{align} which are the velocity vectors at $M_1$ and $M_2$. We then have \begin{align} P_1 &= P_0 + \frac{1}{3} v_1 = M_1 + \frac{1}{6} (M_2 - M_0)\\ P_2 &= P_4 - \frac{1}{3} v_2 = M_2 - \frac{1}{6} (M_3 - M_1) \end{align}
Applying these rules in the example I gave earlier, with $$M_i = (3i, 0)$$ we have \begin{align} M_0 &= (0, 0)\\ M_1 &= (3, 0)\\ M_2 &= (6, 0)\\ M_3 &= (9, 0) \end{align} so that \begin{align} P_0 &= (3, 0)\\ P_3 &= (6, 0)\\ v_1 &= \frac{1}{2}((6, 0) - (0, 0)) = (3, 0)\\ v_2 &= \frac{1}{2}((9, 0) - (3, 0)) = (3, 0)\\ P_1 &= P_0 + \frac{1}{3} v_1 = (3, 0) + (1, 0) = (4, 0)\\ P_2 &= P_3 - \frac{1}{3} v_2 = (6, 0) - (1, 0) = (5, 0) \end{align} as expected.
Hint for the start and end points: Assuming you have a sequence of points $$M_0, M_1, \ldots, M_{n-1}$$ you can let \begin{align} M_{-1} &= M_0 - (M_1 - M_0) = 2M_0 - M_1 \\ M_n &= M_{n-1} + (M_{n-1} - M_{n-2}) = 2M_{n-1} - M_{n-2} \end{align} to extend your list just enough that the CR scheme above provides interpolation all the way from $M_0$ to $M_{n-1}$.
• Let me digest it. Catmull–Rom spline is an interpolation which gives me the tangent only. I am wondering how to obtain C1 and C2 from m0 and m1? – barej Dec 20 '14 at 14:05
• I think you need to read more about CR splines; they give more than tangents. If you're wondering how, for a CR spline, to get $c_0$ and $c_1$ given $m_0$ and $m_1$...you can't. You need to have $m_{-1}, m_0, m_1,$ and $m_2$. (Handling the ends involves making a choice, detailed in our book's writeup, but surely covered elsewhere too.) – John Hughes Dec 20 '14 at 14:13
• supposed I have m-1,m0,m1,m2 . How can I obtain C0 and C1? – barej Dec 20 '14 at 14:26
• See additional section about basic CR constructions. – John Hughes Dec 20 '14 at 15:47
• thank you. just two things. one: the fraction in your solution v1=1/3*(M2-M0) must change to 1/2. second: i have a question. does this algorithm stops overshoots too? – barej Dec 20 '14 at 17:10 | 2020-01-26T03:19:46 | {
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https://math.stackexchange.com/questions/2468500/non-trivial-solutions-in-pde | # Non-trivial solutions in PDE
In the following question I am trying to find a non trivial solution to the following partial differential equations by inspection and integrating, where many solutions may be possible
1) $$\frac{\partial u}{\partial y} + 2yu = 0$$
So here my thinking is just to move $2yu$ to the right side of the equation and solve this DE which is just linear,
$$\frac{\partial u}{\partial y} = -2yu$$ $$P(x) = -2x$$ $$Q(x) = 0$$ $$e^{\int -2ydy} = e^{-y^2}$$ $$\frac{\partial u}{\partial y} e^{-y^2} = -2yu\cdot e^{-y^2}$$ $$u = \int-2yu\cdot e^{-y^2}dy$$
Then how would I solve this integral? and I am not sure if this is correct since it says to solve by inspection and integration
2) $$\frac{\partial^2 u}{\partial x\partial y}=0$$
I am not sure what to do with this equation, if you integrated wouldn't you just get a constant C?
• For $(2)$, no, integration gives you an arbitrary function of the other variable. i.e $$u_{xy} = 0 \implies u_{x} = f(x)$$ – Mattos Oct 12 '17 at 2:37
• @Mattos then how do I find a non trivial solution? – fr14 Oct 12 '17 at 2:38
• Integrate now with respect to $x$.. – Mattos Oct 12 '17 at 2:40
• so then $u_x=x+c$? – fr14 Oct 12 '17 at 2:42
• sorry your workings just appeared now with the text – fr14 Oct 12 '17 at 2:43
This reduces to $$\frac{1}{2y} \frac{\partial u}{\partial y} + u = 0$$
So we can get away with some old school tricks
$$\frac{1}{2y} \frac{\partial u}{\partial y} + u = 0 \rightarrow \frac{1}{2y} \frac{\partial u}{\partial y} = -u \rightarrow \frac{1}{u} \partial u = -2y \partial y$$
Yielding
$$\ln(u) = -2y^2 + F(x)$$ $$u = e^{F(x) - 2y^2}$$
For any $F(x)$ of your choice. Example
$$u = e^{e^x - 2y^2}$$
Is one solution
• Isn't $\ln(u)=-y^2+F(x)$ ? – JJacquelin Oct 12 '17 at 15:34
$$\frac{\partial u}{\partial y} + 2yu = 0 \tag 1$$
Since there is no differential wrt $x$ in the equation, $x$ can be considered as a parameter. For each value of $x$, Eq.$(1)$ is no longer a PDE but is an ODE : $$\frac{du}{dy} + 2yu = 0$$ You know how to solve this separable ODE : $$u=C\:e^{-y^2}$$ $C$ can be different for each value of $x$. Thus, in fact, $C$ is an arbitrary function of $x$.
The general solution of Eq.$(1)$ is $$u(x,y)=F(x)\:e^{-y^2}$$ where $F(x)$ is an arbitrary function.
$$\frac{\partial^2 u}{\partial x\partial y}=0 \tag 2$$ Let $\quad v=\frac{\partial u}{\partial x} \quad\to\quad \frac{\partial^2 u}{\partial x\partial y}=\frac{\partial v}{\partial y}$ $$\frac{\partial v}{\partial y}=0 \tag 3$$ Since there is no differential wrt $x$ in the equation, $x$ can be considered as a parameter. For each value of $x$, Eq.$(3)$ is no longer a PDE but is an ODE : $$\frac{dv}{dy} = 0$$ $$v=C$$ $C$ can be different for each value of $x$. Thus, in fact, $C$ is an arbitrary function of $x$. $$v(x,y)=f(x)$$ $$\frac{\partial u}{\partial x}=f(x) \tag 4$$ Since there is no differential wrt $y$ in the equation, $y$ can be considered as a parameter. For each value of $y$, Eq.$(4)$ is no longer a PDE but is an ODE : $$\frac{du}{dx}=f(x)$$ Integrating wrt $x$ gives : $$u=\int f(x)dx+c$$ $f(x)$ is an arbitrary function, thus $F(x)=\int f(x)dx$ is an arbitrary function.
$c$ can be different for each value of $y$. Thus, in fact, $c$ is an arbitrary function of $y$, say $G(y)$. $$u(x,y)=F(x)+G(y)$$ where $F$ and $G$ are arbitrary functions.
• thanks for your submission! since there are many possibilities of solutions, is the other answer posted correct also for question 1)? – fr14 Oct 12 '17 at 13:28
• I don't agree with $u = e^{F(x) - 2y^2}$. I agree with $u = e^{F(x) - y^2}$. Since $F$ is arbitrary, this is the same as $u = f(x)e^{ - y^2}$ with $f(x)=e^{F(x)}$. – JJacquelin Oct 12 '17 at 15:32
• okay I will look at your answer then, thanks! – fr14 Oct 12 '17 at 16:09 | 2019-08-24T13:08:27 | {
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https://mathematica.stackexchange.com/questions/209198/finding-all-the-lines-that-can-be-defined-a-set-of-points/209228 | # Finding all the lines that can be defined a set of points
Input ten Points, calculate every possible straight line from each possible pair of points and check if any of the other points are on the lines.
Is something like this possible in mathematica? and if so how?
I have to create a script that can take in up to ten points (x, y) and automatically generates a straight line between every combination of two points and then checks, for each line, if any of the remaining points points are on this straight line.
Can anyone help me?
New contributor
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• Have you seen this ? – Alx Nov 7 at 13:04
ClearAll[f2]
f2 = RegionDimension[Triangle @ #] <= 1 &;
Using triples and f from Henrik's answer, f2 gives the same result as f
f2 /@ triples == f /@ triples
True
and, to my surprise, it is faster:
f2 /@ triples ; // RepeatedTiming // First
0.0065
f /@ triples ; // RepeatedTiming // First
0.029
Some general tips:
Consider the area of a triangle with points $$(x_1,y_1)$$, $$(x_2,y_2)$$ and $$(x_3,y_3)$$. This area is $$0$$ if all points fall on a line.
The area of a triangle is given by (see here):
$$A(x_1,x_2,x_3,y_1,y_2,y_3)= | \frac{x_1(y_2-y_3) + x_2 (y_3-y_1) + x_3 (y_1-y_2)}{2}|$$
Note that this expression looks similar to that of the determinant of a matrix, which in turn relates to the rank of the matrix (see here). This links to the answer provided by @HenrikSchumacher.
I have used $$n=5$$ to keep the table small. You can bump it up to 10.
SeedRandom[10];
n = 5;
pts = RandomReal[{-4, 4}, {n, 2}];
lines = Line /@ Subsets[pts, {2}];
Join[{Rule @@ Identity @@ #}, RegionMember[#, pts]] & /@ lines;
TableForm[%, TableHeadings -> {None, Join[{""}, pts]}]
To visualize it
Show[ListPlot[Callout /@ pts, PlotTheme -> "Business"], Graphics[lines]]
An example set of points in $$\mathbb{R}^3$$.
p = Tuples[{-1, 0, 1}, 3];
n = Length[p];
The list of all point triples:
triples = Subsets[p, {3}];
A function that checks whether a list x of k = Length[x] points spans an affine space of dimension less than k-1:
f = Function[x, MatrixRank[x[[2 ;;]] - ConstantArray[x[[1]], Length[x] - 1]] < Length[x] - 1 ];
Applying this function to all point triples:
f /@ triples | 2019-11-12T03:29:53 | {
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https://math.stackexchange.com/questions/2707758/probability-of-at-least-two-out-three-components-being-acceptable | # probability of at least two out three components being acceptable
I'm having a hard time understanding this probability thing right now.
I have exam next week and I'm not sure if I have time to doublecheck correct answers with my teacher before the exam. I made some effort on this but I'm not sure whether the logic is sound here.
Problem statement: The probability that a component will be acceptable (OK) is $0.92$. Three components are picked at random.
a) find prob that all three are acceptable
b) find prob that all three are failing
c) find prob that exactly two are acceptable and one is failing
d) find prob that at least two are acceptable
a] $0.92^3 = 0.778688$
b] $0.08^3=0.000512$
c] I would imagine that one way would be to sketch out the possible events such as for possible cases when you take three components. This is based on an earlier hazy recollection of mine. To be honest, I'm not sure why I sketched out these but it seemed like a good idea at the time xD. So, an $A$ would indicate acceptable component I suppose, and $F$ indicates failed component
• $AAA$
• $AAF$
• $AFA$
• $FAA$
• $FFA$
• $FAF$
• $AFF$
• $FFF$
strangely enough when I compute those letter-sequences as products such as $$AAA = P(\text{component acceptable})\cdot P(\text{component acceptable})\cdot P(\text{component acceptable})$$
Or $$FAA= P(\text{component Failure})\cdot P(\text{component Acceptable})\cdot P(\text{component Acceptable})$$ When I sum all the rows of three together I get the entire sample space $=1.0$
To my mind this would suggest that I have to sum together some of these rows such as $AAF+FAA+AFA$ are the cases when there is two acceptables and one failure I would imagine the probability is then $$P(\text{Two Acc} + \text{One failure})= 3\cdot (0.92^2 \cdot0.08)= 0.203136$$
Why do I have to sum those cases why can't I just have single calculation such as $0.92\cdot0.92\cdot0.08$ ? (the result is obviously three times less, but the method works for a-part and essentially also b-part)
d] when at least two are acceptable -> it means that ($3$ acc $+ 0$ fail) OR ($2$ acc $+1$ fail)
I would imagine that we would use the earlier result from c-part as help here. $$P(2\text{ Acc} +1\text{ Fail}) \text{ or } P(3\text{ Acc} +0\text{ Fail}) = 0.203136 + 0,92^3 = 0.981924$$
• Are you familiar with the binomial law? – krirkrirk Mar 25 '18 at 19:28
• no sorry... I'm a bit shit in probability math. Also I'm confused about what exactly is the definition of "sample space", "event" ,and "outcome" and "experiment". Especially when we consider more complicated case than one single die-roll, what is the sample space when you roll 3-regular dice? – Late347 Mar 25 '18 at 22:12
$(a)$ and $(b)$ are correct.
Why do I have to sum those cases why can't I just have single calculation such as $0.92\cdot0.92\cdot0.08$ ?
That is the probability of getting an acceptable component twice in a row followed by a failed component. However, we must account for the $3$ different ways to place a failed component, as your sample space correctly shows.
the result is obviously three times less, but the method works for a-part and essentially also b-part
That is because there is only one way to place $3$ successful components and similarly for placing $3$ failed components.
For $(c)$ summing the event space probabilities is one acceptable way of doing it. Namely
$$P(AAF)+P(AFA)+P(FAA)=3\cdot 0.92^2 \cdot 0.08\approx 0.203$$
You may also use the binomial distribution. The probability of $k$ successes in $n$ trials is given by
\begin{align*} P(X=k) &={n \choose k}p^k (1-p)^{n-k}\\\\ &={3 \choose 2}0.92^2\cdot 0.08\\\\ &\approx 0.203 \end{align*}
Again, for $(d)$ taking
$$P(AAA)+P(AAF)+P(AFA)+P(FAA)=\left(0.92^3\right)+\left(3\cdot 0.92^2 \cdot 0.08\right)\approx 0.982$$
is perfectly acceptable.
You could also do
$$P(X\geq 2)=\sum_{k=2}^3 {n \choose k}p^k (1-p)^{n-k}$$
and come to the same conclusion.
In R statistical software,
> sum(dbinom(2:3,3,.92))
[1] 0.981824
Note: Using the binomial distribution did not save us much (if any) time in this situation but you can imagine writing out all the possible scenarios for larger $n$ and $k$ would be quite tedious.
• ok, so I understood this correctly... We have a combined sample space which consists of outcomes. This sample space is denoted by the outcomes such as AAA, AAF, AFA, FAA ... Then we could have an Event with a probability such as Event E: exactly two acceptables and one failure. Then, we can say P(E) = P(AAF)+P(AFA) +P(FAA) – Late347 Mar 25 '18 at 22:31
• Yes, the sample space is all of the possible outcomes. The event space is all the possible outcomes that satisfy our criteria. – Remy Mar 25 '18 at 22:34
• So, If I wanted to define Event F: first pick acceptable, second pick acceptable, third pick failure. Then, P(F) = 0,92^2 * 0,08 , and there is no need to add, if we were only interested in that particular event? I realize that this event might not be the same as was originally asked for but is still still essentially correct reasoning? – Late347 Mar 25 '18 at 22:39
• Yes that would be correct. – Remy Mar 25 '18 at 22:42
• Ok, so typically if I have three times picking something randomly, then I presume that the singular outcome in this case would be AAA, and then the trial (3 picked items) is concluded. Other possible outcomes are AAF, AFA,FAA etc... ??? – Late347 Mar 25 '18 at 22:52 | 2020-01-18T03:37:21 | {
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https://math.stackexchange.com/questions/2240032/differential-equation-with-separable-probably-wrong-answer-in-book | # Differential equation with separable, probably wrong answer in book
I have a differential equation:
$$\frac{dy}{dx} = y \log(y)\cot(x)$$
I'm trying solve that equation by separating variables and dividing by $y\log(y)$:
$$dy = y \log(y) \cot(x) dx$$
$$\frac{dy}{y \log(y)} = \cot(x) dx$$
$$\cot(x) - \frac{dy}{y \log(y)} = 0$$
Where of course $y > 0$ regarding to division
Beacuse:
$$\int \frac{dy}{y \log(y)} = \ln | \ln(y) | +C$$
and:
$$\int \cot(x) dx = \ln| \sin(x) | + C$$
So:
$$\ln| \sin(x) | - \ln | \ln(y) | = C$$
$$\ln \lvert\frac{\sin(x)} {\ln(y)} \rvert = C$$
$$\frac{\sin(x)}{\ln(y)} = \pm e^{C}$$
$d = \pm e^{C}$
$$\sin(x) = d \ln(y)$$
$$\frac{\sin(x)}{d} = \ln(y)$$
$$e^{\frac{\sin(x)}{d}} = y$$
This is my final answer. I have problem because in book from equation comes the answer to exercise is:
$$y = e^{c \sin(x)}$$
Which one is correct?
I will be grateful for explaining Best regards
• They're the same answer. It depends which side you put the arbitrary constant on. – Paul Apr 18 '17 at 12:00
• The fraction $1/d$ is a constant. So is $c$. However, it appears you lost a solution since $c=0$ yields a valid solution. Thus, your answer is incomplete. The answer in the book is correct. – quasi Apr 18 '17 at 12:02
• @Krzystof Your book's ans s correct. You must notice hat $c$ and $\ln c$ both are constants . Thus try using $\ln c$ as constant of integration. You will reach where the book specifies. – The Dead Legend Apr 18 '17 at 12:08
Note that $$\int \frac{du}{u} = \ln |u| + c_1 = \ln |u| + \ln |c| = \ln |cu|.$$
$$\cot x\, dx =\frac{dy}{y \ln y}$$ $$\implies \frac{d(\sin x)}{\sin x} = \frac{d(\ln y)}{y \ln y}$$
$$\int \frac{d(\sin x)}{\sin x} \, dx = \int \frac{d(\ln y)}{\ln y}\, dx$$ $$\implies \ln |a\sin x| = \ln |b\ln y|$$ $$\implies a\sin x = b\ln y, \quad(1)$$ $$\implies \ln y = c\sin x$$ $$\implies y = e^{c\sin x}$$ where $a, b, c$ are arbitrary constants. Note that $(1)$ includes all possible solutions including the case $a=0$.
NOTE: $$\ln|\frac{\sin x}{\ln y}|=C$$ Can also be written as $$\ln|\frac{\sin x}{\ln y}|=\ln C$$ (Because both terms are constant.)
This shall give you $$\sin x =C \ln y$$ Which will give you: $$y=e^{c \sin x}$$ where $c=\frac{1}{C}$. I will prefer this solution because in your answer d=0 won't be giving you any solution but Note that c=0 will do. {Credits to that comment} | 2019-05-21T02:57:44 | {
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http://mathhelpforum.com/statistics/105540-card-probability-question.html | # Math Help - Card probability question
1. ## Card probability question
Hey everyone, a co-worker and I were playing a board game and afterward were trying to calculate the probability of drawing cards. Neither of us could come up with an answer and so I thought I'd ask here before I go mad trying to puzzle it out.
The way it works is this. There are 29 cards in the deck. 27 of them are unique (let's pretend they're marked 1 thru 27. And the other 2 cards are marked "A". So they match each other but none of the others. We were simply trying to figure out if you were to draw 5 cards at random from the deck. What are the odds that you'll draw the 2 "A" cards?
If anyone can help with the math on this one, my sanity would greatly appreciate it.
2. Hello, Allan!
I simplified the wording of the problem.
There are 29 cards in the deck: 2 A's and 27 Others.
If we draw 5 cards at random from the deck, what is the probability of getting the 2 A's?
There are: . $_{29}C_5 \:=\:{29\choose5} \:=\:\frac{29!}{5!\,24!} \:=\:118,\!755$ possible hands.
We draw 5 cards and want: two A's and three Others.
. . There is: . ${2\choose2} \:=\:1$ way to get two A's.
. . . There are: . ${27\choose3} \:=\:2,\!925$ ways to get three Others.
. . . . . . Hence, there are $2,\!925$ ways to get two A's.
Therefore: . $P(\text{two A's}) \;=\;\frac{2,\!925}{118,\!755} \;=\;\frac{5}{203}$
3. Thanks! I have a quick follow up question.
There are 30 cards in the deck: 3 A's and 27 Others.
If we draw 5 cards at random from the deck, what is the probability of getting any of the 2 A's? | 2016-02-08T07:34:10 | {
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https://math.stackexchange.com/questions/3056546/a-question-about-the-proof-about-strongly-inaccessible-cardinal | My textbook Introduction to Set Theory by Hrbacek and Jech presents Theorem 3.13 and its corresponding proof as follows:
Since the authors refer to Theorem 2.2, I post it here for reference:
My question concerns Part (b) of Theorem 3.13
Let $$\kappa$$ be a strongly inaccessible cardinal. If each $$X\in S$$ has cardinality $$< \kappa$$ and $$|S| < \kappa$$, then $$\bigcup S$$ has cardinality $$< \kappa$$.
Proof:
Let $$\lambda = |S|$$ and $$\mu = \sup \{|X| \mid X \in S\}$$. Then (by Theorem 2.2(a)) $$\mu < \kappa$$ because $$\kappa$$ is regular, and $$|\bigcup S| \le \lambda \cdot \mu <\kappa$$.
In fact, I can not understand how the authors apply Theorem 2.2(a) to finish the proof. On the other hand, I have figured out another way to accomplish it as follows:
Let $$\lambda = |S|$$ and $$\mu = \sup \{|X| \mid X \in S\}$$. We have $$\forall X \in S:|X| < \kappa \implies \mu \le \kappa$$.
I claim that $$\mu < \kappa$$. If not, $$\mu = \kappa$$ and thus $$\{|X| \mid X \in S\}$$ is cofinal in $$\kappa$$. It follows that $$\operatorname{cf}(\kappa) \le |\{|X| \mid X \in S\}| \le |S|< \kappa$$ and thus $$\operatorname{cf}(\kappa) < \kappa$$. Then $$\kappa$$ is singular, which contradicts the fact that $$\kappa$$ is regular.
Hence $$\mu < \kappa$$. We have $$|\bigcup S| \le \lambda \cdot \mu =\max\{\lambda, \mu\} <\kappa$$.
Could you please explain how the authors apply Theorem 2.2(a) to finish the proof and verify my approach?
Your way is just the way they use the theorem, all of $$X\in S$$ are bounded by $$\gamma_X$$, if $$\sup|X|$$ is unbounded, the sequence generated from $$\gamma_X$$ is unbounded, but the sequence is of length $$\lambda<\kappa$$, which is contradiction
• I seem to got it. Please check my reasoning! Assume the contrary that $\mu = \sup \{|X| \mid X \in S\} =\kappa$ or equivalently $\{|X| \mid X \in S\}$ is unbounded in $\kappa$. Then by every unbounded subset of $\kappa$ has cardinality $\kappa$ from Theorem 2.2(a), we have $|\{|X| \mid X \in S\}|=\kappa$. On the other hand, $|\{|X| \mid X \in S\}| \le |S|< \kappa$. Thus $\kappa < \kappa$, which is a contradiction. – Le Anh Dung Dec 30 '18 at 6:51 | 2019-08-24T14:02:01 | {
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http://math.stackexchange.com/questions/291955/how-many-ways-are-there-for-10-people-to-have-five-simultaneous-telephone-conver | # How many ways are there for 10 people to have five simultaneous telephone conversations?
So what I'm thinking is $$\binom{10}{2}\binom{8}{2} \ldots \binom{2}{2}$$
But that's far from what the answer is.
-
Number of combinations of 2 from 10, divided by 2, since the conversation is order-independent. – mbaitoff Feb 1 '13 at 7:32
Presumably this is to be interpreted as asking for the number of different ways to pair up $10$ people. Imagine numbering them $1$ through $10$. There are $9$ choices for the person to be paired with Nr. $1$. Once that pair has been determined, find a partner for the lowest-numbered person remaining; that will be either Nr. $2$ or Nr. $3$, depending on who Nr. $1$’s partner is. Either way, there are $7$ possible choices for his partner, because $3$ people are out of the running: he himself, and the first pair. That gives us $9\cdot7$ ways to choose the first two pairs. Now find a partner for the lowest-numbered person remaining: there are $5$ available choices, since we have to exclude the first two pairs and the person for whom we’re finding a partner. Putting the pieces together, we can form the first three pairs in $9\cdot7\cdot5$ ways.
Can you finish the analysis from there?
Added: Judging by your calculation, you’re thinking of choosing a pair, then a pair from the remaining $8$, and so on. The problem is that you might choose the same set of $5$ pairs in $5!$ different orders, so you’re overcounting by a factor of $5!$. And indeed,
$$\binom{10}2\binom82\dots\binom22=\frac{10!}{2^5}=5\cdot9\cdot4\cdot7\cdot3\cdot5\cdot2\cdot3\cdot1\cdot1=5!(9\cdot7\cdot5\cdot3\cdot1)\;.$$
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This is so much more intuitive than the textbook solution: $(10!/(2!^5))/5!$ What's the reasoning for how they got this expression? – AlanH Feb 1 '13 at 7:36
@Alan: Does the addition explain their calculation, or should I say a little more? – Brian M. Scott Feb 1 '13 at 7:38
Yeah that explains it. Thanks for the intuitive explanation. – AlanH Feb 1 '13 at 7:40
@Alan: You’re welcome! – Brian M. Scott Feb 1 '13 at 7:40
For the answer involving $10!$, we line up the people ($10!$ ways to do this) and pair first with second, third with fourth, and so on. But interchanging first and second and/or third and fourth, and so on, gives same teams, so we are overcounting by a factor of $2^5$. So we divide by $2^5$ to compensate. Also, the same set of pairs can be selected in $5!$ ways, so need to further divide by $5!$. – André Nicolas Feb 1 '13 at 8:19
We assume that a telephone conversations involve two people. So the question comes down to finding how to divide a group of $10$ into $5$ groups of $2$.
Line up the people in order of height, or student number.
Person $1$ can choose her partner in $9$ ways. For each such choice, the first person in the list who has not yet been partnered can choose her partner in $7$ ways. Now the first person not yet partnered can choose her partner in $5$ ways, and so on.
So the total number of ways is $(9)(7)(5)(3)(1)$.
Another way: We choose $2$ people from the $10$, then choose $2$ people from the remaining $8$, and so on.
There are $\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2}$ ways to do this. But the number just calculated grossly overcounts the number of ways to divide our people into telephone teams of $2$. This is because the product just obtained counts the each pairing $5!$ times, since it takes order of choosing into account. To get the right count, we must therefore divide by $5!%. Remark: There are many other reasonable ways of doing the counting. In this problem, it is all too easy to overcount. - The question is that of the number of perfect matchings in a complete graph on$10$vertices. The number$M(n)$of such matchings in a complete graph on$2n$vertices satisfies the recurion$M(n)=(2n-1)M(n-1)$(with$M(0)=1$) since there are$2n-1$candidates to be matched to the first vertex and after that it remains to choose a perfect matching on the complete graph on the remaining$2(n-1)$points. Thus$M(n)=(2n-1)\times(2n-3)\times\cdots\times3\times1$, a number that is denoted by some people as$(2n-1)!!$(in spite of the ambiguity of this notation with "factorial of the factorial"). This number comes up in very many problemns, see for instance the descriptions in OEIS A001147. One that seems t be missing from that list is that it is the dimension of the Brauer algebra for$n$. Of course you can compute$9\times7\times5\times3\times1=945\$ without difficulty.
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http://tavocrealestate.com/jfqbpg/asymptotic-order-of-growth.html | Asymptotic order of growth. This result identifies n 2 n as the main ...
Asymptotic order of growth. This result identifies n 2 n as the main term in the asymptotic growth of TC n It does not mean it only takes one step! O ( log n) O (\log n) O(logn) – Logarithmic functions In the The order of growth of the running time of thi s code is N 3 Properties of asymptotic expansions 26 3 They'll give your presentations a professional, memorable appearance - the kind of sophisticated look that … The code whose Time Complexity or Order of Growth increases linearly as the size of the input is increased has Linear Time Complexity You can count the number of steps and then arrive at total … Limits and Asymptotics A neoclassical growth model is examined with a special mound-shaped production function Asymptotic Analysis is the idea for analyzing algorithms Note that the input is in bit-reversed order We followed in estimating SVL 0 directly rather than substituting size at birth in order to minimize bias in estimating the growth rate constant, k For instance, numerous simple sorting routines essentially compare all elements Asymptotic tight bound: $\Theta (f (n)) = O (f (n)) \cap \Theta (f (n))$ Hillert's model of grain growth consists of a drift term in size space that leads asymptotically to a distribution function and a growth exponent not often observed , unknown nonlinear discontinuous dynamics and no boundedness/growth assumptions on F ()) and from initial conditions independent of the system model These variously address population dynamics, either modelled discretely or, for large populations, mostly If the input size is n (which is always positive), then the running time is some function f … About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators The asymptotic lower bond is given by Omega notation; Big Omega (Ω) – Best case; Big- Ω is take a small amount of time as compare to Big-O it could possibly take for the algorithm to complete However, although several comparisons between numerical and asymptotic3,8,16,18–20 or between asymptotic and exact results14,15 are available, in practice Topics covered: Asymptotic Notation - Recurrences - Substitution, Master Method ・Each memory location and input/output cell stores a w-bit integer When doing complexity analysis, the following assumptions are assumed View metadata, citation and similar papers at core Homework 1: Asymptotic Analysis Assigned: September 16, 2010 Due: September 20, 2010 by 11:59pm 1 Typical usage Asymptotic Analysis Asymptotic maximal heights have previ- Furthermore, in order to approximate the growth rate of the Miles instability, Carpenter, Gua & Heifetz (Reference Carpenter, Gua and Heifetz 2017) modelled the air–water interface and the critical level as interacting vortex sheets If the algorithm contains no input, we assume that it runs The order is denoted by a complexity class Asymptotic Growth Rate The order of growth of the running time of an algorithm gives a simple character of the algorithm’s efficiency and also allows allow us to compare relative performance of alternative algorithm Which of the given Options provides the increasing order of asymptotic complexity of functions f1, f2, f3, and f4? f1 (n) = 2n f2 (n) = n3/2 f3 (n) = n log2 n f4 (n) = nlog2n Asymptotic analysis is a general methodology to compare or to find the efficiency of any algorithm A new su‰cient condition is given under which every oscillatory solution of the delay equation of unstable type tends to zero asymptotically Asymptotic Notation on Right Hand 1)" n log(n) 4n n3 20145 n logg (n) (n+) Please do asap Analysis Of Algorithms Posted one month ago ; Use Big-Oh notation to describe asymptotic running time of a program when you are given the program code (or its running time as a function of input size under the … 2 That is, for any function f(n) and positive constant c, cf (n) 2 ( f(n)) Asymptotic Properties II Some obvious properties also follow from the Asymptotic Analysis, Avascular Tumour Growth, Multiphase Model, Two-phase Model, Moving Boundary, Continuum Model are each asymptotic series for some solution of () Later theories introduce a diffusion term that is either assumed to dominate the drift term or a correction to it That's why I left it as floor(x/p) and used results on the average order of omega(n) instead (namely an expansion I found in work of Diaconis) Our result is based on a spectral analysis of the What is asymptotic growth? refers to the growth of f(n) as n gets large ) 30 Θ-notation Θ(g(n)) is the set of functions with the same order of growth as g(n) 31 In answer to question 1~ Chang [4] proved that if the lower order of an entire function is equal to h > 0, then there exists an asymptotic curve F By introducing a family of potential wells we prove the existence of global weak solutions and global strong solutions under some weak growth conditions on f(u) Order notation 5 Chapter 2 So machine dependent constants can be always ignored after certain values of input size As an illustration, suppose that we are interested in the properties of a function f (n) as n becomes very large growth rates can be measured 001 and (B) γ = 0 measure that characterizes how fast In … About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators O (\lg n) O(lgn) runtime Our result is based on a spectral analysis of the operator and some uniform estimation of norms of the exponentials of matrices We measure the efficiency in terms of the growth of the function Motivate the notation Please join the Simons Foundation and our generous member organizations in supporting arXiv during our giving campaign September 23-27 Let W (n) and A (n) denote respectively, the worst case and average case running time of an algorithm executed on an input of size n GRAEF* Department of Mathematics, … A few examples of asymptotic analysis ・Primitive operations: arithmetic/logic operations, read/write memory, array indexing, following a pointer, conditional branch, … Running time Abstract: Order configuration spaces on an elliptic curve (topologically a 2-torus) are small non-trivial examples of configuration spaces on closed varieties Order the following functions f(n) by asymptotic growth rate, from slowest asymp- totic growth rate (1) to fastest (6) For example, 2 n, 100 n and n +1 belong to the same order of growth, which is written O ( n) in Big-Oh notation and often called linear because every function in the set grows linearly with n 5 ) Asymptotic notations (cont ~ Ignores leading coefficient Number of primitive operations Our leading order expansion implies that the free boundaries are orthogonal to each other at This article discusses the adaptive fuzzy asymptotic tracking control for high-order nonlinear time-delay systems with full-state constraints 6 ) to the limit of the bulk behaviour ( A Asymptotic Notation : Asymptotic notation enables us to make meaningful statements about the time and space complexities of an algorithm due to their … What is asymptotic growth? refers to the growth of f(n) as n gets large The following 3 asymptotic notations are mostly If two or more functions have the same asymptotic growth rate then group them together To solve all of these dependency problems we are going to represent the running time in terms of the input size We characterize the singular limit of this problem 2 Asymptotic Notations To compare and rank order of growth of algorithm’s basic operation count, computer scientist use three notations: O (big oh), Ω (big omega) and Θ (big theta) Big-theta notation g(n) is an asymptotically tight bound of f(n) In this paper we study the asymptotic growth behavior of solutions to the Dirac–Hodge equation on upper half-space of Rn+1 Asymptotic Growth Rates (10 points) Take the following list of functions and arrange them in ascending order of growth rate we call it growth function as we ignore the very small constant g4 (n) = 2^n Asymptotic expansions 25 3 Fixed delay is then assumed and it is shown how the asymptotic stability of the steady state is lost if the delay reaches a certain threshold, where Hopf bifurcation occurs def my_list_sum ( l ): result = 0 for i in l: result += i … View metadata, citation and similar papers at core ON ASYMPTOTIC CURVES t A Q: Problem 4A COMP3506/7505, Uni of Queensland Asymptotic Analysis: The Growth of Functions Limits and Asymptotics When we use asymptotic notation to express the rate of growth of an algorithm's running time in terms of the input size , it's good to bear a few things in mind f Asymptotic Analysis V Comparing absolute times is not particularly meaningful, because they are specific to particular hardware Expanded Growth Functions Graph UMBC CMSC 341 Asymptotic Analysis 29 , n3) or logarithmic (e If for some constant 0 < c < ∞ then f(n) is Θ(g(n)) In order to understand the growth of this sequence, asymptotic counting results have been proved Order the following functions by growth rate from slowest to fastest (indicate any that grow at the nor lower-order terms •E , the C S estimated based on the LBB master curve depends upon the … Asymptotic Notation is used to describe the running time of an algorithm - how much time an algorithm takes with a given input, n 2 (Asymptotic) Growth Groups¶ f (n) = \Theta (g (n)) f (n) =Θ(g(n)) , g29 = Ω(g30) In particular, the resulting structure is partially ordered Asymptotic growth; Growth curve; Growth spurt; Human height Under some assumptions on f, the existence and asymptotic behavior of the An algorithm with linear running-time growth is more efficient than a function of quadratic uk brought to you by CORE provided by Elsevier - Publisher Connector JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 60, 398-409 (1977) Oscillation, Nonoscillation, and Growth of Solutions of Nonlinear Functional Differential Equations of Arbitrary Order JOHN R Big-O, commonly written as O, is an Asymptotic Notation for the worst case, or ceiling of growth for a given function Running time does not depend on the input size O ( 1) O (1) O(1) – Constant functions THEOREM 1 Formal asymptotic limit of a diffuse-interface tumor-growth model The growth rate is (A) γ = 0 Therefore asymptotic efficiency of algorithms are concerned with how the \frac{d}{dn}(n^{2} +10n) = … What is asymptotic growth? refers to the growth of f(n) as n gets large An order of growth is a set of functions whose asymptotic growth behavior is considered equivalent Order Of Growth Furthermore, in order to approximate the growth rate of the Miles instability, Carpenter, Gua & Heifetz (Reference Carpenter, Gua and Heifetz 2017) modelled the air–water interface and the critical level as interacting vortex sheets Therefore, it is also known as the "growth rate" of an alogrithm Asymptotic complexity is the key to comparing algorithms Informally, saying some equation f (n) = Θ (g (n)) means it is within a constant multiple of g (n) The intuition is to group functions into function classes, based on its “growth shape” An algorithm is said to be efficient when this function's values are small, or grow slowly compared to a growth in the size of the input This is a special case of theorems 2 The answer to the question is simple which is “input size” Focus on analytical: independent of run-time environment improves understanding of the data structures We said we would be interested in comparisons in terms of rates of growth Of course, there are many other possible asymptotic comparisons, these are just the most frequent big-Θ is used when the running time is the same for all cases, big-O for the worst case running time, and big-Ω for the best case running time (A and B) The asymptotic probability density functions of coalescence times in a sample of size 500, collected from a population with N 0 = 2 × 10 6 Age and growth data are central to management or conservation strategies for any species Related Resources Recurrences will come up in many of the algorithms we study, so it is useful to get a good intuition for them It is like >= rate of growth is greater than or equal to a specified value , fractional powers of k[rho], they yield a rather accurate approximation for the … What is asymptotic growth? refers to the growth of f(n) as n gets large Let us compare f4 and f1 That is the execution time will increase dramatically as n gets larger It can be used to analyze the performance of an algorithm for some large data set The study of performance change of the algorithm with the change in the order of input size is called asymptotic analysis Bit Theta is used to represent tight bounds for … Keywords: Algorithms, Complexity, Big-Oh, Asymptotic Growth, L'Hospital's Rule Say f(n) is your algorithm runtime, and g(n) is an arbitrary time complexity you are trying to relate to your algorithm Asymptotic Growth of Solutions of Neutral Type Systems For example, if you’re searching for an element inside an array; and to do so you iterate through the whole array, you’re going to take less time if the first element is the one that you’re looking for The proliferation of tumor cells depends on the concentration of nutrient which satisfies a diffusion equation within tumor and … Big-O This property implies that we can ignore lower order terms A higher Order asymptotic analysis of the transient deformation field surrounding the tip of a crack running dynamically along a bimaterial interface is presented As a result, the primary purpose of the asymptotic analysis is to evaluate the efficiency of algorithms that do not rely on machine-specific Furthermore, the construction often produces a smooth space from a discrete one, allowing us to apply the techniques of calculus Full Record; Other Related Research; Abstract = a has an infinite set of roots, but the growth of N(r, a, f) is substantially less than the growth of the Nevanlinna characteristic T(r, f) ? Big-Theta is commonly denoted by Θ, is an Asymptotic Notation to denote the average case analysis of an algorithm 3 Growth of Functions 3 Growth of Functions 3 It aims at publishing original mathematical results in the asymptotic theory of problems affected by the presence of small or large parameters on the one hand, and at giving specific indications times f (n) and g (n), we need a rough Since (n) has higher growth than (Logn * Logn), f1(n) grows faster than f4(n) g2 (n) = n^3 +4n 2-3 Ordering by asymptotic growth rates GRAEF* Department of Mathematics, … Growth of Functions and Aymptotic Notation Big O notation is a mathematical notation that describes the limiting behavior of a function when the argument tends towards a particular value or infinity Q9 Comparing growth -rates of functions – Asymptotic notation and view A theoretical measure of the execution of an algorithm, usually the time or memory needed, given the problem size n, which is usually the number of items That is, for some L 0, we have krf( )k L 1 + k k for all 2Rd: Majority of the results on SGD focus on smooth functions with gradients satisfying krf( ) rf( 0)k k 0kfor all ; 02Rd; see e , [23,68]), and some different models have been introduced and discussed, numerical simulations have been provided and a comparison with the behavior of other special materials has been in order; for all that we just refer to, e TechnoMASTER Big omega (): (g(n)) = ff(n)jthere exist positive constants cand n 0 0 such that 0 cg(n) f(n) for all n n 0g { (g(n)) is set of functions whose growth g(n) { g(n) represents a lower bound on f(n)’s growth {best(g(n)) represents a lower limit for all inputs to f(n) 4 To continue getting our minds around asymptotic analysis, here are a few examples The advantages and dangers of ignoring constants were discussed near the beginning of this section For exam-ple: “This is an order n 2 algorithm” really means that the function describing the behavior of the algorithm is in Θ n 2 Arrange the following functions in ascending order according totheir asymptotic growth with respect to the o-notation and proveyour results with the help of the definitions, the growth hierarchyand the calculation rules of the O-calculus As a result, the n 2 term limits the growth of f (n) Transcript file_download Download Transcript The theta notation defines exact asymptotic behavior and bounds a function from above and below We typically ignore small values of n, since we are usually interested in estimating how slow the program will be on large inputs Asymptotic Efficiency of an algorithm is defined by the order of growth of running time of an algorithm Rank the following functions by order of growth; that is, find an arrangement g1, g2, Suppose that an algorithm took a constant amount of time, regardless of the input size Our results reveal a 5% increase in the adult survivorship is necessary in order to achieve a positive growth We carefully develop the notations which measure the asymptotic growth of functions 995 Ignore lower order forms; GRAEF* Department of Mathematics, … 6n^2 vs 100n+300 • To compare two algorithms with running Both of these algorithms are asymptotically same (order of growth is nLogn) In the previous article – performance analysis – you learned that algorithm executes in steps and each step takes a “ constant time “ Asymptotic Order Notation Burton Rosenberg September 10, 2007 Introduction We carefully develop the notations which measure the asymptotic growth of functions How do we categorize this algorithm’s e ciency in relation to the size of the input? Big-O: in this case what’s the … Definition In asymptotic analysis, our goal is to compare a function fpnqwith some simple function gpnqthat allows us to understand the order of growth of fpnqas n approaches in nity In … In mathematical analysis, asymptotic analysis, also known as asymptotics, is a method of describing limiting behavior The order of growth of running time of an algorithm is a convenient indicator that allows us to compare its performance with alternative algorithms and gives simplified information regarding how Class 1: Exponential (or higher than polynomial) f 5 = n! f 6 = (lgn)! = ( nlglgn) since lgf We look at large enough n such that only the order of growth of t(n) is relevant The asymptotic behavior of a function f(n) (such as f(n)=c*n or f(n)=c*n 2, etc In this paper we study the initial boundary value problem for a class of fourth order strongly damped nonlinear wave equations utt −Δu+Δu−αΔut = f(u) The equation is read, “f of n is theta g The journal Asymptotic Analysis fulfills a twofold function The algorithm’s lower bound is represented by Omega notation Here are the common running times in order of increasing growth rate You have also some allowed operations, for example, The Big O notation, the theta notation and the omega notation are asymptotic notations to measure the order of growth of algorithms when the magnitude of inputs increases 1 of [1, Chapter 5] It is used to study how the running time of an algorithm grows as the value of the input or the unknown variable increases 1 Asymptotic notation 3 The efficiency of a given algorithm can be checked using these notations Three notations are used to calculate the running time complexity of an algorithm: 1 In particular, we present the sufficient conditions for asymptotic stability of tumor-free equilibrium Now, you might wonder what "asymptotic" means Hayman [6]) ) Ω - notation 28 Ω(g(n)) is the set of functions with larger or same order of growth as g(n) 29 Such “asymptotic estimates” are less precise than an exact formula, but they still provide useful information about the Nov 8, 2015 at 7:39 Let us assume that to be C(n) View Exploration_ Asymptotic Notations_ ANALYSIS OF ALGORITHMS (CS_325_400_U2022) Erik Demaine, Prof For example, it is poor usage to say that the function 2 Using the asymptotic symmetry technique we may prove the following Let's start with something easy " Exponential growth is the most-feared growth pattern in computer science; algorithms that grow this way are basically useless for anything but very small problems Asymptotic Analysis 30 none Asymptotic Order of Growth Upper bounds In such asymptotic analysis, we are interested in whether the function scales as exponential (e In this tutorial you will learn about the big-o notation, theta or Omega notation Does Asymptotic Analysis always work? Asymptotic Analysis is not perfect, but that’s the best way available for analyzing algorithms b Solution We also discuss the stability properties of … T(n) grows exponentially 1 Overview In this lecture we discuss the notion of asymptotic analysis and introduce O, Ω, Θ, and o notation GRAEF* Department of Mathematics, … are satisfied, w is an asymptotic value of /(z) Recent Questions in Management - Others We use big-Θ notation to asymptotically bound the growth of a running time to within constant factors above 34 Asymptotic Analysis is not perfect, but that’s the best way available for analyzing algorithms " That is, N Asymptotic Analysis is a measure of an algorithm’s order of growth (input size) , 10n), polynomial (e One way to compare f1 and f4 is to take Log of both functions For example, if the function f (n) = 8n 2 + 4n – 32, then the term 4n – 32 becomes insignificant as n increases Circumstantial evidence suggests that male whale sharks (Rhincodon typus) grow to asymptotic sizes much smaller than those predicted by age and growth studies and consequently, there may be sex-specific size and growth patterns in the species A function f(x) is said to be growing faster than g(x) if Modified 8 years, 4 months ago Order of growth of Log(f1(n)) is (n) and order of growth of Log(f4(n)) is (Logn * Logn) arXivLabs: experimental projects with community collaborators Perturbation methods 9 2 The Denjoy–Carleman–Ahlfors Theorem asserts that if f has n distinct asymptotic values, then the rate of growth of f is at least order n/2, mean type In particular, for any polynomial p(n) with degree k, p(n) 2 O (nk) 4 We then turn to the topic of recurrences, discussing several methods for solving them I have the following functions that I need to rank in increasing order of Big-O complexity: ( log n) 3, 10 n, n log n, n n, n 4 + n 3, ( 2 To solidify these newfound skills, we introduce the language of "big-O" as a means of 7 % increase in cub survivorship is necessary in order to achieve positive growth rate; a figure certainly within reach of concerted conservation efforts ~ Ignores lower-order terms We tested … The efficiency is measured with the help of asymptotic notations We introduce a new method and attempt to analyze some obvious, and some not so obvious functions, in terms of their The rate of growth(aka Order of Growth) describes how the running time increases when the size of input increases g5 (n) = 3 ^ (3 * log (base 3) n) g6 (n) = 10^n By means of the theory of processes on time-dependent spaces, asymptotic a priori estimate and the technique of operator decomposition and the existence and asymptotic regularity of time-dependent attractors are, respectively, … The order of growth of the running time of an algorithm, defined in Chapter 1, gives a simple characterization of the algorithm's efficiency and also allows us to compare the relative performance of alternative algorithms 1) n ⋅ n 2, 3 n, 2 n ⋅ n 3, n! + n, n n For each of the following program fragments, give an analysis of the running time (Big-Oh will $\endgroup$ – alpoge Later on, there are many researches which give better estimates of hd(M){h_{d}(M)} Danielle Hilhorst, Johannes Kampmann, Thanh Nam Nguyen; In order to model growth of biological systems numerous models have been introduced For each of the following stakeholder groups give an example of an eco-efficiency 0000001 * n^2 has the higher asymptotic growth since it will overtake 500000 * … Furthermore, in order to approximate the growth rate of the Miles instability, Carpenter, Gua & Heifetz (Reference Carpenter, Gua and Heifetz 2017) modelled the air–water interface and the critical level as interacting vortex sheets Now we have a way to characterize the running time of binary search in all cases 33 I'm trying to … About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators In this paper, we study the stability properties of a finite difference scheme for a model where the density evolves down pressure gradients and the growth rate depends on the pressure and possibly nutrients Informally, O(g(n)) is the set of all functions with a smaller or same order of growth as g(n) Let u 2 0 be a C2+a solution of However, their cohomology is not well-understood: it can be described using the Kriz model, but its Betti numbers are unknown The asymptotic theory of extreme order statistics provides in some cases exact but in most cases approximate probabilistic models for random quantities when the extremes govern the laws of interest (strength of materials, floods, droughts, air pollution, failure of equipment, effects of Expression 1: (20n 2 + 3n - 4) Expression 2: (n 3 + 100n - 2) Now, as per asymptotic notations, we should just worry about how the function will grow as the value of n (input) will grow, and that will entirely depend on n2 for the … 1 (1978) in order to make it pass through the origin and fit infants adequately, but we did not succeed As above, we assumed that growth was restricted to 27 April– 17 October CSC 611 -Lecture 2 Asymptotic Notations •A … Asymptotic bounds and limits Proposition The age-structured matrix model of the Serengeti Plain cheetahs by Crooks, et al This is a general method for integrals along the real axis of the form 1 and 4 Meaning of Asymptote Answer: I’m assuming by asymptotic growth rate you mean which function has the largest gradient as n → infinity The amount of time, storage, and other resources necessary to assess the efficiency of an algorithm are well known Partition your list into equivalence classes such that functions Big-oh notation: Big-oh is the formal method of expressing the upper bound of an algorithm's running time Topics: Mathematics and Statistics, SERIES, partial differential equations, ORDER, central index, Valiron's inequalities, maximum term, asymptotic growth, monogenic GRAEF* Department of Mathematics, … Section 4: Asymptotic expansions of integrals 4 What is the asymptotic growth rate of the product of divisor function up to n [duplicate] Ask Question Asked 6 years, Elliptic Equations with Critical Sobolev Growth LUIS A rìMultiplying by g(n) yields 1/2 c · g(n) ≤ f(n) ≤ 3/2 c · g(n) for all n ≥ n 0 The function f(n) is said to be "asymptotically … About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators Using linear algebra and robustness techniques, we analyze this previously published model and the robustness of its conclusion For instance, let’s see this code which returns the sum of a list Xu's solution may be well applied to the case that the flow velocity, U_∞, has the same order of magnitude as the growth speed of dendrite tip, U Instructors: Prof 989 P Problem 3P: Ordering by asymptotic growth ratesa Asymptotic analysis of an algorithm refers to define the mathematical bounds of its run time performance The methodology has the applications across science By Ahmed Tarek We consider a diffuse-interface tumor-growth model which has the form of a phase-field system In particular, we prove that for any n ‚ 1 and 0 < fi • n, there exists a subharmonic function u in the Rn+1+ satisfying the growth con-dition of order fi: u(x) • x¡fin+1 for 0 < xn+1 < 1, such that the Hausdorfi dimension of the asymptotic set S ‚6 , g30 of the functions satisfying g1= Ω(g2), g2 =Ω(g3), However, it does … Taking the first three rules collectively, you can ignore all constants and all lower-order terms to determine the asymptotic growth rate for any cost function The growth patterns above have been listed in order of increasing "size T(n) is Ω(f(n)) if there exist constants c > 0 and n0 ≥ 0 such that for all n ≥ n0 we have T(n) ≥ c · f(n) g The letter O is used because the rate of growth of a … World's Best PowerPoint Templates - CrystalGraphics offers more PowerPoint templates than anyone else in the world, with over 4 million to choose from This order of growth, in turn, is determined by the basic operation count To conclude asymptotic analysis is a means of measuring the performance of an algorithm based on its input Debabala Swain “ 2 10 ” is the constant time so, it is … Asymptotic Notation 16 Common Rates of Growth In order for us to compare the efficiency of algorithms, we need to know some common growth rates, and how they compare to one another Answer: The study of change in … Asymptotic Equivalence 漸近的同等性 | アカデミックライティングで使える英語フレーズと例文集 Solving this equation in the same way as (1 Menu We introduce another smoothness condition which generalizes the concepts of Very regular growth' and 'perfectly regular growth' in the sense that T(r,/) is compared not only with rp (0^p<l/2) but also with rp(r): there exist a proxi-mate order p(r) (p(r)-^p) and two constants c l9 c 2 such that More precisely, from the previous specification one gets that the first coefficients of T ( z) are Asymptotic Notation is a way of comparing function that ignores constant factors and small input sizes g3 (n) = 2n log (base 2) n Function Order of Growth: (20 points) List Asymptotic analysis is input bound, which means that we assume that the run time of the algorithms depends entirely upon the size of the Input to the algorithm rìBy definition of the limit, for any ε > 0, there exists n 0 such that for all n ≥ n 0 Zabolotskii, “Asymptotic properties of meromorphic and δ-subharmonic functions of slow growth,” Candidate's Dissertation, Physicomathematical Sciences, … [Category: Asymptotic Order of Growth] Consider the following eight functions of n: 2n 100 n nyn n log(n) (1 However, one can derive approximate formulas, or “asymptotic estimates” each function grows about the stationary points to second order and performing the integral yields , log 2n), for example If the array size doubles, so does the run-time n) Asymptotic Equivalence 漸近的同等性 | アカデミックライティングで使える英語フレーズと例文集 We investigate a model equation in the crystal growth, which is described by a level-set mean curvature flow equation with driving and source terms In theoretical analysis of algorithms it is common to estimate their complexity in the asymptotic sense, i Lower-order terms and constants • Lower order terms of a function do not matter since lower-order terms are dominated by the higher order term That is, T ( n) f ( n) is only required for all n n0, for some n0 such that for all $n ≥ N$, $c f (n) ≤ g (n) ≤ d f (n)$ Show transcribed image text Order the following functions by asymptotic growth rate, from smallest to largest: n, n lg n, 500n, lg n, 2″, 2n2 996 Biometrics, December 1988 17 1 o 16 1 - ~86 -c 81 0 A good rule of thumb is: the slower the asymptotic growth rate, the better the algorithm (although this is often not the … (Asymptotic) Growth Groups¶ This module provides support for (asymptotic) growth groups The following are common rates of growth 2 Standard notations and common functions Chap 3 Problems Chap 3 Problems 3-1 Asymptotic behavior of polynomials 3-2 Relative asymptotic growths 3-3 Ordering by asymptotic growth rates Asymptotic Growth Rates We have talked about comparing data structure implementations | using either an empirical or analytical approach For example, we know that p n “is asymptotic to” nlogn, i pdf from CS 325 at Sukkur Institute of Business Administration, … Asymptotic Equivalence 漸近的同等性 | アカデミックライティングで使える英語フレーズと例文集 CSE-245 Algorithms Asymptotic Notation f Analyzing Algorithms • Predict the amount of resources required: • memory: how much space is needed? • computational time: how fast the algorithm runs? • FACT: running time grows with the size of the input • Input size (number of elements in the input) – Size of an array, polynomial degree Its solution furnishes displacement potentials which are used to evaluate explicitly the The growth of any function depends on how much the running time is increasing with the increase in … Asymptotic Analysis and Recurrences 2 Solution to Problem 3 Finally, … Suppose (Mn,g){(M^{n},g)} is a Riemannian manifold with nonnegative Ricci curvature, and let hd(M){h_{d}(M)} be the dimension of the space of harmonic functions with polynomial growth of growth order at most d Indeed, Taylor series are a perfect tool for understanding limits, both large and small, making sense of … The order of growth for varying for varying input size of n is as given below 1 CAFFARELLI Institute of Advanced Study BASILIS GIDAS Brown University A Taylor series may or may not converge, depending on its limiting (or "asymptotic") properties Here the matrices , are diagonal, may be taken as the identity matrix, and may be taken to equal ; denotes the matrix exponential and For a smooth solution with compactly supported initial datum, its velocity support is shown to grow like $(t+1)^{\\frac{2}{15}}\\ln^{\\frac{8}{15}}(t+2)$ (5 pts) Take the following list of functions and arrange them in ascending order of growth rate 3a: Order by asymptotic growth rates Bang Ye Wu CSIE, Chung Cheng University, Taiwan September 24, 2008 First we simplify some of them, and classify them into exponential, poly-nomial, and poly-log functions GRAEF* Department of Mathematics, … Let f denote a function, meromorphic in C 32 Asymptotic notations are mathematical tools to represent time complexity of algorithms for asymptotic analysis Viewed 476 times 2 1 $\begingroup$ How can I calculate the Asymptotic Rate of growth of a function, for instance like: Simplification We are only interested in the growth rate as an “order of magnitude” Lower bounds The exercises establish several facts about commonly encoun-tered order sets, and relationships among … As pointed out in the previous section, the efficiency analysis framework con-centrates on the order of growth of an algorithm’s basic operation count as the principal indicator of the algorithm’s efficiency Given the following functions i need to arrange them in increasing order of growth a) $2^{2^n}$ b) $2^{n^2} We already noted that while asymptotic categories such as Θ(n 2) are sets, we usually use "=" instead of "∈" and write (for example) f(n) = Θ(n 2) to indicate that f is in this set (n)) Graph of Growth Functions UMBC CMSC 341 Asymptotic Analysis 28 logarithmic linear quadratic n-log-n cubic exponential Order the following functions by asymptotic order of growth (lowest to highest) | 2n | 3log n | 27+1 |10lo82 "| 10l0810 7" |2100 |n9º|n * 2" Expert Solution 3 A typcial exmaple is any algorithm that makes use of two loops, for instance an insertion sort By means of the Fourier transform we … View metadata, citation and similar papers at core Notably, Gromov used asymptotic cones in his proof that finitely generated groups of polynomial growth are virtually nilpotent Asymptotic normality and consistency of a two-stage generalized least squares estimator in the growth curve model It, however, will not give a good approximation when U_∞ >> U This paper addresses the question of how to invest in a robust growth-optimal way in a market where the instantaneous expected return of the underlying process is unknown 2) Following is another way to compare f1 and f4 We do not repeat the entire proof … •Order of growth –The leading term of a formula –Expresses the behavior of a function toward infinity –Ignores machine depending constants and looks at growth of T(n) as n ® µ CSC 611 -Lecture 2 Asymptotic Order Notation Burton Rosenberg September 10, 2007 Introduction In computer science in the analysis of algorithms, considering the performance of algorithms when applied to very large input datasets Winner of the Standing Ovation Award for “Best PowerPoint Templates” from Presentations Magazine That is, if function g(n) immediately follows function f(n) in your list, then it should be the case that f(n) is O(g(n)) Ignoring lower-order terms is reasonable when performing an asymptotic analysis • Compare functions in the limit, that is, asymptotically! 2 The question of when a deficient value of f, in the sense of Nevanlinna, is an asymptotic value has recently received some attention (see e Big O is a member of a family of notations invented by Paul Bachmann, Edmund Landau, and others, collectively called Bachmann–Landau notation or asymptotic notation When r < 0 and a reduction in the growth rate per capita is present, the growth curve is asymptotic to zero leading to population extinction Interestingly, a new type of nonlinear phenomenon in terms of asymptotic speed of solutions appears which is very sensitive to the shapes of source … th prime It is the measure of the longest amount of time Indeed, Taylor series are a perfect tool for understanding limits, both large and small, making sense of such methods as that of l'Hopital The reason is the order of growth of Binary Search with respect to the input size logarithmic but the order of growth of Linear Search is linear Our results show only an 7 is O(4 All functions with the leading term n2 belong to O Continuous time scales are assumed and a complete steady state and stability analysis is presented Laplace Pf Answer (1 of 8): See, an algorithm’s efficiency is determined by the order of growth of that algorithm Rank the following functions by order of growth; that is, find an arrangement Lecture … Asymptotic expansions of integrals 29 Chapter 4 In order to analyze these type of equations one can use the methodology of Flajolet and Sedgewick (Analytic Combinatorics Book) Functions in asymptotic notation food additives, etc [28, 1] 1)" n log(n) 4n n3 20145 n logg (n) (n+) Please do asap Analysis Of Algorithms Posted 3 days ago The Y-axis represents the ratio of asymptotic crack length with the thickness ( C S / t ), i ( log n) 3 < 10 n < n log n < n n < n 4 + n 3 < ( 2 Solutions for Chapter 3 To compare and rank such orders of growth, computer scientists use three notations: O (big oh), (big omega), and (big theta) Henceforth, we will describe the running time of an algorithm only in the asymptotical (i Landau's symbol comes from the name of the German number theoretician Edmund Landau who invented the notation Notes on Scale Kurt Schmidt (Skipjack Solutions) Program Growth, Asymptotic Behavior November 30, … The often-rediscovered Watson’s lemma[1] gives an asymptotic expansion for certain Laplace transforms, valid in half-planes in C Our purpose in this article is to study the asymptotic behavior of undamped evolution equations with fading memory on time-dependent spaces ) , linear search Skip Navigation However, for large inputs where the size of inputs grows without bound become the main concern for the increase in running time of an algorithm Arrange in increasing order of asymptotic complexity there exists some positive real constants$c$and$d$This module provides support for (asymptotic) growth groups In fact, this method produces exact solutions in cases not treated here; see [1, Chapter 4] 1 In addition, this gives us justi cation for ignoring constant coe cients If f(N) ~ c g(N) for some constant c > 0, then the order of growth of f(N) is g(N) An Asymptotic Notations is the notation which represent the complexity of an algorithm 2), we get the nonzero solutions y= 1 1 2 "1=2 + O("): The corresponding solutions for xare x= 1 "1=2 1 2 + O "1=2 The dominant balance argument illustrated here is useful in many perturbation The asymptotic complexity is a function f ( n) that forms an upper bound for T ( n) for large n An algorithm may not have the same performance for different types of inputs In order to get a more precise result we must understand the radial singular solutions of (1 We obtain a precise norm estimation of semigroup generated by the operator corresponding to the system in question Asymptotic Notation Running time of an algorithm, order of growth Worst case Running time of an algorith increases with the size of the input in the limit as the size of the input increases without bound , if there's no input to the algorithm, it is concluded to work Asymptotic analysis Having characterised an asymptotic limit of the perturbations, we compare it to its numerical counterpart and to the time-dependent solution profiles in order to analytically obtain a condition for instability Its estimator is usually asymptotically normal, but it is non-standard when (X t) t … Asymptotic notations describe the function’s limiting behavior Charles Leiserson Singular perturbation problems 15 Chapter 3 We say that the running time is "big-O of " or just "O of Asymptotic Notation in Equations A good rule of thumb is: the slower the asymptotic growth rate, the better the algorithm (although this is often not the whole story) A notation for “the order of” We’d like to measure the efficiency of an algorithm • Determine mathematically the resources needed There is no such a computer which we can refer to as a standard to measure computing time We introduce “asymptotic” notation • An asymptotically superior algorithm is often Among remaining, n^(3/2) is next Program Growth, Asymptotic Behavior November 30, 2020 25/28 Rogaway Asymptotic Growth Rates Similarly to the previous matching of asymptotics, this is done by comparing the limit of ( A The adverse effect caused by unkno … asymptotic population growth rate is most sensitive to adult survivorship Give an example of a single nonnegative function • Order of growth – The leading term of a formula – Expresses the behavior of a function toward infinity 3 Asymptotic Notations • A way to describe behavior of functions in the limit – How we indicate running times of algorithms – Describe the running time of an algorithm as n grows to ∞ Big Theta Notation 2) is the relative stress factor (X) and it depends upon the magnitude of the range of membrane and bending stress variation for a cyclic condition as defined in Eq I've been looking through several examples online but I have no idea how to do this, it just seems … 3-3 Ordering by asymptotic growth rates We assume acquaintance with the standard notation of the Nevanlinna theory ([[5] Chapter I) which we use without further mention With the increase in the input size, the performance will change if it is a linear, a quadratic or an exponential function Enable new issue alert f (n) = Θ (g (n)) iff there are three positive constants c1, c2 and n0 such that 0 Asymptotic Rate of Growth The optimal investment strategy is identified using a generalized version of the principal eigenfunction for an elliptic second-order differential operator, which depends on the covariance structure of the … long-term dynamic data on height growth; however, if diameter growth is nonasymptotic while height is asymp-totic, it should also be possible to obtain an estimate of average asymptotic height on the basis of static height-diameter relationships For example, say there are two sorting algorithms that take 1000nLogn and 2nLogn time respectively on a machine Superlinear scaling results in crises called ‘singularities’, where population and energy demand tend to infinity in a finite amount of time, which must be avoided by ever more frequent ‘resets’ … Q8 A fractional-order tumor-immune interaction model with immunotherapy is proposed and examined What is Asymptotic Analysis then? Asymptotic Analysis is a way to determine the order of growth for this particular algorithm for a particular case This notation provides an upper bound on a function which ensures that the function never grows faster than the upper bound The asymptotic running time of an algorithm is defined in terms of functions ) refers to the growth of f(n) as n gets large lim n→∞ n→∞ f(n)/g(n) = 1 ⇒ f ∼ g lim n→∞ f(n)/g(n)$= ∞ ⇒ f = O(g) lim n→∞ f(n)/g(n) = 0 ⇒ f = o(g) lim n→∞ n→∞ f(n)/g(n) = ∞ ⇒ f = ω(g) Therefore, skill with limits can be helpful in working out asymptotic relationships A fast DCT algorithm can be derived by transposing the signal-flow Complexity measures instead focus on asymptotic growth, Growth factors act at different levels in the differentiation process, and we consider their action on the mortality rate (apoptosis) of the proliferating cell population Asymptotic vs convergent series 21 3 2/15 Asymptotic Notations • The efficiency analysis framework concentrates on the order of growth of an algorithm’sbasic operation count as the principal indicator of the algorithm’s • To compare and rank such orders of growth, computer scientists use three notations:(big oh), (big omega), and (big theta)efficiency of f/g into facts about the asymptotic relationship between f and g Basically, it tells you how fast a function grows or declines Chapter 1 file_download Download Video In this paper, we … The asymptotic probability density functions of coalescence times in the history for two parameter settings n lglgn -32m)g(2'Ign Ig (n3) smallest largest Justify your answer mathematically by showing values of c and no for each pair of functions that are adjacent in your ordering ; Question: 1 There are three different notations: big O, big Theta (Θ), and big Omega (Ω) In this paper, we are concerned with the following fractional Choquard equation with critical growth: where s ∈ ( 0, 1), N > 2 s, μ ∈ ( 0, N), 2 s ∗ = 2 N N − 2 s is the fractional critical exponent, V is a steep well potential, F ( t) = ∫ 0 t f ( s) d s SVL A is the population mean asymptotic length, k is a growth rate constant and t is age in growth days This asymptotic growth pattern in the larval stage resulted in the narrow ranges in TLs in spite of the wide range of ages of the larvae caught by boat seiners in the coastal waters ; Use Big-Oh notation to describe asymptotic running time of a program when you are given the program code (or its running time as a function of input size under the … asymptotic behavior of functions Also called sub … Chapter 3: Asymptotic Notation - Orders of Growth (3) 3 In general, just the order of the asymptotic complexity is of interest, i , big-O) form, which is also called the The first asymptotic, uniformly valid expansion solution was obtained by Xu (1994) in the limit of the Prandtl number Pr → ∞ A Generalized Set Theoretic Approach for Time and Space Complexity Analysis of Algorithms and Functions Asymptotic Equivalence 漸近的同等性 | アカデミックライティングで使える英語フレーズと例文集 1 GRAEF* Department of Mathematics, … The notation describes asymptotic tight bounds 5 Classifying Functions by Their Asymptotic Growth Rates 55 The terminology commonly used in talking about the order sets is imprecise Want to see the full answer? Check out a sample Q&A here We study the relation between the growth of a subharmonic func-tion in the half space Rn+1+ and the size of its asymptotic set lim (n-> infinite) f(n) / g(n) = infinite _OR_ lim (n-> infinite) g(n) / f(n) = zero Direct Way to calculate About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators and some nonnegative integer $N$ The Cartesian product of growth groups is again a growth group Furthermore we give the asymptotic behaviour of … Assumption 2 Common order-of-growth classiÞcations 42 int count = 0; for (int i = 0; i < N; i++) Engineering Computer Science Data Structures and Algorithms in Java Asymptotic growth rate: According to asymptotic growth rate, the given functions are ordered as follows: 2 10 ,2 logn , 3n+100logn, 4n, nlogn, 4n logn+2n, n 2 +10n, n 3 , and 2 n Explanation: The above is ordered from least to greatest based on the asymptotic functions 8 Download The local and global asymptotic stability of some equilibrium points are investigated After reading this chapter and engaging in the embedded activities and reflections, you should be able to: Understand and appreciate why we do asymptotic analysis using Big-Oh notation Regular perturbation problems 9 2 , to estimate the … Furthermore, in order to approximate the growth rate of the Miles instability, Carpenter, Gua & Heifetz (Reference Carpenter, Gua and Heifetz 2017) modelled the air–water interface and the critical level as interacting vortex sheets " We use big-O notation for asymptotic upper bounds, since it bounds the growth of the running time from above for large enough input sizes g 2 (n) = n 4/3 Fuzzy-logic systems and a separation principle are utilized to relax growth assumptions imposed on unknown nonlinearities Partition your list into equivalence classes such that f(n) and g(n) are in the same class if and only if Asymptotic analysis is input bound i \frac{d}{dn} (4n*ln2n) = 4 + 4ln2n 2 g 1 (n) = 2 n The efficiency is measured using asymptotic notations We establish the well-posedness of solutions and study the asymptotic speed There, take a look at the … Furthermore, in order to approximate the growth rate of the Miles instability, Carpenter, Gua & Heifetz (Reference Carpenter, Gua and Heifetz 2017) modelled the air–water interface and the critical level as interacting vortex sheets In order to be perfectly precise, we should measure the time of an algorithm in the form of n 1 c 1 + n 2 c 2 + n 3 c 3 + ::: 005 0000001 * n^2 is the better running time, and it is for small input sizes but asymptotic growth is about what happens when you increase n arbitrarily List the following functions in non-descending order of asymptotic growth rate 1) n ⋅ n 2 < 2 n ⋅ n 3 < 3 n < n! + n < n n Now, to compare, contrast and rank the order of growth of an algorithm, we use three no The paper continues the research into asymptotic behaviour of solutions to equations with singularit carried out in a series of articles [2], [5], [6] and so on [5] suggests the asymptotic population growth rate is most sensitive to adult If your answer is that you need more information, such as the resources required to implement/run each algorithm, their tradeoffs other than their runtime, etc, you are a true engineer! If you told your intern the runtime of these algorithms can all be the same, then you truly In the case of Theorem 2, since we found the invariant measure involved, therefore we are able to compute the asymptotic growth rate of the f n de ned in (6) A key point using the growth curve model to fit data is determining the degree of polynomial profile form The existence, uniqueness, and nonnegativity of the solutions are proved Introduction In this paper, we discuss the use of a new approach in dealing with complexity rankings of functions in an Algorithm Analysis Course It provides us with an asymptotic upper bound for the growth rate of the runtime of an algorithm It is likewise inappropriate to include constant factors and lower order terms in the big-Oh notation T ( z) = z + z 3 + z 5 + 2 z 7 + 4 z 9 + 8 z 11 + 17 z 13 + 39 z 15 + 89 z 17 + 211 z 19 After proving the existence and uniqueness of viscosity solutions for the eigenvalue problem, we perform an asymptotic expansion in terms of small correlations and obtain semi-analytical approximations of the free boundaries and the optimal growth rate Related Papers We … About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators An asymptotic methodology is used to reduce the problem to one of the Riemann-Hilbert type Ideally, all three directions should be cross-tested, leading to a comprehensive picture of transient nucleation rìThus, f(n) is Θ(g(n)) by definition, with c Overview I Study a way to describe the growth of functions in the limit { asymptotic e ciency I Focus on what’s important (leading factor) by abstracting lower-order terms and constant factors I Indicate running times of algorithms I A way to compare \sizes" of … What is asymptotic growth? refers to the growth of f(n) as n gets large We consider asymptotic behaviors of classical solutions and weak solutions to the three-dimensional Vlasov--Poisson system in the plasma physics case Tight bounds This is the goal of the next several slides This paper shows that the lower order drift term alone determines asymptotic grain growth behavior Asymptotic Equivalence 漸近的同等性 | アカデミックライティングで使える英語フレーズと例文集 Asymptotic Notations and Case Analysis The actual value of the running time of an algorithm depends, many times, on the type of the input Ex Previous studies show that city metrics having to do with growth, productivity and overall energy consumption scale superlinearly, attributing this to the social nature of cities A good rule of thumb is: the slower the asymptotic growth rate, the better the algorithm (although this is often not the … Models of computation: word RAM Word RAM The efficiency of an algorithm is given by the order of growth in run time with respect to input size Asymptotic analysis is based on the idea that as the problem size grows, the complexity will eventually settle down to a simple proportionality to some known function Rank the following functions by order of growth; g1 (n) = n The order on the product is determined as follows: Cartesian factors with respect to the same variable are ordered lexicographically If f(n) = n2 + 3n, then as n becomes very large, the term 3n becomes insignificant compared to n2 Q: 1 rìChoose ε = ½ c > 0 Here’s the plot of running time for Linear Search with respect to input size We consider a differential system of neutral type with distributed delay Asymptotic series 21 3 For example, let hbe a smooth function on (0;+1) all whose derivatives are of polynomial growth, and expressible for small x>0 as h(x) = x g(x) About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How YouTube works Test new features Press Copyright Contact us Creators Using asymptotic analysis, we can very well conclude the best case, average case, and worst case scenario of an algorithm With running times In Asymptotic analysis it is considered that an algorithm 'A1' is better than algorithm 'A2' if the order of growth of the running time of the 'A1' is lower than that of 'A2' n 00 0 10 20 0 10 20 In this paper we study a nonlinear free boundary problem for the growth of radially symmetric tumor with a necrotic core Motivated by a general theory of finite asymptotic expansions in the real domain for functions f of one real variable, a theory developed in a previous series of papers, we present a detailed survey on the classes of higher-order asymptotically-varying functions where “asymptotically” stands for one of the adverbs “regularly, smoothly, rapidly, exponentially” 2 $\begingroup$ @GHfromMO: Oh Asymptotic Growth Rates – “Big-O” (upper bound) f(n) = O(g(n)) [f grows at the same rate or slower than g] iff: There exists positive constants c and n 0 such that f(n) ≤c g(n) for all n ≥n 0 f is bound above by g ¾Note: Big-O does not imply a tight … Asymptotic Notation 1 Growth of Functions and Aymptotic Notation • When we study algorithms, we are interested in characterizing them according to their efficiency It is a technique of representing limiting behavior arXivLabs is a framework that allows collaborators to develop and share … Modeling the dynamics of tumor growth has recently become an important issue in applied mathematics (see, e However, their minimal model does not yield the dependence of the maximum growth rate on the physical parameters A long-standing problem asks whether this conclusion holds for entire functions having n distinct asymptotic (entire) functions, each of growth at most order 1/2, minimal type , the The letter O was chosen by Bachmann to … Asymptotic analysis of an algorithm refers to defining the mathematical boundation/framing of its run-time performance In [5], asymptotic expansions for solutions to equations In order to further characterise the solution y 1 of we we must match the asymptotic at with the bulk asymptotic in a transition region where both approximation are feasible, namely for with We can say that the running time of binary Laplace’s Method In the last section we derived Stirling’s approximation by an approach known that is known as ‘Laplace’s Method’ However, we have now found a completely different seven-parameter asymptotic curve that passes This idea is incorporated in the Big Oh,'' Big Omega,'' and … [Category: Asymptotic Order of Growth] Consider the following eight functions of n: 2n 100 n nyn n log(n) (1 f(n) is O(g(n)), if for some real constants c … according to the authors' best knowledge, asymptotic stability (even without funnel constraints) for 2nd-order systems has not been guaranteed in the related literature under these assumptions (i Growth of Functions and Aymptotic Notation The main purpose of this and other so-called asymptotic notations is to describe the behavior of mathematical functions, by comparing their “orders of growth” ac 30 Under the action of growth factors, proliferating and nonproliferating hematopoietic stem cells differentiate and divide, so as to produce blood cells The analysis that is conducted for large values of input size is called as asymptotic analysis Let n be the size of input to an algorithm, and k some constant Suppose you have an array of n three-digit integers, and that the integers are not necessarily stored in a meaningful order already Such groups are equipped with a partial order: the elements can be seen as functions, and the behavior as their argument (or arguments) gets large (tend to $$\infty$$) is compared ISSN 0921-7134 (P) ISSN 1875-8576 (E) Impact Factor 2021: 0 This is also referred to as the asymptotic running time For a weak solution with bounded initial velocity … Its order of growth is proportional to n² • Constants (multiplied by highest order term) do not matter, since they do not affect the asymptotic growth rate • All logarithms with base b >1 belong to (lg n) since Cutler/Head 1 This is the approach taken in the present study See Solution Putting asymptotic notation in equations lets us do shorthand manipulations during analysis Models of computation: word RAM Word RAM Growth groups are used for the calculations done in the asymptotic ring 100% of your contribution will fund improvements and new initiatives to benefit arXiv's global scientific community O(500n) = It may seem like 0 Thus, $\Theta (f (n))$ is the set of complexity functions $g (n)$ for which , p n ∼ nlogn, and, somewhat more precisely that p n = nlogn+O(n) For small inputs or large enough inputs for the order of growth of execution time, we can find out the more efficient algorithm among all other algorithms with the help of asymptotic notation Based on the stability results, we prove the scheme to be asymptotic preserving (AP) in the incompressible limit ((√n)^(5))*2^(n)3 log n√( n log (√n))((n!) / ((n − 3)!)) * 2^n20^(n^20) Expert Answer Answer to Arrange the following functions in ascending … 2 Big O notation is an asymptotic notation that measures the performance of an algorithm by simply providing the order of growth of the function T(n) is O(f(n)) if there exist constants c > 0 and n0 ≥ 0 such that for all n ≥ n0 we have T(n) ≤ c · f(n) The first and in fact still best such result was obtained in , where the authors showed that there exist constants 0 < c 1 < c 2 such that for all n: (c 1 n) 2 n ≤ TC n ≤ (c 2 n) 2 n On the other hand, the asymptotic behavior of the long-term growth rate ρ deserves more attention Although (10) and (11) only contain the leading order terms of the asymptotics, and the asymptotic decomposition is carried out by using the inverse powers of m, i Colding and Minicozzi proved that hd(M){h_{d}(M)} is finite Asymptotic Equivalence 漸近的同等性 | アカデミックライティングで使える英語フレーズと例文集 As an example construct we asymptotic solutions of Laplace’s equation on a manifold with a second order caspidal singularity The outline of this paper is as Little oh (o): O (n) O(n) runtime The above condition allows for non-smooth In this paper, we study the asymptotic behavior of solutions of the first order delay di¤erential equations of unstable type First, let us recall the notion of a limit Slow growth and therefore long duration of the metamorphosing stage could be influential in determining the cumulative total mortality in the early life stages of of f/g into facts about the asymptotic relationship between f and g Function Order of Growth: (20 points) List the 5 functions below in nondecreasing asymptotic order of growth E O (2^n) O(2n) runtime We discuss asymptotic equality , asymptotic tightness , asymptotic upper bounds O and o, and asymptotic lower bounds and ! The gradient of the objective function fhas at most linear growth a T(n) is Θ(f(n)) if T(n) is both O(f(n)) and Ω(f(n)) 1 (Linear growth) asymptotic growth rate, one has to know the analytical closed-form of the invariant measure (see Section 2 below) associated with the random recurrence It is pronounced "Big Oh of 2 to the n In order to view the full content, Eremenko FUNCTIONS Ask Question Asked 9 years, 7 months ago Read and learn for free about the following article: Asymptotic notation • We are usually interesting in the order of growth of the running time of an algorithm, not in the exact running time The growth curve model is a useful tool for studying the growth problems, repeated measurements and longitudinal data Big Theta: 2 • Hint: use rate of growth onder de 31° +70° 2** 2 lotn6n + 50 loy n 25 2n los n + 3M Show transcribed image text 1 Polynomial Growth • Many algorithms that we encounter will have polynomial growth • In a log-log chart, the asymptotic slope of the line corresponds to … Abstract ECS 20 – Fall 2021 – P The X-axis of the LBB master curve (Fig 3 + 8 The asymptotic cone therefore captures much of the large scale geometry of the metric space e I approached this by first finding the gradient of each function 1 … Imp points Regarding Asymptotic Analysis This causes GrowthGroup ('x^ZZ * log (x)^ZZ') and GrowthGroup ('log (x)^ZZ * x^ZZ') to | 2022-08-09T07:30:22 | {
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https://docs.sciml.ai/dev/modules/PolyChaos/multiple_discretization/ | Multiple Discretization
This tutorial shows how to compute recurrence coefficients for non-trivial weight functions, and how they are being used for quadrature. The method we use is called multiple discretization, and follows W. Gautschi's book "Orthogonal Polynomials: Computation and Approximation", specifically Section 2.2.4, and Example 2.38.
Suppose we have the weight function
$$$\forall t \in [-1,1], \gamma \in [0,1]: \quad w(t;\gamma) = \gamma + (1-\gamma) \frac{1}{\sqrt{1-t^2}},$$$
and we would like to solve
$$$\int_{-1}^{1} f(t) w(t;c) \mathrm{d}t = \sum_{\nu=1}^{N} f(\tau_\nu) w_\nu$$$
by some quadrature rule. We will see that ad-hoc quadrature rules will fail to solve the integral even for the simplest choice $f \equiv 1$. However, finding the recurrence coefficients of the underlying orthogonal polynomials, and then finding the quadrature rule will do just fine.
Let us first try to solve the integral for $f \equiv 1$ by Féjer's rule.
using PolyChaos, LinearAlgebra
γ = 0.5;
int_exact = 1 + pi / 2; # exact value of the integral
function my_w(t, γ)
γ + (1 - γ) * 1 / sqrt(1 - t^2)
end
N = 1000;
nodes, weights = fejer(N);
int_fejer = dot(weights, my_w.(nodes, γ))
print("Fejer error:\t$(abs(int_exact - int_fejer))\twith$N nodes")
Fejer error: 0.00034489625618583375 with 1000 nodes
Clearly, that is not satisfying. Well, the term $\gamma$ of the weight $w$ makes us think of Gauss-Legendre integration, so let's try it instead.
function quad_gaussleg(N, γ)
a, b = rm_legendre(N)
nodes, weights = golubwelsch(a, b)
end
int_gaussleg = dot(weights, γ .+ (1-γ)/sqrt.(1 .- nodes.^2))
print("Gauss-Legendre error:\t$(abs(int_exact-int_gaussleg))\twith$N nodes")
Gauss-Legendre error: 1.5692263158654436 with 1000 nodes
Even worse! Well, we can factor out $\frac{1}{\sqrt{1-t^2}}$, making the integral amenable to a Gauss-Chebyshev rule. So, let's give it anothery try.
function quad_gausscheb(N, γ)
a, b = rm_chebyshev1(N)
nodes, weights = golubwelsch(a, b)
end
int_gausscheb = dot(weights, γ .+ (1-γ)*sqrt.(1 .- nodes.^2))
# int=sum(xw(:,2).*(1+sqrt(1-xw(:,1).^2)))
print("Gauss-Chebyshev error:\t$(abs(int_exact - int_gausscheb))\twith$(length(n)) nodes")
Gauss-Chebyshev error: 4.112336369210823e-7 with 7 nodes
Okay, that's better, but it took us a lot of nodes to get this result. Is there a different way? Indeed, there is. As we have noticed, the weight $w$ has a lot in common with Gauss-Legendre and Gauss-Chebyshev. We can decompose the integral as follows
$$$\int_{-1}^1 f(t) w(t) \mathrm{d}t = \sum_{i=1}^{m} \int_{-1}^{1} f(t) w_i(t) \mathrm{d} t,$$$
with
\begin{align*} w_1(t) &= \gamma \\ w_2(t) &= (1-\gamma) \frac{1}{\sqrt{1-t^2}}. \end{align*}
To the weight $w_1$ we can apply Gauss-Legendre quadrature, to the weight $w_2$ we can apply Gauss-Chebyshev quadrature (with tiny modifications). This discretization of the measure can be used in our favor. The function mcdiscretization() takes the $m$ discretization rules as an input
function quad_gaussleg_mod(N, γ)
nodes, weights = quad_gaussleg(N + 1, γ)
nodes, γ*weights
end
nodes, weights = quad_gausscheb(N + 1,γ)
return nodes, (1-γ)*weights
end
N = 8
a, b = mcdiscretization(N, [n -> quad_gaussleg_mod(n, γ); n -> quad_gausscheb_mod(n, γ)])
nodes, weights = golubwelsch(a, b)
int_mc = sum(w)
print("Discretization error:\t$(abs(int_exact-int_mc))\twith$(length(n)) nodes")
Discretization error: 8.881784197001252e-16 with 7 nodes
Et voilà, no error with fewer nodes. (For this example, we'd need in fact just a single node.)
The function mcdiscretization() is able to construct the recurrence coefficients of the orthogonal polynomials relative to the weight $w$. Let's inspect the values of the recurrence coefficients a little more. For $\gamma = 0$, we are in the world of Chebyshev polynomials, for $\gamma = 1$, we enter the realm of Legendre polynomials. And in between? That's exactly where the weight $w$ comes in: it can be thought of as an interpolatory weight, interpolating Legendre polynomials and Chebyshev polynomials. Let's verify this by plotting the recurrence coefficients for several values of $\gamma$.
Γ = 0:0.1:1;
ab = [ mcdiscretization(N, [n -> quad_gaussleg_mod(n, gam); n -> quad_gausscheb_mod(n, gam)]) for gam in Γ ];
bb = hcat([ab[i][2] for i in 1:length(Γ)]...);
b_leg = rm_legendre(N)[2];
b_cheb = rm_chebyshev1(N)[2]
bb[:,1]-b_cheb
8-element Vector{Float64}:
0.0
-2.7755575615628914e-16
2.220446049250313e-16
-3.608224830031759e-16
1.6653345369377348e-16
-2.220446049250313e-16
3.885780586188048e-16
7.216449660063518e-16
bb[:,end] - b_leg
8-element Vector{Float64}:
0.0
1.1102230246251565e-16
-5.551115123125783e-17
2.7755575615628914e-16
1.1102230246251565e-16
1.1102230246251565e-16
1.6653345369377348e-16
-5.551115123125783e-17
Let's plot these values to get a better feeling.
using Plots
plot(Γ, bb', yaxis=:log10, w=3, legend=false)
zs, os = zeros(N), ones(N)
scatter!(zs, b_cheb, marker=:x)
scatter!(os, b_leg, marker=:circle)
xlabel!("Gamma")
ylabel!("Beta")
The crosses denote the values of the β recursion coefficients for Chebyshev polynomials; the circles the β recursion coefficients for Legendre polynomials. The interpolating line in between stands for the β recursion coefficients of $w(t; \gamma)$. | 2022-07-06T15:54:09 | {
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https://math.stackexchange.com/questions/2561139/why-doesnt-101-mod-13-10 | # Why doesn't -101 mod 13 = 10?
Following the formula found here (r = a - (a/b)b) I have been successfully finding the remainder for several problems, but when I try to solve -101 mod 13, I get the answer 10 even though it should apparently be 3.
101/13 = 7.769, ignore everything after the .
13 * 7 = 91
101 - 91 = 10
Am I missing something obvious here?
• The positive remainder of (for example) $-15$ when divided by $13$ is $11$, because the multiple of $13$ less than or equal to $-15$ is $-26$. – Joffan Dec 11 '17 at 5:57
• The obvious thing you missed is the negative sign. – Matthew Leingang Dec 11 '17 at 5:59
$$-101=-13\cdot7-10=-8\cdot13+3$$
So, the minimum positive remainder is $3$
Well in modular arithmetic if $a \equiv b \pmod m$ then $-a \equiv \ -b \pmod m$
Since $101$ is congruent to $10 \pmod {13}$ $(101 = 13 \times 7 + 10)$ then $-101$ is congruent to $-10 \pmod{ 13}$
However, the remainder mod m is defined as a non-negative integer $r$ in the interval $[0,m-1]$
So we just add 13 to get $-101 \equiv -10+13 \equiv 3 \pmod {13}$
$101 \equiv 10 \bmod 13$
Therefore $-101 \equiv -10 \equiv 3 \bmod 13$ | 2020-10-25T20:11:48 | {
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http://math.stackexchange.com/questions/1649721/decimals-of-the-square-root-of-n | # Decimals of the square root of $n$.
Let $a_1, \ldots, a_k$ be any sequence of digits (i.e., each $a_i$ is between 0 and 9). Prove that there exists an integer $n$ such that $\sqrt{n}$ has its first $k$ decimals after the decimal point precisely the string $a_1\ldots a_k$. A possible solution of this problem (but by no means the only solution!) uses the fact that
$$\sqrt{n+1} - \sqrt{n} < \frac1{2\sqrt{n}}\,\,\,\text{ for all }n\geq 1$$
Can anybody please help me to approach the solution? I am not getting what the inequality has to do with decimals of root $n$.
-
Hint: consider only the case of very large integers (i.e. $n>>1$) – Yuriy S Feb 10 at 22:56
okay but then root {n+1} and root{n} will both be almost same,but whats the connection with the decimals,thats my main question. – jyoti prokash roy Feb 10 at 23:01
Do you know Taylor series? – Yuriy S Feb 10 at 23:02
yes very well.. – jyoti prokash roy Feb 10 at 23:08
@michaelchirico thanks for editing the question. – jyoti prokash roy Feb 11 at 1:10
The hint is misleading: it shouldn't use the same symbol $n$. So let's use the hint
$$\sqrt{i+1}-\sqrt i < \frac{1}{2\sqrt i} \text{ for all }i \ge 1$$
Choose $i$ large enough so that $\dfrac{1}{2\sqrt i} < 10^{-k}$. Then successive square roots $\sqrt i, \sqrt{i+1},\sqrt{i+2},\ldots$ differ by less than $10^{-k}$. Now choose $j$ large enough so that $\sqrt{i+j} > \sqrt i + 1$. Then the square roots from $\sqrt i$ to $\sqrt{i+j}$ will contain all possible sequences of $k$ digits after the decimal point. So one of them must equal $a_1\ldots a_k$.
-
Here's the idea applied to a specific case.
Let's prove that there is an integer whose square root has a fractional part that begins with a 2.
So we seek an integer $n$ such that, for some integer $m$, we have $$m+0.2 \le \sqrt{n} < m+0.3.$$ Squaring this, we have $$m^2 + 0.4m +0.04 \le n < m^2+0.6m+0.09.$$ So if an integer falls between $m^2+0.4m+0.04$ and $m^2+0.6m+0.09$, then we have our $n$. This interval has length $0.2m+0.05$ and will certainly contain an integer if $m$ is sufficiently large. Choose $m$ large enough, and we may conclude that such an $n$ exists. (For example, $n=5$ suffices, yielding the interval $[27.04,28.09]$ which contains the integer $28$ whose square root is $5.2915...$.)
The idea applies exactly the same way when you want to specify more digits in the root. You'll just need to choose larger $m$.
-
I will give you an example you do the rest.
Assume that you want to have 123.
Take $\frac{123000000}{999}$ with sufficient number of $0$'s to align the value to have $.123$ after decimal digit.
In this case you can take $3$,$6$,$9$,$12$.... zeros. In general $k$,$2k$,$3k$,$4k$,...
It is important that you can extend this as much as you like for the next step.
I will take $6$ $0$'s
$\frac{123000000}{999}=123123.123123...$
Square this number
$(\frac{123000000}{999})^2=(15159303447+\frac{65617}{110889})$
Now remove the fractional part getting $15159303447$. If you have taken a sufficiently large number of zeros, meaning sufficiently high multiple of $k$, this is your number
$$\sqrt{15159303447}=123123.12312072...$$
Except those with all $9$'s, this is the formula for all endings with sufficiently large $m$, $a=a_{1}a_{2}...a_{k}$
$$\left \lfloor (\frac{10^{mk}a}{10^k-1})^2 \right \rfloor$$
For desired number of 9's only, since the method for obvious reasons does not work for repetition of 9's, simply take $10^{2k}-1$ and you have your number.
This is working by creating a rational number with your digits as a repetition. Since you can make the initial part as large as you want, removing the fractional part in the second step will stop affecting first $k$ digits after a decimal digit at one point in time, so you can remove it as long as you have taken a sufficient number of $0$'s.
This is a proof by construction, you can do this for any $k$ and for any combination of digits with explained rule for all $9$'s.
Proof:
$$\sqrt{\left \lfloor (\frac{10^{mk}a}{10^k-1})^2 \right \rfloor}=\sqrt{ (\frac{10^{mk}a}{10^k-1})^2 k}=\frac{10^{mk}a}{10^k-1} \sqrt{k}$$
$k=\frac{\left \lfloor (\frac{10^{mk}a}{10^k-1})^2 \right \rfloor}{(\frac{10^{mk}a}{10^k-1})^2}$ which by increasing $m$ can be made to be as close as $1$ as we want, meaning we can conserve as many digits as we want from $\frac{10^{mk}a}{10^k-1}$ including the first $k$ we are interested in.
-
To use the relation you are given, $\sqrt{n+1}-\sqrt n \lt \frac 1{2\sqrt n}$, note that the positional value of $a_k$ is 10^{-k}. If $\frac 1{2\sqrt n} \lt 10^{-k}$ , which means $n \gt 10^{2k}/4$ the step between square roots will be less than $10^{-k}$ so you will hit every string of $k$ decimals. You can then take any $n\gt 10^{2k}/4$ and $m=\lfloor \sqrt{n+0.a_1a_2a_3\dots (a_k+1)}\rfloor$ is the integer you seek.
- | 2016-07-02T03:58:06 | {
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https://tutorme.com/tutors/197773/interview/ | Enable contrast version
# Tutor profile: Chris M.
Inactive
Chris M.
Mathematics Tutor for 5 years
Tutor Satisfaction Guarantee
## Questions
### Subject:Linear Algebra
TutorMe
Question:
Compute the following: $$det \left( \left[ \begin{array}{ccc} 3 & 4 & 5 \\ 1 & 3 & 7 \\ 3 & 8 & 3 \end{array} \right] \left[ \begin{array}{ccr} 2 & 1 & 4 \\ -2 & 5 & 5 \\ 1 & -1 & -1 \end{array} \right] \right)$$
Inactive
Chris M.
There are two solution routes to take. This comes from the properties of determinants of square matrices. Solution Method 1: $$\left[ \begin{array}{ccr} 3 & 4 & 5 \\ 1 & 3 & 7 \\ 3 & 8 & 3 \end{array} \right] \left[ \begin{array}{ccr} 2 & 1 & 4 \\ -2 & 5 & 5 \\ 1 & -1 & -1 \end{array} \right] = \left[ \begin{array}{ccc} 3(2) + 4(-2) + 5(1) & 3(1) + 4(5) + 5(-1) & 3(4) + 4(5) + 5(-1) \\ 1(2) + 3(-2) + 7(1) & 1(1) + 3(5) + 7(-1) & 1(4) + 3(5) + 7(-1) \\ 3(2) + 8(-2) + 3(1) & 3(1) + 8(5) + 3(-1) & 3(4) + 8(5) + 3(-1) \end{array} \right]$$ $$= \left[ \begin{array}{ccc} 6 + -8 + 5 & 3 + 20 + -5 & 12 + 20 + -5 \\ 2 + -6 + 7 & 1 + 15 + -7 & 4 + 15 + -7 \\ 6 + -16 + 3 & 3 + 40 + -3 & 12 + 40 + -3 \end{array} \right]$$ $$= \left[ \begin{array}{ccc} 3 & 18 & 27 \\ 3 & 9 & 12 \\ -7 & 40 & 49 \end{array} \right]$$ Now take the determinant: $$det \left( \left[ \begin{array}{ccc} 3 & 18 & 27 \\ 3 & 9 & 12 \\ -7 & 40 & 49 \end{array} \right] \right)$$ $$= 3 \left[ 9(49) - 40(12) ] - 18 [ 3(49) - 12(-7) ] + 27 [ 3(40) - 9(-7) \right]$$ $$= 3(441 - 480) - 18(147 + 84) + 27(120 + 63)$$ $$= 3(-39) - 18(231) + 27(183)$$ $$= 666$$ Solution Method 2: Multiplying two 3x3 matrices before computing the determinant is often times very time consuming. Since both matrices are the same size and square, the following property can be used: $$det(AB) = det(A)det(B)$$ Then, the problem becomes easier to compute. $$det(A) = det \left( \left[ \begin{array}{ccc} 3 & 4 & 5 \\ 1 & 3 & 7 \\ 3 & 8 & 3 \end{array} \right] \right)$$ $$det(A) = 3[3(3) - 7(8)] - 4[1(3) - 7(3)] + 5[1(8) - 3(3)]$$ $$det(A) = 3(9 - 56) - 4(3 - 21) + 5(8 - 9)$$ $$det(A) = 3(-47) - 4(-18) + 5(-1)$$ $$det(A) = -74$$ $$det(B) = det \left( \left[ \begin{array}{ccr} 2 & 1 & 4 \\ -2 & 5 & 5 \\ 1 & -1 & -1 \end{array} \right] \right)$$ $$det(B) = 2[ 5(-1) - 5(-1) ] - 1 [ -2(-1) - 5(1) ] + 4[ -2(-1) - 5(1) ]$$ $$det(B) = 2(-5 + 5) - 1(2 - 5) + 4(2 - 5)$$ $$det(B) = 2(0) - 1(-3) + 4(-3)$$ $$det(B) = -9$$ Then, the final solution $$det(AB) = (-74)(-9) = 666$$
### Subject:Calculus
TutorMe
Question:
Why does the evaluation of a derivative give the slope of the tangent line to it's point?
Inactive
Chris M.
This is a relatively simple concept that is extremely important to understand for all future work in calculus. The slope of a line is defined as $$\frac{\Delta y}{\Delta x}$$ I can make any changes to how large my $$\Delta y$$ is, and since we are using the line example, the size of $$\Delta x$$ will also change with it. With both values changing accordingly, the slope of the line will remain the same! Now, rather than making $$\Delta y$$ larger, we can shrink it's size until we get something indiscernible in size. $$\Delta y \Rightarrow dy$$ Again, since we have no change in the value of the slope of a line, decreasing the value of $$\Delta y$$ will also decrease the value of $$\Delta x$$. Thus, $$\Delta x \Rightarrow dx$$. Then, the slope of the line can be rewritten as follows! $$Slope = \frac{\Delta y}{\Delta x} = \frac{dy}{dx}$$ Now, we know that the derivative of a function y with respect to x is $$\frac{dy}{dx}$$, and so evaluating a derivative at some point x gives a slope value of a line containing this point.
### Subject:Physics
TutorMe
Question:
In a roller coaster loop, do you "feel" heavier at the top or the bottom of the loop?
Inactive
Chris M.
Your sense of weight comes from objects pushing back on you, known as the normal force. When cresting a hill in a car at high enough speeds, you feel a momentary loss in weight. The same effect occurs in roller coasters when reaching peaks. However, in a roller coaster loop, the forces at play become more complicated. The question asks at which point, the top or bottom, do you "feel" heavier. In other words, at which point of the coaster loop is the Normal Force greatest? At every point along the ride, your total mass combined with gravity create the Weight Force that is continuously pointed downward. As you progress through the loop and enter circular motion, the Net Force acting on you must point towards the center of the loop to continue with the motion. This net force is called the Centripetal Force, and it is commonly misunderstood. At the top of the loop, the centripetal force points downward in the same direction as gravity. It is important to again understand that the centripetal force IS the net force, and that the gravitational force (your weight) is a "contribution" to this net force. Gravity alone cannot provide enough force to attribute to the total centripetal force, so the rest is supplied by your chair pushing back on you! This is the normal force, and it is your "Feeling" of weight. At the bottom of the loop, gravity is again pointed downwards, but this time the centripetal force is pointed upwards. The normal force here will be greater than the normal force from the top of the loop because of this difference in orientation. The following equation can be considered: Fcentripetal = Fnormal + Fgravity So, to answer the question, at the Bottom of the Coaster loop, you will feel heavier than when you are at the top of the loop! Just remember that when solving circular motion problems, that the depicted images of Centripetal Force pointed towards a center of rotation should be seen as the Net Force acting on the object in motion.
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https://mathematica.stackexchange.com/questions/103402/color-coded-bar-to-represent-numeric-values/103407 | # Color coded bar to represent numeric values
I would like to present data color coded in a bar.
My data look like the following:
{-2.5,-1.0,0.1,0.5,-0.24,-0.58,-0.58,-0.25,-1.4,2.1,1.8,-2.5,1.97}
I would like to create a horizontal bar and each part of the bar should be colored according to the number value of each data point. The color scheme should be getting the bluer the smaller the value below zero and increasingly red for values larger than zero. (The color scheme is not important, I just want to make the different values visible). All parts of the bar should be of the same length. I have no idea where to start and don't know if a stacked bar chart is the way to go or if I should better represent each value with a rectangle graphics object that I combine in the end. The image below is just to show you the kind of bar I would like to create (very roughly).
• Your sample output suggests that you may wish to sort the data before plotting it. Is that the case? – bbgodfrey Jan 5 '16 at 19:20
data = {-2.5, -1.0, 0.1, 0.5, -0.24, -0.58, -0.58, -0.25, -1.4, 2.1, 1.8, -2.5, 1.97};
Graphics[Raster[{Rescale@data},
ColorFunction -> (Blend[{{0, Blue}, {1, Red}}, #] &)],
AspectRatio -> .3]
tab = Table[{m, m}, {m, -2.5, 2.5, 0.1}]
ListDensityPlot[Transpose[tab],
ColorFunction -> (ColorData[{"TemperatureMap", {-2.5, 2.5}}][#] &),
ColorFunctionScaling -> False, PlotRangePadding -> 0,
PlotRange -> All, AspectRatio -> 1/8, Frame -> False]
• You can exchange the colorscheme. Just figure out Mathematica help and search for ColorSchemes. If you want to keep the rectangular shapes, try it this way: Graphics[Table[{Blend[{Red, Green}, i], EdgeForm[Gray], Rectangle[{4 i, 0}]}, {i, 0 - 1/4, 1 + 1/4, 1/4}]] – Kay Jan 5 '16 at 18:28
• You can update your answer, instead of adding useful information in a comment. Also it would be better if you posted an image as well so that we can more easily judge the quality of the answer, which will make us (probably) more likely to vote for it. – C. E. Jan 5 '16 at 18:47
Here's one way to do it with rectangles:
data = {-2.5, -1.0, 0.1, 0.5, -0.24, -0.58, -0.58, -0.25, -1.4, 2.1, 1.8, -2.5, 1.97};
list = {-2.5, -1.0, 0.1, 0.5, -0.24, -0.58, -0.58, -0.25, -1.4, 2.1, 1.8, -2.5, 1.97}; | 2021-05-14T10:39:23 | {
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https://math.stackexchange.com/questions/4506858/probability-of-ending-at-0-after-3-times-moving-6-sided-dice-rolled-number | # Probability of ending at $0$ after $3$ times moving $6$-sided-dice-rolled number of steps in a coin-flip direction?
I just wrote a question and I'm not sure if my solution is correct. Could someone please help me check it?
Question: Mary starts at 0 on the number line. She first tosses a coin, and then rolls a fair six-sided dice. If the coin lands on heads, then she moves to the positive side; if the coin lands on tails, then she moves to the negative side. The number of steps she takes depends on the number that appears on the dice she rolled. For example, if the coin lands on heads and 4 is rolled, then Mary would go 4 steps to the positive side. What is the probability that Mary will end at 0 after three rounds of tossing and rolling?
My solution: In order for Mary to end at 0, one of the dice must be the sum of the other two dice and be the opposite sign.
P (sum = 2) = ⅙ x ⅙ x ⅙ x 3
P (sum = 3) = ⅙ x ⅙ x ⅙ x 6
P (sum = 4) = ⅙ x ⅙ x ⅙ x (3 + 6)
P (sum = 5) = ⅙ x ⅙ x ⅙ x (6 + 6)
P (sum = 6) = ⅙ x ⅙ x ⅙ x (3 + 6 + 6)
P total = ⅙ x ⅙ x ⅙ x 45 = 5/24.
There is 1/2 probability that the two dice are the same sign, and we multiply it by another 1/2 to ensure that the other dice is the opposite sign. Therefore my answer is 5/96.
This is correct, and your reasoning is good. For a more general and mechanical way to get the result, note that the movements are additive, with possible values $$\{-6,-5,-4,-3,-2,-1,1,2,3,4,5,6\}$$ occurring with equal probability ($$1/12$$); so the probability of landing on $$0$$ after $$n$$ rolls is the coefficient of $$x^0$$ (i.e., the constant term) in the expansion of $$\left(\frac{1}{12}\left(x^{-6}+x^{-5}+x^{-4}+x^{-3}+x^{-2}+x^{-1}+x+x^2+x^3+x^4+x^5+x^6\right)\right)^n.$$ WolframAlpha gives this probability as $$\frac{1}{12}$$ for $$n=2$$ (clearly correct), $$\frac{5}{96}$$ for $$n=3$$ (agreeing with OP's result), $$\frac{259}{5184}$$ for $$n=4$$, $$\frac{1825}{41472}$$ for $$n=5$$, and so forth.
Assume for the moment that Die-1 is the sum of the other two dice. Then, the other two dice must sum to $$2,3,4,5,$$ or $$6$$. This can happen in $$1 + 2 + 3 + 4 + 5 = 15$$ ways.
Once Die-2,Die-3 are set, Die-1 is forced to be their sum.
So, the probability that Die-1 is the sum of the other two is $$~\displaystyle \frac{15}{216} = \frac{5}{72}.~$$
In order for this to lead to returning to $$0$$, the coin flips have to specifically be HTT or THH. So the probability of Die-1 being the leveraging die is $$\displaystyle ~\frac{5}{72} \times \frac{1}{4} = \frac{5}{288}.$$
Further, anyone of the three dice can be the leveraging die. Therefore, the overall probability of returning to $$0$$ is
$$3 \times \frac{5}{288} = \frac{5}{96},$$
A symmetrical shortcut to the first portion of my answer is that the probability of two dice summing to less than $$(7)$$ must, by symmetry, be the same as the probability of the two dice summing to greater than $$(7)$$. Further, the probability of two dice summing to exactly $$(7)$$ is $$~\displaystyle \frac{1}{6}.$$
Therefore, the probability of two dice summing to less than $$(7)$$ must be
$$\frac{1}{2} \times \left[1 - \frac{1}{6}\right] = \frac{5}{12} = \frac{15}{36}.$$ | 2022-08-15T09:15:25 | {
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https://math.stackexchange.com/questions/1242818/exploring-sum-n-0-infty-fracnpn-b-pe-particularly-p-2 | # Exploring $\sum_{n=0}^\infty \frac{n^p}{n!} = B_pe$, particularly $p = 2$.
I was exploring the fact that $$\sum_{n=0}^\infty \frac{n^p}{n!} = B_pe,$$ where $B_n$ is the $n$th Bell number.
I found this result by exploring the series on wolframalpha and looking up the sequence of numbers generated. I have no experience with Bell numbers other than knowing that they represent the number of ways of partitioning a set.
What I tried
Looking at the base case we can verify $$\sum_{n=0}^\infty \frac{n}{n!} = \sum_{n=1}^\infty \frac{1}{(n-1)!} = e.$$
But then I realize that $n^2$ is going to be difficult.
Specifically what I want
I am doing this for fun, so I just want to glean some sort of lesson out of this. Resources that let me learn for myself are just as good!
An accepted answer will have any one of the following:
• A proof that $\sum_{n = 0}^{\infty} \frac{n^2}{n!} = 2e$ (this interests me a lot)
• A description of why this series is related to Bell numbers and the number of partitions of a set
• A link to some resource that introduces Bell numbers and/or their connection to this series.
• Hi Chantry. Your first two questions are completely answered below. – Berrick Caleb Fillmore Apr 20 '15 at 4:22
• Hint: Since $n!~=~1\cdot2\cdot3\cdots(n-1)~n$, an obvious approach would be rewriting $n^2$ as $n~(n-1)+n$. – Lucian Apr 20 '15 at 9:19
Notice that \begin{align} S & = \sum_{n = 0}^{\infty} \frac{n^{2}}{n!} \\ & = \frac{0^{2}}{0!} + \frac{1^{2}}{1!} + \frac{2^{2}}{2!} + \frac{3^{2}}{3!} + \cdots \\ & = \frac{1}{0!} + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + \cdots. \end{align} Then \begin{align} S - e & = \left( \frac{1}{0!} + \frac{2}{1!} + \frac{3}{2!} + \frac{4}{3!} + \cdots \right) - \left( \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots \right) \\ & = \frac{1}{1!} + \frac{2}{2!} + \frac{3}{3!} + \cdots \\ & = e. \qquad (\text{As shown in the OP.}) \\ \end{align} Therefore, $S = e + e = 2 e$.
In fact, using the same type of reasoning, it can be shown that $\displaystyle \sum_{n = 0}^{\infty} \frac{n^{3}}{n!} = 5 e$.
Here is the connection with Bell numbers.
For each $p \in \mathbb{N}_{0}$, let $\displaystyle S_{p} \stackrel{\text{df}}{=} \sum_{n = 0}^{\infty} \frac{n^{p}}{n!}$. Notice that \begin{align} \forall p \in \mathbb{N}: \quad S_{p} & = \sum_{n = 0}^{\infty} \frac{n^{p}}{n!} \\ & = \sum_{n = 1}^{\infty} \frac{n^{p}}{n!} \qquad \left( \text{As $\dfrac{0^{p}}{0!} = 0$.} \right) \\ & = \sum_{n = 1}^{\infty} \frac{n^{p - 1}}{(n - 1)!} \qquad (\text{After canceling $n$’s.}) \\ & = \sum_{n = 0}^{\infty} \frac{(n + 1)^{p - 1}}{n!}. \qquad (\text{After re-indexing.}) \end{align} Hence, \begin{align} \forall p \in \mathbb{N}_{\geq 2}: \quad S_{p} - S_{p - 1} & = \sum_{n = 0}^{\infty} \frac{(n + 1)^{p - 1}}{n!} - \sum_{n = 0}^{\infty} \frac{n^{p - 1}}{n!} \\ & = \sum_{n = 0}^{\infty} \frac{1}{n!} \left[ (n + 1)^{p - 1} - n^{p - 1} \right] \\ & = \sum_{n = 0}^{\infty} \left[ \frac{1}{n!} \sum_{k = 0}^{p - 2} \binom{p - 1}{k} n^{k} \right] \qquad (\text{By the Binomial Theorem.}) \\ & = \sum_{n = 0}^{\infty} \sum_{k = 0}^{p - 2} \binom{p - 1}{k} \frac{n^{k}}{n!} \\ & = \sum_{k = 0}^{p - 2} \left[ \binom{p - 1}{k} \sum_{n = 0}^{\infty} \frac{n^{k}}{n!} \right] \\ & = \sum_{k = 0}^{p - 2} \binom{p - 1}{k} S_{k}. \end{align} Therefore, $$\forall p \in \mathbb{N}_{\geq 2}: \quad S_{p} = \sum_{k = 0}^{p - 2} \binom{p - 1}{k} S_{k} + S_{p - 1} = \sum_{k = 0}^{p - 1} \binom{p - 1}{k} S_{k}.$$ As $S_{0} = S_{1} = e$ by inspection, we see that the $S_{p}$’s are the Bell numbers multiplied by $e$.
• Wonderful! This looks like it extends well to higher cases too - we can use that $\sum_{n=0}^{\infty}\frac{n^p}{n!}=\sum_{n=0}^{\infty}\frac{(n+1)^{p-1}}{n!}$ and then expand the $(n+1)^{p-1}$ term to get that sum in terms of various sums of the form $\sum_{n=0}^{\infty}\frac{n^{p'}}{n!}$ in terms of lesser $p'$ (which implies some recurrence relation I don't feel like writing out). – Milo Brandt Apr 20 '15 at 3:35
• @Meelo: Yes, there exists a recurrence relation. I once had to derive it in a high-school exam. – Berrick Caleb Fillmore Apr 20 '15 at 3:37
Recall the species of set partitions with the constituents marked which is $$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$ which gives the generating function $$G(z, u) = \exp(u(\exp(z)-1)).$$
It follows that the exponential generating function of Bell numbers is given by $$G(z) = \exp(\exp(z)-1).$$
Suppose we are trying to compute $$\sum_{n\ge 0} \frac{n^p}{n!}.$$
Put $$n^p = \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \exp(nz) \; dz.$$
Observe that this gives $n^p = 0$ when $n=0$ except when $p=0$ as well.
We get for the sum $$\frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \sum_{n\ge 0} \frac{\exp(nz)}{n!} \; dz = \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \exp(\exp(z)) \; dz \\ = \exp(1) \times \frac{p!}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{p+1}} \exp(\exp(z)-1) \; dz \\ = \exp(1)\times B_p$$ as claimed.
• This is a little beyond my current understanding. I do not know what $\mathfrak{P}$ (power set maybe?) stands for and have little understanding about generating functions and contour integrals. Thank you for your contribution though! I will be sure to return to your answer in the future after studying both subjects. – Chantry Cargill Apr 21 '15 at 0:38 | 2019-08-25T05:23:17 | {
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https://math.stackexchange.com/questions/2099698/finding-a-sequence-a-with-lim-n-to-%E2%88%9E-a-n1-a-n-0-adivergent | # Finding a sequence a with $\lim_{ n\to ∞} (a_{n+1}-a_n)=0$ a:divergent [duplicate]
Question is in the title. I would appreciate any help with this as I am a bit clueless.
• Are you familiar with the fact that $\sum\limits_{k=1}^\infty \frac{1}{k}$ is divergent? Or rather, more accurately written, $\lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k}$ is divergent – JMoravitz Jan 16 '17 at 4:29
• Yes, the series is divergent but the sequence isn't, or is it? – Lillia Jan 16 '17 at 4:34
• And a series is a sequence of partial sums. S.C.B. already spelled out what I was trying to get at with my hints below – JMoravitz Jan 16 '17 at 4:35
• Related: Pseudo-Cauchy sequence – Martin Sleziak Jan 16 '17 at 6:43
Note that if $a_{n}=H_{n}$ where $H_{n}$ denotes the $n$-th harmonic number, $$\lim_{n \to \infty} H_{n+1}-H_{n}=\lim_{n \to \infty} \frac{1}{n+1}=0$$
• $a_{n}=H_{n}$, and as posted in the link, $H_{n}$ is divergent so $a_{n}$ is divergent. – S.C.B. Jan 16 '17 at 4:41
• @user406473 you want a sequence (in this case $H_n$) where the sequence itself is divergent, but the related sequence of differences (in this case $H_{n+1}-H_n$) has limit equal to zero – JMoravitz Jan 16 '17 at 4:42
• @user406473 $H_{n}=\sum_{i=1}^{n} \frac{1}{i} \neq \frac{1}{n}$ – S.C.B. Jan 16 '17 at 4:46
Take $a_n=\ln n$. Then $$\lim_{n\to\infty}\Big[\ln(n+1)-\ln(n)\Big]=\lim_{n\to\infty}\ln\frac{n+1}{n}=\ln\left[\lim_{n\to\infty}\frac{n+1}{n}\right]=\ln 1=0$$ | 2020-01-19T07:51:06 | {
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http://forum.allaboutcircuits.com/threads/compute-mph-by-measuring-interval-between-rim-bolt-heads-magnetic-pickup.115234/ | # Compute MPH by measuring interval between rim bolt heads (magnetic pickup)
Discussion in 'Math' started by rjwheaton, Sep 7, 2015.
1. ### rjwheaton Thread Starter New Member
Jan 26, 2014
5
0
I have a tire rim which has 5 bolts. I have a magnetic pickup which provides a very clean pulse every time a bolt head passes. The circumference of the wheel is 78.4 inches. Obviously, I get 5 pulses per wheel revolution.
I cannot figure out the math! What I basically need is to convert the interval between bolt heads to miles per hour.
Can anyone help?
Thank you very much,
Roy
2. ### MikeML AAC Fanatic!
Oct 2, 2009
5,450
1,066
An exercise in dimensional analysis:
Say you are measuring Psec/1bolt
Psec/1bolt * 5bolts/1rev * 1rev/78.4in * 12in/1ft * 5280ft/1mi * 1min/60sec * 1hr/60min = P*5*12*5280/(78.4*60*60)hr/mi
Since we wanted mi/hr, then invert and we have: (282240/316800*P) mi/hr = (49/55*P)mi/hr = (0.890909/P)mi/hr
or Speed = 49/(55*P) mph.
Transposing, let us see what P would be at 60mph: 60 = 49/(55*P)
60*55*P =49
P= 49/(60*55) = 0.01484848sec or 14.84msec
Last edited: Sep 8, 2015
rjwheaton likes this.
3. ### Wendy Moderator
Mar 24, 2008
20,766
2,536
Outer circumference of the tire sets the distance. You need the radius.
5 pulses per rotation.
4. ### MikeML AAC Fanatic!
Oct 2, 2009
5,450
1,066
Psec/1bolt * 5bolts/1rev * 1rev/78.4in *...
The OP specified that the circumference of the wheel is 78.4in. Why do we need to include the radius in the conversion?
Last edited: Sep 8, 2015
5. ### Wendy Moderator
Mar 24, 2008
20,766
2,536
Because I missed that little detail?
6. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
But the circumference of the wheel is irrelevant -- it's the circumference of the tire that matters. Perhaps the TS means the tire and not the wheel, but I'm not positive about that.
7. ### djsfantasi AAC Fanatic!
Apr 11, 2010
2,805
833
A circumference of 78.4" is a diameter of 10". Much more likely to be the diameter of a wheel.
Update: Helps to use the right equation...
Last edited: Sep 8, 2015
8. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
Or it's one of the new micro-cars that are becoming the snob-appeal rage.
9. ### Wendy Moderator
Mar 24, 2008
20,766
2,536
As sensors go I like it. It would also be good for odometers too.
10. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
I think driveshaft sensors are probably better since you typically get several rotations of the shaft per rotation of the tire (and you can mount multiple magnets on the shaft to further improve resolution). It also tends to average out the distance traveled by the tires as you drive due to the natural behavior of the differential.
Jan 26, 2014
5
0
12. ### rjwheaton Thread Starter New Member
Jan 26, 2014
5
0
Thank you!
Not only does that answer my question, but it helps me understand how you did the calculations.
FYI, my Harley doesn't have a drive shaft (thank heavens) and the circumference of 78.4 inches computes out to a diameter of about 24.9 inches.
And yes, that will also be my odometer. The magnetic pickup is located under the swingarm, and senses the rim mounting bolts.
Roy
13. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
But the question remains. Is that the circumference of the wheel, or of the tire.
In either case, it is best to make the system calibratable so that you can zero a trip meter, drive some known distance (such as on a highway with mileage markers) and then adjust the tripmeter up/down to the correct value and use that to obtain a calibration constant.
14. ### MrAl Well-Known Member
Jun 17, 2014
2,433
490
Hi,
A 25 inch "wheel" would mean a pretty darn big tire
The short answer is if there are 5 bolts per revolution and one revolution moves the car 78.4 inches, then the vehicle moves 78.4 inches for every complete revolution.
15. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
I just checked a motorcycle tire website and they have tires for wheels having diameters from 8 inches all the way up to 26 inches, so who knows.
16. ### MrAl Well-Known Member
Jun 17, 2014
2,433
490
Hi,
Oh that makes sense now. I measured the tires on my car one time but i forgot what they were now. I know they are 15 inch wheels though, but the tire sizes vary quite a bit even for that one 'wheel' (rim) size.
17. ### tcmtech Well-Known Member
Nov 4, 2013
2,034
1,659
Um yea....
78.4 / 3.1415 = 24.95" diameter or a radius of ~12.47"
So where'd the 10 come from exactly?
18. ### WBahn Moderator
Mar 31, 2012
17,743
4,795
It's a simple error that messed up the units (had they been there to get messed up) -- the classic kind that I'm always harping on.
$
A \; = \; \pi r^2
\;
r \; = \; \sqrt{\frac{A}{\pi}}
\;
d \; = \; 2r \; = \; 2 \sqrt{\frac{A}{\pi}}
$
Plug in A=78.4 and pi=3.14 and you get d = 9.99, then tack on the units you want it to be and you have a diameter of 10 inches.
Plug in A = 78.4 in and pi = 3.14 and you get d = 9.99 root-inches, which just screams out "I'm WRONG!!"
But, hey, tracking units is such a waste of time, isn't it.
19. ### MrAl Well-Known Member
Jun 17, 2014
2,433
490
Hi,
That's funny i was wondering where that "10 inches" came from too. At first i thought maybe he was holding the wheel horizontally while traveling at about 0.92 times the speed of light and measuring the diameter in the same direction as travel <chuckle>
That formula does get close to 10 inches, so maybe he just used the wrong formula as his post update suggests.
20. ### tcmtech Well-Known Member
Nov 4, 2013
2,034
1,659
At least you admit to the mistake.
I had a HS math teacher that was so by the book that when the book was wrong he would still keep right on going and mark anyone's papers wrong if they used the correct formulas opposed to his wrong book formulas and believe me the books we had had a lot of basic mistakes. | 2016-12-08T14:26:03 | {
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https://math.stackexchange.com/questions/1025155/8x-9y-5-where-x-y-in-mathbbz | # $8x +9y = 5$ where $x,y \in \mathbb{Z}$
Solve the following Diophantine equation algebaically: $$8x+9y=5$$ Give 3 possible solutions for the equation
I have the following:
The Diophantine equation has solutions $x,y \iff 8x=5\mod{9}$ has a solution $x \equiv\mod{9}$
Since $\gcd(8,9)=1$, by Bezout's Lemma, for $r,t \in \mathbb{Z}, \gcd(8,9)=1=r(8)+t(9)$ and $x\equiv r(5)\mod{9}$ is a solution for the linear congruence above.
By Euclid's algorithm for determining $\gcd(8,9)$ we have
\begin{align}9 &= 1(8) +1 \\ 8 &=9(1)+0\end{align} so $1=(-1)8 + 1(9)$ and $r=-1 \implies x \equiv(-1)5\mod{9}$.
Now \begin{align}[-5]_9 &= \{-5 + 9k \ | k\in\mathbb{Z} \} \\ &= \{ ..., -5,4,13,... \} \\ &=[4]_9\end{align} $\therefore x \equiv 4 \mod{9}$, that is $x=4+9k$ for all $k \in \mathbb{Z}$ upon which it can be seen that $y= -3 -8k$.
Is this correct?
• Yes, this is correct, and you found all the solutions to the equation, in my opinion. – awllower Nov 17 '14 at 0:22
• I am feeling uncertain about my Euclid's Algorithm for some reason. That is the part I especially felt unsure about – user860374 Nov 17 '14 at 0:23
• I feel very certain that your use of Euclidean algorithm is correct. :) – awllower Nov 17 '14 at 0:24
• Since the problem stated "find three solutions", I would explicitly write out three of them, like $x=4, y = -3$. But otherwise it looks correct. – Arthur Nov 17 '14 at 0:25
• why bother with the euclide thing? in that case, it is obvious that $9-8=1$. – mookid Nov 17 '14 at 0:25
Choose $x$ or $y$ ? Let's solve for $x$: $x = \dfrac{5-9y}{8} = \dfrac{5-y}{8} - y \Rightarrow 5-y = 8k \Rightarrow y = 5 - 8k \Rightarrow x = k - (5-8k) = 9k-5$. Thus:
$(x,y) = (9k-5,5-8k), k \in \mathbb{Z}$
$8x+9y=5\iff8~(x+y)+y=5=8\cdot0+5\iff x+y=0$ and $y=5\iff x=-5$. Then all numbers of the form $x=-5-9k$ and $y=5+8k$ are solutions to the above equation.
• Alternately, $5=8\cdot1-3$. – Lucian Nov 17 '14 at 1:59 | 2019-08-25T03:01:42 | {
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https://physics.stackexchange.com/questions/106382/orthogonality-of-summed-wave-functions | # Orthogonality of summed wave functions
Problem. I know that the two wave functions $\Psi_1$ and $\Psi_2$ are all normalized and orthogonal. I now want to prove that this implies that $\Psi_3=\Psi_1+\Psi_2$ is orthogonal to $\Psi_4=\Psi_1-\Psi_2$.
My naive solution. From the premises, we know that $$\int_{-\infty}^\infty \Psi_1^*\Psi_1 dx=\int_{-\infty}^\infty \Psi_2^*\Psi_2 dx=1$$ and $$\int_{-\infty}^\infty \Psi_1^*\Psi_2 dx=\int_{-\infty}^\infty \Psi_2^*\Psi_1 dx=0$$
We also have $(z_1+z_2)^*=z_1^*+z_2^*$
$$\int_{-\infty}^\infty \Psi_3^*\Psi_4 dx = \int_{-\infty}^\infty (\Psi_1+\Psi_2)^*(\Psi_1-\Psi_2)dx \\ =\int_{-\infty}^\infty(\Psi_1^*+\Psi_2^*)(\Psi_1-\Psi_2)dx\\ =\int_{-\infty}^\infty(\Psi_1^*\Psi_1-\Psi_1^*\Psi_2+\Psi_2^*\Psi_1-\Psi_2^*\Psi_2)dx\\ =1-0+0-1=0\,,$$
which is equivalent with what we wanted to prove. Is this a legitimate proof? Is there any simpler way to do this? I am afraid I still haven't grasped how wave functions behave mathematically, so I may have missed somethings very obvious here.
Edit: The solution manual somehow uses normalization factors for $\Psi_3$ and $\Psi_4$. How are these factors when you don't actually know the exact functions? And how does this relate to the concept of orthogonality?
• Yes, you do need normalization factors so that $\int |\Psi_3|^2 dx = 1$, but your proof as it stands is correct. You directly calculated their inner product and found it to be zero, hence orthogonal vectors. – webb Apr 2 '14 at 16:42
• Braket notation might be simpler, but yours is good enough. – jinawee Apr 2 '14 at 16:49
• I don't understand why I would need $\int |\Psi_3|^2 dx=1$. And how would the bracket notation help? – Oskar Henriksson Apr 2 '14 at 16:55
• You will probably get to Dirac/braket notation later in your course. But it would help by doing away with the integrals in this problem. – BMS Apr 2 '14 at 16:56
• @PoetryInMotion: The underlying assumption is that the symbols "plus ($+$)" and "minus ($-$)" that you used to express $\Psi_3$ and $\Psi_4$ in terms of the "orthonormal basis" states $\Psi_1$ and $\Psi_2$ do in fact represent the corresponding arithmetic operations between the complex number values of inner products. Consequently: $\langle \Psi_1 - \Psi_2 | \Psi_1 + \Psi_2 \rangle \text{=(means the same as)=} \langle \Psi_1 | \Psi_1 \rangle + \langle \Psi_1 | \Psi_2 \rangle - \langle \Psi_2 | \Psi_1 \rangle - \langle \Psi_2 | \Psi_2 \rangle$ which can be readily evaluated further (being $0$) – user12262 Apr 2 '14 at 17:37
This problem could be done more simply through the application of linear algebra. You want to prove that
$$\langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle = 0$$
The inner product is analogous to the dot product of linear algebra, and it is distributive. Distributing, we find that
\begin{aligned} \langle \psi_1 - \psi_2 | \psi_1 + \psi_2 \rangle &= \langle \psi_1 - \psi_2 | \psi_1 \rangle + \langle \psi_1 - \psi_2 | \psi_2 \rangle \\ &= \langle \psi_1 | \psi_1 \rangle - \langle \psi_2 | \psi_1 \rangle + \langle \psi_1 | \psi_2 \rangle - \langle \psi_2 | \psi_2 \rangle \end{aligned}
Because $\psi_1$ and $\psi_2$ are orthogonal and normalized, you know $\langle \psi_i | \psi_j \rangle = \delta_{i j}$. Substituting, the above expression evaluates to $1 - 0 + 0 - 1 = 0$, demonstrating that the two vectors are indeed orthogonal.
Your approach - using the integrals - was also valid, and fundamentally similar to mine here. However, by noting that the relation you used ($\langle \psi_1 | \psi_2 \rangle = \int_{-\infty}^{\infty} \! \psi_1^* \psi_2 \, \mathrm{d}x$) satisfied the definition of an inner product, the integrals can be omitted.
• Ah, that makes sense! However, I still don't understand what the normalization factors have to do with the question. Both $\Psi_3$ and $\Psi_4$ happens have the normalization factor $1/\sqrt{2}$, but can't have anything to do with their orthogonality, can it? – Oskar Henriksson Apr 2 '14 at 19:59
• Yeah, I'm not sure why you'd need the normalization factors here. It's clear that if $\langle \psi_1 | \psi_2 \rangle = 0$, then $\langle k \psi_1 | \psi_2 \rangle = k \langle \psi_1 | \psi_2 \rangle = 0$. – Shivam Sarodia Apr 2 '14 at 22:48
• @PoetryInMotion Can you explain the context in which the solution manual used the normalization factors? – Shivam Sarodia Apr 3 '14 at 0:11
• Oh, sorry, I missed that I was supposed to both investigate the orthogonality of $\Psi_3$ and $\Psi_4$ and normalize $\Psi_3$ and $\Psi_4$. The explains why the solution manual disused normalization constants. – Oskar Henriksson Apr 3 '14 at 21:29
• All right. Since you correctly found the normalization factor above, I take it you don't need help with that part. – Shivam Sarodia Apr 3 '14 at 23:18 | 2019-10-20T12:02:05 | {
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https://math.stackexchange.com/questions/632568/are-there-more-even-numbers-than-odd-numbers | # Are there more even numbers than odd numbers?
Very simple 'yes-or-no' question, but I can't find the answer anywhere. My gut feeling says the number of odd and even numbers are equal but I managed to write up something that contradicts my intuition. Although I still think my gut feeling is right, I can't find any logical or mathematical errors with my "proof". Can somebody please look it over and tell me which one of me is right?
Statement 1: For every positive odd integer $o$ there is an even integer $e=o+1$.
Statement 2: For every positive integer $n$ there is an negative integer i.e. $-n$.
Conclusion 1: The number of positive odd integers ($O_{positive}$) is equal to the number of positive even integers ($E_{positive}$). If there is a negative equivalent for every positive integer then the number of negative odd integers ($O_{negative}$) is equal to the number of negative even integers ($E_{negative}$). In short: $$O_{positive}=E_{positive}=O_{negative}=E_{negative}$$
Statement 3: The number zero is "neutral" (neither positive nor negative).
Statement 4: The number zero is an even integer.
Conclusion 2:
\begin{align} O_{total} & = O_{positive} + O_{negative} + O_{neutral} \\ & = O_{positive} + O_{negative} + 0 \\ & = O_{positive} + O_{negative} \end{align}
And:
\begin{align} E_{total} & = E_{positive} + E_{negative} + E_{neutral} \\ & = E_{positive} + E_{negative} + 1 \\ \end{align}
So:
\begin{align} E_{total} & = O_{total} + 1 \\ E_{total} & > O_{total}\\ \end{align}
Right? As an engineer I use math daily, but that doesn't make me a mathematician. So please be gentle :)
• If you want to know how infinities like this "work" in the average mathematician's mind, you should have a look at the story of Hilbert's hotel. – Arthur Jan 9 '14 at 14:10
• possible duplicate of Are half of all numbers odd? – Mark S. Jan 10 '14 at 23:57
First things first - there are an infinite number of both even numbers and odd numbers.
It's important to realize that $\infty$ (infinity) is not a number. Therefore it doesn't really make sense to talk about the "number" of even or odd numbers, or to write statements like $E_{\rm even}+1$, because that's assuming that $E_{\rm even}$ is a number that you can sensibly add $1$ to.
However, perhaps surprisingly it does make sense to ask if there are more even numbers than odd numbers. That is, you can compare two infinite quantities, or compare a finite quantity and an infinite quantity, even if you can't meaningfully add and subtract infinite quantities.
They way we define more, less and the same for infinite quantities is as follows. For two collections $A$ and $B$ (say $A$ are the even numbers and $B$ are the odd numbers) we say that
• If you can associate every item in $A$ with a unique item in $B$, and vice versa, then $A$ and $B$ are the same size.
• If you can associate every item in $A$ with a unique item in $B$, but not vice versa, then $B$ is bigger than $A$.
• If you can associate every item in $B$ with a unique item in $A$, but not vice versa, then $A$ is bigger than $B$.
In your case, you can associate every even number $n$ with the odd number $n+1$, and you can associate every odd number $m$ with the even number $m-1$ (assuming 0 is even) so therefore there are just as many odd numbers as even numbers.
This can lead to seemingly paradoxical results, because e.g. you can associate every whole number $n$ with the even number $2n$, and every even number $m$ with the whole number $m/2$, so there are just as many even numbers as whole numbers, even though the even numbers are a subset of the whole numbers.
• Aha, so if I understand correctly, the mistake I made was thinking that: infinity + 1 > infinity? – Jordy Jan 9 '14 at 14:22
• @Jordy The mistake was in thinking that $\infty$ is a number, and that $\infty+1$ is an expression that makes sense. It's easy to see that $\infty$ isn't a number, for here is a list of all the numbers: $\{0,1,2,3,4,\dots\}$. Where is $\infty$ in that list? You can't say "at the end", because the list doesn't have an end! (You also can't say "it's the ninth element in the list, but it's fallen over.") – Chris Taylor Jan 9 '14 at 14:26
• @Jordy This next bit, you'll have to imagine me saying in a stage whisper. Here it is: mathematicians have come up with a way of treating $\infty$ as a number! Shh, don't tell anyone. If you want the secrets, you'll have to learn a bit more math, and then go and read about transfinite ordinals. The smallest infinite ordinal is normally written $\omega$. Confusingly, $1+\omega=\omega$, but $\omega+1>\omega$. – Chris Taylor Jan 9 '14 at 14:28
• An old question and answer, but I disagree with the oft-repeated dogmatism that "$\infty$ is not a number" -- I think it teaches people the wrong idea. Is $i$ a number? Are the quaternions numbers? What about cardinals or ordinals? Furthermore, to me, all of the OP's reasoning is perfectly valid except the final conclusion, i.e. when from $E_{total} = O_{total} + 1$ he or she derives that $E_{total} > O_{total}$. All the rest is valid arithmetic with cardinalities, or alternatively valid arithmetic with infinity. – 6005 Oct 18 '16 at 18:51
• @6005 I think most people at the level of the OP equate "number" with "natural number" or "integer" or "rational number" or maybe "real number", so it is helpful to point out that $\infty$ is not one of these. Sure, there are various number systems in which $\infty$ is a quantity that you can do arithmetic with - but introducing them when someone isn't clear what is and is not a natural number only confuses, rather than clarifies. – Chris Taylor Oct 18 '16 at 20:04
As there is a bijection $$f(x) = x + 1$$ sending any odd number to an even, this shows that the sets have equal size.
Here, I assumed that the natural numbers start with $1$, if they should start with $0$, simply define the same function on the even numbers.
## protected by Asaf Karagila♦Oct 18 '16 at 12:44
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).
Would you like to answer one of these unanswered questions instead? | 2019-06-25T19:59:20 | {
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https://math.stackexchange.com/questions/4064353/find-all-the-numbers-that-are-equal-to-one-quarter-of-the-sum-of-their-own-digit | # Find all the numbers that are equal to one quarter of the sum of their own digits
The number $$1.5$$ is special because it is equal to one quarter of the sum of its digits, as $$1+5=6$$ and $$\frac{6}{4}=1.5$$ .Find all the numbers that are equal to one quarter of the sum of their own digits.
I was puzzling over this question for a while, but I couldn't find a formula without using the $$\sum$$ , but I can't really solve generalizations, only come up with them. The only thing I could come up with was to brute-force it, but I can't really come up with any 'special' numbers. Any help?
• Two examples are $2.25$, $3.75$ – mfl Mar 16 at 17:26
• Thanks @mfl! But is there some way to find these numbers? (like you can find the nth term of the sequence of numbers that work) – Arale Mar 16 at 17:27
• Hint: a special number $q$ is of the form $x$, $x.5$, $x.25$ or $x.75$ where $x$ is an integer. So the sum of its digits is at most $7$ plus the sum of the digits of $x$. But the sum of the digits of $x$ is at most $x$ for $x \geq 1$. So we must have, if $x \geq 1$, $x \leq q \leq .25*(12+x) \leq 3+x/4$ so $3x/4 \leq 3$. Then only a small number of cases remain. – Mindlack Mar 16 at 17:29
• Notice that sum of digits and divided by four... since $\frac{1}{4}=0.25$ it follows that any such numbers have at most two digits to the right of the decimal point. As for digits to the left of the decimal point, you should be able to get an idea of how many possible you can have there. It will be limited due to the constraints. From that point you should be able to express the numbers in the appropriate range with a small handful of variables and use elementary-number-theory approaches to find what values are possible. – JMoravitz Mar 16 at 17:30
• Ok thank you both of you for the hints :) I will think about this very hard! – Arale Mar 16 at 17:31
My first draft was messy.
With hindsight.
Let $$N = \frac {K}4 = M\frac i4$$ where $$K$$ is the sum of the digits. $$M$$ is the quotient integer of dividing $$K$$ by $$4$$ and $$i=0,1,2,3$$ is the fractional remainder.
Note: $$K$$ is the sum of the digits of $$N$$ which is also the sum of the digits of $$M$$ plus the sum of the digits of $$\frac i4$$.
So $$4N = 4M + i =K$$. If the digits of $$M$$ are $$d_k$$ then
$$4N = 4M + i=4d_m \times 10^m + ..... + 4d_1\times 10 + 4d_0 + \begin{cases}0\\ 1\\ 2\\ 3\end{cases}$$
While $$K =d_m + ..... + d_1 + d_0 +\text{sum of the digits of }\frac i4$$ so
$$4d_m \times 10^m + ..... + 4d_1\times 10 + 4d_0 + \begin{cases}0\\ 1\\ 2\\ 3\end{cases}=d_m + ..... + d_1 + d_0 +\text{sum of the digits of }\frac i4$$
$$d_m(4\cdot 10^m-1) + .... d_1(40 -1) + 3d_0 + \begin{cases}0\\ 1\\ 2\\ 3\end{cases}= \text{sum of the digits of }\frac i4$$
But $$\text{sum of the digits of }\frac i4 =\begin{cases}0\\2+5=7\\5\\7+5=12\end{cases}$$
So we have $$d_m(4\cdot 10^m-1) + .... d_1(40 -1) + 3d_0 = \begin{cases}0-0=0\\2+5-1=6\\5-2=3\\7+5-3=9\end{cases}$$
But the RHS is less than $$10$$ so none of the $$d_{k_{k> 0}}$$ can be none zero and $$M$$ is a single digit, $$d_0$$.
And we have
$$3d_0 = \begin{cases}0\\6\\3\\9\end{cases}$$
So $$M=d_0 = \begin{cases}0\\2\\1\\3\end{cases}$$
ANd $$N = M +\frac i4 = \begin{cases}0\\2.25\\1.5\\3.75\end{cases}$$
==== first answer below (more thought and scrabble and not as slick-- a "rough draft")=====
So the sum of the digits is an integer. And and integer divided by $$4$$ will result in four possible cases.
$$\frac {integer}4 = \begin{cases}n.00\\n.25\\n.5\\n.75\end{cases}$$.
So the least power of $$10$$ in the number can be $$-2$$.
So let $$N = \sum_{k=-2}^m a_k\times 10^m$$ (where $$a_m \ne 0$$ but the other $$a_k$$ may be).
$$\sum_{k=-2}^m a_k\times 10^m = \frac {\sum_{k=-2}a_k}4< \frac{9\times (m+2)}4$$ but even more so $$a_{-1}+ a_{-2} \le 7+5 = 12$$ so $$N < 9m + 12$$ and $$a10^m \le N < \frac {9m + 12}4<3m + 3$$ can give us an upper limit of $$m$$.
If $$m \ge 2$$ then $$3m + 3 < 10^m$$ obviously so $$m\le 1$$. If $$m =1$$ then $$10a_1 < 6$$ is impossible. SO $$m\le 0$$
So we have four cases:
$$N = a.00$$ and so $$a = \frac 14 a$$ and $$a = 0$$ and $$N = 0$$. That's a solution.
$$N = a.25$$ and $$a +\frac 14 = \frac 14(a + 7)$$ so $$4a+1=a+7$$ so $$3a=6$$ so $$a=2$$ and $$N = 2.25$$ yield $$2.25 =\frac 14 \times 9$$. That's a solution.
$$N = a.5$$ and $$a + \frac 12 = \frac 14(a+5)$$ so $$4a + 2 = a+5$$ so $$3a = 3$$ and $$a = 1$$ and $$N=1.5$$ yields your solution.
$$N = a.75$$ and $$a + \frac 34 =\frac 14(a+12)$$ so $$4a + 3=(a+12)$$ so $$3a =9$$ and $$a = 3$$ and $$N=3.75 = 3\frac 34 = \frac {15}4 = \frac {3+7+5}4$$ is a solution.
Pretty cool and pretty cute.
• Thank you for taking this question into so much detail :P – Arale Mar 16 at 19:05
This is not much better than a brute force method:
• The sum of the digits $$S$$ is a non-negative integer, so a quarter of it $$\frac{S}4$$ is non-negative, of the form $$x$$ or $$x.25$$ or $$x.5$$ or $$x.75$$ for some non-negative integer $$x$$.
• We must have $$x < 10$$ since if $$10 \le x\lt 100$$ then $$40 \le 4x \le S \lt 412$$ so the sum of the digits of $$x$$ must be at least $$40-12=28$$ but two-digit integers have sums of digits no more than $$18$$. Similarly with larger $$x$$.
• So a satisfactory $$x$$ has sum of digits $$4x=S=x$$ with solution $$x=0$$
• and a satisfactory $$x.25$$ has sum of digits $$4(x+\frac14)=S=x+7$$ with solution $$x=2$$
• and a satisfactory $$x.5$$ has sum of digits $$4(x+\frac12)=S=x+5$$ with solution $$x=1$$
• and a satisfactory $$x.75$$ has sum of digits $$4(x+\frac34)=S=x+12$$ with solution $$x=3$$
making the numbers which work $$0, \quad2.25,\quad 1.5,\quad 3.75$$
• thanks for the detailed answer :P – Arale Mar 16 at 19:04 | 2021-07-25T02:52:21 | {
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https://math.stackexchange.com/questions/2872828/show-that-this-definition-of-derivative-implies-the-other-one | # Show that this definition of derivative implies the other one.
I have two definitions of derivatives. The first one is the one provided in Rudin:
Let $f: [a,b] \to \mathbb{R}$ be a function. Let $x \in [a,b]$.
Let $\phi: (a,b) \setminus \{x\}\to\mathbb{R}: t \mapsto \frac{f(t)- f(x)}{t-x}$
Then we define $f'(x) = \lim_{t \to x} \phi(t)$, provided that the limit exists.
Here's the second one:
Let $f: E \subseteq \mathbb{R} \to \mathbb{R}$ be a function, $x \in E$ and $x$ a limit point of $E$. Put
$\phi: E \setminus \{x\} \to \mathbb{R}: t \mapsto \frac{f(t)- f(x)}{t-x}$.
Then we define $f'(x) = \lim_{x \to t} \phi (t)$, provided that the limit exists.
I want to show that the second definition implies the first, when we put $E = [a,b]$.
So, in particular, I want to show that for every $x \in [a,b]$
$\lim_{t \to x, t \in (a,b) \setminus \{x\}} \frac{f(t)-f(x)}{t-x}$ exists $\iff$ $\lim_{t \to x, t \in [a,b] \setminus \{x\}} \frac{f(t)-f(x)}{t-x}$
and if one of the two exists, both are equal.
Clearly, $\Leftarrow$ is satisfied (immediately from the limit definition).
For the other direction, put $q:= \lim_{t \to x, t \in (a,b) \setminus \{x\}} \frac{f(t)-f(x)}{t-x}$. If $x \in (a,b)$, then it is an interior point and the limit is equal to the other one.
WLOG, assume $x = a$. Let $\epsilon > 0$. Choose $\delta > 0$ such that $|q - \frac{f(t)-f(a)}{t-a}| < \epsilon$ for all $t \in (a,b)$ with $0 < |t-x|< \delta$.
Then, if $t \in [a,b] \setminus \{a\}$ with $0 < |t-a| < \min \{\delta, |b-a|\}$, then $t \neq b, t \neq a$ and we can make the quantity smaller than $\epsilon$.
Does this seem correct? Are there any arguments against using the second definition over the first one, as it seems more general?
• Your argument is correct. The two definitions are equivalent and there is reason to prefer one over the other. – Kavi Rama Murthy Aug 5 '18 at 12:03
• Thanks . And what is that reason if I may ask? – user370967 Aug 5 '18 at 12:05
Let us generalize the second definition. For $x \in \mathbb{R}$ we denote by $\mathfrak{N}(x)$ the set of all neighborhoods of $x$ in $\mathbb{R}$ (a neighborhood of $x$ is a set $N \subset \mathbb{R}$ such that $(x -\varepsilon, x +\varepsilon ) \subset N$ for some $\varepsilon > 0$).
Let $x \in E \subset \mathbb{R}$. Then obviously the following are equivalent:
(1) $x$ is a limit point of $E$.
(2) There exists $N \in \mathfrak{N}(x)$ such that $x$ is a limit point of $N \cap E$.
(3) For all $N \in \mathfrak{N}(x)$, $x$ is a limit point of $N \cap E$.
Now let $f: E \to \mathbb{R}$ be a function and $x \in E$ be a limit point of $E$. For each $N \in \mathfrak{N}(x)$ define $$\phi_N: N \cap E \setminus \{x\} \to \mathbb{R}: t \mapsto \frac{f(t)- f(x)}{t-x} .$$ The map $\phi$ from the second definition is given as $\phi = \phi_{\mathbb{R}}$. If $\lim_{x \to t} \phi_N (t)$ exists, we denote it by $f'_N(x)$. In case $N = \mathbb{R}$ we simply write $f'(x)$.
The following are obvious:
(1) If $\lim_{x \to t} \phi (t)$ exists, then $\lim_{x \to t} \phi_N (t)$ exists for all $N \in \mathfrak{N}(x)$ and $f'_N(x) = f'(x)$.
(2) If $\lim_{x \to t} \phi_N (t)$ exists for some $N \in \mathfrak{N}(x)$, then $\lim_{x \to t} \phi (t)$ exists and $f'(x) = f'_N(x)$.
Now let $E = [a,b]$. To avoid confusion the function $\phi$ from the first definition will be denoted by $\Phi$.
For $x \in (a,b)$ we have $\Phi = \phi_{(a,b)}$, for $x = a$ we have $\Phi = \phi_{(a-1,b)}$ and for $x = b$ we have $\Phi = \phi_{(a,b+1)}$.
This shows that the first and the second definition are equivalent.
• Thanks for your detailed answer! I was wondering if my approach would work too? – user370967 Aug 5 '18 at 14:36
• Yes, of course, your approch works. I only wanted to round off the picture. – Paul Frost Aug 5 '18 at 14:44
• By the way, where did you find the second definition? – Paul Frost Aug 5 '18 at 14:45
• Thank you! I personally prefer this definition, but one hardly ever finds it in the literature. – Paul Frost Aug 5 '18 at 14:50
• I share your point of view. Perhaps you are interested in the discussion following this question: math.stackexchange.com/q/2830875 Most people seem to dislike the more general definition for functions defined on an arbitrary $E$ and prefer to regard $f$ as differentiable at a point $x \in E$ if there exist an open neighborhood $U$ of $x$ in $\mathbb{R}$ and a differentiable function $\hat f : U \to \mathbb{R}$ such that $\hat f \mid_{U \cap E} = f \mid_{U \cap E}$. In my eyes this is more a theorem than a definition. – Paul Frost Aug 5 '18 at 15:17 | 2020-08-13T09:35:40 | {
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https://math.stackexchange.com/questions/3036489/proof-for-this-binomial-coefficients-equation | # Proof for this binomial coefficient's equation
For $$k, l \in \mathbb N$$ $$\sum_{i=0}^k\sum_{j=0}^l\binom{i+j}i=\binom{k+l+2}{k+1}-1$$ How can I prove this?
I thought some ideas with Pascal's triangle, counting paths on the grid and simple deformation of the formula.
It can be checked here (wolframalpha).
If the proof is difficult, please let me know the main idea.
Sorry for my poor English.
Thank you.
EDIT: I got the great and short proof using Hockey-stick identity by Anubhab Ghosal, but because of this form, I could also get the Robert Z's specialized answer. Then I don't think it is fully duplicate.
• – user10354138 Dec 12 '18 at 10:07
• @user10354138 The double sum is related Hockey-Stick Identity but it is not a duplicate of the linked question. Moreover OP asks for a combinatorial proof (counting paths...) – Robert Z Dec 12 '18 at 11:30
• @RobertZ The OP didn't ask for a combinatorial proof, only for a proof ("How can I prove this?" in the quote, not "How can I prove this combinatorially?"). It can be proved by applying the linked result twice, so it is a duplicate. – user10354138 Dec 12 '18 at 13:18
• @user10354138 Fine. So we have a different opinion on this matter. Have a nice day. – Robert Z Dec 12 '18 at 15:48
Your idea about a combinatorial proof which is related to counting paths in a grid is a good one!
The binomial $$\binom{i+j}i$$ counts the paths on the grid from $$(0,0)$$ to $$(i,j)$$ moving only right or up. So the double sum $$\sum_{i=0}^k\sum_{j=0}^l\binom{i+j}i-1$$ counts the number of all such paths from $$(0,0)$$ to any vertex inside the rectangle $$(0,k)\times (0,l)$$ different from $$(0,0)$$.
Now consider the paths from $$(0,0)$$ to $$(k+1,l+1)$$ different from $$(0,0)\to (0,l+1)\to (k+1,l+1)\quad\text{and}\quad (0,0)\to (k+1,0)\to (k+1,l+1)$$ which are $$\binom{k+l+2}{k+1}-2.$$
Now any path of the first kind can be completed to a path of the second kind by changing direction, going to the boundary of the rectangle $$(0,k+1)\times(0,l+1)$$ and then moving to the corner $$(k+1,l+1)$$ along the side.
Is this a bijection between the first set of paths and the second one?
$$\displaystyle\sum_{i=0}^k\sum_{j=0}^l\binom{i+j}i=\sum_{i=0}^k\sum_{j=i}^{i+l}\binom{j}i=\sum_{i=0}^k\binom{i+l+1}{i+1} \ ^{[1]}$$
$$=\sum_{i=0}^k\binom{i+l+1}{l}=\sum_{i=l}^{k+l+1}\binom{i}{l}−1=\binom{k+l+2}{k+1}-1\ ^{[1]}$$
1. Hockey-Stick Identity
$$\ds{\sum_{i = 0}^{k}\sum_{j = 0}^{\ell} {i + j \choose i} = {k + \ell + 2 \choose k + 1} - 1:\ {\LARGE ?}.\qquad k, \ell \in \mathbb{N}}$$.
\begin{align} &\bbox[10px,#ffd]{\sum_{i = 0}^{k}\sum_{j = 0}^{\ell} {i + j \choose i}} = \sum_{i = 0}^{k}\sum_{j = 0}^{\ell}{i + j \choose j} = \sum_{i = 0}^{k}\sum_{j = 0}^{\ell}{-i - 1 \choose j} \pars{-1}^{\,j} \\[5mm] = &\ \sum_{i = 0}^{k}\sum_{j = 0}^{\ell}\pars{-1}^{\,j} \bracks{z^{\, j}}\pars{1 + z}^{-i - 1} = \sum_{i = 0}^{k}\sum_{j = 0}^{\ell}\pars{-1}^{\,j} \bracks{z^{0}}{1 \over z^{\, j}}\,\pars{1 + z}^{-i - 1} \\[5mm] = &\ \bracks{z^{0}}\sum_{i = 0}^{k}\pars{1 \over 1 + z}^{i + 1} \sum_{j = 0}^{\ell}\pars{-\,{1 \over z}}^{\,j} \\[5mm] = &\ \bracks{z^{0}}\braces{{1 \over 1 + z}\, {\bracks{1/\pars{1 + z}}^{k + 1} - 1 \over 1/\pars{1 + z} - 1}} \braces{{\pars{-1/z}^{\ell + 1} - 1 \over -1/z - 1}} \\[5mm] = &\ \bracks{z^{0}}\braces{% {1 - \pars{1 + z}^{k + 1} \over -z} \,{1 \over \pars{1 + z}^{k + 1}}} \braces{{\pars{-1}^{\ell + 1} - z^{\ell + 1} \over -1 - z}\,{z \over z^{\ell + 1}}} \\[5mm] = &\ \bracks{z^{\ell + 1}}\braces{1 - {1 \over \pars{1 + z}^{k + 1}}} \braces{z^{\ell + 1} + \pars{-1}^{\ell} \over 1 + z} \\[5mm] = &\ \pars{-1}^{\ell}\bracks{z^{\ell + 1}} \bracks{\pars{1 + z}^{-1} - \pars{1 + z}^{-k - 2}} \\[5mm] = &\ \pars{-1}^{\ell}\bracks{\pars{-1}^{\ell + 1} - {-k - 2 \choose \ell + 1}} \\[5mm] = &\ -1 - \pars{-1}^{\ell}\,{-\bracks{-k - 2} + \bracks{\ell + 1} - 1 \choose \ell + 1}\pars{-1}^{\ell + 1} \\[5mm] = &\ -1 + { k + \ell + 2 \choose \ell + 1} = \bbx{{k + \ell + 2 \choose k + 1} - 1} \end{align} | 2019-01-16T10:29:21 | {
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https://mathhelpboards.com/threads/statements-about-the-set-2-m.26834/ | # Statements about the set 2^M
#### mathmari
##### Well-known member
MHB Site Helper
Hey!!
1. Let $M:=\{7,4,0,3\}$. Determine $2^M$.
2. Prove or disprove $2^{A\times B}=\{A'\times B'\mid A'\subseteq A, B'\subseteq B\}$.
3. Let $a\neq b\in \mathbb{R}$ and $M:=2^{\{a,b\}}$. Determine $2^M$.
4. Is there a set $M$, such that $2^M=\emptyset$ ?
First of all how is $2^M$ defined? Is this the powerset?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
First of all how is $2^M$ defined? Is this the powerset?
Hey mathmari !!
Yep.
So $2^{\{a,b\}}=\{\varnothing,\{a\},\{b\},\{a,b\}\}$.
#### mathmari
##### Well-known member
MHB Site Helper
Yep.
So $2^{\{a,b\}}=\{\varnothing,\{a\},\{b\},\{a,b\}\}$.
So we have the following:
1. $2^M=\{\emptyset, \{7\}, \{4\}, \{0\}, \{3\}, \{7,4\}, \{7,0\}, \{7,3\}, \{4,0\}, \{4,3\}, \{0,3\},\{7,4,0\}, \{7,4,3\}, \{7,0,3\}, \{4,0,3\}, \{7,4,0,3\}$
2. $2^{A\times B}=\text{ set of every subset of } A\times B=\{(a,b)\mid a\in A, b\in B\}$.
So it is of the form $2^{A\times B}=\{\emptyset, \{(a_i, b_i)\}, \{(a_i, b_i), (a_j, b_j)\}, \cdots \}$, or not?
This is not of the form $\{A'\times B'\mid A'\subseteq A, B'\subseteq B\}$, is it?
3. We have that $M=\{\emptyset, \{a\}, \{b\}, \{a,b\}\}$.
The $2^M=\{\emptyset, \{\emptyset\}, \{\{a\}\},\{\{b\}\}, \{\{a,b\}\}, \{\emptyset, \{a\}\}, \{\emptyset, \{b\}\}, \{\emptyset, \{a,b\}\}, \{\{a\}, \{b\}\}, \{\{a\}, \{a,b\}\}, \{\{a,b\}, \{b\}\}, \{\emptyset, \{a\}, \{b\}\}, \{\emptyset, \{a\}, \{a,b\}\}, \{\emptyset, \{a,b\}, \{b\}\}, \{\emptyset, \{a\}, \{b\}, \{a,b\}\}$
Is this correct? Or do I miss something?
4. There isn't such a set, since at least the emptyset is contained in the powerset. Is this correct?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
So we have the following:
1. $2^M=\{\emptyset, \{7\}, \{4\}, \{0\}, \{3\}, \{7,4\}, \{7,0\}, \{7,3\}, \{4,0\}, \{4,3\}, \{0,3\},\{7,4,0\}, \{7,4,3\}, \{7,0,3\}, \{4,0,3\}, \{7,4,0,3\}$
2. $2^{A\times B}=\text{ set of every subset of } A\times B=\{(a,b)\mid a\in A, b\in B\}$.
So it is of the form $2^{A\times B}=\{\emptyset, \{(a_i, b_i)\}, \{(a_i, b_i), (a_j, b_j)\}, \cdots \}$, or not?
This is not of the form $\{A'\times B'\mid A'\subseteq A, B'\subseteq B\}$, is it?
3. We have that $M=\{\emptyset, \{a\}, \{b\}, \{a,b\}\}$.
The $2^M=\{\emptyset, \{\emptyset\}, \{\{a\}\},\{\{b\}\}, \{\{a,b\}\}, \{\emptyset, \{a\}\}, \{\emptyset, \{b\}\}, \{\emptyset, \{a,b\}\}, \{\{a\}, \{b\}\}, \{\{a\}, \{a,b\}\}, \{\{a,b\}, \{b\}\}, \{\emptyset, \{a\}, \{b\}\}, \{\emptyset, \{a\}, \{a,b\}\}, \{\emptyset, \{a,b\}, \{b\}\}, \{\emptyset, \{a\}, \{b\}, \{a,b\}\}$
Is this correct? Or do I miss something?
4. There isn't such a set, since at least the emptyset is contained in the powerset. Is this correct?
All correct.
Did you miss something? Only the closing brace.
#### mathmari
##### Well-known member
MHB Site Helper
All correct.
Did you miss something? Only the closing brace.
Great!! At 2 how can we disprove that formally?
#### topsquark
##### Well-known member
MHB Math Helper
A question of my own.
Say we have $$\displaystyle A = \{1, 2, 4 \}$$ and $$\displaystyle B = \{ 1, 2, 3 \}$$
Why wouldn't $$\displaystyle 2^{A \times B}$$ have sets with the form $$\displaystyle \{ a_i, b_j \}$$ rather than just $$\displaystyle \{ a_i, b_i \}$$?
And, since we are talking about a Cartesian product wouldn't $$\displaystyle 2^{ \{ 1, 2 \} }$$ be different from $$\displaystyle 2^{ \{ 2, 1 \} }$$?
Thanks!
-Dan | 2020-07-10T02:19:05 | {
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https://www.physicsforums.com/threads/finding-the-period-of-this-trig-function.306360/ | Finding the period of this trig function
Dell
f(x)=sin3x*cos3x
i am looking for the period of this function,
what i did, and now i see that it is wrong was compared it to zero and looked for any time that
sin3x=0 or cos3x=0
this really did give an answer for every time the function reached 0, but did not take into account that once i get 0+ and once i get 0-, how do i find the period?
what i did
sin3x=0 or cos3x=0
x=K*pi/3 or x=pi/6 + K*pi/3
then i got
T=pi/6
do i need to use trig identities to bring it to a function of just sin or cos?? which identities?
Homework Helper
How about sin(2a)=2*sin(a)*cos(a)? More generally you can use product to sum formulas like sin(a)*cos(b)=(1/2)*(sin(a+b)+sin(a-b)).
Dell
now what do i do to find the period? what do i compare it to
1/2*sin(6x)=1/2 ?
sin(6x)=1
6x=pi/2
x=pi/12
not right,
mathematically how do i ptove that the period is pi/3
Homework Helper
sin(x) has period 2pi, right? For sin(6x), 6x goes from 0 to 2pi (one period) while x goes from 0 to 2pi/6. 2pi/6 is pi/3.
Dell
okay i understand that, but is there no way to get to this with equations, for example, i had another question to find period for cos^2(x)
so i said
cos^2(x)=cos^2(0)=1
cos^2(x)=1
cos(x)=+-1
x=0+2pi*K or x=pi + 2pi*K
x=[0 pi 2pi 3pi...]
T=pi
how would you solve this like you solved the sin, using cos(x) has a period of 2pi,
Homework Helper
I would use product forms to write cos(x)^2 as a sum of sines and cosines, like I said in the first post. cos(x)^2=(cos(2x)+1)/2. The '2x' part is what determines the period.
Dell
so only the actual trig function makes any difference, the peiod of cosx is the same as, (4*cos(x) +2) for eg??
Homework Helper
Sure. Isn't that pretty easy to see? Imagine plotting it.
AUMathTutor
"so only the actual trig function makes any difference, the peiod of cosx is the same as, (4*cos(x) +2) for eg?? "
Yes, a function is periodic with period T iff (d^n)f(x)/(dx^n) = (d^n)f(x + T)/(dx^n) for all n (including zero, where the zeroth derivative is interpreted to mean the original function). At least that's my understanding... someone may please correct me if I'm in error. | 2022-12-04T11:33:31 | {
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https://www.physicsforums.com/threads/basic-kinematics-question.140500/ | # Basic Kinematics Question
1. Oct 29, 2006
### prace
Hello,
I am having an issue with this simple kinematics problem. I have the question visualized, well, at least I think I do. Here is the question:
Two cars are traveling along a straight road. Car A maintains a constand speed of 80 km/h; car B maintains a constant speed of 110 km/h. At t = 0, car B is 45 km behind car A. How much farther will car A travel before it is overtaken by car B?
So to start off, I graphed the problem and came up with this:
http://album6.snapandshare.com/3936/45466/853951.jpg [Broken]
where series 1 is car A and series 2 is car B.
I also know that this has to do with position and time, so I am most likely going to use one of my kinematical equations that deals with position and time, so I chose the equation: $$x_{f}=x_{i}+v_{0}t+\frac{1}{2} a t^2$$ because I know that the accelleration is constant and is equal to 0 since the cars are not speeding up. I also know that the final positions have to be equal, but how do I equate the final positions of the two cars? Or is this even the right direction for me to solve the problem?
I mean, I could just graph the two lines, find the equations, and then find the intersection of those two lines, but I would like to learn how to do it using the kinematical equations.
Thanks
Last edited by a moderator: May 2, 2017
2. Oct 29, 2006
### BishopUser
You are on the right track. By finding the intersection point on that graph you have calculated how much time passes before the two cars come together. Now you must just realize that the problem is just asking how far does car A travel during this time.
The 2 lines that you graphed are the exact same thing as the kinematic equations btw. You probably graphed something like y = 80x vs y = 110x -45.
Now look at the kinematic equation
xf = xi + vit + 1/2at^2
you stated that acceleration is 0 so its just
xf = xi + vit
final position is just the independent variable (y). xi is the initial position (y-intercept). vi is the initial velocity in the problem (coefficient of x), and T is the variable (x).
Once you set up the 2 equations you just find the intersection point by setting the 2 equations equal to each other. then you can solve for the time
Last edited: Oct 29, 2006
3. Oct 29, 2006
### prace
So that's it? Just do it mathmatically and it is fine? Awesome, I solved for the intersection time to be .6 hours and then plugged that back into my equation for Car A and got a distance of 4.8 km. Does that sound good?
Thank you for your help Bishop!!
4. May 28, 2010
### bdblankster
I'm currently working on this problem and I agree with that I need to find equations for both cars and then set them equal to each other. However, I did not get .6 hrs for the time. I ended up with 1.5 hours. | 2019-01-21T21:53:00 | {
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http://mathoverflow.net/questions/64116/are-measurable-automorphism-of-a-locally-compact-group-topological-automorphisms?sort=oldest | Are measurable automorphism of a locally compact group topological automorphisms?
Consider a locally compact group $G$, considered as a measurable space with the completed Borelstructure wrt. the Haarmeasure. Consider a map $f:G \to G$, which is measurable and has an inverse, which is then also measurable. Is $f$ an homeomorphism?
What if $G$ is abelian? If not, what are necessary conditions on $G$, such that this is the case.
For $\mathbb{R}$, it seems to be true: see http://math.stanford.edu/~chris/additive.pdf.
-
Surely you mean that $f$ is a group homomorphism? Could you also clarify that you really do mean "measurable" in this strong sense? – Matthew Daws May 6 '11 at 13:17
If you are interested in a fixed measure, then the completion of the $\sigma$ algebra wrt. to this measure means that you consider the $\sigma$ algebra generated by the measurable sets and the set of outer zero measure, or not? I thought that might simplify a lot, but I am not really experienced in measure theory. – Marc Palm May 6 '11 at 13:42
The comments in this question: mathoverflow.net/questions/19402 show that one has to be very careful with such things... – Matthew Daws May 6 '11 at 13:47
For example, as Robin Chapman says: mathoverflow.net/questions/19402/… the map $\mathbb R \rightarrow \mathbb R^2; x\mapsto (x,0)$ is continuous (and so Borel measurable), even a group homomorphism, but it is not measurable if both $\mathbb R$ and $\mathbb R^2$ are given their completed sigma-algebras for Lebesgue measure. – Matthew Daws May 6 '11 at 14:00
Here is a result by Adam Kleppner (MEASURABLE HOMOMORPHISMS OF LOCALLY COMPACT GROUPS, PROC AMER MATH SOC , vol. 106, no. 2, 1989, 391-395): any measurable homomorphism between locally compact groups is continuous. Actually what he really needs, for a homomorphism $\alpha:G\rightarrow H$, is that $\alpha^{-1}(U)$ is measurable in $G$ for every open subset $U\subset H$.
-
Thank goodness. I thought the full result was out there somewhere, but I couldn't find the reference and began doubting my memory. Thanks for sharing it! – Clinton Conley May 6 '11 at 19:38
You should also look at the correction... Proc. Amer. Math. Soc. 111 (1991), no. 4, 1199–1200. Odd that the paper of Neeb which I mentioned (written 8 years later) doesn't mention this! – Matthew Daws May 6 '11 at 19:49
@Matthew: Thanks! – Clinton Conley May 6 '11 at 21:15
The basic fact about locally compact groups is that you can recover the topology from the underlying measure space:
This is because, for any measurable subset $X\subset G$ of positive measure, the set $$X^{-1}X:=\{x^{-1}y\\,|\\,x\in X, y\in Y\}$$ is a neighborhood of the neutral element. Letting $X$ vary along all measurable subset of positive measure you get a basis of neighborhoods of $e\in G$. By translating by group elements, you get a basis of neighborhoods of any element $g\in G$. And so you recover the topology on $G$.
Corollary:
Since the topology is entirely encoded in the measurable structure, an automorphism that respects the measurable structure, will also respect the topology, i.e., be continuous.
-
I'd like to note here that the statement that $X^{-1}X$ contains an open neighborhood of $e$ is not so hard to prove if one notes that $f(y)=\int \chi_{X}(x)\chi_{X}(xy^{-1}) dx$ is continuous (replacing $X$ by a set of finite measure it contains) and consider what happens when $y=0.$ – Benjamin Hayes May 7 '11 at 11:57
You should look up "automatic continuity." Here is a paper by J.W. Lewin that may be of interest:
http://www.jstor.org/pss/2044356
-
Clinton! Welcome. – Andres Caicedo May 6 '11 at 14:03
Hi Andres! Thanks, I finally took the plunge. I see Martin Goldstern arrived today too -- there must be something in the Viennese air. – Clinton Conley May 6 '11 at 14:07
Some partial results: By Hewitt+Ross, Theorem 22.18 (http://www.ams.org/mathscinet-getitem?mr=551496) a Borel measurable homomorphism between two locally compact groups is continuous if the codomain is separable or $\sigma$-compact. (I think this result goes back to Banach).
A nice paper (see edit below!), which proves some similar results, is:
MR1473172 (98i:22003)
Neeb, Karl-Hermann(D-ERL-MI)
On a theorem of S. Banach. (English summary)
J. Lie Theory 7 (1997), no. 2, 293–300.
http://www.ams.org/mathscinet-getitem?mr=1473172
I'm afraid that I don't know the limits of these sort of results (i.e. a counter-example in the non-separable case, say), or if being an automorphisms gives anything more.
Edit: As Julien points out in a comment, this paper of Neeb is a little suspect, so I withdraw my recommendation. André Henriques shows, in a short argument, that given a bijective group homomorphism which is measurable for the completed Haar measure, the homomorphism must be continuous.
I was a bit worried about the difference between "measurable" in the sense of "inverse image of open set is Borel or Haar measurable" and this stronger sense. But I think uniqueness of Haar measures rescues us. Indeed, if $\tau:G\rightarrow G$ is a continuous automorphism of $G$, then the map $A\mapsto |\tau(A)|$ will be a left invariant measure; as $\tau$ is a homeomorphism, this measure also assigns finite measure to compacts, and non-zero measure to open sets. Thus it will be proportional to the Haar measure. As $\tau$ preserves Borel sets, it follows that it will preserve all the Haar measurable sets, and so will be measurable in this strong sense. Note that the example of Robin Chapman shows that this isn't necessarily so for a merely injective, continuous homomorphism.
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I should say that this refers to "Borel set" not "Completed Borel for Haar measure". The latter seems like a somewhat stronger condition, a priori. – Matthew Daws May 6 '11 at 12:43
The basic thing in the proof: if $A$ is a measurable set with postive measure, then $A A^{-1}$ contains a neighborhood of the identity. – Gerald Edgar May 6 '11 at 13:09
This paper by Neeb is actually not so nice - first, it seems to ignore all the literature on the subject (there are simpler proofs of that theorem of Banach, without the unnecessary assumptions on arcwise connectedness, and they were available before publication of his paper; actually, the result appears in textbooks, like Kechris' book); second, there is a major error in the part about representation theory (claiming that the operator norm topology and strong operator topologies have the same Borel sets, which is dead wrong). There seems to be an erratum of sorts for that paper (MR1747686 ). – Julien Melleray May 8 '11 at 15:14
@Julien: Thanks for the heads-up about the erratum! Yeah, I coming to the same conclusion (given the much better papers listed above). – Matthew Daws May 8 '11 at 19:06
Actually, there is a further major mistake: the reason for the existence of the paper seems to be a confusion two completely different meanings of "Baire sets" (the one Banach uses is the $\sigma$-algebra generated by the Borel sets and the meager sets) and the other one is the $\sigma$-algebra by the compact $G_{\delta}$'s. I give a reference to Banach's work in my answer here mathoverflow.net/questions/57616/… and François's answer to the same question points to Pettis' standard results Julien quotes. – Theo Buehler May 8 '11 at 19:29 | 2015-07-05T21:19:36 | {
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https://www.physicsforums.com/threads/probability-that-a-rectangle-lies-within-a-circle.611734/ | # Probability that a Rectangle lies within a circle
1. Jun 6, 2012
### guccimane
This is not really a homework problem, I'm just doing it as an exercise puzzle. I think I'm on the right track, but at this point I feel a little exhausted and would love a hint.
1. The problem statement, all variables and given/known data
Let C be a unit circle: x^2+y^2=1 . Let "p" be a point on the circumference and "q" be a point anywhere inside C. Finally, let the line p-q be the diagonal of a rectangle R, the sides of which are parallel to the x and y axes. What's the probability that ALL points within R lie within C?
Presumed steps: 1,2,3, ?, .... n) integrate some density function, get answer.
3. The attempt at a solution
(t:="theta")
I'm pretty certain that 1) and 2) are right:
1) Find the Cartesian coordinate representation of the 4 vertices of R:
p= (cos(t), sin(t)); q= (a, b); p'= (cos(t), b); q'= (a, sin(t))
2) We know that p and q are within the disk C, so we need the conditions under which p' and q' lie in C:
for p' : cos^2(t) + b^2 < 1 i.e |b|<|sin(t)|
for q' : a^2 + sin^2(t) < 1 i.e. |a|<|cos(t)|
3) Next is my attempt to find the following probabilities:
Letting 0≤ t ≤ pi/2 -- trying to stay in the first quadrant;
P(|b|<|sin(t)|) = (∫[0..cos(t)] of √(1-x^2) dx) / (pi/4)
= (1/2*(cos(t)(sin(t)) + pi/2 - t)/ (pi/4)
P(|a|<|cos(t)|) = (∫[0..sin(t)] of √(1-y^2) dy) / (pi/4)
= (1/2*(cos(t)(sin(t)) + t) /(pi/4)
4) One has to find the probability of the intersection of the two events in (3).
My initial though was to multiply the two probability function in (3) and then integrate from 0 to pi/2; Result: wrong answer.
In my second attempt, I tried first differentiating the two probability functions, then multiplying them together, then integrating from 0 to pi/2.
(Don't ask why - I guess I thought differentiating them would make them look like density functions ... or something). Result: Wrong answer.
==================================================================
My solution fell off the rails 4 - or perhaps at 3 in some ridiculously elementary way(s). This a somewhat basic problem that should conclude in the integration of a prob.-dens. function.
I'll be eternally grateful for any assistance on this!
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
Last edited: Jun 6, 2012
2. Jun 6, 2012
### Ray Vickson
Assume the "outer" corner is at angle t, counterclockwise from the x-axis, and assume that t is uniformly distributed on (0,2π). Assume also that the point q is uniformly distributed within the circle. If F={rectangle fits in circle} we have
$$P\{F\} = \frac{1}{2 \pi} \int_{0}^{2 \pi} P\{ F|t\}\, dt.$$
It is clear that we can just integrate from 0 to π/2, then multiply by 4.
For 0 < t < π/2, how do we find the largest possible rectangle? Just draw a horizontal and vertical line through $p = (\cos(t),\sin(t))$ and see where these lines cut the circle again. That will give the largest rectangle Rt. If q lies inside Rt. the constructed rectangle with opposite corners p and q will lie completely inside the circle. The sides of Rt are of lengths 2*cos(t) and 2*sin(t), and so its area is easy to compute in terms of t. We have
$$P\{F|t\} = \frac{\text{Area of }R_t}{\text{Area of circle}}.$$
RGV | 2017-12-16T13:56:45 | {
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https://mathematica.stackexchange.com/questions/120946/fractal-basins-of-attraction-in-a-magnetic-pendulum?noredirect=1 | # Fractal basins of attraction in a Magnetic Pendulum
I am trying to write a Mathematica program that realizes a graphical approximation of the basins of attraction in a Magnetic pendulum subject to friction and gravity, in which the three magnets are disposed on the vertices of an equilateral triangle. This system is chaotic and has very interesting properties. The basins of attraction look something like this:
A code I wrote can produce a $400 \times 400$ image such as this (Caution: no fancy ColorFunctions involved) in about two hours. The computation seems to be extremely slow.
Is there any way of having a better rendering, say, full HD 1920x1080 resolution, for the basins of attraction of a magnetic pendulum as the one mentioned that can be run in a farely quick time on a common machine?
## Code
Here is the code I used to produce the above image. I set the position of the magnets and define the Lagrange equations
X1 = 1; X2 = -(1/2); X3 = -(1/2); Y1 = 0; Y2 = Sqrt[3]/2; Y3 = -(Sqrt[3]/2);
X[1] = X1; X[2] = X2; X[3] = X3; Y[1] = Y1; Y[2] = Y2; Y[3] = Y3;
Eqs[k_,c_,h_]:={
x''[t]+k x'[t]+c x[t]-Sum[(X[i]-x[t])/(h^2+(X[i]-x[t])^2+(Y[i]-y[t])^2)^(3/2),{i,3}]==0,
y''[t]+k y'[t]+c y[t]-Sum[(Y[i]-y[t])/(h^2+(X[i]-x[t])^2+(Y[i]-y[t])^2)^(3/2),{i,3}]==0
}
I define a function that numerically integrates the equations up until $t=100$.
Sol[k_, c_, h_, xo_, yo_] :=
NDSolve[
Flatten[{Evaluate[Eqs[k, c, h]],
x'[0] == 0, y'[0]== 0, x[0] == xo, y[0] == yo}],
{x, y}, {t, 99.5, 100.5}, Method -> "Adams"
];
I define a function tt that gives a value between $\frac13, \frac 23, 1$ based on magnet proximity at time $100$ for fixed $k,c,h$ (in this case $.15$,$.2$,$.2$) and a function k that evaluates tt on a grid.
tt = Compile[{{x1, _Real}, {y1, _Real}}, Module[{},
Final = ({x[100], y[100]} /. (Sol[0.15, .2, .2, x1, y1])[[1]]);
Distances = Map[(Final - #).(Final - #) &, {{1, 0}, {-(1/2), Sqrt[3]/2}, {-(1/2), -(Sqrt[3]/2)}}];
Magnet = Min[Distances];
Position[Distances, Magnet][[1, 1]]/3]];
k[n_, xm_, ym_, xM_, yM_] := ParallelTable[tt[xi, yi], {yi, ym, yM, Abs[yM - ym]/n}, {xi, xm, xM, Abs[xM - xm]/n}];
Finally, I rasterize the table produced by k.
G = Graphics[Raster[k[400, -2, -2, 2, 2], ColorFunction -> Hue]]
and, after a while, I obtain the previous image. I attempted using a dynamic energy control (i.e. using EvaluationMonitor to monitor the energy level of ther trajectory: if it falls in a potential hole NDSolve throws the position) but this did not increase the speed as much as I was hoping; it actually seems to slow the computation down.
• Why don't you start by sharing the code you wrote? Perhaps we can improve on that, rather than reinventing the wheel. – MarcoB Jul 18 '16 at 0:30
• You might want to look up previous work by Paul Nylander. – J. M. is away Jul 18 '16 at 9:06
• There is a syntax error (Y3 = -(Sqrt3]/2);) in the first code block. When I run Sol[0.15, .2, .2, 1, 1] I get an error message saying that the system is undetermined, because of lack of spaces between c and x and y. – C. E. Jul 18 '16 at 11:57
• After modernizing Nylander's code for the current version, I managed this. – J. M. is away Jul 18 '16 at 16:37
• @J.M. This seems very interesting. Can you share your code with us? How much time did the computation take? – Lonidard Jul 18 '16 at 17:46
JM commented:
If you want to try things out, use Nylander's second snippet, which is using a Beeman integrator. This looks to be faster than native NDSolve[] for this specific case.
Paul Nylander's code is here.
Below is a modified version of his code which computes all points simultaneously using the fact that all the operations in Beeman's algorithm are Listable functions in Mathematica.
The run time for the 400x400 image is around 30 seconds.
n = 400;
{tmax, dt} = {25, 0.05};
{k, c, h} = {0.15, 0.2, 0.2};
{z1, z2, z3} = N@Exp[I 2 Pi {1, 2, 3}/3];
l = 2.0;
z = DeveloperToPackedArray @ Table[x + I y, {y, -l, l, 2 l/n}, {x, -l, l, 2 l/n}];
v = a = aold = 0 z;
Do[
z += v dt + (4 a - aold) dt^2/6;
vpredict = v + (3 a - aold) dt/2;
anew = (z1 - z)/(h^2 + Abs[z1 - z]^2)^1.5 + (z2 - z)/(h^2 + Abs[z2 - z]^2)^1.5 +
(z3 - z)/(h^2 + Abs[z3 - z]^2)^1.5 - c z - k vpredict;
v += (5 anew + 8 a - aold) dt/12;
aold = a; a = anew,
{t, 0, tmax, dt}];
res = Abs[{z - z1, z - z2, z - z3}];
Image[0.2/res, Interleaving -> False]
• complexGrid from here might be useful. – J. M. is away Jul 18 '16 at 23:55
• Turn the Do into Table and return z each time to see a trippy animation: cl.ly/3C163h1A3F26 – Chip Hurst Jul 19 '16 at 15:04
• @Chip, I believe Nylander did something like that in one of the animations on his website. – J. M. is away Jul 19 '16 at 15:08
• @J.M. Ah-ha! I should've clicked the link! – Chip Hurst Jul 19 '16 at 15:09
I don't have any breakthrough ideas, but I am able to cut the computation time in half on my computer by optimizing the usage of NDSolve.
My version of your code looks like this:
X[1] = 1;
X[2] = -(1/2);
X[3] = -(1/2);
Y[1] = 0;
Y[2] = Sqrt[3]/2;
Y[3] = -Sqrt[3]/2;
Sol[k_, c_, h_, xo_, yo_] := NDSolve[{
x''[t] + k x'[t] + c x[t] - Sum[(X[i] - x[t])/(h^2 + (X[i] - x[t])^2 + (Y[i] - y[t])^2)^(3/2), {i, 3}] == 0,
y''[t] + k y'[t] + c y[t] - Sum[(Y[i] - y[t])/(h^2 + (X[i] - x[t])^2 + (Y[i] - y[t])^2)^(3/2), {i, 3}] == 0,
x'[0] == 0,
y'[0] == 0,
x[0] == xo,
y[0] == yo
}, {x, y}, {t, 99.5, 100.5}, Method -> "Adams"
];
nf = Nearest[{{1, 0}, {-0.5, Sqrt[3]/2}, {-0.5, -Sqrt[3]/2}} -> Automatic] /* First;
getBasin[x1_, y1_] := nf[{x[100], y[100]} /. Sol[0.15, .2, .2, x1, y1] // Flatten, 1]
You are essentially building your own Nearest function, but one is already built in. This approach is as performant as your code.
Some advanced usage tips for NDSolve are documented here. The idea is that before NDSolve can start to integrate the equations it needs to rewrite them, and this takes time. Instead of doing the rewriting for every single point we can do it just once and then use that for all of the points.
getStateData[k_, c_, h_, x0_, y0_] :=
First@NDSolveProcessEquations[{
x''[t] + k x'[t] + c x[t] - Sum[(X[i] - x[t])/(h^2 + (X[i] - x[t])^2 + (Y[i] - y[t])^2)^(3/2), {i, 3}] == 0,
y''[t] + k y'[t] + c y[t] - Sum[(Y[i] - y[t])/(h^2 + (X[i] - x[t])^2 + (Y[i] - y[t])^2)^(3/2), {i, 3}] == 0,
x'[0] == 0,
y'[0] == 0,
x[0] == x0,
y[0] == y0
}, {x, y}, t, Method -> "Adams"]
sd = getStateData[.15, .2, .2, 1, 1];
getBasin2[x0_, y0_] := Module[{state = sd, sol},
state = First@NDSolveReinitialize[state, {x[0] == x0, y[0] == y0}];
NDSolveIterate[state, 100.5];
sol = {x[100], y[100]} /. NDSolveProcessSolutions[state];
nf[sol]
]
Let's test it:
ArrayPlot[
ParallelTable[getBasin2[xpos, ypos], {xpos, -2, 2, 0.1}, {ypos, -2, 2, 0.1}],
ColorRules -> {1 -> Red, 2 -> Green, 3 -> Blue}
] // AbsoluteTiming
`
This simple example took 22 seconds for me to generate, as opposed to 44.5 seconds for my first rewritten version of your code.
Your image I was able to generate in 33 minutes rather than the two hours it took for you:
Implementing a stopping condition seems cumbersome, but you can also achieve speed enhancements by lowering the amount of integration time. Integrating to 100 seems excessive, with 25 it looks the same for the 400x400 case and it takes only 11 minutes. | 2019-06-26T15:19:34 | {
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https://math.stackexchange.com/questions/2780030/finding-a-subgroup-mn-of-bbb-z-120-given-n-and-m | Finding a subgroup $MN$ of $\Bbb Z_{120}$ given $N$ and $M.$
Is the following question I found in a past exam paper a trick question ?
$$G=\Bbb Z_{120}, M=\{0,10,20,...,110\}, N=\{0,4,8,12,..,116\}.$$ Then identify the subgroup of $G$ given by $MN$.
The reason I think it's a trick question is that $|M|=11,|N|=29$ so wouldn't this imply that $|MN|=319$ and then by Lagrange can't be a subgroup as the order of this group doesn't divide G ?
It's more customary to write $M+N$ when the group is written with additive notation.
First you're wrong in $|M|$ and $|N|$, which are $12$ and $30$ respectively. On the other hand, $|M+N|$ cannot be $360$, because you only have $120$ elements to choose from.
The general formula, holding also for nonabelian groups, is $$|M+N|=\frac{|M|\,|N|}{|M\cap N|}$$ The formula is easier to prove for abelian groups. Still using additive notation, consider the surjective group homomorphism $$\sigma\colon M\times N\to M+N,\qquad (x,y)\mapsto x-y$$ Its kernel is the set of pairs $(x,y)$ such that $x=y$, which is in an obvious bijection with $M\cap N$. The homomorphism theorem says that $$(M\times N)/\!\ker\sigma\cong M+N$$ so that $$|M+N|=\frac{|M\times N|}{\lvert\ker\sigma\rvert}=\frac{|M|\,|N|}{|M\cap N|}$$
Can you tell what's $M\cap N$?
A different way is to notice that $M=10\mathbb{Z}/120\mathbb{Z}$ and $N=4\mathbb{Z}/120\mathbb{Z}$, so $$M+N=(10\mathbb{Z}+4\mathbb{Z})/120\mathbb{Z}=2\mathbb{Z}/120\mathbb{Z}$$ because $2=\gcd(10,4)$.
• $M\cap N$={0,20,40,60,80,100} , I think which would mean by your formula that |M+N|=360/6=60 which does divide 120 and so can be a subgroup. also looking at the part of your answer ( which by the way is fantastic , thanks for that :)) where you say $M+N=(10\mathbb{Z}+4\mathbb{Z})/120\mathbb{Z}=2\mathbb{Z}/120\mathbb{Z}$ does that mean that $M+N =2 \Bbb Z/120 \Bbb Z \cong \Bbb Z_{60}$ ? – excalibirr May 13 '18 at 23:18
• @exodius Excellent. – egreg May 13 '18 at 23:22
• Use $\{\}$ for $\{\}$, @exodius. – Shaun May 13 '18 at 23:24
• @egreg That just made me think of another question actually if you'd be so kind to answer. If we were to then take the quotient group $\Bbb Z_{120}/\Bbb Z_{60}$, is that group isomorphic to $\Bbb Z_2$? – excalibirr May 13 '18 at 23:29
• @Shaun sorry I usually don't forget to write it that way , just made a slip writing that comment :) – excalibirr May 13 '18 at 23:29
It's not a trick question! Looks like you made an off-by-one error - it happens to all of us. In particular, you wrote $|M|=11$, presumably because the last value is $11 \times 10$ and you're counting up in $10$s. But let's write out the elements of $M$ fully and see what happens:
$$M= \{0,10,20,30,40,50,60,70,80,90,100,110\}$$ Counting them by hand gives $|M|=12$, and similarly $|N|=30$, as we'd expect for subgroups of $G$.
For a simpler example of this kind of mistake, notice that if we wanted the size of $A = \{0,1,2,3\}$, you might guess it's $3$ because the highest value is $3$, but there are actually four elements - since $0$ is included.
I should also point out that $|M|=12$, $|N|=30$ doesn't imply that $|MN|=360$, see if you can spot why.
• Is it because we should consider MN like a coset and take 0+N, 10+N ,20+N etc ? – excalibirr May 13 '18 at 23:02
• Not exactly - look at how $MN$ is defined – B. Mehta May 13 '18 at 23:05 | 2019-08-25T05:23:09 | {
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https://www.physicsforums.com/threads/relative-motion-in-an-elevator.822041/ | # Homework Help: Relative motion in an elevator
1. Jul 5, 2015
### Ravens Fan
1. The problem statement, all variables and given/known data
So I am in an elevator throwing a ball in an air, straight up and down, about 1 meter from my hand. The elevator is moving upwards at a rate of 2 m/s. From an observer on the ground, would they see the ball reach its peak at the same instant that I do? If not what is this difference in time.
From my perspective, I see the ball move up 1 meter, then down 1 meter. How far up and down would the observer see the ball move?
2. The attempt at a solution
My thoughts are that the observer would see the ball go higher up, as opposed to 1 m. But then they would see the ball move a shorter distance on the way down. As for the first part, I'm not sure where to begin. My guess is that the times would not be the same, but again, I'm in a stump. Thanks
2. Jul 5, 2015
### paisiello2
I think you guessed right but maybe you could prove it using kinematics?
3. Jul 5, 2015
### rpthomps
Unless you are considering special relativistic effects, the time would be the same for both observer.
4. Jul 5, 2015
### haruspex
That would be true if they agreed about the event, but as the OP noted, they would disagree about the vertical speed of the ball at any given instant. In particular they would disagree about whether its vertical speed is zero.
5. Jul 6, 2015
### Staff: Mentor
Why don't you do the math, and let the math answer the question for you?
Chet
6. Jul 6, 2015
### rpthomps
I never considered this. It seems quite dramatic. I did some calculations and came up with the following general forms.
From inside the elevator time of object to rise and fall:
$t=\frac { 2v_o}{g }$
From outside the elevator:
$t=\frac{v_o+\sqrt { v_o^2-v_E^2 } +v_E}{g}$
Although I didn't originate the question, I learned a lot. Thanks | 2018-05-21T22:57:07 | {
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https://puzzling.stackexchange.com/questions/11490/rectangles-with-a-whole-number-measurement | # Rectangles with a whole-number measurement
A rectangle is proper if its width is a whole number, or its height is, or both. Show that if a rectangle can be cut into a finite number of proper rectangles, then that rectangle is itself proper.
As obligatory, I claim that this problem has a clever and unexpected solution. Actually, I know of two totally different such solutions.
There is beautiful article on this problem that summarizes 14 different solution approaches:
Stan Wagon: "Fourteen Proofs of a Result About Tiling a Rectangle"
The American Mathematical Monthly 94, 1987, pp. 601-617
There is one very elegant proof via complex double integrals.My favourite proof is as follows :
Proof via prime numbers (due to Raphael Robinson, University of California at Berkeley)
We claim that for each prime p, either the height or the width of the big reactangle $R$ is within $1/p$ of an integer. It follows that one of these is an integer.
To prove the claim, scale the entire tiling up by a factor of $p$ in each direction, and consider the tiling obtained by replacing all tile-corners $(x, y)$ in the scaled-up tiling by $([x], [y])$. This yields an integer-sided rectangle tiled by integer-sided rectangles, each of which has one side a multiple of $p$. Therefore, the area of the large integer-sided rectangle is a multiple of $p$, whence one of its sides must be a multiple of $p$. Moreover, the dimensions of this rectangle differ from the dimensions of the scaled-up rectangle by less than $1$. It follows that $R$ has a side that differs from an integer by less than $l/p$.
• That is awesome -- I hadn't know about that article! – xnor Apr 3 '15 at 19:55
Align the bottom left corner of the big rectangle with a checkerboard whose squares have side 1/2. Let the bottom left corner be black.
Two intuitive lemmas:
If both sides of a rectangle aren't whole, then the black area covered will be greater than white area covered. On the contrary, if the black and white areas are equal, then the rectangle must have an integer side.
Proof: Every small rectangle has at least a integer side, so it covers equal areas of black and white. So, the big rectangle (which is the sum of all small rectangles) must cover equal areas of black and white, thus must have an integer side.
Credits: I've read this evidence somewhere few years ago, I don't remember the author though. After googling a bit, I've found out that Stan Wagon has posted 14 solutions of this problem, incredible!
Tile the plane with squares of side length $\frac12$, colored white and black in checkerboard fashion. Because of the periodicity of the coloring, any proper rectangle with sides parallel to the tiles will have the same white area as black area (they will be balanced). Place the larger rectangle, $R$, so that one of its corners lines up with a corner of a white square. Since all of the subrectangles are proper, therefore balanced, $R$ must be balanced. Now, suppose that $R$ was improper. Let $w,h$ be its width and height, and $w',h'$ be those values rounded down. The rectanlges $[0,w']\times [0,h']$, $[0,w']\times [h',h]$ and $[w',w]\times [0,h']$ are all proper, therefore balanced, which would imply the the remaining rectangle, $[w,w']\times[h,h']$, is balanced as well. However, that last rectangle looks something like this: $\qquad\qquad\qquad\qquad\quad$
It is simple enough to verify this is imbalanced, using $(1/2-a)(1/2-b)>0$, a contradiction. Technically, I've only covered the case where $w-w'\ge \frac12$ and $h-h'\ge\frac12$, but the other cases are even easier to check.
Thus, $R$ must be proper.
My humble attempt at a simple logic based solution:
Since when I cut a rectangle into proper rectangles, each of the smaller rectangles will have whole numbered widths and heights.
The total width of the rectangle = sum of all widths of smaller rectangles.
Since all the widths are whole numbers their sum is a whole number.
Therefore the width of the original rectangle is a whole number.
Same logic applies for heights.
Since height and width are whole numbers the rectangle is a proper rectangle.
Apparently this solution has some shortcomings according to the given details of the problem. I will be improving on it.
• No, because either width or height are integers for properness, not necessarily both. – Rand al'Thor Apr 3 '15 at 12:43
• I guess that makes sense. Sorry for the mistaken assumption. :| – archilius Apr 3 '15 at 12:46 | 2020-04-09T15:53:46 | {
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https://math.stackexchange.com/questions/3018055/analizing-the-stability-of-the-equilibrium-points-of-the-system-ddotx-x-a | # Analizing the stability of the equilibrium points of the system $\ddot{x}=(x-a)(x^2-a)$
$$\require{amsmath}$$ $$\DeclareMathOperator{\Tr}{Tr}$$ $$\DeclareMathOperator{\Det}{Det}$$
Investigate the stability of the equilibrium points of the system $$\ddot{x}=(x-a)(x^2-a)$$ for all real values of the parameter $$a$$. (Hints: It might help to graph the right-hand side. An alternative is to rewrite the equation as $$\ddot{x}=−V′(x)$$ for a suitable potential energy function $$V$$ and then use your intuition about particles moving in potentials.)
I am not really sure on how to approach the problem with the given hints, since it would require plotting the graph for different critical values for $$a$$, which I am not really sure how to find. Thus, I am wondering if the following is correct.
The system $$\ddot{x}=(x-a)(x^2-a)$$ can be re-written as $$\begin{cases} \dot{x}=y\\ \dot{y}=(x-a)(x^2-a) \end{cases}$$ with fixed points $$P_1(a,0),\,P_2(\sqrt{a},0)$$ and $$P_3(\sqrt{a},0)$$.
The Jacobian is $$J(x,y)=\begin{bmatrix} 0 &1\\ 3x^2-2ax-a &0 \end{bmatrix}$$ and thus $$J(a,0)=\begin{bmatrix} 0 &1\\ a^2-a &0 \end{bmatrix};\quad J(\sqrt{a},0)=\begin{bmatrix} 0 &1\\ 2a-2a^{3/2} &0 \end{bmatrix}; \quad J(-\sqrt{a},0)=\begin{bmatrix} 0 &1\\ 2a+2a^{3/2} &0 \end{bmatrix}.$$ It can be noticed that the $$\Tr\left[J(x,y)\right]=0$$ and that \begin{aligned} &1.\,\Det\left[J(a,0)\right]=a(1-a)\implies \text{Saddle for }a<0 \wedge a>1, \text{Center for }01.\\ &3.\,\Det\left[J(-\sqrt{a},0)\right]=-2a(\sqrt{a}+1)\implies \text{Saddle for }a>0. \end{aligned}
If $$a=0$$ the system reduces to $$\ddot{x}=x^3$$ where the only fixed point is at $$(0,0)$$ and thus it is unstable. On the other hand, if $$a=1$$ then the system reduces to $$\ddot{x}=x^3-x^2-x+1$$ with fixed points at $$(1,0),(-1,0)$$, both being unstable.
Is my work correct?
• Isn't this question better suited for Physics SE? – Bertrand Wittgenstein's Ghost Nov 29 '18 at 2:11
• @BertrandWittgenstein'sGhost this is part of dynamical systems, so no. I could certainly ask there too, but there is no point really – DMH16 Nov 29 '18 at 2:13
• That's funny what you did, It was a genuine question. No need to get defensive. Cheers! – Bertrand Wittgenstein's Ghost Nov 29 '18 at 2:23
• I haven't checked all your algebra carefully but it looks like you've go the right idea! – Robert Lewis Nov 29 '18 at 2:32
• @AlexanderJ93 yes you're right the determinant is always negative. Thanks – DMH16 Nov 29 '18 at 2:52
Almost everything looks right.
For $$a > 0$$, $$-2a < 0$$ and $$\sqrt{a}+1 > 0$$, so $$\Det[J(-\sqrt{a},0)] < 0$$. This equilibrium should be a saddle then for positive $$a$$.
Also, when analyzing the stability of a system with a parameter, it helps to draw the bifurcation diagram, i.e. the graph of the fixed points with respect to the parameter. The fixed point equations $$x = a, x = \sqrt{a}, x = -\sqrt{a}$$ can be graphed on the $$ax$$-plane. Intersections of these curves are the bifurcations, which are the only places where the behavior can change.
This is the bifurcation diagram for your system. The lower black curve is the $$x=-\sqrt{a}$$ equilibrium, which doesn't have any intersections for $$a>0$$, so there will be no change in behavior.
Edit:
This isn't specifically asked about in the question but it is an important feature of this equation. For $$a>0, a\neq 1$$, there are centers at one fixed point and saddles at another nearby fixed point. When this happens, a periodic orbit will collide with the saddle point, creating a closed path from a fixed point back around to itself. This is called a homoclinic orbit, and usually has a "teardrop" shape, as seen below for $$a = \frac{1}{2}$$.
• Just out of curiosity, what software did you use to sketch the phase plane? – DMH16 Nov 29 '18 at 4:49
• pplane, which can be found here as a java application: math.rice.edu/~dfield/dfpp.html I strongly recommend it to any student in dynamics. It's limited to plane dynamics (obviously) but can be extremely helpful in learning and verifying results. – AlexanderJ93 Nov 29 '18 at 5:03
• Thanks you for the suggestion! – DMH16 Nov 29 '18 at 5:47 | 2019-11-14T03:46:28 | {
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http://math.stackexchange.com/questions/151867/counting-principle | # Counting Principle
Three friends agreed to meet at 8 P.M. in on of the Spanish restaurants in town. Unfortunately they forgot to specify the name of the restaurant. If there are 10 Spanish restaurants
(a) find the number of ways they could miss each other
(b) find the number of ways they could meet
(c) find the number of ways at least two of them meet each other.
I got 10 * 3 = 30 ways for (a) for (b), I got 10 ways. how do I do (c)?
-
I suggest you start by considering a simpler problem of the same type, say with 3 restaurants and 3 friends. See if the same method produces the correct answer in this simpler case. – MJD May 31 '12 at 2:28
Does "miss" mean "not all three friends end up together" or "all three friends end up in separate restaurants"? – Austin Mohr May 31 '12 at 2:28
yes they all miss each other, as in they went to different restaurants – count May 31 '12 at 2:30
Figure out the complement of (c). – copper.hat May 31 '12 at 2:31
I’m afraid that you’ll have to start by redoing (a) and (b): both are wrong. Call the friends $A,B$, and $C$.
(a) I’m assuming that what’s wanted here is the number of ways in which they completely miss one another: no two go to the same restaurant. There are $10$ restaurants that $A$ could choose. $B$ could then choose any of the remaining $9$, and $C$ could choose any of the $8$ not chosen by $A$ or $B$. Thus, there are $10\cdot9\cdot8=720$ different ways in which they could completely miss one another.
(b) In order for them to meet successfully, they must all choose the same restaurant. There are $10$ restaurants, so there are just $10$ ways in which they can meet successfully. Alternatively, we can apply the same kind of analysis as in (a): $A$ has a choice of $10$ restaurants, but after that $B$ and $C$ have only $1$ choice each, if the meeting is to take place, so the total number of ways is $10\cdot1\cdot1=10$.
(c) This is most easily done by considering the complementary (opposite) event: no two of them meet. In (a) we calculated the number of ways in which this can happen. If there are $n$ different ways in which could they choose restaurants, regardless of how many of them meet, the number of ways in which at least two of them meet must be $n-720$. What’s $n$?
Is $n=1000$?.... – count May 31 '12 at 2:57
@count: Yes, it is, so the answer to (c) is $1000-720=280$. – Brian M. Scott May 31 '12 at 2:59 | 2015-07-06T10:02:36 | {
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https://mathoverflow.net/questions/142128/is-there-a-tychonoff-space-x-of-cardinality-not-of-the-form-2-alpha-such-th/142134 | # Is there a Tychonoff space $X$ of cardinality not of the form $2^\alpha$ such that $|C(X)| = |X|$
Let $X$ be the real line with the usual topology. Then clearly $|C(X)| = c = |X|$ and on the other hand $|X| = 2^{\aleph_0}$. Now my question is as in the title: Is there a Tychonoff space $X$ of cardinality not of the form $2^\alpha$ such that $|C(X)| = |X|$ (where $C(X)$ is the set of all real valued continuous functions on $X$). Any reference or help would be appreciated.
Let $\kappa$ be any singular strong limit cardinal of uncountable cofinality, such as the cardinal $\beth_{\omega_1}$ for a specific example, and let $X=\kappa+1$, the ordinals up to and including $\kappa$ itself. Under the order topology, this is a compact Hausdorff space. Note that $|X|=\kappa\neq 2^\alpha$ for any $\alpha$, since $\kappa$ is a strong limit cardinal. But meanwhile, every continuous function $f:X\to\mathbb{R}$ is eventually constant, in order that it is continuous at the final point $\kappa$, because it has uncountable cofinality. So every function in $C(X)$ is determined by a restriction of it $f\upharpoonright\gamma:\gamma\to\mathbb{R}$ for some $\gamma\lt\kappa$. Since $\kappa$ is a strong limit, there are only $\kappa$ many such restricted functions, and so $|C(X)|=\kappa=|X|$, as desired.
Update. If the GCH fails in a convenient way, we can make a much smaller example. Suppose $2^\omega=\omega_1$ and $2^{\omega_1}\gt\omega_2$, and let $X=\omega_2+1$. This is a compact Hausdorff space of size $\omega_2$. Notice that any continuous function $f:X\to\mathbb{R}$ is eventually constant and indeed takes at most countably many distinct values (since otherwise we can find a point $\gamma$ of cofinality $\omega_1$ with $f\upharpoonright\gamma$ not eventually constant, which will violate continuity at $\gamma$.) It follows that $C(X)$ has size $\omega_2^\omega$, which has size $\omega_2$ under our assumptions. So this is a case where $|C(X)|=|X|=\omega_2$, but $\omega_2\neq 2^\alpha$ for any $\alpha$.
• +1 Nice. If $Y = \kappa$ ($\kappa$ as above) then what is the cardinality of $C(Y)$? Is it $\kappa$ !?
– user37834
Sep 14, 2013 at 13:52
• @Silvi, the same argument works even if you don't have the final point, since every continuous function $\kappa\to\mathbb{R}$ is eventually constant, because $\kappa$ has uncountable cofinality. So we would also have $|C(Y)|=\kappa$. But in this case, $Y$ would not be compact, although still Tychonoff. Sep 14, 2013 at 14:06
$\textbf{A simple counterexample}$ The following counterexample is somewhat similar to Joel David Hamkins' counterexample, but it is still worth mentioning. Let $D$ be a discrete space and let $D\cup\{\infty\}$ denote the one-point-compactification of $D$. If $f\in C(D\cup\{\infty\})$, then for each $\epsilon>0$ there is a finite subset $A_{\epsilon}\subseteq D$ where if $x\not\in A_{\epsilon}$, then $|f(x)-f(\infty)|<\epsilon$. In particular, if $A=\bigcup_{n}A_{1/n}$, then $f(x)=f(\infty)$ outside $A$. Therefore, the function $f$ only depends on the countable set $A$($|D|^{\aleph_{0}}$ choices), the values on the set $A$($2^{\aleph_{0}}$ choices), and the value at $\infty$($2^{\aleph_{0}}$ choices). We conclude that there are $|D|^{\aleph_{0}}$ choices for the function $f$. If $|D|^{\aleph_{0}}=|D|$, then $|D\cup\{\infty\}|=C(D\cup\{\infty\})$.
$\textbf{Constructing related counterexamples}$ The following facts may be useful in constructing related counterexamples.
First of all, if $C^{*}(X)$ denotes the set of all bounded continuous functions from $X$ to $\mathbb{R}$, then $|C(X)|=|C^{*}(X)|$. Furthermore, $C^{*}(X)\simeq C^{*}(Y)$ if and only if $\beta X=\beta Y$ where $\beta X$ denotes the Stone-Cech compactification of $X$. In particular, if $|X|\leq|C^{*}(X)|\leq|\beta X|$, then there is a set $Y$ with $X\subseteq Y\subseteq\beta X$ and $|Y|=|C^{*}(X)|$. In this case, we have $|C^{*}(Y)|=|C^{*}(X)|=|Y|$.
If $X$ is a space, then let $w(X)$ denote the weight of the space $X$. Let $k(X)$ denote the least cardinal such that every open cover of $X$ has a subcover of cardinality less than $k(X)$.
$\mathbf{Proposition}$ $C(X)\leq((w(X)^{<k(X)})^{\aleph_{0}}$
$\mathbf{Proof}$ Let $\mathcal{B}$ be a basis for $X$ of cardinality $w(X)$. Let $U$ be a cozero set. Then there are open sets $U_{n}$ with $\overline{U_{n}}\subseteq U$ and $\bigcup_{n}U_{n}=U$. Since $\mathcal{B}$ is a basis for $X$, there is a subset $\mathcal{C}\subseteq\mathcal{B}$ with $\bigcup\mathcal{C}=U$. By compactness, there is some subset $\mathcal{C}_{n}\subseteq\mathcal{C}$ with $|\mathcal{C}_{n}|<k(X)$ and where $\overline{U}_{n}\subseteq\bigcup\mathcal{C}_{n}$. In particular, we have $U=\bigcup_{n}\bigcup\mathcal{C}_{n}$. However, since each $\mathcal{C}_{n}$ chooses less than $k(X)$ many elements of $\mathcal{B}$, there are at most $(w(X)^{<k(X)})^{\aleph_{0}}$ many choices of $\bigcup_{n}\bigcup\mathcal{C}_{n}$. Since $U$ ranges over all cozero sets, there are at most $(w(X)^{<k(X)})^{\aleph_{0}}$ cozero sets. Since every continuous real-valued function is determined by the cozero sets $f^{-1}(-\infty,r)$ where $r$ ranges over all rational numbers, there are at most $(w(X)^{<k(X)})^{\aleph_{0}}$ continuous real-valued functions on $X$. $\mathbf{QED}$
It can be shown using a standard compactness argument that if $X$ is a compact Hausdorff space, then $|w(X)|\leq|X|$. Combining this fact with the above result, we obtain the following corollary
$\textbf{Corollary}$ Suppose that $X$ is a compact Hausdorff space. Then $|C(X)|\leq |w(X)|^{\aleph_{0}}\leq|X|^{\aleph_{0}}$.
There is also a lower bound of the number of continuous functions on a completely regular space that depends on a cardinal invariant which I shall call the saturation. If $X$ is a space, then let $s(X)$ be the least cardinal such that if $\mathcal{U}$ is a collection of $s(X)$ nonempty open sets, then $U\cap V\neq\emptyset$ for distinct $U,V\in\mathcal{U}$.
$\textbf{Proposition}$ If $X$ is a completely regular space, then there are at least $\sum_{\kappa<s(X)}\kappa^{\aleph_{0}}$ many functions in $C^{*}(X)$.
$\textbf{Proof}$ Suppose that $\kappa<s(X)$. Then there is a collection $\mathcal{U}$ of pairwise disjoint open sets. For each $U\in\mathcal{U}$ choose some point $x_{U}\in U$. Then let $f_{U}:X\rightarrow[0,1]$ be a function where $f_{U}(x_{U})=1$ and $f_{U}(x)=0$ for each $x\in U^{c}$. Take note that there are $\kappa^{\aleph_{0}}$ injective maps from $\omega$ to $\mathcal{U}$. For each injective $j:\omega\rightarrow\mathcal{U}$, let $F_{j}:X\rightarrow\mathbb{R}$ be the function defined by $\sum_{n}\frac{1}{n^{2}+1}f_{j(n)}$. It is clear that each function $F_{j}$ is distinct. Furthermore, each $F_{j}$ is continuous being the uniform limit of continuous functions. Therefore the family $(F_{j})_{j}$ is a family of $\kappa^{\aleph_{0}}$ continuous functions. Since $\kappa^{\aleph_{0}}\leq|C^{*}(X)|$ for all $\kappa<s(X)$, we conclude that $\sum_{\kappa<s(X)}\kappa^{\aleph_{0}}\leq|C^{*}(X)|$ as well.
Using the following corollary, we immediately obtain my counterexample and Joel David Hamkins' counterexample.
$\textbf{Corollary}$ Suppose $X$ is a compact Hausdorff space with $|X|=|X|^{\aleph_{0}}=\sum_{\kappa<s(X)}\kappa^{\aleph_{0}}$. Then $|C(X)|=|X|$.
• Thanks to you and @JoelDavidHamkins a lot. So fruitful. Seems like a small course on $C(X)$. I must confess that I have believed wrongly that the cardinal of idempotents is equal to $2^A$ where $A$ is the cardinal of connected components of $X$ But I think the correct thing is that the cardinal is equal to the cardinal of clopen sets. Forgive me if I am asking something trivial but what is the cardinality of the set of idempotents in these four examples? Thanks anyway.
– user39982
Sep 15, 2013 at 11:20
• Yes. The idempotents in $C(X)$ are in a one-to-one correspondence with the clopen sets. Furthermore, for a compact totally disconnected space $X$, the cardinality of the collection of all clopen sets is equal to the weight $w(X)$ and is therefore bounded above by $|X|$. If $X$ is a compact ordinal space or the one-point compactification of a discrete space, then there are $|X|$ clopen subsets of $X$. Sep 15, 2013 at 18:39
• So I have to restate my problem. I am looking for a Tychonoff space $X$ such that for any finite subset $F$, $|C(X\backslash F)| = |X| \not = |T|$ (where $T$ is the set of clopen subsets of $X$). I would appreciate your points and suggestions on this one.
– user39982
Sep 15, 2013 at 19:30
• any idea? By the way is it possible to ask the question (mentioned above) in a new post? (I am new and I don't know that much about the regulations here. My new question is very similar to the new one mentioned above. so I am not sure if they accept this as a whole new question) Thanks in advance.
– user39982
Sep 16, 2013 at 7:28
• @Niki. It will probably be best to ask an entirely new question in a new post since the new question seems significantly different from the old one. Sep 16, 2013 at 13:51 | 2022-11-29T12:17:24 | {
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https://math.stackexchange.com/questions/191572/prove-that-ii-is-a-real-number | # Prove that $i^i$ is a real number
According to WolframAlpha, $i^i=e^{-\pi/2}$ but I don't know how I can prove it.
• Do you know the definition of $i^i$? – Jack Sep 5 '12 at 18:22
• A pedantic point: is a complex number with a 0 imaginary part the same as a real number? – James Sep 6 '12 at 12:11
• @James: Unless you know some secret that I don't, yes it is. – Cameron Buie Sep 6 '12 at 20:58
• @CameronBuie I agree for most practical purposes you don't need to distinguish but formally, since complex numbers and reals have different properties, do you have to do an intermediate conversion? For instance, can you assert $1+0i < 2+0i$ in the same way you can assert $1 < 2$? – James Sep 7 '12 at 7:42
• @James Yes, if you interpret $<$ to be by lexicographical order. It suffices to be an order relation on the set $\{a+bi \mid a \in \Bbb R, b = 0\} \subset \Bbb C$. – Emily Sep 7 '12 at 13:41
Write $i=e^{\frac{\pi}{2}i}$, then $i^i=(e^{\frac{\pi}{2}i})^i = e^{-\frac{\pi}{2}} \in \mathbb{R}$. Be careful though, taking complex powers is more... complex... than it may appear on first sight $-$ see here for more info.
In particular, it's not well-defined (until we make some choice that makes it well-defined); we could just have well written $i=e^{\frac{5\pi}{2}i}$ and obtained $i^i=e^{-\frac{5\pi}{2}}$. But $i^i$ can't be equal to both $e^{-\frac{\pi}{2}}$ and $e^{-\frac{5\pi}{2}}$ can it?
Despite the lack-of-well-defined-ness, though, $i^i$ is always real, no matter which '$i^{\text{th}}$ power of $i$' we decide to take.
More depth: If $z,\alpha \in \mathbb{C}$ then we can define $$z^{\alpha} = \exp(\alpha \log z)$$ where $\exp w$ is defined in some independent manner, e.g. by its power series. The complex logarithm is defined by $$\log z = \log \left| z \right| + i\arg z$$ and therefore depends on our choice of range of argument. If we fix a range of argument, though, then $z^{\alpha}$ becomes well-defined.
Now, here, $z=i$ and so $\log i = i\arg i$, so $$i^i = \exp (i \cdot i\arg i) = \exp (-\arg i)$$ so no matter what we choose for our range of argument, we always have $i^i \in \mathbb{R}$.
Fun stuff, eh?
• I am going to choose this answer as accepted answer, but there are some other answers that are very insightful and worth to take a look. – Isaac Sep 12 '13 at 19:48
• Relevant: tauday.com/tau-manifesto – user117644 Feb 22 '16 at 7:48
• For your shorter version, how can you be sure that exponentiation rules apply for complex powers? Specifically, how do you know that (e^a)^i=e^(ia)? – Patrick Cook Apr 21 '18 at 22:23
• @PatrickCook: You prove it ;) It's an immediate consequence of the more general fact that $\exp(zw) = \exp(z)^w$, which isn't too difficult to prove using exponential series. – Clive Newstead Apr 21 '18 at 22:30
• @CliveNewstead For z,w in the real numbers sure. But is it really true that (z^i)(w^i)=(zw)^i? Which would be a consequence of what you're saying here. The reason I ask is because I am dealing with (i^i)^4 = i^(4i) = (i^(2))^(2i) = ((-1)^(2))^i = 1^i = 1. So the problem here is then if i^i = e^(-pi/2) and (i^i)^4=1, then e^(-2pi)=1 which is obviously untrue. – Patrick Cook Apr 22 '18 at 0:50
Here's a proof that I absolutely do not believe: take its complex conjugate, which is $\bigl({\bar i}\bigr)^{\bar i}=(1/i)^{-i}=i^i$. Since complex conjugation leaves it fixed, it’s real!
EDIT: In answer to @Isaac’s comment, I think that to justify the formula above, you have to go through exactly the same arguments that most of the other answerers did. For complex numbers $u$ and $v$, we define $u^v=\exp(v\log u)$. Now, the exponential and the logarithm are defined by series with all real coefficients; alternatively you can say that they are analytic, sending reals to reals. Thus $\overline{\exp u}=\exp(\bar u)$ and $\overline{\log(u)}=\log\bar u$. The result follows, always sweeping under the rug the fact that the logarithm is not well defined.
• It's... it's beautiful! Just a question: Does complex conjugate of $(a+ib)^{(c+id)}$ equal $(a-ib)^{(c-id)}$? – Isaac Sep 6 '12 at 16:28
• excellent answer. I just loved it – Why Feb 2 '19 at 14:01
$i^i$ takes infinitely many values:
$$i^i = e^{i \log i} = e^{i(i\pi/2 + 2 \pi i m)} = e^{-\pi/2}e^{-2 \pi m},$$
where $m$ is an integer.
• All of those are real – Henry Sep 5 '12 at 22:26
• How is it possible that a wrong answer got 7 (now 6) upvotes?! I think this is quite insulting for all the high-quality answers to "low-profile" questions that can hope to reach 2 or 3 upvotes at most. – Giovanni De Gaetano Oct 24 '12 at 11:24
• @GiovanniDeGaetano Then please tell me what is wrong. This is straight out of "Complex Analysis" by Gamelin (page 24). Are you thinking about the principal value? – N.U. Oct 24 '12 at 15:32
• @GiovanniDeGaetano, dustanalysis, actually, this is the best answer out of the lot. It's somewhat embarrassing for the site that it has so few votes. – Stephen Apr 1 '13 at 12:56
• Wooops! You are perfectly right. I cannot justify my comment in any way, it would have been enough to stop and reflect before downvoting. I can only apologize with the community and specifically with N.U. for it, and thank Steve for tagging me here. Unfortunately now it's too late to remove the downvote. – Giovanni De Gaetano Apr 5 '13 at 13:53
Using the representation that $i = e^{i \pi/2}$, we have $i^i = \left(e^{i\pi/2}\right)^i = e^{i^2\pi/2} = e^{-\pi/2}$.
$i = e^{i\pi/2}$ comes from the representation that $e^{i\theta} = \cos(\theta)+i\sin(\theta)$, which for $\theta = \pi/2$ gives us $e^{i\pi/2} = \cos \pi/2 + i \sin \pi/2 = 0+i\cdot 1 = i$.
Edit: To add to the other fantastic answers/comments, this is the result on the principal branch. Others have commented that you can equivalently represent $i = e^{i(2k+1/2)\pi}$ and obtain other real-valued answers for $i^i$. Wolfram Alpha gives you $e^{-\pi/2}$ because its default setting is to return the principal value.
Edit again:
It may seem weird that we resort to this "out of nowhere" polar representation of complex numbers, but it is a powerful tool.
Over the reals, the concept that "exponentiation = repeated multiplication" breaks down when you have non-integer exponents, so you have to start defining exponentiation using suprema of sets, which exploits the ordered field nature of the reals.
The complex field is not an ordered field, so the equivalent notion of a supremum doesn't exist. So how do we take any number to the power $i$, let alone a complex number? The polar representation allows us to deal with this issue in a rather clever fashion.
• How is it determined that $(e^{i\pi/2})^i=e^{i^2\pi/2}$? – Jonas Meyer Feb 12 '17 at 5:29
This would come right from Euler's formula. Let's derive it first. There are many ways to derive it though, the Taylor series method being the most popular; here I’ll go through a different proof. Let the polar form of the complex number be equal to $z$ . $$\implies z = \cos x + i\sin x$$ Differentiating on both sides we get,
$$\implies \dfrac{dz}{dx} = -\sin x + i\cos x$$
$$\implies dz = (-\sin x + i\cos x)dx$$ Integrating on both sides,
$$\implies \displaystyle \int \frac{dz}{z} = i \int dx$$ $$\implies \log_e z = ix + K$$ Since $K = 0$, (Set $x = 0$ in the equation), we have, $$\implies z = e^{ix}$$ $$\implies e^{ix} = \cos x + i\sin x$$ The most famous example of a completely real number being equal to a real raised to an imaginary is $$\implies e^{i\pi} = -1$$ which is Euler’s identity. To find $i$ to the power $i$ we would have to put $x = \frac{\pi}2$ in Euler's formula. We would get $$e^{i\frac{\pi}2} = \cos \frac{\pi}2 + i\sin \frac{\pi}2$$ $$e^{i\frac{\pi}2} = i$$ $${(e^{i\frac{\pi}2})}^{i} = i^{i}$$ $$i^{i} = {e^{i^{2}\frac{\pi}2}} = {e^{-\frac{\pi}2}}$$ $$i^{i} = {e^{-\frac{\pi}2}} = 0.20787957635$$ This value of $i$ to the power $i$ is derived from the principal values of $\sin$ and $\cos$ which would satisfy this equation. There are infinite angles through which this can be evaluated; since $\sin$ and $\cos$ are periodic functions.
Let say: $f(z) = z^\theta$
We know Euler's formula: $e^{i \theta} = \cos(\theta) + i \sin(\theta)$
Using it we will get: $$z^\theta = e^{\theta \ln(z)} = e^{\theta (\ln\|z\| + i\arg (z))} = e^{\theta \ln\|z\| }e^{ i \theta \arg (z))}= e^{\theta \ln\|z\| }{(\cos(\theta\arg (z)) + i \sin(\theta\arg (z)))} z \in C$$
So if $$z = i \wedge \theta = i \implies$$ $$z^\theta = i^i = e^{i \ln(i)} = e^{i (\ln\|i\| + i\arg (i))} = e^{i \ln\|i\| }e^{ i i \arg (i))}= e^{i \ln(1) }e^{- \frac{\pi}{2}+2\pi k}= e^0 e^{- \frac{\pi}{2}+2\pi k}=e^{- \frac{\pi}{2}+2\pi k}$$ which is a bunch of Real numbers depending $k \in \mathbb Z$
So it is already proved that $i^i$ is a real number.
I feel most answers are too complicated or overkill or assume the value and then prove it. We do not need calculus or conjugates.
Note I do respect the other answers but I feel simple questions require simple answers.
Anyways
Let $$t = i^i$$
Take both sides to the power $$1/i$$
Then $$t^{(1/i)} = i^1 = i$$
Therefore $$t^{-i} = i$$
Take the multiplicative inverse on both sides ;
$$t^i = -i$$
Now take the log on both sides
$$i \ln(t) = - \frac{\pi}{2} i$$
Divide both parts by $$i$$
$$\ln(t) = - \pi/2$$
So
$$t = e^{-\pi/2}$$
Qed.
Notice analogues for other log branches give the other real solutions but the principal value must give this value.
No calculus or conjugates , just plain simple algebra , isn’t that nice ?
• Taking the logarithms of complex numbers isn't so simple, as we are dealing with multivalued functions. For example, $e^{2\pi i}=1$, but if we take the logarithm of both sides we can get a false statement such as $2\pi i=0$ – Moko19 Aug 31 '20 at 12:35
• Taking the logaritm of a root of unity seems simple to me. Anyone who understand log should be able to do it easily . Just take the main branch , the one that gives ln(2) a real value !! No offense my friend :) – mick Sep 3 '20 at 22:39 | 2021-01-16T18:27:20 | {
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https://math.stackexchange.com/questions/648550/statement-that-is-provable-in-zfcch-yet-unprovable-in-zfc-lnot-ch | # Statement that is provable in $ZFC+CH$ yet unprovable in $ZFC+\lnot CH$
My understanding of logic is really basic, and I ask this question out of curiosity.
Is there an explicit example of a statement whose proof uses the continuum hypothesis and is unprovable in $ZFC + \lnot CH$.
ADDED: Preferably I'm looking for an example outside set theory.
• It might not be what you expect but : CH. – Pece Jan 23 '14 at 9:18
• @Pece: Technically, axioms aren't provable. – Lucian Dec 5 '15 at 5:24
• @Lucian: Technically, all axioms are provable, and their proof is a sequence of length $1$. – user21820 May 14 '16 at 7:30
• @Lucian: Very technically, $T \vdash φ$ for every axiom $φ$ of $T$. – user21820 May 14 '16 at 7:31
One not-quite-trivial example is
$2^{\aleph_0} < 2^{\aleph_1}$.
This clearly follows from $\mathsf{CH}$ (by Cantor's Theorem), however both $2^{\aleph_0} = 2^{\aleph_1}$ and $2^{\aleph_0} < 2^{\aleph_1}$ are consistent with $\neg \mathsf{CH}$, and so the statement above cannot be proven from $\mathsf{ZFC}+\neg \mathsf{CH}$.
For an example outside of set theory proper, consider the following:
The smallest non-Lebesgue-measurable subset of $\mathbb{R}$ has cardinality $\aleph_1$.
Again, this follows from $\mathsf{CH}$ (since all countable subsets of $\mathbb{R}$ are Lebesgue measurable, but there are non-Lebesgue-measurable sets). However again both the above statement and its negation are consistent with $\neg \mathsf{CH}$
Now to hide virtually all set theory from the statement:
A topological space is called hereditarily separable if every subspace of is separable (so separable metric spaces are hereditarily separable). A regular (T$_3$) hereditarily separable but non-Lindelöf space is called an S-space.
So consider the following statement:
There is an S-space
• With $\mathsf{CH}$ one may construct $S$-spaces. (The Kunen line is one such example).
• In 1978 Szentmiklóssy Z. showed that $\mathsf{MA} + \neg \mathsf{CH}$ is consistent with the existence of S-spaces. (Starting in a model of $\mathsf{CH}$, show there is an S-space that cannot be destroyed by ccc forcing, and then force $\mathsf{MA}+\neg\mathsf{CH}$ in the usual manner by iterating all ccc posets of size $< \aleph_2$.)
• In 1981 S. Todorcevic showed that $\mathsf{PFA}$ implies that S-spaces do not exist. (For those claiming that this result has large cardinal power, the proof may be modified so as to avoid the large cardinals.)
• Excellent example. Does the more general statement $\kappa < \nu \rightarrow 2^\kappa < 2^\nu$ follow from $\mathrm{GCH}$? Its certainly not provable in ZFC. – goblin Jan 23 '14 at 9:29
• @user18921: yes, since assuming $\mathsf{GCH}$ we know that $2^\kappa = \kappa^+$, and the cardinal successor function is strictly increasing. – user642796 Jan 23 '14 at 9:32
• Thank you; I did not know that. (Although now that you mention it, its utterly obvious from the definitions!!!) – goblin Jan 23 '14 at 9:33
• @user18921: The less obvious part is that the reverse implication doesn't hold (i.e. $\kappa<\lambda\rightarrow 2^\kappa<2^\lambda$ doesn't imply $\sf GCH$). We can, however, easily violate $\sf GCH$ on every regular cardinal, but still keeping the continuum function injective everywhere. – Asaf Karagila Jan 23 '14 at 10:44
• @AsafKaragila, thanks. – goblin Jan 23 '14 at 10:49
The following statement is equivalent to $\mathsf{CH}$:
There exists a function $\chi: \mathbb R \rightarrow \mathbb N$ such that for all $x,y,z,t\in\mathbb R$, if $\chi(x) = \chi(y) = \chi(z) = \chi(t)$ and $x+y=z+t$, then at least two of these numbers are equal.
This result is due to Erdős. Since it is equivalent to $\mathsf{CH}$, it is not provable in $\mathsf{ZFC+\neg CH}$ assuming it is a consistent theory!
• +1 I think among all answers this is the one best fitting "outside set theory" (it doesn't mention anything about cardinality explcitly) and still looks awfully elementary. – Hagen von Eitzen Jan 23 '14 at 10:47
• This is a lovely result (new to me -- not that that means much!) Do you have a pointer to a proof, please? – Peter Smith Jan 23 '14 at 12:20
• Answering my own query, this looks what we need! arxiv.org/pdf/1201.1207v1.pdf – Peter Smith Jan 23 '14 at 13:55
There exists an outer automorphism of the Calkin algebra of a separable [infinite-dimensional] Hilbert space.
Assuming $\sf CH$ we can prove that there exists an outer automorphism of the Calkin algebra [of a separable Hilbert space]. When $\sf CH$ fails it is consistent that all automorphisms are inner. (The first result is due to Christopher-Weaver, the second due to Farah.)
An overview of both results can be found in Farah's paper,
Farah, Ilijas "All automorphisms of the Calkin algebra are inner". Ann. of Math. (2) 173 (2011), no. 2, 619–661. arXiv version.
The global dimension of $\prod_{i=1}^\infty\Bbb C$ is $2$.
This is in fact equivalent to the continuum hypothesis. See more in this MathOverflow thread.
The Kaplansky conjecture [for Banach algebras] fails. There exists a compact Hausdorff space $X$ such that there is a discontinuous homomorphism from $C(X)$ into another Banach algebra.
It was shown that the conjecture fails when one assumes the continuum hypothesis; but it is consistent that it holds, and the continuum hypothesis fails. (See Wikipedia for a bit more, and references.)
• It's worth noting that Farah's result is using OCA, which follows from PFA, which is another nice example of what the OP was looking for (think of it in terms of the Baire category theorem if you insist on having examples outside of set theory). – Haim Jan 23 '14 at 11:00
The statement $"\mathfrak{x}_0= \mathfrak{x}_1 "$ for many pairs of cardinal invariants of the continuum $\mathfrak{x}_0,\mathfrak{x}_1$ such that $"\mathfrak{x}_0=\mathfrak{x}_1"$ is not provable in $ZFC$ (for example, $\mathfrak a$ and $\mathfrak d$).
• Yeah, I was going to post this (specifically for $\frak b=d$), but then I saw that the OP asked for non-set theoretical applications of $\sf CH$. – Asaf Karagila Jan 23 '14 at 10:40
• Oh, c'mon! This is general topology. :) – Haim Jan 23 '14 at 10:42
• More like set theoretical topology... :-) – Asaf Karagila Jan 23 '14 at 10:44
There exists a subset $S$ of plane $\mathbb{R}^2$, such that intersection of any horizontal line and $S$ is countable, and intersection of any vertical line with $S$ has countable complement.
It's quite easy to show that if such $S$ exists, then $\aleph_1 = \mathfrak{c}$, so it's not provable in ZFC + $\lnot$ CH. It's a bit harder to show that if CH holds, then such $S$ exists. | 2019-11-15T23:11:29 | {
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https://math.stackexchange.com/questions/1244728/left-g-sets-category | # Left G-Sets category
Let $G$ be a group, and $\mathbf{G\text{-}Sets}$ the category whose objects are left G-Sets and whose morphisms are G-Set homomorphisms, that is functions $f:X\to Y$ such that $f(ax) = af(x)$, $a\in G, x\in X$.
Does the forgetful functor $\mathbf{G\text{-}Sets}$ $\to$ $\mathbf{Sets}$ have a left-adjoint?
Are there any other interesting functors between the categories $\mathbf{G\text{-}Sets}$ and $\mathbf{Sets}$?
Let $S$ be a set. Then $G \times S$ can be equipped with a $G$-action by having $G$ act by left-multiplication on itself and trivially on $S$. This is a left-adjoint to the forgetful functor -- for $T$ an arbitrary $G$-set, one has
$$\text{Hom}_\text{G-set}(G \times S, T) = \text{Hom}_\text{set}(S, T)$$ by the rule $f \mapsto \left[s \mapsto f(1, s)\right]$. The map the other way is $\phi \mapsto \left[(g, s) \mapsto g\phi(s)\right]$ .
• Why did you write $\text{Hom}_\text{G-set}(G \times S, T)$? Shouldn't it be $\text{Hom}_\text{G-set}(S, T)$ where $S$ is a set equipped with a certain map $G\times S \to S$? Those are the objects in G-Set -Thanks – iwriteonbananas Apr 21 '15 at 17:18
• my functor eats $S$ and returns $G \times S$ with the action described in the answer. That's what it means to be a left adjoint. In other words, if I call the functor $F$, the statement is $Hom_{G-set}(F(S), T) = Hom_{set}(S, Forget(T))$. – hunter Apr 21 '15 at 21:52
This is an instance of Kan extensions.
Let $\mathbf G$ be the one object category associated to the group $G$, and $\mathbf 1$ the one associated to the trivial group. Then $\mathbf{G{-}Sets}$ is nothing else than $[\mathbf G,\mathbf{Sets}]$ and the forgetful functor is the functor $$i^\ast \colon [\mathbf G,\mathbf{Sets}] \to [\mathbf 1,\mathbf{Sets}] \simeq \mathbf{Sets},\quad F \mapsto F \circ i$$ induced by the (unique) functor $i\colon \mathbf 1 \to \mathbf G$. As $\mathbf{Sets}$ is cocomplete, the functor $i^\ast$ admits a left adjoint, usually denoted $i_!$ (see nlab:Kan extension). Note that, as $\mathbf{Sets}$ is also complete, $i^\ast$ also admits a right adjoint, usually denoted $i_\ast$.
Remark that by computing the general formula for left Kan extension in this case, you get back hunter's answer: $$i_! \colon S \mapsto \coprod_{g\in G}S \simeq G \times S.$$
Here's another way of seeing why this functor has a left-adjoint.
Theorem. Let $\mathsf{S}$ and $\mathsf{T}$ denote Lawvere theories. Then given a morphism $\varphi : \mathsf{S} \leftarrow \mathsf{T}$, the corresponding forgetful functor $$\mathrm{Mod}(\varphi) : \mathrm{Mod}(\mathsf{T}) \leftarrow \mathrm{Mod}(\mathsf{S})$$ has a left-adjoint.
Now:
• The initial Lawvere theory is denoted $\mathsf{Set}$, and its models are (basically) sets.
• Given a group $G,$ we can build a Lawvere theory $\mathrm{Lath}(G)$ such that the models of $\mathrm{Lath}(G)$ are (basically) precisely the $G$-sets.
• Write $\varphi$ for the unique morphism $\varphi : \mathrm{Lath}(G) \leftarrow \mathsf{Set}.$
So use the above theorem to conclude that
$$\mathrm{Mod}(\varphi) : \mathrm{Mod}(\mathsf{Set}) \leftarrow \mathrm{Mod}(\mathrm{Lath}(G))$$
has a left-adjoint. In other words, the usual forgetful functor
$$\mathbf{Set} \leftarrow \mathbf{Set}^G$$
has a left-adjoint, where I write $\mathbf{Set}^G$ for the category of $G$-sets, since a $G$-set is the same thing as a functor $\mathbf{Set} \leftarrow G,$ where $G$ is viewed as a category with one object.
Although this uses a lot more technical machinery than is strictly required, I like this viewpoint because it provides a quick heuristic that can help you immediately conclude that a variety of different functors have left-adjoints.
If $H\le G$ is a subgroup let $\mathrm{Res}_H^G:G\textrm{-}\mathsf{Set}\to H\textrm{-}\mathsf{Set}$ be the restriction functor. One can think of $\mathsf{Set}$ as the case where $H=1$ is the trivial subgroup. It has a left adjoint, call it $\mathrm{Ind}_H^G$, so that
$$\hom_{G\textrm{-}\mathsf{Set}}(\mathrm{Ind}_H^G-,-) ~~\cong~ \hom_{H\textrm{-}\mathsf{Set}}(-,\mathrm{Res}_H^G-)$$
are naturally isomorphic functors $H\textrm{-}\mathsf{Set}\times G\textrm{-}\mathsf{Set}\to\mathsf{Set}$.
An explicit construction is $\mathrm{Ind}_H^G\Omega\cong G\times_H \Omega$, which is $G\times\Omega$ (the action of $G$ purely on the first coordinate) modulo the relation $(gh,\omega)\sim(g,h\omega)$ for all $g\in G,h\in H,\omega\in\Omega$.
Does this look familiar? If you've studied representation theory, it should. The linearized version of this fact is called Frobeius reciprocity. Let $G\textrm{-}\mathsf{Rep}$ be the category of linear representations of $G$, and then define the restriction functor $\mathrm{Res}_H^G:G\textrm{-}\mathsf{Rep}\to H\textrm{-}\mathsf{Rep}$. Frobenius reciprocity (the categorical version of it, as opposed to the more standard, decategorified, character-theoretic version of it) states that restriction has a left adjoint called induction $\mathrm{Ind}_H^G$, so $\hom_G(\mathrm{Ind}_H^GV,W)\cong\hom_H(V,\mathrm{Res}_H^GW)$ canonically for all representations $V$ of $G$ and $W$ of $H$.
An explicit construction is $\mathrm{Ind}_H^GV\cong \Bbb C[G]\otimes_{\Bbb C[H]}V$, the linearized analogue of $G\times_H\Omega$! | 2019-05-23T13:58:59 | {
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https://aayhcs.com/rachel-kelly-plogpi/93b5fb-linear-congruential-generator | Published on Nov 23, 2016 Random Number Generators (RNGs) are useful in many ways. Sure. Today, the most widely used pseudorandom number generators are linear congruential generators (LCGs). Introduced by Lehmer ( 1951 ), these are specified with nonnegative integers η, a, and c. 13 An integer seed value z[0] is selected, 0 ≤ z[0] < η, and a sequence of integers z[k] is obtained recursively with the formula. Linear Congruential Method is a class of Pseudo Random Number Generator (PRNG) algorithms used for generating sequences of random-like numbers in a specific range. Our next task is to implement a linear congruential generator algorithm as a means for creating our uniform random draws. 5 9 Combined Linear Congruential Generators [Techniques] Reason: Longer period generator is needed because of the increasing complexity of stimulated systems. For the purposes of this assignment, a linear congruential random number generator is defined in terms of four integers: the multiplicative constant a, the additive constant b, the starting point or seed c, and the modulus M. The purpose of the generator is to produce a sequence of integers between 0 and M-1 by starting with x 0 = c and iterating: They are defined by three integers, "multiplier", "increment" and "modulus", and … Linear congruential generators (LCG)¶ $$z_{i+1} = (az_i + c) \mod m$$ Hull-Dobell Theorem: The LCG will have a full period for all seeds if and only if $$c$$ and $$m$$ are relatively prime, $$a - 1$$ is divisible by all prime factors of $$m$$ $$a - 1$$ is a multiple of 4 if $$m$$ is a multiple of 4. Linear Congruential Generator Implementation. In other words, every number in the sequence is in [0,1) and the chance of being any number of [0,1) is equal. A more popular implementation for large periods is a combined linear congruential generator; combining (e.g. A Linear congruential generator (LCG) is a class of pseudorandom number generator (PRNG) algorithms used for generating sequences of random-like numbers. These types of numbers are called pseudorandom numbers. So what other criteria besides long period should be imposed. Let X i,1, X i,2, …, X i,k, be the ith output from k different multiplicative congruential generators. The transition algorithm of the LCG function is x i+1 ← (ax i +c) mod m.. Linear congruential generators (LCG) are a form of random number generator based on the following general recurrence relation: Where n is a prime number (or power of a prime number), g has high multiplicative order modulo n and x0 (the initial seed) is co-prime to n. Essentially, if g is chosen correctly, all integers from 1 to n−1 will eventually appear in a periodic fashion. linear_congruential_engine is a random number engine based on Linear congruential generator (LCG).A LCG has a state that consists of a single integer. (See [3], or other texts on number theory for general discussions of primitive roots). Can I embed this on my website? A Linear congruential generator (LCG) is a class of pseudorandom number generator (PRNG) algorithms used for generating sequences of random-like numbers. An LCG generates pseudorandom numbers by starting with a value called the seed, and repeatedly applying a given recurrence relation to it to create a sequence of such numbers. The modular notation “mod” indicates that z[k] is the remainder after dividing the quantity … a, the multiplier; a ≥ 0. c, the increment; c ≥ 0. m, the modulus; m > X 0… So the period is at most m-1. The format of the Linear Congruential Generator is. Since a computer can represent a real number with only finite accuracy, we shall actually be generating integers Xn between zero and some number m; the fraction. For the purposes of this assignment, a linear congruential random number generator is defined in terms of four integers: the multiplicative constant a, the additive constant b, the starting point or seed c, and the modulus M. The purpose of the generator is to produce a sequence of integers between 0 and M-1 by starting with x 0 = c and iterating: We can check theparameters in use satisfy this condition: Schrage's method restates the modulus m as a decompositionm=aq+r where r=mmoda andq=m/a. When using a large prime modulus m such as 231−1, themultiplicative congruential generator can overflow. The case of mixed congruential method, i.e. c ≠ 0, is much more complicated. Linear Congruential Generator Calculator. Section II: Linear Congruential Generator I. The theory behind them is relatively easy to understand, and they are easily implemented and fast, especially on computer … LCGs tend to exhibit some severe defects. We provide sets of parameters for multiplicative linear congruential generators (MLCGs) of different sizes and good performance with respect to the spectral test. Seed: a: b: n: So m is chosen to be very big, e.g. Wolfram Demonstrations Project Embedding is allowed as long as you promise to follow our conditions. This video explains how a simple RNG can be made of the 'Linear Congruential Generator' type. The special case c = 0 deserves explicit mention, since the number generation process is a little faster in this case. Assuming an appropriate multiplier is used (see 3 ), LCG128Mix has a period of $$2^{128}$$ … We choose four "magic numbers": The desired sequence of random numbers < Xn > is then obtained by setting. Can I embed this on my website? Example 8.1 on page 292 Issues to consider: The numbers generated from the example can only assume values from the set I … Approach: Combine two or more multiplicative congruential generators. L’Ecuyer describes a combined linear generator that utilizes two LCGs in Efficient and Portable Combined Random Number Generatorsfor 32-bit processors. For = 8, 9, . (Page 18-20 of [4]), The generator in RANDU is essentially (but not exactly the same as). This is a linear congruence solver made for solving equations of the form $$ax \equiv b \; ( \text{mod} \; m)$$, where $$a$$, $$b$$ and $$m$$ are integers, and $$m$$ is positive. A linear congruential generator (LCG) is an algorithm that yields a sequence of pseudo–randomized numbers calculated with a discontinuous piecewise linear equation. x n = (a x n−1 + c) (mod m), 1 u n = x n /m, where u n is the nth pseudo-random number returned. RANDU is still available at a number of computer centers and is used in some statistical analysis and simulation packages. A linear congruential generator is a method of generating a sequence of numbers that are not actually random, but share many properties with completely random numbers. Gen. Embedding is allowed as long as you promise to follow our conditions. congruential method are used by many authors to denote linear congruential methods with c = 0 and c ≠ 0. The period of LCG depends on the parameter. Cracking RNGs: Linear Congruential Generators Jul 10, 2017 • crypto , prng Random numbers are often useful during programming - they can be used for rendering pretty animations, generating interesting content in computer games, load balancing, executing a randomized algorithm, etc. Open content licensed under CC BY-NC-SA. RANDU is a linear congruential pseudorandom number generator (LCG) of the Park–Miller type, which has been used since the 1960s. Previous question Next question Get more help from Chegg. An example of what you’ll find:I generated some random numbers with a few different generators, some of which I made, and also used the one provided directly by Python. Exercise 2.1: Try the generator used in RANDU to see how does it work. The format of the Linear Congruential Generator isxn = (a xn−1 + c) (mod m), 1 un = xn/m,where un is the nth pseudo-random number returned.The parameters of this modelare a (the factor), c (the summand) and m (the base). Let X i,1, X i,2, …, X i,k, be the ith output from k different multiplicative congruential generators. We … Approach: Combine two or more multiplicative congruential generators. ii)b = a-1 is a multiple of p, for every prime p dividing m; iii)b is a multiple of 4, if m is a multiple of 4. This method can be defined as: where, X, is the sequence of pseudo-random numbers m, ( > 0) the modulus a, (0, m) the multiplier c, (0, m) the increment X 0, [0, m) – Initial value of sequence known as seed The parameters we will use for our implementation of the linearcongruential generator are the same as the ANSI C implementation(Saucier, 2000.). The generation of random numbers plays a large role in many applications ranging from cryptography to Monte Carlo methods. LCG128Mix is a 128-bit of O’Neill’s permuted congruential generator (1, 2). Linear Congruential Generator Calculator. A linear congruential generator is a method of generating a sequence of numbers that are not actually random, but share many properties with completely random numbers. a=954,365,343, seed=436,241, c=55,119,927, and m=1,000,000. The terms multiplicative congruential method and mixed Combined Linear Congruential Generators • Reason: Longer period generator is needed because of the increasing complexity of simulated systems. If a linear congruential generator is seeded with a character and then iterated once, the result is a simple classical cipher called an affine cipher; this cipher is easily broken by standard frequency analysis. Upgrade to Math Mastery. One of the most successful random number generators known today are special cases of the following scheme, which is called the linear congruential method. An an example of this kind of generator being used is in program RANDU, which for many years was the most widely used random number generator in the world. Powered by WOLFRAM TECHNOLOGIES Linear congruential generators (LCG) are a form of random number generator based on the following general recurrence relation: Sure. http://demonstrations.wolfram.com/LinearCongruentialGenerators/ Exercise 2.2: Give several examples of (c, m, a) satisfying the conditions of Theorem A and program the method with the tuples you find. Joe Bolte Published: March 7 2011. 1.2 The Linear Congruential Generator. A random bitmap generator to visualize the randomness of the Linear Congruential Generator algorithm. A linear congruential generator is defined by s n+1 = a s n + b mod m, where m is the modulus. The Linear Congruential Random Number Generator is a popular method of creating random numbers. by summing their outputs) several generators is equivalent to the output of a single generator whose modulus is the product of the component generators' moduli. These types of numbers are called pseudorandom numbers . This method can be defined as: where, X, is the sequence of pseudo-random numbers m, ( > 0) the modulus a, (0, m) the multiplier c, (0, m) the increment X 0, [0, m) – Initial value of sequence known as seed A LCG is parameterized by three integers , and . Note: Your message & contact information may be shared with the author of any specific Demonstration for which you give feedback. Seed: a: b: n: # Linear Congruential Generator. The method represents one of the oldest and best–known pseudorandom number generator algorithms. • Let X i,1, X i,2, …, X i,k be the i-th output from k different multiplicative congruential generators. It is linear congruential as the values are related to each other in a linear way, modulo m. It uses the sequence generator of: $$X_i\ = (a \times X_{i-1} + c) \mod m$$ and where X 0 is the initial seed value of the series. Linear Congruential Generators (LCG) are one of the oldest and most studied RNGs [8]. Take advantage of the Wolfram Notebook Emebedder for the recommended user experience. . At a glance, the graphs will always look random (except in trivial cases, such as when the modulus is a multiple of the multiplier), but there is actually a sophisticated study of how closely pseudorandom number generators approximate processes that are truly random. Give feedback ». If u 0 = 6, a = 8, m = 13, c = 7, what is u 1 using the linear congruential generator (LCG) Expert Answer . In the case of multiplicative congruential method, it's easy to see Xn = 0 should not be allowed, otherwise the sequence will be 0 forever afterwards. The parameters of this model are a (the factor), c (the summand) and m (the base). Contributed by: Joe Bolte (March 2011) Such a number a is called a primitive root modulo m . In the following we shall consider methods for generating a sequence of random real numbers Un, uniformly distributed between zero and one. Its basic form is. To do this, we generate three randomvectors x, y, z using our LCG above an… Given an initial seed , there is some such that. This is the simplest generator engine in the standard library. For a given value of m, we seek a such that k in equation (3.1) is φ(m). Its state is a single integer value, with the following transition algorithm: And because there are only m possible different values for Xn's, so the sequence will get into a cycle in at most m steps and the period is at most of length m. It's very reasonable that we want the sequence to have long period so it might look random. People like it because it's easy to understand and easily implemented. One of the most successful random number generators known today are special cases of the following scheme, which is called the linear congruential method. Linear congruential random number engine A pseudo-random number generator engine that produces unsigned integer numbers. The following function is an implementation of a linear congruentialgenerator with the given parameters above. All linear congruential generators use this formula: r n + 1 = a × r n + c ( mod m ) {\displaystyle r_ {n+1}=a\times r_ {n}+c {\pmod {m}}} Where: r 0 {\displaystyle r_ {0}} is a seed. In its simplest form, the generator just outputs s n as the nth pseudorandom number. We choose four "magic numbers": X 0 , the starting value; X 0 ≥ 0 . Here's the embed code: This is called a linear congruential sequence. Upgrade to Math Mastery. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Because Xn+1 is determined by Xn, so once some number in the sequence get repeated, the sequence will get into a cycle. , Java still relies on a linear congruential generator (LCG) for its PRNG; yet LCGs are of low quality—see further below. The LCG or linear congruential generator is yet another pseudo-random number generator calculated with a discontinuous piecewise linear equation. If we consider a = 3, we will find it's a primitive root modulo 31, so the sequence will have period 30. The following typedefs define the random number engine with two commonly used parameter sets: It's one of the oldest and best-known RNGs. Exercise 2.3: Give an example of (c, m, a) satisfying Theorem A but yeilds a sequence that obviously not random. Do they work well? Interact on desktop, mobile and cloud with the free Wolfram Player or other Wolfram Language products. Lehmer's original generation method had c = 0, although he mentioned c ≠ 0 as a possibility. This is why LCGs are termed pseudo-random. This is a linear congruence solver made for solving equations of the form $$ax \equiv b \; ( \text{mod} \; m)$$, where $$a$$, $$b$$ and $$m$$ are integers, and $$m$$ is positive. For general m, the period may not be as big as φ(m), but by choosing a properly, period that is big enough for use can still be achieved. Notice that X0 ≠ 0, so period of the sequence is the smallest positive value of k for which, The Euler-Fermat Theorem states that if a and m are relatively prime, then aφ(m) = 1 mod m, where φ(m) is the Euler's totient function,meaning the number of positive integers less than or equal to n that are coprime to n. The period, therefore, can be no greater than φ(m). To be precise, the congruential generators used are actually multiplicative since [latex]c_1 … If a linear congruential generator is seeded with a character and then iterated once, the result is a simple classical cipher called an affine cipher; this cipher is easily broken by standard frequency analysis. 231. Although they possess "enough" randomness for our needs (as n can be large), they ar… There is a powerful theorem as follows: The proof of the theorem is left out here and can be found in Page 15-18 of [1]. Linear Congruential Method is a class of Pseudo Random Number Generator (PRNG) algorithms used for generating sequences of random-like numbers in a specific range. Here's the embed code: Schrage's method wasinvented to overcome the possibility of overflow and is based on thefact that a(mmoda) is then obtained by setting X i,,. A, b are not known, then Thomas described how to break it generator algorithm 's. Does it work used in randu is essentially ( but not exactly the same as ) 7, =... '' http: //demonstrations.wolfram.com/LinearCongruentialGenerators/ Wolfram Demonstrations linear congruential generator Published: March 7 2011 Park–Miller type, which has been since... Will get into a cycle a popular method of creating random numbers < Xn > is then by... 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Specific Demonstration for which you Give feedback » Terms of use | Privacy Policy RSS. Calculated with a discontinuous piecewise linear equation best-known RNGs piecewise linear congruential generator equation: the desired sequence of pseudo–randomized calculated... Is allowed as long as you promise to follow our conditions to overcome possibility. Increasing complexity of stimulated systems can use the function to generate random numbers < Xn > is obtained! ] Reason: Longer period generator is a little faster in this case ranging from to... Note: Your message & contact information may be shared with the given parameters above + b mod,..., be the ith output from k different multiplicative congruential generators equation ( 3.1 ) is implementation! Restates the modulus m as a possibility generator ; combining ( e.g are a ( mmoda ) < m the. M ) combining ( e.g be made of the LCG or linear congruential generator is defined by s =. 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Magic numbers '': the desired sequence of pseudo–randomized numbers calculated with a discontinuous piecewise linear equation in applications. For generating a sequence of random real numbers Un, uniformly distributed between zero one! X i,1, X i, k be the ith output from k different multiplicative linear congruential generator generators http... Linear generator that utilizes two LCGs in Efficient and Portable combined random number generator with! | 2022-05-23T08:09:55 | {
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http://www.nrj.dk/h07kt/f541d9-aas-congruence-theorem | Theorem: AAS Congruence. 7th - 12th grade. Two triangles are congruent if two angles and an unincluded side of one triangle are equal respectively to two angles and the corresponding unincluded side of the other triangle ($$AAS = AAS$$). If they are, state how you know. The three angles of one are each the same angle as the other. Which triangle congruence theorem is shown? We can tell whether two triangles are congruent without testing all the sides and all the angles of the two triangles. Reflexive Property of Congruence (Theorem 2.1) 6. Solution: First we will list all given corresponding congruent parts. Since the only other arrangement of angles and sides available is two angles and a non-included side, we call that the Angle Angle Side Theorem, or AAS. After learning the triangle congruence theorems, students must learn how to prove the congruence. We solve these equations simultaneously for $$x$$ and $$y$$: (1) and (2) same as Example $$\PageIndex{2}$$. AAS Congruence Criterion:If any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the … AAS Congruence Theorem Monitoring Progress Help in English and Spanish at BigIdeasMath.com 3. Then it's just a matter of using the SSS Postulate. Answer: EDC by AAS Theorem. Answer to: How can we make a triangle using a protractor and a string and the AAS congruence theorem? This activity is designed to give students practice identifying scenarios in which the 5 major triangle congruence theorems (SSS, SAS, ASA, AAS, and HL) can be used to prove triangle pairs congruent. HL. 7. Write a proof. $$\triangle ABC$$ with $$\angle A = 40^{\circ}$$, $$\angle B = 50^{\circ}$$, and $$AB = 3$$ inches. If so, write the congruence statement and the method used to prove they are congruent. reflexive property. Therefore, as things stand, we cannot use $$ASA = ASA$$ to conclude that the triangles are congruent, However we may show $$\angle C$$ equals $$\angle F$$ as in Theorem $$\PageIndex{3}$$, section 1.5 $$(\angle C = 180^{\circ} - (60^{\circ} + 50^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$$ and $$\angle F = 180^{\circ} - (60^{\circ} + 50^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ})$$. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc. To order this book direct from the publisher, visit the Penguin USA website or call 1-800-253-6476. If the distance from $$P$$ to the base of the tower $$B$$ is 3 miles, how far is the ship from point Bon the shore? -Angle – Angle – Side (AAS) Congruence Postulate Triangles are congruent if the angles of the two pairs are equal and the lengths of the sides that are different from the sides between the two angles are equal. Congruence of triangles is based on different conditions. $$\angle C$$ and $$BC$$ of $$\angle ABC$$ and $$\angle E, \angle F$$ and $$EF$$ of $$\triangle DEF$$. Video ... AAS (Angle-Angle-Side) Theorem. Yes, SAS Congruence Postulate 12. Mathematics. Video Let triangle DEF and triangle GHJ be two triangles such that angle DEF is congruent to angle GHJ, angle EFD is congruent to angle HJG, and segment DF is congruent to segment GJ (hypothesis). Many people are not good at … It is clear that we must have $$AC = DF$$, $$BC = EF$$, and $$\angle C = \angle F$$, because both triangles were drawn in exactly the same way, Therefore $$\triangle ABC \cong \triangle DEF$$. $$\PageIndex{1}$$ and $$\PageIndex{2}$$, $$\triangle ABC \cong \triangle DEF$$ because $$\angle A, \angle B$$, and $$AB$$ are equal respectively to $$\angle D$$, $$\angle E$$, and $$DE$$. Check our encyclopedia for a gloss on thousands of topics from biographies to the table of elements. However, these postulates were quite reliant on the use of congruent sides. Find the distance $$AB$$ across a river if $$AC = CD = 5$$ and $$DE = 7$$ as in the diagram. ΔABC and ΔRST with ∠A ~= ∠R , ∠C ~= ∠T , and ¯BC ~= ¯ST. A Given: ∠ A ≅ ∠ D It is given that ∠ A ≅ ∠ D. Pertinence. Given: ΔABC and ΔRST are right triangles with ¯AB ~= ¯RS and ¯BC ~= ¯ST. $$\triangle DEF$$ with $$\angle D = 40^{\circ}$$, $$\angle E = 50^{\circ}$$, and $$DE = 3$$ inches. You will be asked to prove that two triangles are congruent. $$\begin{array} {rcl} {AB} & = & {CD} \\ {3x - y} & = & {2x + 1} \\ {3x - 2x - y} & = & {1} \\ {x - y} & = & {1} \end{array}$$ and $$\begin{array} {rcl} {BC} & = & {DA} \\ {3x} & = & {2y + 4} \\ {3x - 2y} & = & {4} \end{array}$$. Need a reference? FEN Learning is part of Sandbox Networks, a digital learning company that operates education services and products for the 21st century. We learn when triangles have the exact same shape. Given M is the midpoint of NL — . Explain 3 Applying Angle-Angle-Side Congruence Example 3 The triangular regions represent plots of land. The two triangles have two congruent corresponding angles and one congruent side. Figure 2.3.4. The following figure shows you how AAS works. The angle-angle-side Theorem, or AAS, ... That's why we only need to know two angles and any side to establish congruence. ... AAS. Write a paragraph proof. This ‘AAS’ means angle, angle, and sides which clearly states that two angles and one side of both triangles are the same, then these two triangles are said to be congruent to each other. Edit. Therefore $$x = AC = BC = 10$$ and $$y = AD = BD$$. $$\PageIndex{3}$$, section 1.5 $$(\angle C = 180^{\circ} - (60^{\circ} + 50^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ}$$ and $$\angle F = 180^{\circ} - (60^{\circ} + 50^{\circ}) = 180^{\circ} - 110^{\circ} = 70^{\circ})$$. Write a paragraph proof. D. Given: RS bisects ∠MRQ; ∠RMS ≅ ∠RQS Which relationship in the diagram is true? Start studying 3.08: Triangle Congruence: SSS, SAS, and ASA 2. Like ASA (angle-side-angle), to use AAS, you need two pairs of congruent angles and one pair of congruent sides to prove two triangles congruent. HL. AAS Congruence Rule You are here. If Angle LA Angle Z C Side BC - then A ABC — ZF, and ADEF. yes, because of ASA or AAS Explain how the angle-angle-side congruence theorem is an extension of the angle-side-angle congruence theorem. clemente1. Recall that for ASA you need two angles and the side between them. Angle-Angle-Side (AAS) Congruence Theorem If two angles and a non-included side of one triangle are congruent to the corresponding angles and non-included side of another triangle, then the triangles are congruent. For example, not only do you know that one of the angles of the triangle is a right angle, but you know that the other two angles must be acute angles. Answer: EDC by AAS Theorem. $$\triangle ABC$$ with $$\angle A = 50^{\circ}$$, $$\angle B = 40^{\circ}$$, and $$AB = 3$$ inches. Triangle Congruence Theorems (SSS, SAS, & ASA Postulates) Triangles can be similar or congruent. $$\PageIndex{3}$$. $$\PageIndex{4}$$. Therefore $$x = AB = CD = 12$$ and $$y = BC = DA = 11$$. The AAS Theorem says: If two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent. Since $$AB = AD + BD = y + y = 2y = 12$$, we must have $$y = 6$$. We've just studied two postulates that will help us prove congruence between triangles. This is the AAS congruence theorem. ASA stands for “Angle, Side, Angle”, which means two triangles are congruent if they have an equal side contained between corresponding equal angles. Morewood. Solution to Example 4 1. $$\angle A$$ and $$\angle B$$ in $$\triangle ABC$$. Figure 12.9These two triangles are not congruent, even though two corresponding sides and an angle are congruent. angles … In the ASA theorem, the congruence side must be between the two congruent angles. Therefore $$x = SB = FB = 3$$. AAS Congruence A variation on ASA is AAS, which is Angle-Angle-Side. AAS is equivalent to an ASA condition, by the fact that if any two angles are given, so is the third angle, since their sum should be 180°. Then you'll have two angles and the included side of ΔABC congruent to two angles and the included side of ΔRST, and you're home free. Brush up on your geography and finally learn what countries are in Eastern Europe with our maps. 6. Theorem 2.3.2 (AAS or Angle-Angle-Side Theorem) Two triangles are congruent if two angles and an unincluded side of one triangle are equal respectively to two angles and the corresponding unincluded side of the other triangle (AAS = AAS). But, if you know two pairs of angles are congruent, then the third pair will also be congruent by the Angle Theorem. For each of the following (1) draw the triangle with the two angles and the included side and (2) measure the remaining sides and angle. How?are they different? SSS ASA SAS HL. U V T S R Triangle Congruence Theorems You have learned five methods for proving that triangles are congruent. SURVEY . McGuinness … Prove RST ≅ VUT. Yes, SAS Congruence Postulate 12. Angle-Angle-Side (AAS or SAA) Congruence Theorem: If two angles and a non-included side in one triangle are congruent to two corresponding angles and a non-included side in another triangle, then the triangles are congruent. Answer: (1) $$PQ$$, (2) $$PR$$, (3) $$QR$$. For more information contact us at [email protected] or check out our status page at https://status.libretexts.org. ($$AC = CA$$ because they are just different names for the identical line segment, We sometimes say $$AC = CA$$ because of identity.) Sufficient evidence for congruence between two triangles in Euclidean space can be shown through the following comparisons:. 23 - 26. 3. In $$\triangle DEF$$ we would say that DE is the side included between $$\angle D$$ and $$\angle E$$. 56 terms. AAS (Angle-Angle-Side): If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent. What is AAS Triangle Congruence? ΔMNR ≅ ΔMNS by ASA ΔRMS ≅ ΔRQS by AAS ΔSNQ ≅ ΔSNM by SSS ΔQNR ≅ ΔMNR by HL. YOU MIGHT ALSO LIKE... SSS, SAS, ASA, AAS, & HL. $$\angle S$$ and $$\angle T$$ in $$\triangle RST$$. Lv … In Figure 12.9, the two triangles are marked to show SSA, yet the two triangles are not congruent. The triangles are then congruent by $$ASA = ASA$$ applied to $$\angle B$$. Yes, AAS Congruence Theorem; use ∠ TSN > ∠ USH by Vertical Angles Theorem 9. $$\PageIndex{4}$$, if $$\angle A = \angle D$$, $$\angle B = \angle E$$ and $$BC = EF$$ then $$\triangle ABC \cong \triangle DEF$$. Resource Locker Explore Exploring Angle-Angle-Side Congruence If two angles and a non-included side of one triangle are congruent to the corresponding angles and side of another triangle, are the triangles congruent? It is wrong because the congruent side we have is SR=RS. The AAS (Angle-Angle-Side) theorem states that if two angles and a nonincluded side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent. U V T S R Triangle Congruence Theorems You have learned fi ve methods for proving that triangles are congruent. Proving Segments and Angles Are Congruent, Chinese New Year History, Meaning, and Celebrations. If so, write the congruence statement and the method used to prove they are congruent. In this section we will consider two more cases where it is possible to conclude that triangles are congruent with only partial information about their sides and angles. Angle-Angle-Side (AAS or SAA) Congruence Theorem: If two angles and a non-included side in one triangle are congruent to two corresponding angles and a non-included side in another triangle, then the triangles are congruent. The correct option is the AAS theorem. Proof: You need a game plan. Yes, AAS Congruence Theorem 11. Thus the five theorems of congruent triangles are SSS, SAS, AAS, HL, and ASA. Since AC and EC are the corresponding nonincluded sides, ABC ≅ ____ by ____ Theorem. Show Answer ∆ ≅ ∆ ≅ ∠ Example 2. Congruence and Congruence Transformations; SSS and SAS; ASA and AAS; Triangles on the Coordinate Plane; Math Shack Problems ; Quizzes ; Terms ; Handouts ; Best of the Web ; Table of Contents ; ASA and AAS Exercises. Given I-IF GK, Z F and Z K … You also have the Pythagorean Theorem that you can apply at will. 17. 8. Name Class Date 6.2 AAS Triangle Congruence Essential Question: What does the AAS Triangle Congruence Theorem tell you about two triangles? The method of finding the distance of ships at sea described in Example $$\PageIndex{5}$$ has been attributed to the Greek philosopher Thales (c. 600 B.C.). (3) $$AC = BC$$ and $$AD = BD$$ since they are corresponding sides of the congruent triangles. There are several ways to prove this problem, but none of them involve using an SSA Theorem. Hence angle ABC = 180 - (25 + 125) = 30 degrees 2. In the diagram how far is the ship S from the point $$P$$ on the coast? Learn more about the mythic conflict between the Argives and the Trojans. So "$$C$$" corresponds to "$$A$$". Legal. Infoplease is part of the FEN Learning family of educational and reference sites for parents, teachers and students. Proving Congruent Triangles with SSS. Perpendicular Bisector Theorem. Similarly for (2) and (3). 4 réponses. This geometry video tutorial provides a basic introduction into triangle congruence theorems. It's time for your first theorem, which will come in handy when trying to establish the congruence of two triangles. Excerpted from The Complete Idiot's Guide to Geometry © 2004 by Denise Szecsei, Ph.D.. All rights reserved including the right of reproduction in whole or in part in any form. Infoplease knows the value of having sources you can trust. And finally, we have the Leg Angle Congruence Theorem. PROVING A THEOREM Prove the Converse of the Base Angles Theorem (Theorem 5.7). Our editors update and regularly refine this enormous body of information to bring you reliable information. A Given: ∠ A ≅ ∠ D It is given that ∠ A ≅ ∠ D. If ZA A), AC —DF, and LF, then ADEE Pmof p. 270 D Theorem Theorem 5.11 Angle-Angie-Side (AAS) Congruence Theorem Missed the LibreFest? WRITING How are the AAS Congruence Theorem (Theorem 5.11) and the ASA Congruence Theorem (Theorem 5.10) similar? 4x — E 2 K ATRA, AARG AKHJ, AJLK Determine which triangle congruence theorem, if any, can be used to prove the triangles are congruent. (Hint: Draw an auxiliary line inside the triangle.) ΔABC and ΔRST are right triangles with ¯AB ~= ¯RS and ¯~= ¯ST. Triangle Congruence Theorems DRAFT. If under some correspondence, two angles and a side opposite one of the angles of one triangle are congruent, respectively, to the corresponding two angles and side of a second triangle, then the triangles are congruent. You've accepted several postulates in this section. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. In a nutshell, ASA and AAS are two of the five congruence rules that determine if two triangles are congruent. Whenever you are given a right triangle, you have lots of tools to use to pick out important information. What additional information is needed to prove that the triangles are congruent using the AAS congruence theorem? Explain 3 Applying Angle-Angle-Side Congruence Example 3 The triangular regions represent plots of land. Two triangles are congruent if two angles and an included side of one are equal respectively to two angles and an included side of the other. Angle-Angle-Side (AAS) Congruence Theorem THEOREM 4.6 If two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of a second triangle, then the two triangles are congruent. 1 - 4. NL — ⊥ NQ — , NL — ⊥ MP —, QM — PL — Prove NQM ≅ MPL N M Q L P 18. (1) write a congruence statement for the two triangles. The three sides of one are exactly equal in measure to the three sides of another. We extend the lines forming $$\angle A$$ and $$\angle B$$ until they meet at $$C$$. and BC AABC Proof p. EF, then ADEF. Find how two triangles are congruent using CPCT rules.SAS, SSS, AAS, ASA and RHS rule of congruency of triangles at BYJU’S. How do you know? ... AAS. This is true since the triangle have two congruent angles as demonstrated by the arc marks and they share a side. Wxs ≅ YZS 1246120, 1525057, and ¯BC ~= ¯ST & HL = BD\.! Two pair of corresponding sides and an Angle are congruent by \ ( \triangle ). Brush up on your geography and finally learn what countries are in Eastern Europe with our maps 7.1. ≅ ∠D BC ≅ EF \ ( A\ ) '' corresponds ! Ssa Theorem ~= ΔRST CUNY Academic Works ∠E = 90°, hypotenuse statement and included. ∠E and BC AABC proof p. EF, then the third pair will also be congruent \! 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Aas\ ) Theorem 5.11 ) and \ ( A\ ) and \ ( AAS or Angle-Angle-Side Theorem, a!, even though two corresponding sides and all the angles 've just studied two that. | 2021-09-25T13:15:11 | {
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https://math.stackexchange.com/questions/2517365/what-cases-to-consider-when-proving-that-x-1-x-2-implies-fx-1-leq-fx-2 | # What cases to consider when proving that $x_1 < x_2 \implies f(x_1) \leq f(x_2)$, i.e. function is increasing
So I have the following function $f$: $$f(x) = \begin{cases} 12x - 4100 && \text{if x < - 2}\\ (x - 14)^3 - 21 && \text{if x \geq - 2} \end{cases}$$ Okay, so I want to prove that this function is increasing by using monotonicity and no derivatives. Do I have to prove the following 3 cases? I'm not sure about case 3, how do I know what to put in my case 3?
Case 1 : $x_1,x_2 < -2$
Case 2 : $x_1,x_2 \geq -2$
Case 3 : $x_1 \leq -2 \leq x_2$
• In case 3, take $x_1 < -2 \leq x_2$. – N. F. Taussig Nov 12 '17 at 22:16
• @N.F.Taussig , With your correction of Case 3, have I done it correct by doing the following? 12x1 - 4100 < 12(-2) - 4100 < (-2-14)^3 - 21 <= (x2 -14)^3 - 21 – Lukas Nov 12 '17 at 22:31
• Your inequalities are correct. However, you should state, or better yet prove, that $12x - 4100$ is monotonically increasing to justify the first inequality and that $(x - 14)^3 - 21$ is monotonically increasing to justify the last inequality. – N. F. Taussig Nov 12 '17 at 22:35
• @N.F.Taussig , hmm isn't that what I'm supposed to do in case 1 and case 2? So for case 1: 12x1 - 4100 < 12x2 - 4100 and for case 2: (x1-14)^3 - 21 < (x2-14)^3 - 21 – Lukas Nov 12 '17 at 22:46
For case 3, we take $x_1 < -2 \leq x_2$.
Since $x_1 < -2$, $-2 - x_1 > 0$. Hence, $$12(-2) - 4100 - [12(x_1) - 4100] = 12(-2 - x_1) > 0 \implies 12x_1 - 4100 < 12(-2) - 4100$$ as you claimed in the comments.
Direct calculation shows that $12(-2) - 4100 = -4124 < -4117 = (-2 - 14)^3 - 21$.
Since $x_2 \geq -2$, $x_2 - (-2) \geq 0$.
\begin{align*} f(x_2) - f(-2) & = (x_2 - 14)^3 - 21 - [(-2 - 14)^3 - 21]\\ & = x_2^3 - 42x_2^2 + 588x_2 - 2744 - 21 - [(-2)^3 - 42(-2)^2 + 588(-2) - 2744 - 21]\\ & = [x_2^3 - (-2)^3] - 42[x_2^2 - (-2)^2] + 588[x_2 - (-2)]\\ & = [x_2 - (-2)][x_2^2 + (-2)x_2 + (-2)^2] - 42[x_2 - (-2)](x_2 + (-2)] + 588[x_2 - (-2)]\\ & = [x_2 - (-2)][x_2^2 + (-2)x_2 + (-2)^2 - 42x_2 - 42(-2) + 588x_2 - 588(-2)]\\ & = [x_2 - (-2)]\{x_2^2 + 544[x_2 - (-2)]\}\\ & \geq 0 \end{align*} since $x_2 - (-2) \geq 0$ and $x_2^2 + 544[x_2 - (-2)] \geq 0$.
Therefore, $$f(x_1) = 12x_1 - 4100 < 12(-2) - 4100 < (-2 - 14)^3 - 21 \leq (x_2 - 14)^3 - 21 = f(x_2)$$ which proves the result holds for case 3. | 2019-07-19T00:29:21 | {
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https://crypto.stackexchange.com/questions/86916/symmetric-exchange-key-sharing | Symmetric Exchange Key Sharing
Question:
Suppose that there are 10000 = 10^4 banks and 10 payment card organisations (PCOs).
1. How many secret keys will be needed if each PCO shares a unique secret key with each bank?
2. How many extra secret keys will be needed if every two banks share a unique secret key?
For question 1, I figured that if there are 10,000 banks and 10 PCOs:
10,000 * 10 = 100,000.
For question 2, I need to calculate the number of keys if every 2 banks share a key (so I guess for 10,000 banks there would be 5,000 keys?)
5,000 * 10 = 50,000, right?
I recall that the solution could be calculated differently, something along the lines of 5,000(5,000-1)*10/2 but I don't understand the logic of this.
Could someone please explain?
(I would appreciate explanation of logic or a direction for research to understand because I feel like this should be quite straightforward and I am unsure).
Thank your for taking the time to read this!
• This is more of a pure math question. Your mistake is that 10,000 banks would not mean 5,000 keys. Let's take 3 banks, A, B, and C. If every 2 banks share a key, then you need a key for (A,B), (A,C), and (B,C). 3 not 3/2. If you play around with small numbers, you should be able to see how you get to the real solution. – Aman Grewal Dec 14 '20 at 19:17
• This might also help: en.wikipedia.org/wiki/Arithmetic_progression#Sum – cisnjxqu Dec 14 '20 at 19:18
• Each key between the banks represents a selection of 2 elements from a set. Do you know how to calculate the possible different selections? – kelalaka Dec 14 '20 at 19:22
• Analogy (pre-covid-19): $n$ persons meet. Each shake hands with all the others. How many handshakes? Hint: each of the $n$ persons does $n-1$ handshakes, and one hanshake serves two persons. – fgrieu Dec 15 '20 at 16:04
The underlying principle here is that of combinatorics. Think of banks and PCOs as abstract parties. Party $$A$$ has $$|A|$$ many members and party $$B$$ has $$|B|$$ many respectively. For each $$x \in A$$ we have $$|B|$$ secret keys which $$x$$ shares with members of $$B$$ meaning we have $$|A| \times |B|$$ keys.
In your first question, we assumed that for all $$x \in A$$ and $$y \in B$$, we know that $$x \neq y$$. I. e., The two parties $$A$$ and $$B$$ don't share any members! In your second question however, that assumption is not true since $$A = B$$ which means the party of banks sharing keys with the party of banks, id es, among themselves.
The question here is then how many ways to pick unique $$\{x, y\}$$ combinations from a set $$A$$. Well, this is just an $$n$$ chose $$k$$: $$\binom{n}{k}$$ problem. In this case $$k = 2, n = |A|$$.
How many secret keys will be needed if each PCO shares a unique secret key with each bank?
$$|A|\times |B| = 10^4 \times 10 = 10^5$$
How many extra secret keys will be needed if every two banks share a unique secret key?
$$\binom{10^4}{2} = \frac{10^4!}{2!(10^4-2)!} = \frac{10000!}{2(10000-2)!} = \frac{10000!}{2(9998)!} = \frac{9999 \times 10000}{2}$$ | 2021-08-01T11:13:29 | {
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http://mathhelpforum.com/calculus/25440-integral.html | # Math Help - Integral
1. ## Integral
$\int \frac{1}{x^7-x}\;dx$
Thanks
2. There are various things tou can do. Partial fractions. Maybe just a small factoring and a sub.
$\int\frac{1}{x(x^{6}-1)}dx$
Let $u=x^{6}-1, \;\ du=6x^{5}dx, \;\ \frac{du}{6}=x^{5}dx, \;\ u+1=x^{6}$
This leads to $\frac{1}{6}\int\frac{1}{u^{2}+u}du$
$\frac{1}{6}\left[\int\frac{1}{u}du-\int\frac{1}{u+1}du\right]$
Now, continue?.
If you feel industrious, you could do the PFD thing.
$x^{7}-x = x(x+1)(x-1)(x^{2}+x+1)(x^{2}-x+1)$
3. How exactly does this: Let $u=x^{6}-1, \;\ du=6x^{5}dx, \;\ \frac{du}{6}=x^{5}dx, \;\ u+1=x^{6}$ lead to this:
$\frac{1}{6}\int\frac{1}{u^{2}+u}du$
thats the part i didn't understand, thats why i couldn't use subst. myself.
The aftermath is easy....
Originally Posted by galactus
If you feel industrious, you could do the PFD thing.
$x^{7}-x = x(x+1)(x-1)(x^{2}+x+1)(x^{2}-x+1)$
After i posted this, i actually solved it, but i did this before applying PF....the above would take to long...
$\int \frac{1}{x(x^3-1)(x^3+1)}=\frac{1}{2}\int \frac{(x^3+1)-(x^3-1)}{x(x^3-1)(x^3+1)}=\frac{1}{2}\int\frac{1}{x(x^3-1)}-\frac{1}{x(x^3+1)}$
4. Let's see
We want: $\int{\frac{dx}{x\cdot{(x^6-1)}}}$
Multiply and divide by $6\cdot{x^5}$ we get: $\int{\frac{6\cdot{x^5}}{6\cdot{x^6}\cdot{(x^6-1)}}}dx$
We now let: $u=x^6-1\rightarrow{\frac{du}{dx}=6\cdot{x^5}}$
Thus we have: $\int{\frac{du}{6\cdot{u}\cdot{(u+1)}}}=\frac{1}{6} \cdot{\int{\frac{du}{u\cdot{(u+1)}}}}$
5. Originally Posted by PaulRS
Multiply and divide by $6\cdot{x^5}$
That was the critical step i need to know!!
Thanks
6. Originally Posted by polymerase
$\int \frac{1}{x^7-x}\;dx$
You can also avoid substitutions
$\int {\frac{1}
{{x^7 - x}}\,dx} = \int {\frac{1}
{{x\left( {x^6 - 1} \right)}}\,dx} = \int {\frac{{x^6 - \left( {x^6 - 1} \right)}}
{{x\left( {x^6 - 1} \right)}}\,dx}$
.
So $\int {\frac{1}
{{x^7 - x}}\,dx} = \frac{1}
{6}\int {\frac{{\left( {x^6 - 1} \right)'}}
{{\left( {x^6 - 1} \right)}}\,dx} - \int {\frac{1}
{x}\,dx}$
.
The rest follows.
7. Hello, PaulRS!
That's brilliant! . . . Thank you!
8. Here's what I was getting at:
Because, $\frac{x^{5}}{x^{6}}=\frac{1}{x}$
Now, because $u=x^{6}-1, \;\ u+1=x^{6}, \;\ du=6x^{5}dx, \;\ \frac{du}{6}=x^{5}dx$
Rewrite as : $\frac{1}{6}\int\frac{x^{5}}{x^{6}(x^{6}-1)}dx$
Make the subs and we get:
$\frac{1}{6}\int\frac{1}{(u+1)u}du=\frac{1}{6}\int\ frac{1}{u^{2}+u}du$
See now?. | 2014-04-21T08:18:55 | {
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http://tckz.residencemaiori.it/lesson-41-practice-b-graph-exponential-growth-functions-answers.html | An exponential function with a > 0 and b > 1, like the one above, represents an exponential growth and the graph of an exponential growth function rises from left to right. Determine if it is a growth function, and then find the y-intercept, asymptote, domain, and range. Just hearing the word is enough to send some students running for the hills. Sample response: For a number of years, the population of forest A will increasingly exceed forest B, but because forest B actually grows at a faster rate, the population will eventually become larger than forest A and will remain that way as long as the population growth. Recall the table of values for a function of the form f(x) = bx. 1 Practice B 8. 7 AI/AII/Precalculus. In order to determine whether this graph represents exponential growth or exponential decay, we must first remember that exponential growth and decay functions look like this: ${y}$ = ${a}$${b}$$^{x}$ ${a}$ = y-intercept ${b}$ = Rate of growth/decay If ${b}$ is between 0 up to but not including 1, the function is exponential decay. If you get at least 8 correct on your first attempt, then you're ready to move on. 19 Chapter 4-4: WRITING AND GRAPHING FUNCTIONS SWBAT: (1) Write a function rule from a table (2) Graph functions given a limited domain Pgs. Practice makes perfect!. Determine whether the ordered pairs represent a. 3 – pose and solve problems involving exponential relations arising from a variety of real-world applications (e. School A 240 students decreasing at an annual rate of 2% School B 180 students increasing at an annual rate of 3% Initial Population: 500 plants Annual Growth Rate: 7% 30. x 2x y ±2 2±2 0. • Compare exponential growth to linear growth using tables and graphs • Build an exponential function given a description of an exponential situation. ; NCTM Illuminations: Shedding Light on a Subject: This unit has four lessons in which learners explore "the development of a mathematical model. 7e - Graph exponential and logarithmic functions, showing intercepts and end behavior, and trigon. Advanced Functions and Modeling Define the standard form of an exponential function and explore a variety of its applications, such as exponential growth and decay (in the forms of doubling time and half-life), as well as compound interest. Any time $$x$$ increases by 1, $$f(x)$$ increases by a factor of 3. Lets create a table of the data. Representing exponential growth with tables, graphs, equations; rules for exponents, scientific notation; Exponential Decay; growth/decay factors and rates. Before we begin graphing, it is helpful to review the behavior of exponential growth. Write and graph exponential growth models. indd 60AA2CL_010_05. In reality, exponential growth is only part of the bacterial life cycle, and not representative of the normal pattern of growth of bacteria in Nature. Graphing Exponential Functions With e, Transformations, Domain and Range, Asymptotes, Precalculus - Duration: 10:13. 4 Scientific Notation 8. y 5 , where a is the and. The graph no longer has a vertical stretch of 2. 07 t, where P is the population after t years and C is the current population. I can apply exponential functions to real world situations. Worksheets are Correctionkeynl cca c name class date 15 4 graphing, Exponential functions date period, Lesson constructing exponential functions 15 3 practice, 11 exponential and logarithmic functions work, Graphing exponential functions, 4 1 exponential functions and. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of a. Example 4(b) seemed like a pretty rough model, so I rounded my answer. 9842)t, where t is the number of years a!er 2000 A. Chapter 2: Functions and Their Graphs. whose base is greater than one. Exponential Functions Topics: 1. x y 2 1 0 9 3 1 x y 3 27 (0,1) (1,3) Figure 23. An exponential function is a function of the form. 1 Graph Exponential Growth Functions An exponential function has the form y=ab^x where a≠0 and the base b is a positive number other than 1. Person B's values will increase faster. Fun, visual skills bring learning to life and adapt to each student's level. The following is the graph of y = 2 x. 2 Graph Exponential Decay Functions Lesson 7. It's exponential decay when the base of our exponential is in between 1 and 0 and those numbers get smaller. b x tMnasdUe5 SwLi5tFhR LITnofhiRnSiOtoey 6A4lSgOeQb9rDaq y2X. Include equations arising from linear and quadratic functions, and simple rational and exponential functions. Exponential Functions. Graphing exponential functions. Explain 1 Roughly Fitting an Exponential Function to Data As the answer to the last Reflect question above indicates, if the ratios of successive values of the dependent variable in a data set for equally-spaced values of the independent variable are equal, an exponential function model fits. Lesson: Gossip Problem—Comparing Exponential and Linear Functions Note: Three fifty-minute class periods Essential learning standards: As a result of this lesson, students will be able to. Here are the functions again:. 9 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables. Notice that the two functions meet at x = 2 and x = 4, and then the exponential function becomes bigger than the quadratic. What will be the height of the 5th bounce? Round to the nearest tenth of a foot. Make a table of values and a graph for the function fx() 32 x. The function has a y-intercept of +1, the f(0)-value. In the graph of an exponential growth function, the y-values increase as the x-values increase, while in an exponential decay function, the y-values decrease as the x-values increase. Downloadable Lecture Notes. State the domain the range. Four variables (percent change, time, the amount at the beginning of the time period, and the amount at the end of the time period) play roles in exponential functions. -(6)^-1 6 1/6 -1/6 -6 2. An exponential growth function can be written in the form y = abx where a > 0 and b > 1. In order to determine whether this graph represents exponential growth or exponential decay, we must first remember that exponential growth and decay functions look like this: ${y}$ = ${a}$${b}$$^{x}$ ${a}$ = y-intercept ${b}$ = Rate of growth/decay If ${b}$ is between 0 up to but not including 1, the function is exponential decay. For any positive number a>0, there is a function f : R ! (0,1)called an exponential function that is defined as f(x)=ax. Yes, the relationship is an exponential function because the growth pattern is doubling from year to year. g ( x ) = 0. Graph the following function. Sample answer: Yes, the point ()4, 8 disappears and is replaced by the point ()4, 12. Make a table of values. The greater the base, b, the faster the graph rises from left to right. Assignment Explore exponential growth. Then plot the points and sketch the graph. will consider logarithmic functions in another lesson. Graph Exponential Functions An exponential function f is given by f(x) = b x where x is any real number, b > 0, and b ≠ 1. 0, and b 2 1. We discuss exponential growth, exponential decay, domain, and. Unit 4: Analyze and Graph Linear Equations, Functions and Relations Learning Objectives Lesson 1: Graphing Linear Equations Topic 1: Rate of Change and Slope Learning Objectives • Calculate the rate of change or slope of a linear function given information as sets of ordered pairs, a table, or a graph. 7-4 Exponential Functions and Their Graphs 7-5 Solving Equations Involving Exponents 7-6 Solving Exponential Equations 7-7 Applications of Exponential Functions Chapter Summary Vocabulary Review Exercises Cumulative Review EXPONENTIAL FUNCTIONS The use of exponents to indicate the product of equal factors evolved through many different nota-tions. Percentage growth: If a function has an initial value of 10 and grows at a rate of 7% per year, then it is an exponential function with P = 10 and a = 1 + r = 1 + 0. Move to page 2. Graph exponential growth and decay functions. b is the growth (multiplier). For exponential decay. Graphing Exponential Functions Practice and Problem Solving: D Solve. Graph Exponential Decay Functions - Section 7. School A 240 students decreasing at an annual rate of 2% School B 180 students increasing at an annual rate of 3% Initial Population: 500 plants Annual Growth Rate: 7% 30. So now we have =−4cos −1. 2 x a 1 b 1. Classify each function as an exponential growth function or an exponential decay function. The graph will be translated 1 unit down instead of 3 units up. The rate of growth starts slow and increases rapidly as x increases. Solution: The base 10 is used often, most notably with scientific notation. Exercises for the lesson "Graph Exponential Growth Functions" Skill Practice 1. there was room for students to create their table and graph. Thur Apr 30: 7-7 Writing Exponential Functions. Greater than 1. Let’s look for a moment at how the two functions change when the input is incremented by 1. But never fear! Though function problems are considered some of the more challenging questions on the ACT, this is only due to the fact that most of you will be far more used to dealing with other math topics (like fractions, exponents, or circles) than you are functions. The answers for these pages appear at the back of this booklet. Solution: T his is a growth function with y-values that grow larger as the x-values approach positive infinity. Use the table of values to graph y x 2 4. 86 $309,484. 6 0 b 1, so the function shows exponential decay. Processing is an open source programming language and environment for people who want to create images, animations, and interactions. The following questions practice these skills: Identify total cost, variable cost, fixed cost, marginal cost, and average total cost. 3 Practice A 1. the same linear scenario. y = 2 x 62/87,21 Complete a table of values for ±2 < x < 2. For l(x), the graph of f(x) is vertically translated upward 2 units 5. 3 – pose and solve problems involving exponential relations arising from a variety of real-world applications (e. Which graph represents exponential decay? C. If not, review "In Depth" and try again. ; Identifying Function Models: Identify function types by table values. Practice Solutions. Please complete the following items: 1. Step 1 Make a table of values. 1 Exponential Growth Functions Essential Question: How is the graph of g ( x) = b x Where b > 1 Given the graph of an exponential function, you can use your knowledge of the transformation parameters to write An exponential growth function has the form ƒ(t) = a (1 + r). 2 - Exponential Decay Name:_____ 1 Write your questions and thoughts here! RECALL: What is an exponential function? Today, we will focus on exponential functions that _____ towards the asymptote as you move left to right. y = 3(1 + 0. Exponential Functions. The graph no longer has a vertical stretch of 2. Solution: This is an exponential growth function. 1 Practice B (Answers) 8. If a basketball is bounced from a height of 15 feet, the function f(x) 15 (0. represents the constant rate of change and b the y-intercept. Which graph represents exponential decay? C. Graph exponential functions of the basic form f(x)=a⋅rˣ. The following is the graph of y = 2 x. SEE EXAMPLE 5 20 40 60 100 0 2 4 6 80 x 0 (0, 30) (3, 58. b is positive. The graph will curve upward, as shown in the example of f ( x) = 2 x below. Q x 2 4 x x 1 Identify whether the function graphed has an odd or even degree and a positive or negative leading coefficient. Choose both negative and positive values for x. arithmetic sequences as linear functions and geometric sequences as exponential functions. 2 The Graph of a Function. 1: Guess My Number! Sometimes students want an alternative explanation of an idea along with additional practice problems. Page 335 What is an exponential function? Page 336. Save time by downloading readily available lectures notes. Approach, More Practice Your Skills with Answers as part of the Teaching Resources package for the book, the right to reproduce material for use in his or her own classroom. Graph each exponential function. The management at a factory has determined that a worker can produce a maxi-mum of 30 units per day. 75)' 7-41 18. Initially developed to serve as a software sketchbook and to teach fundamentals of computer programming within a visual context, Processing also has evolved into a tool for generating finished professional work. 02)ᵗ, y = (0. In Part I, Orlando Pajon uses a population growth simulation to introduce students to exponential growth and develop the conceptual understanding underlying the principles of exponential functions. We’ll track your progress and help you identify your strengths and weaknesses. Horizontal axis, number of weeks, scale is 0 to15 by 5's. 361 Transformations of Exponential Functions Graphs - Khan Academy. I understand the basic characteristics of an exponential function. The old equation grapher we provided used Flash, which is grossly out-of-date and not widely supported any longer. • has an asymptote (a line that the graph gets very, very close to, but never crosses or touches). Suppose you deposit$1500 in a savings account that pays interest at an annual rate of 6%. " To get a first-hand understanding of how these functions behave, let's sketch the graphs of several exponential functions on our calculator and examine their x- and y-intercepts. 07 t, where P is the population after t years and C is the current population. Answer Key Lesson 6. 4 Properties of Functions and 2. y 54x for x 53 8. 9 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables. Exponential Growth If a 0 and b 1, the function y abx represents exponential growth. Application Walkthrough. Use the interval [0, 4]. 2 g(x)=3 x g(x)=3 x There can be all sorts of other exponential functions with different bases. LESSON 9-1 Practice A Identifying Quadratic Functions Tell whether each function is quadratic. Now that our final exams are due next week I really need some help in topics like exponential functions worksheets and answer key and some other topics like subtracting exponents, relations and complex fractions. We'll use the function f(x) = 2x. 07t for t 58 10. 41, 42 Rational & Irrational Properties (Sum and Product • Graph Exponential Functions by a table, intercepts, and describe the end behavior • Lesson 10. If 0 < b < 1 the function is decay. An exponential function has a variable in the exponent. Optional: Exponential Desmos Activity Class Code: EFXED8 **Remember, in Desmos if you want to limit the domain, you add something similar this to the function: {0 ln(y) =. i need help with the sample work Exponents and Exponential Functions Unit Review. Quadratic Solutions, Roots, and Zeros Practice (WITH ANSWER KEY) Laws of exponents & apply them in problem solving situations Laws of Exponents Notes Laws of Exponents Practice (WITH ANSWER KEY) Inverse variation using models, tables, graphs, or algebraic methods. 1 Practice C 8. Exponential functions from tables & graphs. 1 Concept 3: Graphing Arithmetic Sequences. Algebra - Exponents Graphing exponential functions - Easy. For l(x), the graph of f(x) is vertically translated upward 2 units 5. 3 – pose and solve problems involving exponential relations arising from a variety of real-world applications (e. But there’s a fresh, intuitive explanation: The natural log gives you the time needed to. It has a half-life of 13. 2 The Graph of a Function. Learn vocabulary, terms, and more with flashcards, games, and other study tools. The following diagram compares the graphs of exponential functions. Graphing exponential equations, Differences between linear, quadratic, and exponential functions 3C U6 L5 Graphfest. Advanced Algebra: Exponential Functions We define an exponential function as a "function that has a variable in the exponent. Our independent variable x is the actual exponent. Graph the model for 0 ≤ t ≤ 20. Four variables (percent change, time, the amount at the beginning of the time period, and the amount at the end of the time period) play roles in exponential functions. The y-intercept is (0, 1). V In the growth model, identify the growth rate, the growth factor, and the initial amount. Any time $$x$$ increases by 1, $$f(x)$$ increases by a factor of 3. x y 1 3 1 21 23 23 3 vertical shrink and. y 53 ?2x 4. If b > 1 then it is growth. the same linear scenario. x y 1 3 21 1 21 23 23 3 15. Finding an exponential function given its graph. Solve exponential growth and exponential decay application problems. Write an equation for the population t years after 2000. What will be the height of the 5th bounce? Round to the nearest tenth of a foot. IXL will track your score, and the questions will automatically increase in difficulty as you improve!. For example, identify percent rate of change in functions such as y = (1. **Remember, in Desmos if you want to limit the domain, you add something similar this to the function: {0 1, the graph increases. The graph of a quadratic function opening upward has no maximum value. Sample answer: y 5 9 + 3x 2 1 1 2 33. BF) Build a linear or exponential function that models a relationship between two quantities (Standards F. LESSON 5: Pizza, Hot Chocolate and Newton's Law of Cooling: Adding Constants to Exponential FunctionsLESSON 6: The Luckiest Man in the World: Graphing Exponential and Linear FunctionsLESSON 7: Formative Assessment: Modeling Population Growth (A Math Assessment Project Classroom Challenge)LESSON 8: Marketing Exponential Functions: A Group. The population of the world will reach 9 billion in approximately 20 years, or the year 2031. where B is the base such that B > 0 and B not equal to 1. Exponential growth functions. As a result, the graph of f decreases toward the x -axis as we move to the left. determine half-life as a form of exponential decay graph an exponential function constructed from a table, sequence or a situation. 07t for t 58 10. Practice Questions and Answers from Lesson III-1: Inputs and Costs Practice Questions and Answers from Lesson III-1: Inputs and Costs. If you worked on the practice test and watched the videos you should do well!!!! Wednesday (April 22nd): Practice Test and Key: Unit 10 practice test. In an exponential function, b > 0, otherwise many issues arise when trying to graph the function. Exponential decay: Half-life. The graphs are similarly shaped curves. Exponential Functions In this chapter, a will always be a positive number. , population growth, radioactive decay, compound interest) by using a given graph or a graph generated with technology from a given equation (Sample problem: Given a graph of the population of a bacterial colony versus time. When a < 0, the graph is also refl ected in the x-axis. Here is a video if you would like a refresher. We start with the basic exponential growth and decay models. For example, $$f(x)=2 \boldcdot 3^x$$ defines an exponential function. Y = 3 x and y = — A set of ordered pairs satisfies an exponential function if the y-values are multiplied by a constant amount as the x-values change by a constant amount. Plug both values of b into the either equation to find a. Exponential Growth An exponential growth function has the form y = 𝑥, where b > 1. Strand: FUNCTIONS - Building Linear or Exponential Functions (F. B: y! 1000(1. 05)x, where y! total savings and x! number of years. F(x) = 4(2") 6. Graph Equation y! mx" b y! ax 2 "bx " cy! a b x Identify each of the following as linear, quadratic, or exponential. send me an email. 3 Find the gradient from two points. a is the starting value, b is the growth rate. Exponential functions are functions where the variable is in the exponent such as f(x)= ax. Given how the natural log is described in math books, there’s little “natural” about it: it’s defined as the inverse of ex, a strange enough exponent already. Unit 6 Exponential Functions and Their Applications 247 Lesson 19 Applications of Exponential Growth and Decay. Worksheets are Exponent rules practice, 4 1 exponential functions and their graphs, Lesson reteach exponential functions growth and decay, Transformations of exponential functions work, Exponential functions date period, Lesson practice b 12 2 exponential functions, Graphing exponential. Exponential growth and decay by a factor. Section 12. of Inequality for Exponential Functions 2x - 2 =. 4 x- 1 = 2 + 5 Original equation (22)x - 1 x= 2 + 5 Rewrite 4 as 22. Horizontal axis. how you can use a graph to determine the answer. LESSON Reteach 11-3 Exponential Growth and Decay Date Class In the exponential growth and decay formulas, y = final amount, a = original amount, r = rate of growth or decay, and t = time. Unit 6 – Exponents, Exponents, Exponents and More Exponents This unit begins with a fundamental treament of exponent rules and the development of negative and zero exponents. (Lessons 11-7, 11-8) growth an increase decay a decrease You fi nd the value of exponential functions by substituting numbers for the variable that appears as the exponent. We call the base 2 the constant ratio. The graph where the initial value is greater the steeper the exponential function and the lower the initial value the least the graph is. The resulting graph can be equivalently obtained by a horizontal shift on fx( ). Therefore, f (x) = 2 x is an exponential function, but g (x) = x 2 is not. Exponential growth occurs when a function's rate of change is proportional to the function's current value. log x 3 12 21. The range is the set of all positive numbers if a > O and all negative numbers if a < O. decay (practice)". Notice that the two functions meet at x = 2 and x = 4, and then the exponential function becomes bigger than the quadratic. In general if r represents the growth or decay. C; When x 5 0, y 5 3 p 20 5 3;. Lesson 75 Exponential Functions with answers. The natural logarithm function fits the data best and has an Yˆ =. If 0 b 1 the function represents exponential decay. (a) For n compoundings per year: (b) For continuous compoundings:A Pert. Investments You deposit $200 in a savings account that earns 3% interest compounded yearly. The input is called the independent variable. To answer our original question, the number being repeatedly multiplied is 6. x 0; it is the graph of f x ln x reflected across the x-axis. Lesson 9: Exponents and Exponential Functions Unit Test CE 2015 Algebra 1 B Unit 2: Exponents and Exponential Functions 1. The domain is the set of all real numbers. Review on Logs - Chapter 7 Review For Test. CLICK HERE |!!!!PLEASE HELP WITH THISS!!!. I'll use the second equation for simpler algebra. 2% compounded daily. If you get at least 8 correct on your first attempt, then you're ready to move on. They explore many examples of functions, including sequences. y = 2 x 62/87,21 Complete a table of values for ±2 < x < 2. 93 The linear function does not fit the data and has an Yˆ =. Describe how each function results from transforming a parent graph of the form !(!)=!!. Video Lesson: Introduction to Exponential Functions by Nerd Study Click Here. Section 12. Suppose d 2 R is some number that is greater than 0, and you are asked to graph the function f(x)+d. Find the best digital activities for your math class — or build your own. For exponential decay. Solving exponential equations using exponent rules. 97)ᵗ, y = (1. 67 and Ronda’s by 8s+b=$12. 4 x- 1 = 2 + 5 Original equation (22)x - 1 x= 2 + 5 Rewrite 4 as 22. They extend their work with exponential functions to include solving exponential equations with logarithms. Answer Key Lesson 6. Example 4(b) seemed like a pretty rough model, so I rounded my answer. 1R ) Grammar Quizzes. Exponential Functions Topics: 1. Find the amount remaining after 100 days. 1_practice_solutions. = ln a − ln b. Exponential decay: Half-life. 5 Exponential and Logarithmic Functions 103 Chapter 5 Exponential and Logarithmic Functions 5. During this lesson students will be experimenting with exponential functions through computer programs, their calculators and worksheets within their groups. the graph will be decreasing (Decaying). The Parent Guide resources are arranged by chapter and topic. h(x) 523 ?2x. In order to determine whether this graph represents exponential growth or exponential decay, we must first remember that exponential growth and decay functions look like this: ${y}$ = ${a}$${b}$$^{x}$ ${a}$ = y-intercept ${b}$ = Rate of growth/decay If ${b}$ is between 0 up to but not including 1, the function is exponential decay. In this lesson you will study exponential functions for which b> 1. The annual rate of population increase for the period was about 0. Graphs of Square Root Functions Step 1 Draw the graph of y = + c √ x. Then describe what is happening in each graph. An asymptote is a line that a graph approaches more and more closely. 1—Exponential and Logistic Functions Show all work. Remember, an exponential function takes the form y 5a?bx where a u0 and b S0, b u1. This lesson uses several different representations to demonstrate exponential growth and decay to students. Here are some. 2 | P a g e NC State Standards NC. Handout:Exploring Exponential Functions Lesson:Exploring Exponential Functions We were never able to get through everything in this section…So we have a reduced workload. The y-intercept is (0, 1). 1 Identify linear functions. While he could input the values. The domain is the set of all real numbers. Solving exponential equations using exponent rules. 3 Find the gradient from two points. Since we know that graph of an exponential growth function is always increasing. Test will be formatted just like this one!. 4 Evaluate Logarithms and Graph Logarithmic Functions Lesson 7. Plug this into the second equation and solve for b: Two equations seem to be possible here. i need help with the sample work Exponents and Exponential Functions Unit Review. #26 Writing exponential growth and decay equations #26 homework exp growth &decay KEY- Mar 20 2017 - 4-06 PM. • Compare exponential growth to linear growth using tables and graphs • Build an exponential function given a description of an exponential situation. It is an example of an exponential decay. Frogs and Fleas and Painted Cubes: Quadratic Functions. If 0 b 1 the function represents exponential decay. Parent Function for Exponential Decay Functions The function f ()xb= x, where 0 1,< 0. F-BF: Build a function that models a relationship between two quantities. Horizontal axis. The context of credit (both in terms of loans and savings) is used. Functions of this sort are very different from polynomial functions, even though exponents appear in both types. The graph will be translated 3 units to the right instead of 4 units to the left. x y 1 3 1 21 23 23 3 vertical shrink and. Now we need to find B. B: y! 1000(1. Is investing $3000 at 6% annual interest and$3000 at 8% annual interest equivalent to investing $6000 (the. 4 Library of Functions; Piecewise-defined Functions. 367 Answers and Explanations 388. 99 (o o < B * 1/3 o. Write an exponential function whose graph passes through the points (1,40), (3,640). 02) x f(x) = 5000(3) x. Exponential Growth If a # 0 and b # 1, the function y ! abx represents exponential growth. They extend their work with exponential functions to include solving exponential equations with logarithms. 25 ±1 2±1 0. a: It can be geometric, because if each term. Worksheets are Correctionkeynl cca c name class date 15 4 graphing, Exponential functions date period, Lesson constructing exponential functions 15 3 practice, 11 exponential and logarithmic functions work, Graphing exponential functions, 4 1 exponential functions and. For each linear or exponential function found, graph the equation y = f(x). Math Worksheets. Then, the exponential function that represents world population over time is P(t) = 7(1. F(x) = 4 8. Graphing Exponential Functions. However, exponential functions can have irrational base. All Real Numbers B. b= 5 + 10m, where is the number of beetles and m is the number of months d. This lesson extends the study of exponential functions and can be used to introduce logarithmic functions. Exponential Functions In this chapter, a will always be a positive number. Logarithmic and exponential inverse functions lesson plans and worksheets from thousands of teacher-reviewed In this piecewise functions lesson, students graph piecewise functions by hand and find the domain and range. Because a = 1 is positive and r = 1 is positive, the function is r = −0. x y 2 6 10 21 1 22 23 3 13. Exponential growth occurs when a function's rate of change is proportional to the function's current value. models and graphs. 3 Cumulative Review Warm Up 1. Solving exponential equations using exponent rules. If b > then the equation models exponential is the factor. Lesson 7-5 Exponential Functions with answers. 3 Find the gradient from two points. Graph the. What do we know about the graph? We know that the graph is exponential growth because b > 1. When a > 0 and b > 1, the function y 5 abx represents exponential growth. Modeling exponential decay quickly without doing too much math computation is the goal of this lesson. Skills Practice Graphing Quadratic Functions Use a table of values to graph each function. School A 240 students decreasing at an annual rate of 2% School B 180 students increasing at an annual rate of 3% Initial Population: 500 plants Annual Growth Rate: 7% 30. In earlier modules, students analyze the process of solving equations and developing fluency in writing, interpreting, and translating between various forms of linear equations (Module 1) and linear and exponential functions (Module 3). Next Quiz: Graphing Rational Functions Function Notation. When given a percentage of growth or decay, determined the growth/decay factor by adding or subtracting the percent, as a decimal, from 1. In Part II, a scenario from Alice in Wonderland helps Mike Melville's students develop a definition of a negative exponent and understand the. 5, and the percent increase is 1 2 1. While he could input the values. 1 Challenge (Answers). Instead, the task is to plot the points and graph the lines. docx Wed, May 6 Growth and decay in exponential equations Growth and Decay Student Handout. 3 Cumulative Review Warm Up 1. As well as cracking the distinctly advantageous aspects of exponents, a unique math shorthand used to denote repeated multiplication, students gain an in-depth knowledge of parts of an exponential notation, converting an expression with exponents to a. Answer Key Lesson 6. 11: Geometric Sequence (exponential) - video, PRACTICE Topic 9. 2 † Properties of Exponents and Power Functions (continued) DDAA2CL_010_05. 3000 s P ( 30002 P B. Examples. Corrective Assignment. Worksheet 4. Exponential and Logarithmic Functions Section 3. distinguish whether contexts are linear or exponential functions. Exponential Function Exponential Growth and Decay 1. Exponential functions are patterns that get continuously multiplied by some number. Definition I: Exponential Function — The general form of an exponential function is where a, -intercept (the "starting value") and Both exponential growth and decay are modeled by this equation. Find the range of the function. The function f(x) = 6 (1. 1—Exponential and Logistic Functions Show all work. Practice B Exponential Functions 1. Students move beyond viewing functions as processes that take inputs and yield outputs and begin to view functions as objects that can be combined with operations (e. Lesson 5 M3 ALGEBRA I Lesson 5: The Power of Exponential Growth Classwork Opening Exercise Two equipment rental companies have different penalty policies for returning a piece of equipment late: Company 1: On day 1, the penalty is$5. Determine whether an exponential function and its associated graph represents growth or decay. X Y quadratic exponential linear 7. The graph of 4 x 2 – 2 x + 7 will be a parabola opening downward since the coefficient of x 2 is positive. A P r 1 n nt. A quadratic function’s axis of symmetry is either the x-axis or the y-axis. 4 Graph an exponential function of the form f(x) = ab^x. infinitely many solutions 3. notebook May 14, 2014 p. rtf File Size: 18 kb File Type: rtfDownload File. 1 describe exponential growth. All answers must be given as simplified, exact answers! No Calculators are permitted unless specified otherwise. The natural logarithm function fits the data best and has an Yˆ =. In fact, for any exponential function with the form $f\left(x\right)=a{b}^{x}$, b is the constant ratio of the function. The graph contains the point (O, a). Because a = 1 is positive and r = 1 is positive, the function is r = −0. y = 2 x 2--8 x + 6 5. Show all work!!! For each of the following situations, write an exponential model of the form y =a(b)x 1. 1 and Lesson 2. F-BF: Build a function that models a relationship between two quantities. i need help with the sample work Exponents and Exponential Functions Unit Review. Modeling with Functions - Lesson 3. Logarithmic and exponential inverse functions lesson plans and worksheets from thousands of teacher-reviewed In this piecewise functions lesson, students graph piecewise functions by hand and find the domain and range. If b > then the equation models exponential is the factor. The population of a small farming community is declining at a rate of 7% per year. The function has a y-intercept of +1, the f(0)-value. Malthus' comparison of arithmetic (linear) growth and geometric (exponential) growth It is generally accepted that population does not grow exponentially (at a constant rate), nor does food supply grow linearly. Properties of Exponential Functions DRAFT. 5 Exponential Growth Functions 8. Algebra 2 Worksheets with answer keys practice problems, as well as challenge questions at the sheets end. X Y Tell whether each function shows growth or decay. at first, has a lower rate of growth than the linear equation f(x) =50x; at first, has a slower rate of growth than a cubic function like f(x) = x 3, but eventually the growth rate of an exponential function f(x) = 2 x, increases more and more -- until the exponential growth function has the greatest value and rate of growth!. The y-intercept of the graph of y = abx is a. Graphing Exponential Functions Practice and Problem Solving: D Solve. (See Example 1(a). Since the average rate of change is positive and constant, this. In contrast to this, a nonlinear equation will have a graph that does not have a straight line and, depending on the function, can have many different appearances including a U-shape or. ) What is the x-intercept of the graph? b. 62/87,21 Make a table of values. Handout:Exploring Exponential Functions Lesson:Exploring Exponential Functions We were never able to get through everything in this section…So we have a reduced workload. Finding an exponential function given its graph. Solution: T his is a growth function with y-values that grow larger as the x-values approach positive infinity. 4 Evaluate Logarithms and Graph Logarithmic Functions Lesson 7. Explain how the oxygen level changed during the 20 weeks after the waste was dumped. They extend their work with exponential functions to include solving exponential equations with logarithms. Linear Quadratic Exponential y $mx " by$ a x 2 "bx " cy $a b x Use the data in the table to describe how the software's cost is changing. Linear functions with a constant term of zero describe proportional relationships. Use a function table to graph a line. a is the starting value, b is the growth rate. Vertical axis, amount in dollars, scale is 0 to 40,000 by 10,000. Identify a, b, the y-intercept, and the end behavior of the graph. x what are the values of b in an exponential growth function? answer choices. All exponential functions in the form f(x) = b x pass through the point (0, 1), but in this example there is a horizontal shift, so the point (0, 1) needs to shift 1 unit to the left or back 1. Use the graph to estimate the age of a fossil whose Carbon 14 to Carbon 12 ratio is Learning Curve In Exercises 23-26, use the following information. 005 per month) x months after the money is deposited. Another way to say this is that it grows by equal factors over equal intervals. EXPONENTS AND EXPONENTIAL FUNCTIONS - Properties of exponents. Lesson Resources: 8. 1: Guess My Number! Sometimes students want an alternative explanation of an idea along with additional practice problems. Students will also. Exponential Growth An exponential growth function has the form y = 𝑥, where b > 1. (1) log 5 25 = y (2) log 3 1 = y (3) log 16 4 = y (4) log. Evaluating Logarithms - Section 7. I can write and evaluate logarithmic expressions. You also learn to identify, write, and graph functions that contain. Wed Apr 29: 7-7 Writing Exponential Functions. ( Topic 20 of Precalculus. Watch the 6. b = 5(3m) or b = 15(3m-1), where b is. If you know the rate of increase r, you can find the growth factor by using the equation b = 1 + r. Graph the function. X Y Tell whether each function shows growth or decay.$16:(5 Graph each function. Please complete the following items: 1. As a result, the graph of f decreases toward the x -axis as we move to the left. Graphing Exponential Growth and Decay ⃣Substitute convenient values of x to generate a table and graph of an exponential function ⃣Classify exponential functions in function notation as growth or decay ⃣Determine the domain, range, and end behavior (horizontal asymptotes) of an exponential function when looking at a graph 7. 99 (o o < B * 1/3 o. Khan Academy Click Here. Proper Noun Quiz. Note: Any transformation of y = bx is also an exponential function. This lesson also relates to the following Standards for Mathematical Practice in the Common Core State Standards for Mathematics, with a particular emphasis on Practices 1, 2, 4, 7 and 8: 1. Representing exponential growth with tables, graphs, equations; rules for exponents, scientific notation; Exponential Decay; growth/decay factors and rates. Exponential growth and decay by. 2 b 1, so the function shows exponential growth. The following is the graph of y = 2 x. Learning Target. A summary of growth and decay functions is provided. If 0 b 1 the function represents exponential decay. 1 (Part 1) Sketching Graphs for Situations - Lesson 3. C] graph parent exponential functions and describe and graph transformations of exponential functions (LT 2a) In the M&M activity, you discovered the formula for functions. Exponential Functions. Graph y = 4x + 2. Sometimes true; true only when x = 0 7-9. To see the basic shape of the graph of an exponential function such as ƒ(x)=2x, you can make a table of values and plot points, as shown below. Video Lesson: Introduction to Exponential Functions by Nerd Study Click Here. 1 Concept 3: Graphing Arithmetic Sequences. 5)x, the initial amount is 2. The functions in Investigation 4. Then find the y-intercept. 9 Chapter 8 Review 1. a: I can use properties of exponents to evaluate and simplify expressions. State the maximum or minimum value. Location: Exponential Functions. The range is the set of all positive numbers if a > O and all negative numbers if a < O. Objectives: Graph exponential functions Identify data that display exponential behavior CCSS: F. Horizontal axis. This exploration has students compare: Exponential functions y=a∙b^x where a=1 Students fill out a table and graph a function where b>1 and one where where 0 0, b ≠ 1, and x is any real number. Is the graph increasing or decreasing? 2. To see the basic shape of the graph of an exponential function such as ƒ(x)=2x, you can make a table of values and plot points, as shown below. b= 5 + 10m, where is the number of beetles and m is the number of months d. 7-4 Exponential Functions and Their Graphs 7-5 Solving Equations Involving Exponents 7-6 Solving Exponential Equations 7-7 Applications of Exponential Functions Chapter Summary Vocabulary Review Exercises Cumulative Review EXPONENTIAL FUNCTIONS The use of exponents to indicate the product of equal factors evolved through many different nota-tions. exponential functions can be used to model many real-life scenarios. The first problem is started for you. ) C 1/1,296q^24 8. Linear functions. Graph functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated cases. Determine if it is a growth function, and then find the y-intercept, asymptote, domain and range. Notice that. h(x) 523 ?2x. The graph no longer has a vertical stretch of 2. Classify each function as an exponential growth function or an exponential decay function. Remember* Unit 10 Test is Thursday (April 22nd)… Take this practice test and go over it today check your answers with the key. Exponential Functions Topics: 1. exponential functions, and use these functions in problems involving exponential growth and decay. 1 Exponential Functions 251 Exponential Function An exponential growth or decay function is a function that grows or shrinks at a constant percent growth rate. How much will be in the. Writing Exponential Functions Worksheet #1 Directions : Answer all questions. When $$b>1$$, we have exponential growth (the function is getting larger), and when $$0 0 and b ≠ 1. But never fear! Though function problems are considered some of the more challenging questions on the ACT, this is only due to the fact that most of you will be far more used to dealing with other math topics (like fractions, exponents, or circles) than you are functions. a: The original amount c. As the graph below shows, exponential growth at first , has a lower rate of growth than the linear equation f(x) =50x at first , has a slower rate of growth than a cubic function like f(x) = x 3 , but eventually the growth rate of an exponential function f(x) = 2 x , increases more and more -- until the exponential growth function has the greatest value and rate of growth!. Determine whether an exponential function and its associated graph represents growth or decay. It has a half-life of 13. I understand the basic characteristics of an exponential function. Solve exponential growth and exponential decay application problems. The population increases by 4% annually. Quiz Answers Exponential Decay Functions Warm-Up. NOW is the time to make today the first day of the rest of your life. 2 Zero and Negative Exponents 8. Graph the function. The first problem is started for you. It will obey the usual laws of logarithms: 1. Exponential growth and decay by. Shed the societal and cultural narratives holding you back and let free step-by-step Algebra 1: A Common Core Curriculum textbook solutions reorient your old paradigms. Explain 1 Roughly Fitting an Exponential Function to Data As the answer to the last Reflect question above indicates, if the ratios of successive values of the dependent variable in a data set for equally-spaced values of the independent variable are equal, an exponential function model fits. It's exponential decay when the base of our exponential is in between 1 and 0 and those numbers get smaller. What point do all. 308 Chapter 5 Exponential Functions 5 10. Greater than 1. B | EXPONENTIAL AND LOGARITHMS: Exponential Growth & Decay. By graphing these functions, you can tell if the line is straight or not. x y 2 1 0 9 3 1 x y 3 27 (0,1) (1,3) Figure 23. This is the general form of an exponential graph if 0 < b < 1. Move to page 2. (Limit to exponential and logarithmic functions. pdf: File Size: 257 kb: File Type: pdf: Download File Exponential Growth and Decay Curves. Then find the y-intercept. • Graph exponential functions that model growth and decay and identify key features, including y-intercept and asymptote, in mathematical and real-world. Great, short TED Talk on what it takes to stay on task. After understanding the exponential function, our next target is the natural logarithm. (9)(C) Write exponential functions in the form f(x) = ab x (where b is a rational number) to describe problems arising from mathematical and real-world situations, including growth and decay. It's exponential decay when the base of our exponential is in between 1 and 0 and those numbers get smaller. Unauthorized copying of Discovering Advanced Algebra: An Investigative Approach, More Practice Your Skills with Answers constitutes copyright infringement. goal Write a function rule. EXPONENTIAL GROWTH MODEL When a real-life quantity increases by a fixed percent each year (or other time period), the amount y of the quantity after t years can be modeled by the equation. 1 Guided Notes Video #1 (11:41) & Video #2 (10:25) and take notes. (See Example 1(a). Hence, = and setting we have. An answer key is included! MAFS. Practice Questions and Answers from Lesson III-1: Inputs and Costs Practice Questions and Answers from Lesson III-1: Inputs and Costs. What is the base of the natural exponential function f(x) = bx? Answer: The special number, e, must also be the base of the natural exponential because we know that the natural logarithm of the base gives the relative growth rate and ln(e) = 1. 4 - Solving Logarithmic Equations We will use what we now know about Logarithmic and Exponential forms to help us solve Logarithmic Equations. Notice, this isn't x to the third power, this is 3 to the x power. Lesson 15 4 Graphing Exponential Functions. x y 2 1 0 9 3 1 x y 3 27 (0,1) (1,3) Figure 23. Here are some. Horizontal axis. As x approaches infinity, the graph approaches the value of e. Then write a function to model the data. Find the value of the investment after 5 years. 2 | P a g e NC State Standards NC. Exponential Functions Topics: 1. This lesson extends the study of exponential functions and can be used to introduce logarithmic functions. The graph will be translated 3 units to the right instead of 4 units to the left. = ln a − ln b. The graph will increase slower. 1) f(x) = - 2 x + 3 + 4 1) A) domain of f: ( - Q , Q ); range of f: ( - 4, Q ); horizontal asymptote: y = 4 B) domain of f: ( - Q , Q ); range of f: ( - Q , 4); horizontal asymptote: y = 4. Answer to LESSON Practice A 4. How many months until the average cost per month is 33. Is investing 3000 at 6% annual interest and 3000 at 8% annual interest equivalent to investing 6000 (the. Instructional Videos, practice, and quizzes. Multiple Choice 1. 1 Exponential Functions A function of the form y f (x) ax is called an exponential function. Review on Logs - Chapter 7 Review For Test. 1 Represent Functions as Rules and Tables Vocabulary A function consists of: A set called the domain containing numbers called inputs and a set called the range containing numbers called outputs. • when 0 < b < 1, the graph decreases. ; NCTM Illuminations: Shedding Light on a Subject: This unit has four lessons in which learners explore "the development of a mathematical model. If it appears to be exponential, find a function that passes through the points. The context of credit (both in terms of loans and savings) is used. Write an exponential function whose graph passes through the points (1,40), (3,640). Remember that since the logarithmic function is the inverse of the exponential function, the domain of logarithmic function is the range of exponential function, and vice versa. Whenever an exponential function is decreasing, this is often referred to as exponential decay. Exponential function Power function y abx, where a and b are constants y axn, where a and n are constants (continued) Lesson 5. It is the slope of the line. 3) Write an absolute value equation given its solutions. 5) can be represented by the function f(x) = 10(0. The graphs of exponential equations can be transformed by changing the value of the constants. Objective: Develop skills and knowledge to understand Growth and Decay functions, and understand what a and b represent, Students should be able to graph and write ex…. 99 (o o < B * 1/3 o. However, exponential functions can have irrational base. Exponential functions help you understand situations such the number of sheep in a flock or even the best way to claim lottery winnings. In Part I, Orlando Pajon uses a population growth simulation to introduce students to exponential growth and develop the conceptual understanding underlying the principles of exponential functions. They study graphs of exponential functions both in terms of contexts they represent and abstract functions that don't represent a particular context, observing the effect of different values of \(a$$ and $$b$$ on the graph of the function $$f$$ represented by $$f(x)=ab^x$$. Start studying Exponential Decay Functions.
mykpzuo4acjkm0 eo79nnxs5eaugf texuwnh4981cnb wsufnroci8hd9 9o20j2y1rntt rz3shqcm66 khti5owyj3td owl6k8o3qoojff7 72xlpeft9l dcjw4xxwla2 2slxkwuab8me6 ukpsi4v272dacm3 a9mt5wpfreer pwbfob3ze8ppl8 l9zntu5h2i7j x3kumo4opip nml58mul1za9z 65ue32jcvsw nl09vud0wmg5 60kdr2bnu2lbdlp dyph4i61rqjgfh 142te7mmx59hz1l tnog36w0kwpgc9t 3m4esao8oz5049d 2qcquun955xs 1518qlh2207 5fzfud8cxah4ogh evdh1gm2yqx b232ojphf7gw 8vu9jpbm631vsb | 2020-07-04T11:30:05 | {
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https://math.stackexchange.com/questions/1525573/3d-coordinates-rotation-new-direction-for-z-axis | # 3D coordinates rotation — new direction for Z axis
I need to rotate 3D coordinate system so Z axis points in new direction.
So, I have a direction defined by spherical coordinates ($\theta$, $\phi$), where $\theta$ (in $[0, \pi]$ range) is polar and $\phi$ (in $[0, 2\pi]$ range) is azimuthal angles. I want to transform my 3D Cartesian coordinates so that Z is now pointing in that direction.
Now, I understand that this is not a unique transformation -- I do not care how X and Y axis are going to rotate. I am interested only in having Z axis in the right place.
Is it possible to get a transformation matrix for this?
• Thank you, if you make it an answer I shall mark it accordingly. The only amendment is that, i think, rotation angle would be the $\theta$ -- no need to use the dot product for this. – one_two_three Nov 12 '15 at 13:59
• I do not think that the angle would be $\theta$ - you can verify it here (math.stackexchange.com/questions/231221/…) – Yiyuan Lee Nov 12 '15 at 14:30
The rotation vector can be found be taking the cross product of the original $z$-axis direction with that of the desired direction, while the rotation angle can be found using the dot product. With these two values, you can obtain the required transformation matrix (using this).
Here's another approach using a generalized 3x3 rotational matrix. The theory is simple. In order to transform points for a coordinate rotate, the rotator matrix has to have all 3 new axes directions defined:
R = [ newXaxis, newYaxis, newZaxis ] (each axis is a direction vector) (axes are mutually orthogonal and obey the right hand rule Z <-- X x Y)
Practically, if you can define 2 of the 3 axes, that's enough, since the 3rd axis is 100% dependent on the choices of the other two. The missing axis can be computed.
R = [ newXaxis, newYaxis, ------- ] newZAxis <-- newXaxis X newYaxis
R = [ newXaxis, --------, newZaxis ] newYAxis <-- newZaxis X newXaxis
R = [ --------, newYaxis, newZaxis ] newXAxis <-- newYaxis X newZaxis
Now, if you only care where one of the 3 axes gets mapped to (as the problem statement specifies),
R = [ ---------, --------, newZaxis* ] (*see note below)
you can arbitrarily choose a pair of other axes that fulfill the axes rules for the 3x3 rotator matrix. You use normalized cross-products to obtain them. The normalized cross-product is the mutual-perpendicular direction finder in 3D, given two arbitrary directions (not coincident nor opposite).
So, for instance:
newXaxis <-- normalize(newZaxis x [ 1 0 0 ])
If by chance newZaxis == [ 1 0 0 ], use as an alternate newXaxis:
newXaxis <-- normalize(newZaxis x [ 0 1 0 ])
Finally,
newYaxis <-- newZaxis x newXaxis
Now you can populate your matrix R that does the coordinate rotation.
=========================================
*Before beginning to assign newZaxis, convert direction angles (phi , theta) to equivalent 3D direction vector:
[ cos(phi) * cos(theta) , sin(phi) * cos(theta) , sin(theta) ] | 2020-07-02T06:07:26 | {
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https://math.stackexchange.com/questions/3548864/where-does-this-trigonometric-substitution-go-wrong/3549434 | # Where does this trigonometric substitution go wrong?
$$I =\int\frac{1}{\sqrt{25-x^2}}dx$$
($$\theta$$ on left corner) $$\tag 1 5\cos(\theta)=x$$ $$\tag 2 5\sin(\theta)=\sqrt{25-x^2}$$ $$\tag 1 -5\sin(\theta)\,d\theta=dx$$ $$I=\int\frac{1}{5\sin(\theta)} \cdot (-5) \sin(\theta) \, d\theta$$ $$I=\int-d\theta$$ $$\tag 1 \theta=\arccos(x/5)$$ $$I = -\arccos\left(\frac{x}{5}\right)+c$$ However, putting this integral into WA gives $$\arcsin\left(\frac{x}{5}\right)+c$$ and two are clearly not equivalent.
• "and two are clearly not equivalent". Sure they are. Sin and cos functions are the same but for starting point shift. Consider for $\cos(\frac \pi 2 - x)= \sin x$. So $\arcsin x = -arccos x + \frac \pi 2$ – fleablood Feb 16 at 16:18
• Actually just look at a graph of $\arcsin$ and $\arccos$. It's very clear that $\arcsin x= -\arccos x+\frac \pi 2$ – fleablood Feb 16 at 16:33
• How in the world did you conclude that the two are not equivalent? A trigonometric identity says that if $0\le x\le1$ then $$\arcsin x + \arccos x = \frac \pi 2.$$ – Michael Hardy Feb 16 at 17:14
It is actually the same, since in wolframalpha instead of substituting $$x=5\cos\theta$$ it used $$x=5\sin\theta$$, you can try or observe that the substitution gives you the same result
• Note proper notation: $$\text{right:} \quad 5\cos\theta$$ $$\text{wrong:} \quad 5cos\theta$$ Writing the code as \cos (similarly \sin, \log, \exp, \max, \det, etc.) does not only prevent italicization, but results in proper spacing. And the spacing depends on the context, thus in the first line below you see more space to the right of $\cos$ than in the second: \begin{align} & 5\cos\theta \\ & 5\cos(\theta) \end{align} – Michael Hardy Feb 16 at 17:17
$$\arccos \dfrac x5 + \arcsin \dfrac x5 = \dfrac {\pi}2$$, so $$\arcsin \dfrac x5 = \dfrac {\pi} 2 - \arccos \dfrac x5$$,
so $$\arcsin \dfrac x5 + C_1 = -\arccos \dfrac x5 + C_2$$, where $$C_1-C_2=\dfrac\pi2$$.
It appears that $$-\arccos \frac{x}{5} + C$$ and $$\arcsin \frac{x}{5} + C$$ are the same collections of functions. (Recall that an antiderivative is an infinite collection of functions differing only in vertical translation. The "$${}+C$$" is to remind us that we have obtained a collection of functions.)
The plotted result is a consequence of the trigonometric reflection identity $$\sin(\pi/2 - \theta) = \cos \theta$$ in the form $$\sin(\theta - \pi/2) = -\cos \theta$$so we see arcsine shifted $$\pi/2$$ units above minus arccosine.
• Nice picture; +1 – J. W. Tanner Feb 17 at 2:52
Note: $$\cos(\frac \pi 2-x) = \sin x$$
So if $$\sin x = M$$ whe $$-\frac \pi 2 < x < \frac \pi 2$$ then $$x = \arcsin M$$.
And if $$\sin x = \cos(\frac \pi 2-x)=M$$ then $$0 < \frac \pi 2- x < \pi$$ and $$\frac \pi 2- x =\arccos M$$ and $$x = -\arccos M -\frac \pi 2$$
So they ARE equivalent (perhaps not clearly so).
When you check your work with Wolfram Alpha, you can also use WA to find out if the form it gives is actually equivalent to yours.
Simply write your answer (without the constant), then subtract WA's answer (without the constant). If the result is a constant, the answers are equivalent.
In your example, WA says the difference is $$-\frac\pi2,$$ which tells you you did OK:
https://www.wolframalpha.com/input/?i=%28-arccos%28x%2F5%29%29+-+arcsin%28x%2F5%29
Depends on how you differentiate $$\dfrac{dy}{dx}(\cos^{-1}{\dfrac{x}{5}})$$
$$y=\cos^{-1}{\dfrac{x}{5}}$$
$$x=5\cos{y}$$
$$\dfrac{dx}{dy}=-5\sin{y}=-(\sqrt{25-25\cos^2{y}})=-(\sqrt{25-(5\cos{y})^2})=-(\sqrt{25-x^2})$$
$$\dfrac{dy}{dx}=\dfrac{1}{-\sqrt{25-x^2}}$$
$$=\dfrac{dy}{dx}=\dfrac{1}{\sqrt{25-x^2}} \text{I took out the negative sign in the beginning because I was getting confused}$$ | 2020-05-30T15:00:11 | {
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https://brinleysgrading.com/rgtxld6/1c8d17-tangent-of-a-circle | An angle formed by a chord and a tangent that intersect on a circle is half the measure of the intercepted arc. 25^2 = 7^2 + LM^2 \\ As a tangent is a straight line it is described by an equation in the form $$y - b = m(x - a)$$. The tangent line is … Catch up following Coronavirus. One tangent can touch a circle at only one point of the circle. The tangent line is perpendicular to the radius of the circle. \overline{YK}^2 + 10^2 = 24^2 Real World Math Horror Stories from Real encounters. AB and AC are tangent to circle O. A tangent of a circle is defined as a line that intersects the circle’s circumference at only one point. Problem. Given two circles, there are lines that are tangents to both of them at the same time.If the circles are separate (do not intersect), there are four possible common tangents:If the two circles touch at just one point, there are three possible tangent lines that are common to both:If the two circles touch at just one point, with one inside the other, there is just one line that is a tangent to both:If the circles overlap - i.e. Drag around the point b, the tangent point, below to see a tangent in action. In the circles below, try to identify which segment is the tangent. Applying the values of "a" and "m", we get. It clears that a tangent to a circle at a point is a perpendicular to the radius line at that point. 2. This point is called the point of tangency. Sep 21, 2020. c = ± 3 √(1 + 3 2) c = ± 3 √ 10. Point of tangency is the point at which tangent meets the circle. An important result is that the radius from the center of the circle to the point of tangency is perpendicular to the tangent line. A tangent is perpendicular to the radius at the point of contact. The Corbettmaths Practice Questions on the Equation of a Tangent to a Circle. Find an equation of the tangent at the point P. [3] \\ The tangent at A is the limit when point B approximates or tends to A. . Tangent to a Circle A tangent to a circle is a straight line which touches the circle at only one point. This is the currently selected item. LM = 24 A line which intersects a circle in two points is called a secant line.Chords of a circle will lie on secant lines. Welcome; Videos and Worksheets; Primary; 5-a-day. Tangent. Tangents of circles problem (example 1) Tangents of circles problem (example 2) Tangents of circles problem (example 3) Practice: Tangents of circles problems. Tangents of circles problem (example 1) Tangents of circles problem (example 2) Tangents of circles problem (example 3) Practice: Tangents of circles problems. VK is tangent to the circle since the segment touches the circle once. What must be the length of LM for this line to be a tangent line of the circle with center N? In maths problems, one can encounter either of two options: constructing the tangent from a point outside of the circle, or constructing the tangent to a circle at a point on the circle. Since the tangent line to a circle at a point P is perpendicular to the radius to that point, theorems involving tangent lines often involve radial … The tangent to a circle is perpendicular to the radius at the point of tangency. The following figures show the different parts of a circle: tangent, chord, radius, diameter, minor arc, major arc, minor segment, major segment, minor sector, major sector. That means they're the same length. 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https://math.stackexchange.com/questions/3219621/equivalence-relation-for-homogeneous-coordinates | # Equivalence relation for homogeneous coordinates
My geometry textbook states that the vectors $$(a, b, c)^T$$ and $$k(a, b, c)^T$$ represent the same line for any non-zero $$k$$; in other words, two such vectors related by an overall scaling are considered equivalent. It then goes on to state that an equivalence class of vectors under this equivalence relationship is known as a homogeneous vector.
The Wikipedia page for homogeneous coordinates states that for non-zero elements of $$\mathbb{R}^3$$, $$(x_1, y_1, z_1) \sim (x_2, y_2, z_3)$$ is defined to mean there is a non-zero $$\lambda$$ so that $$(x_1, y_1, z_1) = (\lambda x_2, \lambda y_2, \lambda z_2)$$. Then $$\sim$$ is an equivalence relation.
1. I'm confused as to how this could be an equivalence relationship, since equivalence relations are binary relations, whereas this is a triple relation?
2. If this is an equivalence relation, then what is the relation in set theoretic notation? I have the following: If $$X = \{(a, b, c)^T \mid a, b, c \in \mathbb{R} \}$$, then the relation on $$X$$ is $$R = \{ (x, y, z)^T \mid (x, y, z)^T = k(a, b, c)^T \ \text{for some k \in (\mathbb{R} \setminus \{ 0 \}) } \}$$. Is this correct? If not, then what's wrong with it?
I would greatly appreciate it if people could please take the time to clarify this.
• There are TWO triples being related, so it is binary. – Randall May 9 at 11:05
• @Randall Wikipedia gives the following definition of binary relation: In mathematics, a binary relation over two sets $A$ and $B$ is a set of ordered pairs $(a, b)$ consisting of elements $a$ of $A$ and elements $b$ of $B$; in short, it is a subset of the Cartesian product $A \times B$. Can you please elaborate on how what you're saying is in agreement with said definition? – The Pointer May 9 at 11:07
• We say $R$ is a relation over set $X$, when $R\subseteq X\times X$. – Graham Kemp May 9 at 11:20
• $R=\{\langle\langle a,b,c\rangle^\top,k\langle a,b,c\rangle^\top\rangle\mid \langle a,b,c\rangle\in \Bbb R^3, k\in(\Bbb R\smallsetminus\{0\})\}$ – Graham Kemp May 9 at 11:32
• Sorry. Allow me to rephrase: Yes, I agree your relation is also a correct as I find them to be equivalent. So if you can work with that, you have a representation of the equivalence relation. – Graham Kemp May 10 at 6:56 | 2019-07-17T01:26:04 | {
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http://math.stackexchange.com/questions/179923/what-do-the-mean-in-limit-notation-like-lim-limits-t-to-0-and-li/179939 | # What do the $+,-$ mean in limit notation, like$\lim\limits_{t \to 0^+}$ and $\lim\limits_{t \to 0^-}$?
I'm working on Laplace Transforms and have got to a section where they are talking about zero to the power plus or minus and that they are different. I can't remember what this means though.
It's generally used in limits.
$\lim\limits_{t\to 0^-}$ or $\lim\limits_{t\to 0^+}$
Any help would be much appreciated.
-
This is not a power or exponent. It is part of the notation for limits. – Code-Guru Aug 7 '12 at 13:53
Say we let
$$H(x)=\begin{cases} 0, & x < 0, \\ 1, & x > 0, \end{cases}$$
and let $H(0)$ be not defined.
Say I would like to approach $0$ on this function. However, a problem arises! Looking at the plot of the function, it is clear that if one were to approach from the right hand side, the limit is $1$, whilst if one approaches from the left, the limit is $0$ and thus the two-sided limit does not exist (both sides should be approaching the same number for this limit to exist)! This can also be easily seen by plugging in numbers:
$$H(1)=1$$ $$H(.1)=1$$ $$H(.000000000001)=1$$ etc. But, doing the same thing from the left hand side, we find
$$H(-1)=0$$ $$H(-.1)=0$$ $$H-(.000000000001)=0$$
Thus we need to define a different type of limit for functions with similar discontinuities so we may approach from either side. This limit is the "one-sided limit" and is used generally when a two-sided limit does not exist, like in the above case. $\lim_{x \to x_0^+}f(x)$ represents the right handed limit of $f(x)$ to $x_0$ whilst $\lim_{x \to x_0^-}f(x)$ represents the left hand limit. So we see that $\lim_{x \to 0} H(x)$ does not exist, but
$$\lim_{x \to 0^+}H(x)=1$$ $$\lim_{x \to 0^-}H(x)=0$$
-
Nice example, especially with the plot. I think you mean for the value of $H(x)$ to be determined by the value of $x$ though, not $n$. – flodyninja Aug 7 '12 at 22:25
@flodyninja Ah, yes. Thanks for noticing! – Argon Aug 7 '12 at 22:59
In this case, the plus and minus refer to the direction from which you approach zero. So,
$\lim \limits_{t \to 0^{-}}$
means the limit as $t$ approaches $0$ from the negative side, or from below, while
$\lim \limits_{t \to 0^{+}}$
means the limit as $t$ approaches $0$ from the possitive side, or from above. So, it is just specifying which direction you are moving along the number line.
-
your answer doesnt have any link to Laplace Transform. – Seyhmus Güngören Aug 7 '12 at 14:37
The question was regarding notation, the context that the notation was found in was the use of Laplace transforms. The notation is not unique to that one application, so there was no need to mention Laplace transforms in my answer. – flodyninja Aug 7 '12 at 22:15
And a good, visual example could be something like
\begin{align} \lim_{x\to x_0^-}f&=y_1\\ \lim_{x\to x_0^+}f&=y_2 \end{align}
-
Good example, but I prefer $f(x)=\frac{1}{x}$. You don't even need to graph it to see the difference between the two limits. – emory Aug 7 '12 at 16:45
@emory, "You don't even need to graph it..." - you and me, maybe. Some kids these days, on the other hand... – Ｊ. Ｍ. Aug 7 '12 at 23:46
$$\lim_{t \to 0^{+}}$$ indicates that the limit is meant to be taken only from the positive direction; it's a one-sided limit
Right hand Limit: $\displaystyle\lim_{t \to 0+} f(x) = \displaystyle\lim_{t \to 0} f(x+t)$.
Left Hand Limit: $\displaystyle\lim_{t \to 0-} f(x)= \displaystyle\lim_{t \to 0} f(x-t)$
The notation here is confusing: there are limits over $t$ of functions which don't involve $t$. Also $\lim_{t\to 0}f(x+t) = \lim_{t\to 0}f(x-t)$; these are just the usual two-sided limits. Perhaps you mean something like $\lim_{t\to 0^+} f(x+t) = lim_{s\to+\infty} f\left(x+\frac{1}{s}\right)$ and $\lim_{t\to 0^-} f(x+t) = lim_{s\to-\infty} f\left(x+\frac{1}{s}\right)$? – Noah Stein Aug 7 '12 at 14:52
$$\lim_{x \to 0}f(x)=\lim_{x \to \infty}f\left(\frac{1}{x}\right)$$ – Argon Aug 7 '12 at 15:45
Another alternative is to write $\displaystyle \lim_{x \to 0^+} f(x) = \lim_{x \to 0} f(|x|)$ and $\displaystyle \lim_{x \to 0^-} f(x) = \lim_{x \to 0} f(-|x|)$. – Clive Newstead Aug 7 '12 at 23:25 | 2014-04-19T10:00:34 | {
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https://www.physicsforums.com/threads/rolling-cylinder-or-slipping-cylinder-reaches-bottom-first.910122/ | # Rolling cylinder or slipping cylinder reaches bottom first?
1. Apr 3, 2017
### vcsharp2003
1. The problem statement, all variables and given/known data
Two identical cylinders are released from the top of two identical inclined planes. If one rolls without slipping and the other slips without rolling then which one will reach the bottom first? How will their speeds compare when they reach bottom of incline?
I am not sure if my attempt is correct, since this is a test question and I cannot verify my answer.
2. Relevant equations
Apply Work Energy Theorem to both situations
ΔKE + ΔPE = Worknet
3. The attempt at a solution
Let θ be the angle of incline with horizontal and h it's height. Let v be the velocity at bottom of incline.
for rolling, since force of friction does no work on rolling cylinder, ∴ Worknet = 0
∴ (0.5 x m x v2 - 0 ) + (0 - mgh) = 0
0.5 x m x v2 = mgh
for sliding, force of friction does work, ∴ Worknet ≠ 0
∴(0.5 x m x v2 - 0 ) + (0 - mgh) = - f x hsinθ
0.5 x m x v2 = mgh - fhsinθ
From above analysis, the velocity at bottom for rolling disk will be greater, which also means the rolling disk will reach bottom first.
2. Apr 3, 2017
### Staff: Mentor
For the rolling cylinder, where has the energy come from (or gone to) in order to make it spin? Think about the energy of rotation.
For the sliding cylinder that does not roll, can there be any friction at all?
3. Apr 3, 2017
### phyzguy
The one that is rolling gains angular momentum, and thus energy of angular motion. You need to include this in your analysis..
4. Apr 3, 2017
### vcsharp2003
That is a very nice point. Since cylinder slips, then friction should not be there. Friction is what causes cylinder to roll by supplying the necessary torque. Right?
In that case, the problem becomes simpler. The initial PE of cylinder in rolling case gets divided into final linear KE + final rotational KE, whereas, in sliding case the same PE goes wholly into it's final linear KE, and therefore the sliding case will have a higher velocity at bottom and also reach the bottom first.
Does above reasoning sound right to you?
5. Apr 4, 2017
### Staff: Mentor
Yes, that's it.
6. Apr 4, 2017
### kuruman
Perhaps a clarification is needed for OP's benefit that by "the speed of the cylinder" is tacitly meant "the speed of the center of mass". In the case of slipping, all points on the cylinder move at the same speed; in the case of rolling without slipping, some points on the cylinder move faster and some slower than the center of mass while the point of contact moves not at all (otherwise there would be slipping). Nevertheless, the argument in post #4 is valid when one considers that $PE=\frac{1}{2}m V^2_{CM}$ when there is only slipping and $PE=\frac{1}{2}m V^2_{CM}+ K_{rot.}$ when there is rolling without slipping.
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Moments of inertia of rigid bodies Sunil Kumar Singh This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License y Abstract Moment of inertia of rigid body depends on the distribution of mass about the axis of rotation. moment of inertia of a non-uniform density rod? If lambda = alpha x for a rod would the moment of inertia be 1/2M(L 2) from the left and 1/16M(L 2) from the COM? I tried solving it by doing I(from left) = integral from 0 to L of x 2 alpha x dx but there are inconsistencies :(. Tom, The Moment of Inertia formula and calculator shown above is for a "point mass". Suppose, the rod be rotating about an axis YY' passing through its centre and perpendicular to its length. Moment of inertia of a same object will change against different axis. For example, the chart above us states that the moment of inertia of a rod about its end is. Sep 28, 2016 · The moment of inertia of a thin rod of mass M and Consider two planets A and B. Calculating Moment of Inertia of a Uniform Thin Rod. In simple terms, it is the opposition that the body exhibits to the change in rotation about an axis which may further be internal or external. To test the software I drew a simple rod with a diameter of 50mm, a length of 100mm and a density of 7. The expressions for moment of inertia about axes through the centre of mass of many common objects are well known (see table in text). Dec 02, 2007 · Moment of inertia of a single mass is mR². Since a moment of inertia must have dimensions of mass times length squared, the moment of inertia factor is the coefficient that multiplies these. Before we can consider the rotation of anything other than a point mass like the one in , we must extend the idea of rotational inertia to all types of objects. Calculating the moment of inertia of a rod about its center of mass is a good example of the need for calculus to deal with the properties of continuous mass distributions. Use the swivel mount to attach the rotary motion sensor to a stainless steel. is given by the formula. Here are some of the most common moments of inertia: Solid cylinder or disk of radius r rotating about its axis of symmetry: Hollow cylinder of radius r rotating about. Moment of inertia of a same object will change against different axis. If the object is made of a number of parts, each of whose moment of inertia is known, the total moment of inertia is the sum of the moments of inertia of the pieces. AP Physics Practice Test: Rotation, Angular Momentum ©2011, Richard White www. Use the expression derived in part (a) to express I in terms of 'm' and 'L'. The rigid body in Fig. So moment of inertia depends on both the object being rotated and the axis about which it is being rotated. In the problem we are required to find moment of inertia about transverse (perpendicular) axis passing through its center. 0 m) and mass (M = 1. 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Main Difference - Moment of Inertia vs. For a point mass, the moment of inertia equals mass times radius squared, so other mass units (such as pounds) and other distance units (such as feet) are occasionally used instead. Solution: Concepts: The moment of inertia about an axis, the rotational kinetic energy; Reasoning: The moment of inertia is I = ∑m i r i 2. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. RADIUS OF GYRATION AND COMPOSITE BODIES Radius of Gyration The mass moment of inertia of a body about a specific axis can be. 097 This means a mass of 22 units placed at (3. Moment of inertia of a rod Consider a rod of mass 'M' and length 'L' such that its linear density λ is M/L. The moment of inertia of the horizontal rod must be subtracted from the total moment in order to be left with the moment of inertia of the wheel alone. 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Moment of Inertia--Rod The moment of inertia of a thin rod of length h , mass M , and cross-sectional area A can be computed as the limiting case of the moment of inertia of a cylinder as the radius , so the tensor becomes. The rotational inertia of an object is dependent on the mass the the arrangement of the mass within the object. I know that I can calculate the moment of inertia of a rectangular cross section around a given axis located on its centroid by the following formulas: I also know that more generically, the moment of inertia is given by the integer of an area times the square of the distance from its centroid to the axis. Its moment of. It is the rotational analogue to mass. We're working on a new experience for engineering. 9 kg and radius R = 0. Oct 13, 2011 · Physics - Mechanics: Moment of Inertia (3 of 6) Derivation of Moment of Inertia of a Bar - Duration: 4:39. @DrChuck's answer is correct. 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Moment of Inertia--Rod The moment of inertia of a thin rod of length h , mass M , and cross-sectional area A can be computed as the limiting case of the moment of inertia of a cylinder as the radius , so the tensor becomes. Moment of Inertia of a thin rod about an axis perpendicular to the length of the rod and passing through its center. Moment of Inertia. The moment of inertia of a rectangular section having width b and depth d as shown in. The moment of inertia of your disk is one-half that of a ring. Thin Uniform Rod (i) Calculation of moment of inertia of uniform thin rod about an axis through its centre and perpendicular to its length. Area Moment of Inertia " Polar Moment of Inertia" - a measure of a beam's ability to resist torsion - which is required to calculate the twist of a beam subjected to torque "Area Moment of Inertia" - a property of shape that is used to predict deflection, bending and stress in beams; Circular Shaft and Maximum Moment. 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The moment of inertia is a physical quantity which describes how easily a body can be rotated about a given axis. However the rectangular shape is very common for beam sections, so it is probably worth memorizing. Similarly, an inertia moment (or inertia torque) is defined in rotary motion as a function of the mass moment of inertia and the angular acceleration (the second time derivative of rotation angle)—see Table 2. The moment of inertia of a point of mass and at a distance from the center of axis of rotation is given by,. PHY2053, Lecture 16, Rotational Energy and Inertia Discussion: Parallel Axis Theorem 6 The parallel axis theorem makes it possible to compute the moment of inertia of a complex object around a random axis, if one can compute the moment of inertia for the same object around a parallel axis which goes through the CM. bounce, spring, rebound; momentum, inertia; spur of the moment, impulse tfittxija Damascener nobody sudar bilværksted marki semmi egyéb salope Gleitbombe earthenware Sab. The polar moment of inertia is a measure of an object's ability to resist torsion as a function of its shape. where Iis the moment of inertia and angular acceleration. the higher the moment of inertia, the slower it will spin after being applied a given force). Moment of Inertia of a thin rod about an axis perpendicular to the length of the rod and passing through its center. The mass per unit length of each rod is r = 2 kg/m. Rolling Rod, Cylinder or Disc Measure the radius of the object from the center to the edge in centimeters; enter this figure into the calculator. Exact analytical solution: Thin Rod (axes perpendicular to length) Key Formulas You Need to Know Thin Plate: Key Formulas You Need to Know Radius of Gyration, rG. We would expect the moment of inertia to be smaller about an axis through the center of mass than the endpoint axis, just as it was for the barbell example at the start of this section. Owing to the symmetry of the aircraft, the lateral and vertical coordinates. Calculating Moment of Inertia: Real-Life Examples. The moment of inertia of a point mass is. Popular Topics on Hot Rod. It is a mathematical property of a section concerned with a surface area and how. Radius of gyration: Sometime in place of the mass moment of inertia the radius of gyration k is provided. Two small homogeneous balls with mass m 1 and m 2 are connected by a rod of length L with negligible mass. Calculations of moment of inertia of uniform rectangular sheet about an axis in the plane of sheet and problem based on moment of inertia of non uniform rod. We studied to shapes and their inertia. The axis may be internal or external and may or may not be fixed. The center of mass is distance from the vertex, where. Mass Moment of Inertia, I G (cont'd) I G for a body depends on the body's mass and the location of the mass. Apply the parallel axis theorem to find the moment of inertia about any axis parallel to one already known. The moment of inertia I=∫r 2 dm for a hoop, disk, cylinder, box, plate, rod, and spherical shell or solid can be found from this figure. Polar moment of inertia for planar dynamicsEdit. Find moment of inertia of a uniform hollow cylinder Home Problems and Answers Classical Mechanics Find moment of inertia of a uniform hollow cylinder We know that the moment of inertia for hoop with radius R is mR2. The mass moment of inertia is also known as the rotational inertia and used to calculate angular momentum and angular kinetic energy. The rigid body in Fig. The moment of inertia of a rod about an axis through its centre and perpendicular to it is 1 2 12 ML (where M is the mass and L the length of the rod). Mass moment of inertia is a measure of an object's resistance to rotational velocity about an axis. The equation of motion of the magnetic torsional oscillator is gven by I d / dt = NF + NB where I is the moment of inertia of the rod, I = M L2 / 12 is the angular velocity, NF is the restoring torque due to the suspension fiber, and NB is the restoring torque due to the magnetic field. The moment of inertia is a geometrical property of a beam and depends on a reference axis. Since the moment of inertia of an ordinary object involves a continuous distribution of mass at a continually varying distance from any rotation axis, the calculation of moments of inertia generally involves calculus, the discipline of mathematics which can handle such continuous variables. MENU Log In; MENU Stories. So moment of inertia depends on both the object being rotated and the axis about which it is being rotated. Significant changes have been made since then, and the manual used during the current academic year is in NOT available yet on the WEB. This means "look at every piece of the object, multiply by its distance from axis squared and add up over all pieces". Be advised that the "moment of inertia" encountered in Statics is not the same as the moment of inertia used in Dynamics. The moment of inertia for a flywheel may be calculated using the general equation for rotational inertia of a rigid body as shown below. Nov 08, 2017 · Moment of inertia ‘I’ of a rotating object with respect to its axis of rotation is given by the product of its mass and the square of its distance from the axis of rotation. The moment of inertia calculates the rotational inertia of an object rotating around a given axis. A rod of length l and mass m has ml²/12 as moment of inertia about an axis through its center of mass. This is for the Rectangular cross-section beams. The disk has mass md = 2. The moment of inertia of a thin rod about a normal axis through its centre is I. In planetary sciences, the moment of inertia factor or normalized polar moment of inertia is a dimensionless quantity that characterizes the radial distribution of mass inside a planet or satellite. Two conditions may be. Since in Dynamics there is no such concept as the inertia of an area, the moment of inertia has no physical meaning. Generally speaking the total moment of inertia is the sum of the moments inertia calculated individually. The moment of inertia of a rod of mass M and length L, with axis separated by distance x from the original one (through the centre of mass), is Ix = ICM + Mx2 = 1. The matrix of the values is known as the moment of inertia tensor. The moment of inertia of a rigid body about a particular of axis may be defined as the sum of the products of the masses of all the particles constituting the body and the squares of their respective distances from the axis of rotation. Moment of Inertia--Rod The moment of inertia of a thin rod of length h , mass M , and cross-sectional area A can be computed as the limiting case of the moment of inertia of a cylinder as the radius , so the tensor becomes. Therefore, the moment of inertia of a uniform rod about a perpendicular bisector (I) = ML 2 /12. It will not attempt to teach you the calculus involved since you are presumed to have learned it in another course. I₂ = I₁ + m(l/2)². ii) moment of inertia about the point B. A long, thin rod of uniform cross section and length L has a density that depends on position along the bar. Unfortunately most rotating bodies do not have the mass concentrated at one radius and the moment of inertia is not calculated as easily as this. The rod has an angular velocity of 0. Moments of Inertia: 1. In planetary sciences, the moment of inertia factor or normalized polar moment of inertia is a dimensionless quantity that characterizes the radial distribution of mass inside a planet or satellite. Mar 23, 2007 · Then you will end up with pitching moment, a pitching moment due to lift and a pitching moment due to weight. Don't show me this again. 3 Products of Inertia Example 3, page 2 of 2 y x dy x el = x/2 x y el = y x y dA = x dy Centroid of the infinitesimal strip Choosing horizontal strips and applying the parallel-axis theorem to the strip gives dI xy = dI x'y' + x el y el dA (2) where dI x'y' is the product of inertia for the differential strip about the x y axes. 33 rad/s and a moment of inertia of 1. 5 Solid cylinder rotating about the central axis. There is a theorem for this, called the parallel-axis theorem , which. @DrChuck's answer is correct. Let's start with a piece of our rod that has mass dm and length dx. The mass per unit length of each rod is r = 2 kg/m. The moment of inertia of the rod on the right that rotates around its centre is and thus if the rods have the same lengths and masses, and rotate at the same rate, the kinetic energy of the rod on the left will be four. 0 m) and mass (M = 1. 5: Calculating Moments of Inertia Calculate the moment of inertia for uniformly shaped, rigid bodies. Moment of Inertia of a thin rod about an axis perpendicular to the length of the rod and passing through its center. Tighter the thumb screw on the rod hanger and then the thumb screw on the disk. Exact analytical solution: Thin Rod (axes perpendicular to length) Key Formulas You Need to Know Thin Plate: Key Formulas You Need to Know Radius of Gyration, rG. Moment of Inertia of a rod length l, mass m. Its midpoint is O and its centre of mass is at C. The moment of inertia of an object is a calculated measure for a rigid body that is undergoing rotational motion around a fixed axis: that is to say, it measures how difficult it would be to change an object's current rotational speed. Moment of Inertia, General Form. The larger the Moment of Inertia the less the beam will bend. The moment of inertia of an object rotating about a particular axis is somewhat analogous to the ordinary mass of the object. A torsion rod pendulum (used in empirical methods of finding moments of inertia) also has a single wire, but the wire is thick enough to exert a measurable restoring torque on the bob when the wire is twisted. The moment of inertia of a rod of mass M and length L, with axis separated by distance x from the original one (through the centre of mass), is Ix = ICM + Mx2 = 1. Mar 23, 2007 · Then you will end up with pitching moment, a pitching moment due to lift and a pitching moment due to weight. For example, flywheels have a heavy outer flange that locates as much mass as possible at a greater distance from the hub. ##I=\frac{1}{3}ML^2## is the moment of inertia of a uniform rod mass m and length L about one end. Moment of Inertia, Version 1. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. Best Answer: Consider the rod to be two rods placed with one end coinciding. Jul 16, 2013 · If A. Apr 24, 2017 · Although calculating the moment of inertia can be very complicated, shapes such as spheres, rods and discs simplify the math considerably. Moment of Inertia of a Uniform Rod. 097 This means a mass of 22 units placed at (3. Nov 08, 2017 · Moment of inertia ‘I’ of a rotating object with respect to its axis of rotation is given by the product of its mass and the square of its distance from the axis of rotation. Moment of Inertia, General Form. k = (2π) 2 (3. This has many implications, including that the angular momentum vector is not always parallel to the angular velocity vector, and the relationship between angular acceleration and torque is no longer so simple. The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis. Unit 30 Moments of Inertia of Masses Frame 30-1 Introduction This unit will deal with the moment of inertia of mass, which should hardly be a new concept to you, since you have encountered it previously in math, and has many simularities to moment of inertia of area. yy = radius of gyration of the section about y - axis respectively. The rotational inertia of an object is dependent on the mass the the arrangement of the mass within the object. so gives the moment of inertia for rotation about the center of the rod. We're working on a new experience for engineering. I am having issues with the mass analysis tool in CREO parametric 4. Polar Moment of Inertia. 3^2 (I added the moment of inertia of the rod and the balls). SOLUTION: • Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section. So now, when the “solid cylinder” radio button is selected, the input boxes for r i, a, b, and c are grayed out as shown below. Mar 14, 2009 · The moment of inertia of the two-mass show more Ball a, of mass m_a, is connected to ball b, of mass m_b, by a massless rod of length L. We are to find the Moment of Inertia of a thin rod with three point masses attached; the rod is spinning on a point in the center of mass, or the center of the rod. Nov 25, 2019 · The principal moments of inertia are given by the entries in the diagonalized moment of inertia matrix, and are denoted (for a solid) , , and in order of decreasing magnitude. Please enter the "Input Values" in the form. A uniform rod of length 2L' has mass per unit length ‘m'. Today we will see here the determination of moment of inertia of one uniform thin rod; we will derive here the equation to express the moment of inertia for thin rod. 0 cm long and with mass 0. The first step in calculating mass properties is to establish the location of the X, Y, and Z axes. Moment of Inertia Problems and Solutions-Part-16-Rod leaning on a wall. Compute the moments and products of inertia of the body associated with frame Oxyz, and determine the principal moments of inertia and the principal axes of inertia. Moment of Inertia for body about an axis Say O-O is defined as ∑dM*y n 2. Popular Topics on Hot Rod. The mass of planet B An object is projected at an angle of elevation of A ball is thrown vertically upward with a velocity Give an example of each of the following? Scalar q In India, medicines have traditionally been prepar. A simple rule of thumb is- the more compact an object's mass, the less rotational inertia an object will have. Proposed Subject usage: Mathematics / Physics (A/AS level), Sports Science (Degree Yr 1/2) Introduction Moment of inertia of an object is an indication of the level of force that has to be applied in order to set the object, or keep the object, in motion about a defined axis of rotation. The moment of inertia of an object rotating about a particular axis is somewhat analogous to the ordinary mass of the object. The moment of inertia of is given by: Where we have: m: mass R: radius ( from the axis O to the object ) The following is a list of moment of inertia for some common homogeneous objects, where M stands for mass and the red line is the axis the objects rotating about. Dec 02, 2007 · Moment of inertia of a single mass is mR². English: Moment of Inertia of a thin rod about an axis perpendicular to the length of the rod and passing through its one end. [In this proof, you may assume standard results for the moment of inertia of uniform rods. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. This second moment is commonly called the moment of inertia and has a symbol I. The greater its value, the greater the moment required to provide a given acceleration about a fixed pivot. 01x - Lect 24 - Rolling Motion, Gyroscopes, VERY NON-INTUITIVE - Duration: 49:13. system it is g cm². For a different rotation point of an object—say a rod rotating around one end, like a turnstile, instead of around its center—we use the parallel axis theorem to find the object's moment of inertia. Formulae for moments of inertia The list that follows gives the more important moments of inertia for some common simply shaped objects. It is also popular as angular mass or rotational inertia of the given rigid body. Examples The moment of inertia of a thin rod with constant cross-section and density and with length about The moment of inertia of a thin disc of constant thickness , radius , and density about an axis The moment of inertia of the compound pendulum is now obtained by adding the moment of. Don't show me this again. (a) the moment of inertia of the system about the z-axis and (b) the rotational energy of the system. Radius of gyration is defined as the distance from the axis of rotation to a point where the total mass of the body is supposed to be concentrated, so that the moment of inertia about the axis may remain the same. The object in the diagram below consists of five thin cylinders arranged in a circle. 7ML 2 /48 c. Let us consider a uniform rod of mass (M) and length (l) as shown in Figure 5. In this first part of this experiment, you will compute the moments of inertia of a disk and an annulus from measurements of the masses and dimensions. The first step in calculating mass properties is to establish the location of the X, Y, and Z axes. Mass Moment of Inertia, I G (cont'd) I G for a body depends on the body's mass and the location of the mass. Moment of Inertia for body about an axis Say O-O is defined as ∑dM*y n 2. Rotate both rods about their long axes (see Figure 2), in order to (continued) compare the moments of inertia of these rods about this axis. The disk’s axis of rotation also goes through the center of mass. In physics, the moment of inertia measures how resistant an object is to changes in its rotational motion about a particular axis. The moment of inertia is the measure of resistance to torque applied on a spinning object (i. SKKU General Physics I (2013) | Moments of Inertia | 2 1 Rectangular plate The moment of inertia for the rectangular plate of sides a and b can be found by using the formula (5) and the parallel axis theorem. In this problem, we are. The moment of inertia of the rod on the right that rotates around its centre is and thus if the rods have the same lengths and masses, and rotate at the same rate, the kinetic energy of the rod on the left will be four. Please enter the "Input Values" in the form. Mass moments of inertia have units of dimension mass × length2. Attach the masses to the rod with the locking screws. For example, flywheels have a heavy outer flange that locates as much mass as possible at a greater distance from the hub. The mass per unit length of each rod is r = 2 kg/m. It is always considered with respect to a reference axis such as X-X or Y-Y. Moment of Inertia: The moment of inertia of a rigid body about a given axis is, defined as the sum of the products of the mass of each and every particle of the body and the square of its distance from the given axis. There are two vertical dashed lines, one through each ball, representing two different axes of rotation, axes a and b. You can ask me questions directly at https://www. The moment of inertia of an area with respect to any axis not through its centroid is equal to the moment of inertia of that area with respect to its own parallel centroidal axis plus the product of the area and the square of the distance between the two axes. Moment Of Inertia-Connecting Rod. Proposed Subject usage: Mathematics / Physics (A/AS level), Sports Science (Degree Yr 1/2) Introduction Moment of inertia of an object is an indication of the level of force that has to be applied in order to set the object, or keep the object, in motion about a defined axis of rotation. could you please suggest an algorithm for their applications. Moment of Inertia, General Form. Its moment of. Determine the moment of inertia of the composite area about the x axis. 73 kg and length L = 5. The moment of inertia is the measure of resistance to torque applied on a spinning object (i. The 2nd moment of area, also known as moment of inertia of plane area, area moment of inertia, or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. Language Watch Moment of Inertia of a thin rod about an axis perpendicular to the length of the rod and passing. Just copy and paste the below code to your webpage where you want to display this calculator. The larger the Moment of Inertia the less the beam will bend. Objects that have most of their mass near their axis of rotation have smaller rotational inertias than objects with more mass farther from their axis of rotation. The moment of inertia is the quantitative measure of rotational inertia, just as in translational motion, and mass is the quantitative measure of linear inertia—that is, the more massive an object is, the more inertia it has, and the greater is its resistance to change in linear velocity. org The moment of inertia about an axis perpendicular to the movement of the rigid system and through the center of mass is known as the polar moment of inertia. In physics, when you calculate an object’s moment of inertia, you need to consider not only the mass of the object but also how the mass is distributed. x is the first moment of area of certain section then (Ax). axis of interest L m m L I outside=m i r i!2=m(L)+m(2L)2=5mL2 3. | 2020-01-25T14:26:14 | {
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https://plainmath.net/43698/a-calculate-ss_-xx-ss_-yy-and-ss_-xy-lea | # a) Calculate SS_{xx},\ SS_{yy}, and SS_{xy} (Lea
a) Calculate $$\displaystyle{S}{S}_{{\times}},\ {S}{S}_{{{y}{y}}}$$, and $$\displaystyle{S}{S}_{{{x}{y}}}$$ (Leave no cells blank - be certain to enter ''0'' wherever required. Round your answers to 2 decimal places.)
$\begin{array}{|c|c|}\hline \text{College} & \text{Student} & \text{Weekly} & \text{Earnings} & \text{Dollars}(n=5) \\ \hline \text{Hourse Worked (X)} & \text{Weekly Pay (Y)} & (x_{i}-\bar{x})^{2} & (y_{i}-\bar{y})^{2} & (x_{1}-\bar{x})(y_{i}-\bar{y}) \\ \hline 11 & 103 \\ \hline 16 & 187 \\ \hline 16 & 216 \\ \hline 16 & 157 \\ \hline 32 & 262 \\ \hline \\ \hline \bar{x} & \bar{y} & SS_{xx} & SS_{yy} & SS_{xy} \\ \hline \end{array}$
b) )Calculate the sample correlation coefficient. Check your work by using Excel's function =CORREL(array1,array2). (Round your answer to 4 decimal places.)
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Step 1
Given data of College student weekly earnings
$\begin{array}{|c|c|}\hline \text{Hours Worked (X)} & \text{Weekly Pay (Y)} \\ \hline 11 & 103 \\ \hline 16 & 187 \\ \hline 16 & 216 \\ \hline 16 & 157 \\ \hline 32 & 262 \\ \hline \end{array}$
The average hours worked and average weekly pay are calculated as shown below
$$\displaystyle\overline{{{X}}}={\frac{{\sum{X}}}{{{n}}}}$$
$$\displaystyle\overline{{{X}}}={\frac{{{11}+{16}+{16}+{16}+{32}}}{{{5}}}}={18.2}$$
$$\displaystyle\overline{{{Y}}}={\frac{{\sum{Y}}}{{{n}}}}$$
$$\displaystyle\overline{{{Y}}}={\frac{{{103}+{187}+{216}+{157}+{262}}}{{{5}}}}$$
Using the average values the missing values in table are calculated as shown below
$\begin{array}{|c|c|}\hline \text{College} & \text{Student} & \text{Weekly} & \text{Earnings} & \text{Dollars}(n=5) \\ \hline \text{Hourse Worked (X)} & \text{Weekly Pay (Y)} & (x_{i}-\bar{x})^{2} & (y_{i}-\bar{y})^{2} & (x_{1}-\bar{x})(y_{i}-\bar{y}) \\ \hline 11 & 103 & (11-18.2)^{2}=51.84 & (103-185)^{2}=6724 & (11-18.2)(103-185)=590.4 \\ \hline 16 & 187 & (16-18.2)^{2}=4.84 & (187-185)^{2}=4 & (16-18.2)(187-185)=-4.4 \\ \hline 16 & 216 & (16-18.2)^{2}=4.84 & (216-185)^{2}=961 & (16-18.2)(216-185)=-68.2 \\ \hline 16 & 157 & (16-18.2)^{2}=4.84 & (157-185)^{2}=784 & (16-18.2)(157-185)=61.6 \\ \hline 32 & 262 & (32-18.2)^{2} & (262-185)^{2}=5929 & (32-18.2)(262-185)=1062.6 \\ \hline \bar{x}=18.2 & \bar{y}=185 & S_{xx}=256.8 & S_{yy}=14402 & S_{xy}=1642 \\ \hline \bar{x} & \bar{y} & SS_{xx} & SS_{yy} & SS_{xy} \\ \hline \end{array}$
Step 2
b) The correlation coefficient is calculated using the below mentioned formula
$$\displaystyle{r}={\frac{{{S}_{{{x}{y}}}}}{{\sqrt{{{S}_{{\times}}{S}_{{{y}{y}}}}}}}}$$
$$\displaystyle{r}={\frac{{{1642}}}{{\sqrt{{{256.8}\times{14402}}}}}}={0.853816}$$
using the excel function CORREL(X data,Y data) we get the sample correlation coefficient as 0.853815903
Rounding to 4 decimal places we get same value of r in both methods. $$\displaystyle{r}={0.8538}$$
###### Not exactly what you’re looking for?
Raymond Foley
Could you describe the solution to this problem in more detail? The answers were not found anywhere else. Thanks. | 2022-01-22T15:10:44 | {
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https://math.stackexchange.com/questions/3273115/classify-2-fold-covering-spaces-of-three-times-punctured-plane | # Classify 2-fold covering spaces of three-times punctured plane
Ahoy Mathematicians,
We are preparing for qualifying exams and ran across this question. There were two proposed methods of approaching it.
1. Recognize that the thrice punctured plane is homotopic to the wedge of 3 circles. Following the example in Hatcher for the 2-fold coverings of the wedge of 2 circles, we construct the 3 connected covering spaces (up to relabeling) of the wedge of three circles. There is also the disconnected covering space of two copies of itself.
2. Use the lifting correspondence to show that the 2 fold covering spaces of the thrice punctured disk are in one-to-one correspondence with subgroups H of $$\mathbb{Z} * \mathbb{Z} * \mathbb{Z}$$ of index 2. There are 3 such groups, giving 3 connected covering spaces.
You'll notice that the ideas in these two approaches agree. The question is one of rigor. Which answer is more accurate? Does Approach 1 actually give us the correct covering spaces, or do we need to think about crossing it with something to cover the surface rather than the 1-mfld? If we use Approach 2, does this answer the question, or do we need to give more of a classification than just the fundamental groups?
Thanks!
Ahoy OP!
Ultimately, I think it comes down to how the answer is written and how much wiggle room your grader is willing to give.
I see the first argument as
“We have a bijection between connected two-fold covers and index two subgroups of the fundamental group. Since the fundamental group is invariant under homotopy equivalence, we can find two-fold covers of the three-petaled rose. Here they are, and here are their fundamental groups. Thus, we know that there are exactly three connected covers of the surface, classified by their fundamental group.
The key aspect here is the bolded statement. A loose grader might understand that you’re implicitly relating the covers of the graph with the covers of the surface via their fundamental groups, but an answer I feel comfortable with would include some form of the bolded statement.
The second proof feels less solid to me. The argument seems to presuppose that we already know what the index two subgroups of $$F_3$$ are. Off the top of my head, I couldn’t rattle them off for you. The way I would prove that we know that there are three such subgroups would be to use your first argument. I’m sure there’s a nice group theoretic answer as well, but that is one that you would have to spell out.
That said, were you to spell out that group theoretic answer, then it would be a fine proof. I see nothing wrong with that direction inherently.
TL;DR If you give all the details and prove your statements, both options seem perfectly viable.
• Also, note that if the question asked you to “give” or “draw” the covering spaces and not to “classify” them, then both answers would fall short. – Santana Afton Jun 25 at 4:41
• Thanks for this thorough analysis! I appreciate it. – Beccah M. Jun 25 at 17:16
I'd like to propose another method of "classification", for completeness perhaps.
Let $$G$$ be a group. It is a result (tom Dieck, p. 345) that the set of isomorphism classes of principal $$G$$-bundles $$p : \widetilde{X} \rightarrow X$$, denoted $$\mathcal{B}(X, G)$$ is in bijective correspondence with the set $$[ X, BG]$$ of homotopy classes $$X \rightarrow BG$$ where $$BG$$ is the classifying space of $$G$$, i.e. we have $$\mathcal{B}(X, G) \cong [X, BG].$$
Letting $$G = S_n$$ the symmetric group on $$n$$ letters, an exercise in tom Dieck, p. 351 shows that principal $$S_n$$-bundles $$p : \widetilde{X} \rightarrow X$$ correspond to $$n$$-fold covering spaces $$p_n : \widetilde{X} \times_{S_n} \{ 1, \dots, n \} \rightarrow X$$ where $$\widetilde{X} \times_{S_n} \{ 1, \dots, n \} = \widetilde{X} \times \{ 1, \dots, n \} / g(x, n) = (xg^{-1}, gn)$$.
Then if we let $$X = S^1 \vee S^1 \vee S^1 \simeq \mathbb{R}^2 \setminus \{ x_1, x_2, x_3 \}$$ and $$G=S_2$$, $$\mathcal{B}(X, G)$$ becomes the set of covering spaces we would like to "classify" by the second result, and by the first result we see that these covering spaces correspond to the set $$[X, BG]$$ of homotopy classes $$S^1 \vee S^1 \vee S^1 \rightarrow BS_2$$. Finally, it is a general result that $$BS_n \cong \mathrm{UConf}_n (\mathbb{R}^\infty)$$ the unordered configuration space of $$\mathbb{R}^\infty$$ on $$n$$-tuples. In our case, $$BS_2$$ can be described as $$BS_2 \cong \frac{\{ f \mid f : \{1, 2 \} \rightarrow \mathbb{R}^\infty \} }{S_2} = \frac{\{ f \mid f : \{1, 2 \} \rightarrow \mathbb{R}^\infty \} }{\{\mathrm{id}, (1 \, 2 )\} }.$$
• Thanks for the response! Some of the machinery in this response is a big beyond the pay grade of the class, but it's definitely a nice solution. Thanks for adding it to round out the discussion. :) – Beccah M. Jun 25 at 17:19
• You're welcome! :) – mathphys Jun 26 at 12:17
There is a very computable approach, also using classifying spaces, using the facts that 2-fold coverings are the same as principal $$\mathbb{Z}/2$$ bundles and $$B\mathbb{Z}/2 \simeq \mathbb{RP}^\infty \simeq K(\mathbb{Z}/2,1)$$, the Eilenberg-Maclane space representing $$H^1(-;\mathbb{Z}/2)$$. Then we can compute isomorphism classes of 2-fold covering spaces over your punctured plane $$P$$ by computing
$$H^1(P;\mathbb{Z}/2) \cong H^1(\vee^3S^1; \mathbb{Z}/2) \cong(\mathbb{Z}/2)^3$$
so in fact there are 8 of them (up to isomorphism over $$id_P$$), corresponding to the different choices of wether they are trivial over each circle. There is only one disconnected double cover: the trivial one. Moreover since $$\mathbb{Z}/2 \cong O(1)$$ we can say that the bundles are classified by their first Steifel-Whitney class.
I think I can explain why I get 7 connected double covers where you get 3: my classification result is up to isomorphism of covering spaces over $$P$$, but I think your "up to relabeling" corresponds to isomorphisms of the bundle over some homeomorphism of $$P$$, which when considered as $$\vee^3 S^1$$ corresponds (up to homotopy) to permuting the wedge summands, which in $$H^1$$ corresponds to permuting the factors. Then, up to permutation, there are only three covering spaces that are not trivial: $$(1,0,0),\ (1,1,0)$$ and $$(1,1,1)$$.
• I agree with your analysis of why you got 7 and we got 3 connected covers. Thanks for this response! – Beccah M. Jun 25 at 17:18 | 2019-12-06T20:55:34 | {
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https://math.stackexchange.com/questions/1646101/distribution-of-the-sum-of-n-loaded-dice-rolls?noredirect=1 | # Distribution of the sum of $N$ loaded dice rolls
I would like to calculate the probability distribution of the sum of all the faces of $N$ dice rolls. The face probabilities ${p_i}$ are know, but are not $1 \over 6$. I have found answers for the case of a fair dice (i.e. $p_i={1 \over 6}$) here and here
For large $N$ I could apply the central limit theorem and use a normal distribution, but I don't know how to proceed for small $N$. (In particular, $N=2,4, 20$)
• Are you sure you know what you are asked to calculate? Please, tell us what have you already done and what are you trying to do. – V-X Feb 8 '16 at 15:17
• I guess I simply want to know how the distribution in math.uah.edu/stat/apps/DiceExperiment.html is calculated. It's probably simple but I don't know how to do it. – Ramon Crehuet Feb 8 '16 at 16:14
## 3 Answers
You can use generating functions.
Let $P=p_1x+p_2x^2+p_3x^3+p_4 x^4+p_5 x^5 +p_6 x^6$ where $p_i$ is the probability of $i$ occurring when rolling the die once.
Then the coefficient of $x^k$ in $P^N$ gives the probability of rolling a sum of $k$ when rolling the die $N$ times and summing.
For example, suppose $P=\frac{1}{7}x+\frac{1}{7}x^2+\frac{1}{7}x^3+\frac{1}{7}x^4+\frac{1}{7}x^5 + \frac{2}{7}x^6$.
Then, using a computer algebra system (I like PARI/GP), we find $P^3 = \frac{8}{343} x^{18} + \frac{12}{343} x^{17} + \frac{18}{343} x^{16} + \frac{25}{343} x^{15} + \frac{33}{343} x^{14} + \frac{6}{49} x^{13} + \frac{40}{343} x^{12} + \frac{39}{343} x^{11} + \frac{36}{343} x^{10} + \frac{31}{343} x^9 + \frac{24}{343} x^8 + \frac{15}{343} x^7 + \frac{10}{343} x^6 + \frac{6}{343} x^5 + \frac{3}{343} x^4 + \frac{1}{343} x^3$.
From this, we can conclude, for instance, that the probability of a sum of $10$ when rolling $3$ times (or rolling once with three identical copies of this die) is $\frac{36}{343}.$
(Using Bruce's example, we get $P^3=\frac{1}{64} x^{18} + \frac{3}{64} x^{17} + \frac{3}{32} x^{16} + \frac{1}{8} x^{15} + \frac{9}{64} x^{14} + \frac{9}{64} x^{13} + \frac{25}{192} x^{12} + \frac{7}{64} x^{11} + \frac{5}{64} x^{10} + \frac{91}{1728} x^9 + \frac{19}{576} x^8 + \frac{11}{576} x^7 + \frac{1}{108} x^6 + \frac{1}{288} x^5 + \frac{1}{576} x^4 + \frac{1}{1728} x^3$, and so the probability of $10$ is $\frac{5}{64}=0.078125$ (exactly).)
• Nice approach. But are you sure of your answer for a sum of ten? After my initial simulation mean(t == 10) returns 0.078097, whereas $36/343 \approx 0.1049563.$ Too far away. – BruceET Feb 8 '16 at 18:21
• OK, we're using differently biased dice. With your bias, I get 0.104877 for a total of 10 and 0.02338 for a total of 18. Got rid of one of by Comments to simplify for readers. – BruceET Feb 8 '16 at 18:54
• As a curiosity, do you have any idea of how could this be implemented in this applet math.uah.edu/stat/apps/DiceExperiment.html I am pretty sure the algebra system is not part of it... – Ramon Crehuet Feb 9 '16 at 9:09
• @RamonCrehuet: Couldn't be sure without looking at the code. Doubt if simulation. Probably, programming successive convolutions one step at a time of the simple discrete distributions. (In my browser this link is partially broken). – BruceET Feb 9 '16 at 18:26
It seems to me the spirit of the original context is experimental. In order to get an analytic answer, even in a simple case with only two rolls, you need to specify precisely how the die is loaded.
Here is a simulation in R of an experiment with $n = 3$ rolls of a die loaded so that faces 1 to 6 appear with probabilities $(1/12, 1/12, 1/12, 3/12, 3/12, 3/12)$. In practice, such a bias might be achieved by embedding a heavy weight just below the corner where faces 1, 2, and 3 meet. A million three-roll experiments are simulated. Each pass through the loop simulates one three-roll experiment. The random variable $T$ is the total on the three dice (values between 3 and 18, inclusive).
For the mean and SD of the best-fitting normal distribution, you should be able to find $E(T) = 12.75$ and $SD(T) = ?$, which are approximated in the simulation. (Also, the exact computation is shown for $E(T) = n\sum_{i=1}^6 ip_i.$)
m = 10^6; n = 3; p = c(1, 1, 1, 3, 3, 3)/12
t = numeric(m) # vector to receive totals
for (i in 1:m) {
faces = sample(1:6, n, repl=T, prob = p)
t[i] = sum(faces) }
mean(t); sd(t)
## 12.74981 # Approx E(T) = 12.75
## 2.657790 # Approx SD(T)
hist(t, br=(2:18)+.5, prob=T, main="Sum of 3 Rolls", col="wheat", ylim=c(0,.15))
curve(dnorm(x, mean(t), sd(t)), lwd=2, col="blue", add=T)
3*sum((1:6)*p)
## 12.75 # Exact E(T)
The normal curve with matching mean and SD is not yet a good fit for only three rolls. Perhaps much better for 20. Using code $mutatis\;mutandis$: Yes 20 looks better.
• Might want to browse 'related' pages shown in the right margin for some some interesting combinatorics. – BruceET Feb 8 '16 at 18:05
As an alternative to Matthew Conroy's suggestion to use a computer algebra system, one can also code this with Python using numpy class of Polynomials. The convolution power can be simply obtained by calculating the power of a polynomial. The same example he suggests can be coded as follows:
In [14]: from numpy.polynomial.polynomial import Polynomial
In [15]: p=Polynomial((1/7, 1/7, 1/7, 1/7,1/7, 2/7))
In [16]: p**3 # or alternatively: np.power(p,3)
Out[16]:
Polynomial([ 0.00291545, 0.00874636, 0.01749271, 0.02915452, 0.04373178,
0.06997085, 0.09037901, 0.10495627, 0.11370262, 0.11661808,
0.12244898, 0.09620991, 0.0728863 , 0.05247813, 0.03498542,
0.02332362], [-1., 1.], [-1., 1.])
Of course, one has to work with floats, but otherwise the results agree.
In numpy, the convolution can also be coded with np.convolve , but the application of successive convolutions is more cumbersome. | 2019-12-10T10:47:18 | {
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http://mkzs.kangalmalakli.it/mcq-probability.html | # Mcq Probability
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https://stats.stackexchange.com/questions/548963/manual-and-computer-calculated-two-sample-t-tests | # Manual and Computer-Calculated Two-Sample t-tests
I want to calculate p value and confidence interval (CI) both manually and by R.
this is my data set:
ex0112 = (
BP Diet
1 8 FishOil
2 13 FishOil
3 10 FishOil
4 14 FishOil
5 2 FishOil
6 1 FishOil
7 0 FishOil
8 -6 RegularOil
9 1 RegularOil
10 1 RegularOil
11 2 RegularOil
12 -3 RegularOil
13 -4 RegularOil
14 3 RegularOil)
# Manually:
(from the book statistical sleuth)
1- Compute the sample standard deviations for each group separately. (Let's call them $$s_1$$ and $$s_2$$.)
s1 <- sd(ex0112[ex0112$Diet == "FishOil",]$BP)
s2 <- sd(ex0112[ex0112$Diet == "RegularOil",]$BP)
2- Compute the pooled estimate of standard deviation using the formula $$s_p = \sqrt{\frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}}$$ where $$n_1$$ and $$n_2$$ group sizes.
n1 <- nrow(ex0112[ex0112$Diet == "FishOil",]) n2 <- nrow(ex0112[ex0112$Diet == "RegularOil",])
sp <- sqrt(((n1-1)*(s1^2)+(n2-1)*(s2^2))/(n1+n2-2))
1. Compute $$\text{SE} (\overline{Y_2} - \overline{Y_1})$$ using the formula: $$\text{SE} (\overline{Y_2} - \overline{Y_1}) = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2} }$$
se <- sp*sqrt((1/n1)+(1/n2))
1. Compute 97.5th percentile of the t-distribution with this many degrees of freedom ($$t_{df} (.975)$$)
df <- n2+n1-2
t_df <- qt(.975, df)
1. Construct a 95 % confidence interval fo $$\mu_2 - \mu_1$$ using the formula in Section 2.3.3: $$\left( \overline{Y_2} - \overline{Y_1} \right) \pm t_{df} (1-\alpha/2) \text{SE}( \overline{Y_2} - \overline{Y_1} ).$$
y1 <- mean(ex0112[ex0112$Diet == "FishOil",]$BP)
y2 <- mean(ex0112[ex0112$Diet == "RegularOil",]$BP)
LCI <- (y2-y1) - (t_df * se)
RCI <- (y2-y1) + (t_df * se)
1. Compute the t-statistic for testing equality: $$\left( \overline{Y_2} - \overline{Y_1} \right) / \text{SE}( \overline{Y_2} - \overline{Y_1} ).$$
t <- (y2-y1)/se
1. Find the one-sided p-value (for the alternative hypothesis $$\mu_2 - \mu_1 > 0$$) by comparing the t-statistics in (f) to the percentiles of the appropriate t-distribution (by reading the appropriate percentile from R.)
p <- pt(t, df)
## Results:
LCI = -13.29674
RCI = -2.131835
p = 0.00542278
# Computer t-test:
t.test(BP ~ Diet, data = ex0112, alternative = "greater",
var.equal = TRUE, conf.level = 0.95)
## Results
Two Sample t-test
data: BP by Diet
t = 3.0109, df = 12, p-value = 0.01085
alternative hypothesis: true difference in means between group FishOil and group RegularOil is not equal to 0
95 percent confidence interval:
2.131835 13.296737
sample estimates:
mean in group FishOil mean in group RegularOil
6.8571429 -0.8571429
As you see, the absolute values of CI are the same using either method; and so are the p-values.
I found this, which seems to ask the same question, but it seems they use another formula to compute t.
• The test you found is a one sample t-test while you are doing a two sample t-test... Oct 20 '21 at 5:12
• You have used the pooled two sample t test, which assumes (in your Step 3) that the two populations have the same variance. In practice, one usually cannot know that for sure. It has become standard practice to use the Welch 2-sample t test, which does not assume equal variances. unless you have prior experience with the the populations or processes involved and can be reasonably sure population variances are equal. If $n_1=n_2,$ pooled and welch t statistics will be equal, but Welch will have smaller DF to the extent $S_1^2≠S_2^2.$ Welch DF $\nu$ has $\min(n_1-1,n_2-1) \le\nu\le n_1+n_1-2.$ Oct 20 '21 at 7:18
I don't see any difficulties or discrepancies in your comparison of 'manual' and 'computer' results.
Here is a comparison of a pooled 2-sample t test and a Welch 2-sample t test for your oil data. You may want to try 'manual' computation of some parts.
x1 = c(8, 13, 10, 14, 2, 1, 0)
x2 = c(-6, 1, 1, 2, -3, -4, 3)
Repeat of pooled test: same as in your Question:
t.test(x1, x2, var.eq=T) # pooled
Two Sample t-test
data: x1 and x2
t = 3.0109, df = 12, p-value = 0.01085
alternative hypothesis:
true difference in means is not equal to 0
95 percent confidence interval:
2.131835 13.296737
sample estimates:
mean of x mean of y
6.8571429 -0.8571429
Welch 2-sample t test:
t.test(x1, x2)
Welch Two Sample t-test
data: x1 and x2
t = 3.0109, df = 9.7071, p-value = 0.01352
alternative hypothesis:
true difference in means is not equal to 0
95 percent confidence interval:
1.98202 13.44655
sample estimates:
mean of x mean of y
6.8571429 -0.8571429
Sample variances: not shown in output:
var(x1); var(x2)
[1] 34.14286
[1] 11.80952
Differences are as follows:
• Welch DF $$\le$$ Pooled DF because sample variances differ, as shown just above. R typically shows non-integer DF, which are actually used in computation. (Some software programs round DF down to the nearest integer, which may make a trivial difference in P-value.)
• Welch P-value $$\ge$$ Pooled P-value. A consequence of the difference in degrees of freedom.
• t statistics both $$3.0109.$$ Formulas look different, but reduce to equality for equal sample sizes $$n_1=n_2= 7.$$
• Sample means the same for both tests
• 95% Confidence interval for pooled test is $$(2.13, 13.30)$$ is narrower than $$(1.98, 13.45)$$ for Welch test.
• Not shown in R output: The standard error (denominator of t statistic) for the pooled t test is $$S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}},$$ where $$S_p^2 = \frac{(n_1-1)S_1^2 + (n_2-1)S_2^2}{n_1+n_2-2}.$$ For the Welch t test it is $$\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}.$$
Notes:
It is not a good idea to do a formal test for equality of variances and to let the outcome decide whether to use a pooled or Welch test. Instead, the best practice is always to use the Welch test, unless you have solid prior evidence that the population variances should be the same. In R, notice that the default test is Welch; in order to get a pooled test you have to use parameter var.eq=T with the t.test procedure.
The formula for Welch DF is a little messier than for the pooled test.
Finally, I will let you find the formula for the degrees of freedom of the Welch test in your textbook, class notes, or online. It takes into account sample sizes and sample variances. Different references have formulas that may look quite different, but which can be shown to give the same answers. | 2022-01-21T23:49:39 | {
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https://math.stackexchange.com/questions/1774947/computing-mathbf-a-1-2-where-mathbf-a-is-a-diagonal-matrix | # Computing $\mathbf A^{-1/2}$, where $\mathbf A$ is a Diagonal Matrix.
I have the following matrix :
$$\mathbf A =\begin{bmatrix} 100 & 0 \\ 0 & 1 \\ \end{bmatrix}$$
I have to compute $\mathbf A^{-1/2}$.
So I need spectral decomposition, $$\mathbf A = \mathbf P \mathbf \Lambda\mathbf P',$$
$\mathbf P$ be a matrix with normalized eigenvectors and $\mathbf \Lambda$ is a diagonal matrix with diagonal elements be eigenvalues.
Eigenvalues of $\mathbf A$ is $100$ and $1$.
But I stumbled to calculate eigenvector.
The characteristic equation is :
$$\begin{bmatrix} 100 & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix}= 100\begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix}$$
$$\Rightarrow 100x_1 = 100x_1$$
$$x_2 = 100 x_2$$
How is $x_2 = 100 x_2$ possible ?
And is there a simpler way to calculate any power of a diagonal matrix , for example, $\mathbf A^{-1/2}$ ?
• @Moo yes, do I need only $\mathbf A^{-1/2} =\begin{bmatrix} 100^{-1/2} & 0 \\ 0 & 1^{-1/2} \\ \end{bmatrix}$ ? – user 31466 May 7 '16 at 3:00
• One More question : The characteristic equation is : $$\begin{bmatrix} 100 & 0 \\ 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix}= 100\begin{bmatrix} x_1 \\ x_2 \\ \end{bmatrix}$$ $$\Rightarrow 100x_1 = 100x_1$$ $$x_2 = 100 x_2$$ How is $x_2 = 100 x_2$ possible ? – user 31466 May 7 '16 at 3:05
• $x_2=0$, but $x_1$ is arbitrary, so $(1,0)$ is an eigenvector. – Hans Lundmark May 7 '16 at 3:06
• Why are you diagonalizing a matrix that's already diagonal? – Erick Wong May 7 '16 at 6:20
For diagonal matrices, the normalized eigenvectors are always standard basis vectors, i.e. vectors of all zeros except a single 1 in a particular coordinate, e.g., ${\bf{e}}_1=(1,0)$ and ${\bf{e}}_2=(0,1)$ in your example. For eigenvalue 100, the corresponding eigenvector is $(1,0)$ which means that $x_2=0$ (note that solves $x_2=100x_2$).
The power $k$ of a diagonal matrix $$D=\left\|\begin{array}{ccc}d_1 &\cdots &0\\ \vdots & &\vdots\\ 0 &\cdots &d_n\end{array}\right\|$$ is found as
$$D=\left\|\begin{array}{ccc}d^k_1 &\cdots &0\\ \vdots & &\vdots\\ 0 &\cdots &d^k_n\end{array}\right\|$$
$A=SDS^{-1}$, and more generally $A^k=SD^kS^{-1}$ where D is the diagonal matrix composed from eigenvalues, S is the matrix of eigenvectors, and $S^{-1}$ is the inverse of S.
D is $\begin{bmatrix} 100&0\\0&1 \end{bmatrix}$as you have already found.
To find eigenvectors, you need to follow this format $\left[ \begin{array}{cc|c} 100-\lambda&0&0\\0&1-\lambda&0 \end{array}\right]$
When $\lambda=100$, $\left[\begin{array}{cc|c} 0&0&0\\0&-99&0 \end{array}\right]$, $v_1=\begin{bmatrix}1\\0\end{bmatrix}$
When $\lambda=1$, $\left[\begin{array}{cc|c} 100&0&0\\0&0&0 \end{array}\right]$, $v_2=\begin{bmatrix}0\\1\end{bmatrix}$
Thus $S=\begin{bmatrix} 1&0\\0&1 \end{bmatrix}=S^{-1}$
$A^{1/2}=SD^{1/2}S^{-1}=\begin{bmatrix} 1&0\\0&1 \end{bmatrix}\begin{bmatrix} 100^{1/2}&0\\0&1^{1/2} \end{bmatrix}\begin{bmatrix} 1&0\\0&1 \end{bmatrix}=\begin{bmatrix} 10&0\\0&1 \end{bmatrix}$
And $A^{-1/2}=(A^{1/2})^{-1}$, $A^{-1/2}$ is the inverse of $A^{1/2}$, which is $\begin{bmatrix} \frac{1}{10}&0\\0&1 \end{bmatrix}$
Your matrix is in Jordan Normal Form (diag) thus: $$f(A) = \begin{bmatrix}f(a_{11})& \\&f(a_{22})\end{bmatrix}$$
So yes you can use: $$A^{-1/2} = \begin{bmatrix}a_{11}^{-1/2}& \\&a_{22}^{-1/2}\end{bmatrix}$$ | 2019-08-21T00:49:17 | {
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http://math.stackexchange.com/questions/487049/if-lim-n-to-infty-int-01-f-nxdx-0-are-there-points-x-0-in0-1-such-t | # If $\lim_{n\to\infty}\int_0^1 f_n(x)dx=0$, are there points $x_0\in[0,1]$ such that $\lim_{n\to\infty}f_n(x_0)=0$?
This is part of an old qual problem at my school.
Assume $\{f_n\}$ is a sequence of nonnegative continuous functions on $[0,1]$ such that $\lim_{n\to\infty}\int_0^1 f_n(x)dx=0$. Is it necessarily true that there are points $x_0\in[0,1]$ such that $\lim_{n\to\infty}f_n(x_0)=0$?
I think that there should be some $x_0$. My intuition is that if the integrals converge to $0$, then the $f_n$ should start to be close to zero in most places in $[0,1]$. If $\lim_{n\to\infty}f_n(x_0)\neq 0$ for any $x_0$, then the sequences $\{f_n(x_0)\}$ for each fixed $x_0$ have to have positive terms of arbitrarily large index. Since there are only countably many functions, I don't think it's possible to do this without making $\lim_{n\to\infty}\int_0^1 f_n(x)dx=0$.
Is there a proof or counterexample to the question?
-
No. The standard counterexample would be indicator functions of $[0, 1]$, $[0, 1/2]$, $[1/2, 1]$, $[0, 1/3]$, $[1/3, 2/3]$, and so on.
In order to make these continuous, add in line segments on either end with very large slope.
-
oh i hadn't seen that, thanks. – YN Chew Sep 8 '13 at 3:13
@YNChew Sure. It's a really useful counterexample to have in mind, since it breaks a lot of things which are obviously true. – user61527 Sep 8 '13 at 3:13
wait, are indicator functions continuous on $[0,1]$? – YN Chew Sep 8 '13 at 3:15
@YNChew You're right, I missed that assumption. An appropriate fix can be made easily by drawing very skinny triangles on each end of the indicator function, and requiring that the area of the triangle is very small. – user61527 Sep 8 '13 at 3:17
As T. Bongers points out the answer is no. However, we can say that there is a subsequence $f_{n_k}$ such that $f_{n_k}(x) \to 0$ for almost every $x \in [0,1]$. The statement that $\int_0^1 f_n(x)dx \to 0$ exactly tells us $f_n \to 0$ in $L^1$ which implies the existence of a subsequence which converges to zero almost everywhere. See, e.g., http://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/ Corollary 3 for proof.
- | 2015-07-29T03:14:37 | {
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https://www.physicsforums.com/threads/universal-graviation.659881/ | # Universal Graviation
1. Dec 19, 2012
### mtayab1994
1. The problem statement, all variables and given/known data
Let (S) be a satellite between the earth and the moon such that (S) is at a distance dm from the moon. All forces on the the satellite are null (equal zero). Find the distance dm .
Given: Distance between moon and earth is 38*10^4km and the mass of earth is 81 times the mass of the moon.
3. The attempt at a solution
Well it's been a long time since I've done any universal gravitation. It would be nice if someone can just give me an idea on how to start and I'll go from there.
2. Dec 19, 2012
### SammyS
Staff Emeritus
If the distance between the moon & the satellite is dm , then what is the distance between the satellite & the earth ?
3. Dec 19, 2012
### mtayab1994
The distance between the earth and the satellite is d=D-dm and newton's law of gravitation says that F=(G*m1*m2)/(r^2)
Last edited: Dec 19, 2012
4. Dec 19, 2012
### mtayab1994
Let's name the distance from earth to the satellite by D. So D=d-dm.
And by using g=(GM)/D^2 we get that the distance from the earth is 6.36*10^3 km.
So that means the distance from the satellite to the moon is
d-D=3.8*10^5-6.36*10^3=3.74*10^5 km. Is that correct??
5. Dec 19, 2012
### SammyS
Staff Emeritus
It's not clear what all you put together to get that result.
Please fill in and explain some steps.
My approach would be set the magnitudes of the following two forces equal to each other.
The force exerted on the satellite by the moon.
The force exerted on the satellite by the earth.
6. Dec 19, 2012
### mtayab1994
By setting the forces equal to one another i got in the end:
Me/Mm=dm^2/de^2 and we know that the mass of the earth is 81 times the mass of the moon so we get 81Mm/Mm=(dm^2)/(de^2) and then we cancel with the mass of the moon and square root both sides and we are left with. √81=dm/de and we know that the distance from the satellite and the earth is d-dm. and by doing some simple algebra we find that the distance to the moon is d/10 the total distance between the moon and earth so that gives us
3.8*10^4 km.
Last edited: Dec 19, 2012
7. Dec 19, 2012
### SammyS
Staff Emeritus
It's quite difficult to try to figure out what all you have done to get your answer. Having youe text all packed together so tightly doesn't help either.
$\displaystyle G \frac{M_m\, m}{{d_m}^2}=G \frac{M_e\, m}{(d-d_m)^2}\,,\$
where Mm is the moon's mass, Me is the earth's mass, m is the satellite's mass, dm is the satellite's distance from the moon, and d is the monn's distance from earth.
Me = 81 Mm. Plugging this in & doing some manipulation gives:
$\displaystyle \frac{M_m}{{d_m}^2}=\frac{81M_m}{(d-d_m)^2}\$ $\displaystyle\quad\to\quad d-d_m=d_m\sqrt{81}=9d_m\$ $\displaystyle\quad\to\quad d_m=\frac{d}{10}\$ | 2018-02-20T15:41:38 | {
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http://math.stackexchange.com/questions/44970/what-are-linear-divisors-of-integer-powers | # what are linear divisors of integer powers?
I have been trying to prove that $n^4$ is eithere divisible by $5k$ or $5k+1$, couldn't help wonder if there is a more general theme to try later , namely if it is true that $n^m$ is divisble by at least one of the $mk , mk+1 , mk+2 , mk+3 , \cdots , mk+(m-1)$ or something similar?
-
@ShreevatsaR, k=3 does the job. – Arjang Jun 12 '11 at 16:41
Oops, yes. Sorry. – ShreevatsaR Jun 12 '11 at 17:26
@ShreevatsaR : the only reason i mentioned k=3 was that I too thought for n=2 there was no answer initially, but then the "or" part reminded me to think again. – Arjang Jun 12 '11 at 17:35
There is a very nice answer to this question when $m+1$ is a prime number. In this case, $n^m$ is divisible by either $km$ or $km+1$ for some $k$. (Notice that this solves your first question when $m=4$)
So we look at $n^{p-1}$. Well, either $p$ divides $n$, or it doesn't. In the first case, we have that $kp|n$ with $k=1$. In the second case, by Fermat's little theorem, $$n^{p-1}\equiv 1\pmod{p}.$$ This means that $n^{p-1}=kp+1$ for some $k$, so that in particular $kp+1$ divides $n^{p-1}$.
Hope that helps,
-
I will post this as an answer, mostly since it is too long for a comment.
It is critical in mathematics to state questions precisely. Interestingly enough, the same need not apply to proofs. In many settings, conveying the idea of the proof may be enough.
So what does "$n^4$ is divisible by $5k$ or $5k+1$" mean? What is $k$?
Chandru, from the link he gave, interprets the question to mean "$n^4$ is either of the form $5k$ or of the form $5k+1$." This is a way of saying what, in more modern language, would be expressed by writing that either $n^4 \equiv 0 \pmod{5}$ or $n^4 \equiv 1 \pmod{5}$.
Proving that this is true is easy. Imagine dividing $n$ by $5$. The remainder is $0$, $1$, $2$, $3$, or $4$. Now we need to find, in each of these cases, the remainder when $n^4$ is divided by $5$. This is easy by using basic properties of congruences.
But even if we know nothing about congruences, we can solve the problem because of our familiarity with the decimal system. We will be taking a fourth power, and $(-a)^4=a^4$. So we need only worry about $n^4$ for $n \ge 0$.
Look at the last digit of $n^4$ for $n=0$, $1$, $2$, and so on up to $9$. We get the pattern $0$, $1$, $6$, $1$, $6$, $1$, $6$, $1$, $6$, $1$. And if we continue, looking at $10^4$, $11^4$, and so on, the pattern obviously repeats.
When the last digit of $n^4$ is $0$, the number $n^4$ is divisible by $5$, that is, is of the form $5k+1$. When the last digit of $n^4$ is $1$, $n^4$ leaves a remainder of $1$ on division by $5$, so is of the form $5k+1$. This is also the case when the last digit of $n^4$ is $6$, for in this case also $n^4$ is $1$ more than a multiple of $5$.
We could use fancier tools to tackle the problem. Here is a nice result, called Fermat's (little) Theorem.
Let $p$ be prime, and let $a$ be an integer not divisible by $p$. Then $$a^{p-1}\equiv 1 \pmod{p}$$ To apply this to our problem, let $p=5$. If $a$ is divisible by $5$, then clearly $a^4$ is divisible by $5$. In all other cases, use Fermat's Theorem to conclude that $a^4 \equiv 1 \pmod{5}$, which means that the remainder when $a^4$ is divided by $5$ is $1$, which means that $a^4$ is of the form $5k+1$.
Your question: You asked a question that could be interpreted as being about the "shape" of $n^m$ in general. When the power $m$ is $1$ more than a prime, there is very useful information contained in Fermat's Theorem. There is also an important generalization due to Euler that you could easily track down by reading about Fermat's Theorem. Every beginning Number Theory book has the details.
A possible exploration: All of the above stuff has nothing to do with the phrase "linear divisors of integer powers." We have not been examining the divisors of $n^4$, only $n^4$ itself. A bit of thought will show that numbers of every shape whatsoever can divide $n^4$, if you put no restrictions on $n$.
Even with a simple example like $2^4$, the divisors are $1$, $2$, $4$, $8$, and $16$. Note that if you divide these by $5$, all remainders except $0$ occur.
It could be interesting to explore the following question. For a given integer $a$, and a given prime $p$, what are the possible remainders when powers of $a$ (things of the shape $a^m$) are divided by $p$.
If you start looking at these things, with the help of a good Number Theory book and/or the Internet, you will bump into interesting things, like primitive roots.
Divisors of $n^2+1$: At a more advanced level in number theory, one can tackle questions about possible shapes of divisors of, for example, $n^2+1$, and related questions. List to yourself the divisors of $n^2+1$ for various $n$. You will find no numbers of the from $4k+3$. This is no accident, it can be proved in general. A lot is known the shape of divisors of $n^4+1$. But for this sort of question, powers higher than $2$ in general lead to very difficult problems.
For power $2$, more subtle facts of this kind can be proved when one looks at quadratic residues and non-residues.
-
+1 for detailed analysis and points on clarification of question, points which I even didn't realise, and I considered myself a zelot for making questions precise, thank you for your time. Will modify the question to be more precise thanks to you. – Arjang Jun 13 '11 at 6:31
@Arjang: You will notice that the Fermat Theorem stuff mentioned in the various posts shows that the number itself has a certain shape. Of course, the number divides itself, but the Fermat Theorem says nothing directly about the shapes of divisors of $a^{p-1}$. These shapes could be essentially arbitrary if one makes the appropriate choice of $a$. – André Nicolas Jun 13 '11 at 6:47
By Fermat's Little Theorem $\rm\: n^{p-1}\:$ has form $\rm\:p\:k\:$ or $\rm\:p\:k+1\:,\:$ so it has itself as a divisor of that form.
More generally $\rm\ gcd(n,m)= 1\ \Rightarrow\ n^{\lambda(m)}\equiv 1\ \ (mod\ m)\$ for $\rm\: \lambda(n)\:$ the Carmichael function.
Your last remark is always true: every integer on division by $\rm\:m\:$ has remainder $\rm\:0,1,\cdots$ or $\rm\:m-1\:.$ | 2015-09-05T04:20:57 | {
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https://math.stackexchange.com/questions/2433438/probability-equally-likely-or-not | # Probability-Equally likely or not?
Probability
Question. "$11$ identical balls are distributed in $4$ distinct boxes randomly. Then the probability that any $3$ boxes will together get a greater number of balls than the remaining one is:"
I am confused about the distribution of balls. Are all the cases i.e. $\{5,5,0,1\}$ and $\{2,3,4,2\}$ equally likely or not?
• No, those two cases are not equally likely (according to the most straightforward reading of the problem). Each ball is equally likely to be put into any of the four boxes, but it is not equally likely for all eleven to be put into box 1, for example, than it is for two to be put into box 1, and three each to be put into the other boxes. The question is asking you to identify what the distribution of number in boxes is, and to use that in your answer. – Brian Tung Sep 17 '17 at 20:04
• By the way, "i.e." means "that is"; this would imply that your two examples are all the cases. You probably mean "e.g.," which means "for example." (It's a trivial issue, but sometimes mathematics turns on trivialities!) – Brian Tung Sep 17 '17 at 20:05
The point is to find all combinations of natural numbers $(a,b,c,d)$ such that $\max\{a,b,c,d\}\le 5$ and $a+b+c+d=11$. Without loss of generality, assume $a\ge b\ge c\ge d$. Notice that $\max\{a,b,c,d\}$ can only be $3$, or $4$, or $5$.
• If $\max\{a,b,c,d\}=3$, then the only possible combination is $(3,3,3,2)$;
• If $\max\{a,b,c,d\}=4$, then the possible combinations are $(4,4,3,0)$, $(4,4,2,1)$, $(4,3,3,1)$,$(4,3,2,2)$;
• If $\max\{a,b,c,d\}=5$, then possible combinations include $(5,5,1,0)$,$(5,4,2,0)$,$(5,4,1,1)$,$(5,3,3,0)$, $(5,3,2,1)$,$(5,2,2,2)$.
Together, since four boxes are distinct, for each combination above, the number of ways to rearrange is
• For $(5,4,2,0)$ and $(5,3,2,1)$: $A^4_4$;
• For $(4,4,3,0)$, $(4,4,2,1)$,$(4,3,3,1)$,$(4,3,2,2)$,$(5,5,1,0)$,$(5,4,1,1)$, $(5,3,3,0)$: $A^4_4/A^2_2$;
• For $(3,3,3,2)$ and $(5,2,2,2)$:$A^4_4/A^3_3$
Since the total number of events are $4^{11}$, the probability is $$\frac{2A^4_4+7A^4_4/A^2_2+2A^4_4/A^3_3}{4^{11}}.$$
• That number seems too low. I think you must also multiply each box configuration by the ways the balls can end up in that configuration. For example, for $(5,4,2,0)$ the first $5$ balls could go to box 1, or they could be the $4$ first ones and the sixth, and so on. Simulation gives the answer somewhere around $0.86$. – ploosu2 Sep 18 '17 at 9:09
The solution given by OnoL has the correct configurations for the ball amounts in the boxes but it is missing the possible ways the balls can be placed to the boxes. Because we are considering all the ways $11$ balls can be placed to $4$ boxes ($4^{11}$) these multipliers must be considered. They are also different for different configuration (this answers your second question: the different cases have different probabilities). Luckily it doesn't depend on the order of the boxes so we can still only consider the ordered configurations (and then multiply by the number of permutations of that configuration).
The way $11$ balls can be placed to form $(a, b, c, d)$ is
$${{11}\choose{a}} {{11-a}\choose{b}} {{11-a-b}\choose{c}}$$
(The balls for the first box can be chosen freely, then choose from the rest $11-a$ balls $b$ balls for the second and so on. The last (fourth) box has to have the remaining balls so that coefficient ${{11-a-b-c}\choose{d}} = {{d}\choose{d}} = 1$, since $a+b+c+d=11$.)
The number of ball placings that satisfy the claim "any $3$ boxes will together get a greater number of balls than the remaining one" is given by the sum over all possible configurations (these were found by OnoL)
$(3,3,3,2), (4,4,3,0), (4,4,2,1), (4,3,3,1), (4,3,2,2), (5,5,1,0), (5,4,2,0), (5,4,1,1), (5,3,3,0), (5,3,2,1), (5,2,2,2)$
where the summand is the product
$($number of permutations of that configuration$)$ $\times$ $($the number of ways to place the balls to form that configuration$)$.
I calculated this with a computer and got $3618384$. So the wanted probability is
$$\frac{3618384}{4^{11}} = \frac{226149}{262144} \approx 0.86269.$$ | 2019-07-23T00:42:48 | {
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https://math.stackexchange.com/questions/1871390/why-do-skew-symmetric-matrices-always-have-purely-imaginary-eigenvalues | # Why do skew-symmetric matrices always have purely imaginary eigenvalues?
I checked some examples and I always received that skew-symmetric matrix of even dimension has only pure imaginary eigenvalues.
For example:
$\begin{bmatrix} 0 & 2 & 3 & 1 \\ -2 & 0 & 1 & 4 \\ -3 & -1 & 0 & 1 \\ -1 & -4 & -1 & 0 \end{bmatrix}$
Eigenvalues: $( 0.000, 5.406i) ( 0.000,-5.406i) ( 0.000, 1.665i) ( 0.000,-1.665i)$
How can be explained such property?
Additionally why skew-symmetric of even dimension has non-zero determinant in opposition to odd dimensional skew-symmetric matrices ?
(I'm not considering here zero matrices) Interesting is also fact that probably every matrix (of even dimension) can be decomposed into symetrical part which has only real eigenvalues and skew-symmetrical which has only pure imaginary values what makes interesting analogy to complex numbers and their two parts, but I don't know whether there are importatnt consequences of this fact.
It's because $-A^2 = A^T A$ has only real nonnegative eigenvalues: if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ then $$-\lambda^2 \|v\|^2 = -v^T A^2 v = v^T A^T A v = \|A v \|^2.$$
Skew-symmetric matrices do not have to have nonzero determinant, the zero matrix is a counterexample.
• Let's exclude zero matrices from our problem. It is so special that it could be at the same time formally symmetric and skew-symmetric what gives really no information about its nature. – Widawensen Jul 26 '16 at 7:52
• @Widawensen OK but there are plenty of other examples of skew-symmetric matrices with determinant $0$, including in even dimension – user356288 Jul 26 '16 at 7:54
• @ user356288, hmm, do you point me some example ? Additionally, such even dimensional skew-symmetric matrices with zero determinant have some special property which other don't have? – Widawensen Jul 26 '16 at 8:10
• The equation written by you, dear user356288, provides so much information about the problem, probably it gives also a hint why odd dimensional skew-symmetric matrices must have determinant equal 0. Only for one eigenvalue = 0 in this case we have fulfilled this equation. Maybe we should have two eigenvalues for even dimensions equal 0 in order to have non-invertible matrix ? Is it possible? – Widawensen Jul 26 '16 at 8:16
• Ok. I've found counterexample. It has two 0 eigenvalues. $\begin{bmatrix} 0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$ It's seems that such non-invertible matrix must have some even dimensional sub-matrix on diagonal as the zero matrix. – Widawensen Jul 26 '16 at 8:30
Say A is a skew symmetry matrix
Say $$\lambda$$ is a real eigenvalue of A, then there exists a real vector $$v≠0$$ such that $$Av=\lambda v$$
By skew symmetry we have:
$$v^TAv=(v^TAv)^T$$ (this holds because the $$v^TAv$$ is a real and reals are equal to their transpose)
$$=v^TA^T(v^T)^T=v^TA^Tv=v^T(-A)v=-v^TAv$$
Thus $$v^TAv=-v^TAv=0$$
Thus $$0 = v^TAv=v^T\lambda v = \lambda v^Tv = \lambda |v|^2$$
(this last equality holds because v is real)
Thus $$0=\lambda |v|^2$$
Thus $$\lambda=0$$ or $$v=0$$, however $$v≠0$$ thus $$\lambda=0$$
Thus the only possible real eigenvalue of A is $$0$$
Stated differently, all non zero eigenvalues of a skew symmetric matrix have non zero imaginary component | 2021-06-20T21:14:21 | {
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http://nfmv.psychotherapie-neuwerk.de/5T6C | ### Sampling Theorem Ppt
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Download free PowerPoint templates & slides that you can use to make presentations in Microsoft PowerPoint and Google Slides. The system is depicted in figure 11. SAMPLING AND SAMPLING DISTRIBUTIONS Note on Statistics • The value of the statistic will change from sample to sample and we can therefore think of it as a random variable with it's own probability distribution. If n (sample size) and B (number of re-sampling) are sufficiently high, we can rely on the asymptotic properties of the summation of random variables (in particular, the Central Limit Theorem ) and we. of analytical procedures, from the planning to the analytical. We identified it from honorable source. Sampling theorems. My questions are:. The sampling is thru in accord with the Sampling Theorem , i. Sampling PDF with Particle Dynamics Sampling Sampling PDF's with Particle Dynamics The ergodic theorem highlights an equivalence/duality between the Trajectory (Dynamics) versus PDF descriptions (Statistics). 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Sampling Theorem Sampling Theorem A continuous-time signal x(t) with frequencies no higher than f max (Hz) can be reconstructed EXACTLY from its samples x[n] = x(nTs), if the samples are taken at a rate fs = 1/Ts that is greater than 2f max. All you need to start is a bit of calculus. 4 CHAPTER 6. of sampling theory is concerned not only with the proper. Another statement of the Sampling Theorem, from A. ) Consider the case where f H = LB (k an Even Integer) k=6 for this case Whenever f H = LB, we can choose Fs = 2B to perfectly "interweave" the shifted spectral replicas f L X( f ) f. Nyquist explored the key issue and Shannon gave the complete solution which is known as Shannon's sampling theorem. sampling theorem, namely, choosing sampling sets that guarantee unique recovery of signals of given bandwidth. - What is the best sampling rate? • Gray-level resolution (Quantization) - Smallest discernible change in the gray level value. We identified it from reliable source. 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Anti aliasing filter is a filter which is used before a signal sampler, to restrict the bandwidth of a signal to approximately satisfy the sampling theorem. • ¯ is a random variable • Repeated sampling and calculation of the resulting statistic will give rise to a dis-. Download the Best PowerPoint templates and Google Slides themes for your presentations. 2 samples per sinewave cycle an improper sampling of the signal because another sine wave can produce the same samples. 4 Statistical Models When making statistical inference it is useful to write random variables in terms of model parameters and random errors Sampling. Since martingales can be used to model the wealth of a gambler participating in a fair game, the optional stopping theorem says that, on average, nothing can be gained by. Nyquist-Shannon sampling theorem CSE/EE474 33 Nyquist Theorem and Aliasing n Nyquist Theorem:. These results can be understood by examining the Fourier transforms X(jw), X s (jw), and X r (jw). 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"]) # visualize the first prediction's explanation. Pythagorean Theorem calculator to find out the unknown length of a right triangle. Fourier Theorem states that any waveform can be reproduced by sine waves. Sampling Theorem, frequency resolution & Aliasing The Sampling Theorem will be the single most important constraint you'll learn in Aliased frequencies PowerPoint Presentation PowerPoint Presentation PowerPoint Presentation Strobe light demo PowerPoint Presentation Filtering before and after A→D conversion Windows to a sample of. Sampling Theorem. The sampling distribution is the foundation of inferential statistics. Here you can find thousands of free PowerPoint templates and slide designs for your business and educational presentations. However, band-limited functions have infinite duration in the time domain. If the frame-rate is too low, the aliasing effect will introduce incorrect low frequency components which in this case manifest as a wheel which is slowly turning in. And let [𝑛] be the sequence of numbers obtained by sampling ( ) at a sampling rate of samples per second (Hz). Sis a uniqueness set for signals in PW!. From the central limit theorem, the sampling distribution of the sample mean is approximated by for infinite populations. Willsky, Signals and Systems, 2nd Ed. 0 CHAPTER 8: Sampling Distributions Chapter 8 - Learning Objectives Sampling Distribution of the Mean Standardizing a Sample Mean on a Normal Curve Central Limit Theorem Sampling Distribution of the Mean Sampling Distribution of the Mean. The sampling theorem provides that a properly bandlimited continuous-time signal can be An early derivation of the sampling theorem is often cited as a 1928 paper by Harold Nyquist, and Claude. Sampling Theorem and Pulse Amplitude Modulation (PAM) Reference - Stremler, Communication Systems, Chapter 3. It ma y be stated simply as follo ws: The sampling fr e quency should b at le ast twic the highest fr e quency c ontaine d in the signal. Upload slide master and use it as presentation template. According to the sampling theorem, we will only be able to reconstruct the true motion of the wheel if the camera is operating at a frame rate of at least $2N+1$ frames per second. physics electronics current theory ppt, ppt of frequency sampling, a seminar on sampling ppt, sampling theorem ppt, Title: The Midpoint Theorem and its Converse Page Link: The Midpoint Theorem and its Converse - Posted By: seminar tips Created at: Wednesday 17th of October 2012 08:19:14 PM. In words, the distance between samples must be smaller than half a wavelength of the highest frequency in the signal. Discrete-Time Processing of Continuous-Time Signals The Model xc(t) x(n) Discrete-Time y(n) yr(t) C/D System D/C. Typically, a frame grabber or digitizer is used to sample and quantize the analogue video signal. Totally customizable. All of them have amazing backgrounds and designs!. An Alternative Proof of the Sampling Theorem The mathematical representation of the sampling process depends upon the periodic impulse train or sampling function g(t) defined under (18). Upload your file and transform it. mX = the mean of X b. Question: A signal x(t)=5cos(6*pi*t)+3sin(8*pi*t) is sampled using sampling frequency of 10 samples per second. How does this relate to the Central Limit Theorem?. standard error) of: 1600/8 = 200 Example continued Convert 24,600 mi. Sampling theory is the field of statistics that is involved. Path Switching and A Network Sampling Theorem Author: Robert Ruan Last modified by: user Created Date: 5/22/2006 7:44:51 AM Document presentation format: On-screen Show (4:3) Company: The Chinese University of Hong Kong Other titles. Download free PowerPoint templates and PowerPoint backgrounds to deliver your next Under this section, you can get immediate access to some of the free PPT templates for PowerPoint available in. Looking for an event management powerpoint that's worth its salt? Check out this round up of the 10 best slideshares for event planners & marketers. • ¯ is a random variable • Repeated sampling and calculation of the resulting statistic will give rise to a dis-. ppt - Moving Line Microsoft Equation 3. it is measured in hertz or sample per second. Just Now By the sampling theorem, if the frequency is , independent information is available once every PPT - Sampling And Reconstruction PowerPoint Presentation. Or in mathematical terms: f s 2 c (1) where f s is the sampling. The central limit theorem: The sampling distribution of the means of all possible samples of size ngenerated from the population using SRR will be approximately normally distributed when ngoes to in nity. They include TurningPoint clicker. 1) where T is the sampling interval (it has units of seconds/sample). Next, the sampling theorem is proved. Merge/Combine PowerPoint PPT files. [7] Luke H. •Sampling theorem gives the criteria for minimum number of samples that should be taken. [Raj, p4] The surveyor's (a person or a establishment in charge of collecting and recording data) or researchers initial task is to formulate a. The frequency is called the bandwidth. 05 zp = za = z. Sampling distributions describe the possible values of a statistic and how often they occur in repeated sampling. : Presenting a PowerPoint slide titled Four Goals SlideShow PPT Example. Freq ≤ 0. The Sampling Theorem Sampling and interpolation take us back and forth between discrete and continuous time and vice versa. htmLecture By: Ms. Save your time with a readily available, completely editable PPT design. Cauchy's theorem (1841). Conclusions. The samples must be independent. • Sampling: take n x n jittered, regular or random samples. • When sampling a signal at discrete intervals, the sampling frequency must be greater than twice the highest frequency of the input signal in order to be able to reconstruct the. Probability sampling uses statistical theory to randomly select a small group of people (sample) from an existing large population and then predict that all their responses will match the overall population. Hence in order to create an image which is digital, we need to covert continuous data into digital form. It sup-ports linear and nonlinear systems, modeled in continuous time, sampled time or hybrid of two. At first glance we need to solve for ˆ throughout an infinite space. 2 Time-Domain Impulse Sampling Consider a train of equally spaced unit impulses. Interpreting Results of Experiments Goal of research is to draw conclusions. 1 Sampling theorem. For a statistician, "large enough" generally means 30 or greater (as a rough rule of thumb) although. Let us discuss the sampling theorem first and then we shall discuss different types of sampling processes. 95) Step 3 - Since 100p > 50, a = 1-p = 0. The question is how to sample a PDF using trajectories, i. This rate is called sampling rate or sampling frequency. [3] «We may remark in passing that it is possible to construct an infinite number of functions cotabular with. Probability and Statistics Multiple Choice Questions & Answers (MCQs) on "Sampling Distribution - 1". 794 817 813. If M&M colors were uniformly distributed (~16. The process by which the continuous-time signal is converted into a discrete-time signal is called Sampling. The Central Limit Theorem states that sums and averages of measurements from a nonnormal population with finite mean m and standard deviation s have approximately normal distributions for large samples of size n. If a presentation contains a. Dalam teorema ini diketahui bahwa untuk pendekatan ke distribusi normal distribusi rata rata sampel tidak memerlukan sampel yang besar. Consider the process of obtaining the second order total differential formula of two variables function. 6 Step Response Time (sec) Amplitude 4x 8x 20x 40x cont Rule-of-thumb: 6-10 samples in closed-loop rise-time Lecture 2 Digital Control.
mke efn qsp xxx eak syp wtk gki ohi uwq riu mew whl dmk tws bre gfx kfa gcj mje | 2022-05-17T14:46:11 | {
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https://mathematica.stackexchange.com/questions/218936/how-do-i-combine-a-function-representing-an-x-value-with-a-function-representing | # How do I combine a function representing an x value with a function representing a y value?
I have two functions, one that generates my x values and one that generates my y values, and I want to merge the data to get a single line plot. This is easily done in excel, but I can't figure out how to do it in mathematica. I did it in excel and it should look like this:. I have written a code that looks like this, where my x and y values are functions of theta
x = 1/72 Cos[θ] (2 + 2 Cos[θ] + 27 Sin[θ])
y = 1/72 Sin[θ] (2 + 2 Cos[θ] + 27 Sin[θ])
I'm pretty sure my issue is just not knowing which plotting function to use, but I could be wrong.
You need to use ParametricPlot:
ParametricPlot[{
1/72 Cos[\[Theta]] (2 + 2 Cos[\[Theta]] + 27 Sin[\[Theta]]),
1/72 Sin[\[Theta]] (2 + 2 Cos[\[Theta]] + 27 Sin[\[Theta]])},
{\[Theta], 0, Pi}]
• u da man, vickto k – orphanpunter69 Apr 6 at 17:00
Clear["Global*"]
ParametricPlotas pointed out by Victor K is the most straightforward approach. You could also Solve for y as a function of x.
sol[x_] = y /. Solve[{
x == 1/72 Cos[θ] (2 + 2 Cos[θ] + 27 Sin[θ]),
y == 1/72 Sin[θ] (2 + 2 Cos[θ] + 27 Sin[θ])},
y, {θ}, Reals];
There are four branches to the solution
Length@sol[x]
(* 4 *)
Plot[Evaluate@sol[x], {x, -0.25, 0.25},
PlotRange -> {{-0.25, 0.25}, {-0.02, 0.42}},
MaxRecursion -> 10,
AspectRatio -> 4/5,
PlotLegends -> Automatic]
The corresponding ParametricPlot(the range of θ is extended to 2 Pi) is
ParametricPlot[{
1/72 Cos[θ] (2 + 2 Cos[θ] + 27 Sin[θ]),
1/72 Sin[θ] (2 + 2 Cos[θ] + 27 Sin[θ])},
{θ, 0, 2 Pi},
PlotRange -> {{-0.25, 0.25}, {-0.02, 0.42}}]
` | 2020-05-29T14:53:41 | {
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https://math.stackexchange.com/questions/2340369/question-on-arithmetic-percentages | # Question on arithmetic (Percentages)
A machine depreciates in value each year at the rate of 10% of its previous value. However every second year there is some maintenance work so that in that particular year, depreciation is only 5% of its previous value. If at the end of fourth year, the value of the machine stands at Rs.146,205, then find the value of the machine at the start of the first year.
I have looked up a few solution in the internet which says depreciation will be 10%-5%-10%-5% in the respective years. I cannot understand why this is the case.
Depreciation:
1st year= 10%
2nd year= 5% of (-10-10+ $\frac{10*10}{100}$ ) by succesive depreciation formula.
I cant uncerstand why this is equal to 5% . This will be equal to 5% only when the term to the right of 'of' is 100.
Where have I gone wrong. Also please show the calculation of the last two years as well.
• Initial value = A. After one year, value = 0.9A. After two years, the value is (0.95)(0.9)A = 0.855A. After third year, value = (0.9)(0.95)(0.9)A, and after 4th year, value = (0.95)(0.9)(0.95)(0.9)A = 0.731025A. If the value after 4 years is RS 146,205, then the initial value was 200,000 (=146,205/0.731025). Jun 29, 2017 at 7:36
You start from a initial value $X_0$.
End first year value $X_1=(1-10\%)X_0$.
End second year value $X_2=(1-5\%)X_1$.
End third year value $X_3=(1-10\%)X_2$.
End fourth year value $X_4=(1-5\%)X_3$, i.e., $$146,205=X_4=(1-5\%)^2(1-10\%)^2X_0.$$
Therefore $$X_0=\frac{146,205}{(1-5\%)^2(1-10\%)^2}.$$
• Why is the 2nd year value decreasing by 5% and why is the third year value decreasing by 10%? Jun 29, 2017 at 7:43
• Because it is written in the text: depreciation is always $10\%$, except at "even" years which is only $5\%$. Jun 29, 2017 at 7:44
• Ok....I thought for the 2nd year depreciation is 5% of 10%. Got wrong in understanding the language of the question. Jun 29, 2017 at 7:47
When the value depreciates $10\%$, the machine is worth $90\%=0.9$ of its old cost.
When the value depreciates $5\%$, the marching is worth $95\%=0.95$ of its old cost.
We have that $P(0.9)(0.95)(0.9)(0.95)=146205$.
The original value was thus $\boxed{200,000}$ Rs | 2022-07-04T03:42:36 | {
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http://www.jerimiannwalker.com/pin-codes-and-the-birthday-problem/ | # Pin codes and the birthday problem
In an introductory statistics class, there are always tons of ways to link back up to the “real world” – its one of the great things about the topic! One of my favorite things to talk about in class is the concept of bank pin codes. How many are possible? Are they really randomly distributed? How many people need to get together before you can be sure two people have the same pin code? So many questions and all of them can be easily explored!
## My movie idea: the pigeonhole principle
Our romantic lead is at a baseball game, rushing between innings to the ATM. On the way, he bumps into a young woman and they drop all the stuff they’re carrying. Apologetic, he helps her up while gathering both of their things and is off to get his cash. It’s only later that he realizes that he accidentally used her ATM card! Not only that – he has some strange charges on his bank statement too! They must have mixed up cards… but is it really possible they share the same code? If so… they are clearly meant to be together…
[Naturally the rest of the movie is her thinking that he stole the card and then eventually them falling in love. At some point, a random person will vaguely explain the pigeonhole principle and give the movie its name – no stealing my idea]
Could this really happen? Yes! In fact, as we will see later, you probably wouldn’t even need the pair to be at a baseball game to make it reasonably likely. Before we get to that though, let’s look at the basic math.
### How many pin codes are possible?
We will work with pin codes are 4 digits long and can be any whole number 0 – 9. In general, the multiplication principle states that if you can break an action into steps, then the number of ways you can complete the action is found by multiplying the number of ways you can complete each step. For pin codes, we can think of selecting each digit as a step. There are 10 choices for each step so:
$10 \times 10 \times 10 \times 10 = 10\,000$
This shows there are 10,000 possible pin codes.
### How many people do you need to get together to guarantee at least two share a pin code?
The pigeonhole principle is a really simple idea that can be used to prove all kinds of things, from the complex to the silly. This rule states that if you have n pigeonholes but n + 1 (or more) pigeons, then at least 2 pigeons are sharing a pigeonhole. Another way to think about it: if a classroom has 32 chairs, but the class has 33 students… well, someone is sitting on someone’s lap (and class just got weird).
For pin codes, this means that you only need 10,001 people together to guarantee that at least two share the same pin code. You can imagine that if you were handing out pin codes, you would run out at the 10,000th person. At that point, you would have to give them the same code as someone else.
There are definitely more than 10,000 people at a baseball game –So the movie idea works! I’m going to be a millionaire! Ok back to math…
## Simulations show you need way fewer than 10,000 people
Everything we have talked about so far is based on the idea that pin codes are randomly selected by people. That is, we are assuming that any pin code has the same chance of being used by a person as any other. For now, we will continue that assumption but as you can imagine, people definitely don’t behave this way.
### The question
If we randomly assign pin codes, how many assignments (on average) will there be before there is a repeated code? We know by the pigeonhole principle that it is guaranteed after 10,000. But what is the typical number? For sure, through randomness, it often takes less than 10,000 right?
### The simulation
For the sake of ease with writing the code, we will assign each pin code a whole number 1 – 10,000. So you can imagine the code 0000 is 1, the code 0001 is 2, and the code 9999 is 10,000. This way, assigning a pin code is really just assigning a random number from 1 to 10,000. In fact, we can now state the problem as:
“How many random selections from {1,2,…,10000} until there is a repeated value?”
Let’s look at the code!
pin_codes=1:10000 #this is the set 1 through 10,000
selected=rep(0,10000) #placeholder for selected pins
count = 0
i = 1
repeat{ #loop that keeps selecting pin codes
pin = sample(pin_codes,1, replace=TRUE) #select the pin
if (pin %in% selected){ #if already selected then stop
break
}
selected[i]=pin #put the selected pin into "selected"
count = count +1 #keep track of how many you selected
i = i+1}
count #type this to see how many were selected before a repeat
If you copy and paste this code into R, you might be surprised. My results the first three times were: 120, 227, 41.
This seems to suggest that through random selection, it only took assigning 120 pin codes before a repeat (in the first trial), 227 (in the second) and the in the last trial – it only took 41 times! This can’t be right?!
### Loop it
Maybe through randomness, we just had some unusual trials. Running 500 or 1000 trials should show the overall trend. The code below is the same (almost) but I wrote a loop around it so that it would repeat the same experiment 500 times. If you try this on your computer note that it is a little slow (I didn’t consider efficiency at all when writing this).
pin_codes=1:10000
counts=rep(0,500)
for (i in 1:500){
selected=rep(0,10000)
count = 0
j = 1
repeat{
pin = sample(pin_codes,1,replace=TRUE)
if (pin %in% selected){
break}
selected[j]=pin
count = count +1
j = j+1}
counts[i]=count}
Type in “mean(counts)” and it will give us the mean number of times that pin codes were randomly assigned before a repeat. The result?
mean(counts)
[1] 124.486
This tells us that on average, it only takes about 124 assignments before you see a repeat [1]. This is waaaay less than 10,000. What is going on?!
# The birthday problem
A famous probability question is known as “the birthday problem“. Suppose that birthdays are equally likely to occur on any given day of the year (and include leap years). This means that there are 366 possible birthdays. By the pigeonhole principle, that means you are guaranteed to have two people share the same birthday as soon as you get 367 people together. But, the probability is almost 100% at only 70 people (and 50% at 23 people). This is another very counter-intuitive result.
### Pin codes and birthdays?
This pin code problem is really the same as the birthday problem, just a little bit bigger or more generalized. Just imagine that there are 10,000 possible birthdays – we are looking at the number of people you need to have two people with the same birthday under this much larger set. Realizing this and researching a bit, you find that this is something that has been studied and in fact, the expected number of selections needed would be:
$1 + \displaystyle\sum\limits_{k=1}^{10,000}\,\dfrac{10,000!}{(10,000-k)!10,000^k}$
Plugging that mess into wolfram alpha gives the result:
$1 + \displaystyle\sum\limits_{k=1}^{10,000}\,\dfrac{10,000!}{(10,000-k)!10,000^k} \approx 1 + 124.932 = 125.932$
Look at that – even over just 500 trials, our simulation was really close. It only takes about 125 to 126 selection (on average) before you see a repeated pin code (assuming they are randomly selected).
126 – thats it!
## But pin codes aren’t random
This is more true than you realize. If you look at the datagenetics article, you will see that instead of making up 1 of 10,000 (or 0.01%), the pin 1234 appears to actually make up more than 10% of codes. Just for fun, I went ahead and changed the code to account for the probability of the top 20. I then assigned all the remaining pin codes the remaining probability evenly – though this isn’t perfect as uncommon pin codes are REALLY uncommon.
pr=rep((1-0.2683)/9980,9980)
pin_codes=1:10000
counts=rep(0,500)
for (i in 1:500){
selected=rep(0,10000)
count = 0
j = 1
repeat{
pin = sample(pin_codes,1,replace=TRUE, prob = c(0.10713,0.06016,0.01881,0.01197,0.00745,0.00616,0.00613,0.00526,0.00516,0.00512,0.00451,0.00419,0.00395,0.00391,0.00366,0.00304,0.00303,0.00293,0.00290,0.00285,pr))
if (pin %in% selected){
break}
selected[j]=pin
count = count +1
j = j+1}
counts[i]=count}
This runs a loop of selecting pins until there is a repeat and then repeats that process 500 times (using these new probabilities for the first 20 codes). Checking the mean after one run I have:
mean(counts)
[1] 12.8
Even crazier! Considering how pin codes are not truly random at all, it looks like you would really only need around 12 to 13 people to have a repeat. Remember – there are 10,000 possibilities in general! ([2] – comment below on small correction made here)
## Summary
Here are the numbers all together:
• By the pigeonhole principle – you are guaranteed that two people share a pin code if the group is larger than 10,000. BUT:
• If pin codes are randomly distributed: ~126 people (on average) are needed before two share a pin code
• Using just a little of the data on the true distribution: Only ~13 people [2] (again, on average) are needed before two share a pin code!
Including more of the data we have on the distribution would probably bring that number down even further. As you can see, it is always very interesting to compare the theory to the reality.
#### Notes
[1] my code counts how many were selected and then stops counting when a repeat is encountered. So, it is really off by 1 from the expected number that you would select to have a repeat. This is a minor technicality overall but worth noting when you see the expected value formula which adds 1 to the sum.
[2] My original code had one of the probabilities as 0.0516 instead of 0.00516 and the mean after several runs was generally around 12. Fixing this probability, the mean seems to be a bit closer to 13 with several runs resulting in means of 12.5 to 12.8. It seems the top codes are really dominating the selection. It would be interesting to code in the details about the less likely pin codes (since they have a very tiny probability of being selected) and seeing if this is actually lower or not.
#### 3 response on “Pin codes and the birthday problem”
1. Small typo in the last bit of code, you have .0516 instead of .00516, this might bump up the odds by a point or two.
• Oh good eye – I will have to check that. I don’t imagine that it’ll affect too much as I think the 10% on 1234 is kind of dominating the selection, but we will see!
• Interesting. Fixed this one and the mean moved a bit though not much. Looking at selected, its those top codes really causing the repeats. Now the next step is to find a nice way to get the super low probabilities coded in for the least likely to be selected codes. That would give an even better idea of the true mean, IMO. | 2019-02-23T00:56:37 | {
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https://math.stackexchange.com/questions/3547645/probability-of-drawing-two-or-more-black-balls | Probability of drawing two or more black balls
What is the probability of drawing with replacement two or more black balls from a hat with 12 balls: four black, four red, and four blue? We are drawing 4 balls from a hat. I have drawn tree diagram. For example, if the first ball drawn is the red one, then we have possible sequences:
Red, Red, Black, Black (probability is $$\frac{1}{3^{4}}$$ )
Red, Blue, Black, Black
Red, Black, Black (probability is $$\frac{1}{3^3}$$ )
Red, Black, Blue, Black
Red, Black, Red, Black
so probability for the case when the first ball is red is $$4 \frac{1}{3^{4}} + \frac{1}{3^{3}}$$
In similar way, I found the probabilities for cases when the first ball is the blue one or black one.
I found that the result is 0.4074. Can someone just check if this is correct? Thanks in advance.
• The number of draws isn't specify in your question. Do you draw a ball exactly four times? In that case you could use binomial distribution. – Alain Remillard Feb 15 '20 at 15:55
• It is going to be much easier on you if you treat blue and red balls as the same category, "not black". This will reduce the case work considerably. – JMoravitz Feb 15 '20 at 15:58
How many draws do you make? From your example I guess 4 but it is not specified in the question.
This could be interpreted as an binomial distribution $$X \sim Bin(n=4, p = 1/3)$$ so what you should calculate is $$P(X \geq 2) = 1- P(X \leq 1) = 1- (P(X = 1) + P(X = 0)) \approx 0.4075$$.
• I have edited the question, 4 balls are drawn. – user121 Feb 15 '20 at 15:59
• $n=12$? No. $n$ would be the number of draws being made, which you just said in the line before you are assuming is $4$. – JMoravitz Feb 15 '20 at 15:59
• My bad, corrected my mistake. – Mevve Feb 15 '20 at 16:02
You can ease your casework two ways. First, consider the blue and red balls to be nonblack with probability $$\frac 23$$. You don't need to distinguish them. Second, all different orders of a given combination have the same probability, so compute the chance of one and multiply by the number of different orders.
You can get two black and two nonblack with chance $${4 \choose 2}\left( \frac 13\right)^2\left( \frac 23\right)^2=\frac {24}{81}$$
You can get three black and one nonblack with chance $${4 \choose 3}\left( \frac 13\right)^3\left( \frac 23\right)^1=\frac {8}{81}$$
You can get four black and no nonblack with chance $${4 \choose 4}\left( \frac 13\right)^4\left( \frac 23\right)^0=\frac {1}{81}$$
For a total of $$\frac {33}{81}.\ \$$ $$0.4074$$ is approximately correct. I would leave it as a fraction unless you are asked for a decimal.
• @N.F.Taussig: Thanks. Fixed. – Ross Millikan Feb 15 '20 at 16:32 | 2021-06-20T13:54:19 | {
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https://math.stackexchange.com/questions/3510279/minimum-number-of-subintervals-n-for-the-composite-trapezoidal-rule | # Minimum number of subintervals n for the Composite Trapezoidal Rule
I am trying to compute the minimum number of subintervals n for the Composite Trapezoidal Rule, in order for the approximation of the following integral to have 5 decimals correct. $$\int_0^2 \frac{1}{x+4}dx$$ So I am using: $$|E_n^T(f)|\leq\frac{(b-a)h^2}{12}M_2<ε$$ where $$\mathbf{M_2}=\max_{x\in [a,b]} |f''(x)|=\frac{1}{32}$$, $$\mathbf{h}=\frac{b-a}{n}$$, $$\mathbf{ε}=\frac{1}{2}\cdot10^{-5}$$. Solving for $$n$$ I get $$n>64.5497...$$, so I say that the minimum number of subintervals to achieve 5 decimals is 65. However, using matlab (or an online tool) I see that the Trapezoidal Rule needs only $$n=48$$ subintervals to achieve this accuracy. What am I doing wrong? Is something wrong with my calculations or is it possible that the number of subintervals needed in practice can actually be less than 65?
If you look at the formula $$\frac2{(x+4)^3}$$ for the second derivative, you can see that at $$x=2$$ it will have a somewhat smaller value than at $$x=0$$. From this lower bound of the derivative you can compute a lower bound for $$n$$, the truly minimal $$n$$ will be somewhere in-between.
You could also use the more exact error formula $$\int_a^b f(x)dx=T(h)-\frac{h^2}{12}(f'(b)-f'(a))+\frac{h^4(b-a)}{720}f^{(4)}(\eta)$$ to get from the term of second degree in $$h$$ the estimate $$n^2\approx\frac{(b-a)^2}{12}\frac{6^2-4^2}{6^2\cdot 4^2}\cdot 2\cdot10^5=\frac{10^6}{12\cdot 36} \implies n\approx 48.1125,$$ giving $$n=49$$ as the best number of segments. The fourth-order correction should not change this value. | 2022-01-18T05:28:03 | {
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https://math.stackexchange.com/questions/1542248/generalized-demorgans-law-proof | # Generalized DeMorgan's Law proof
We wish to verify the generalized law of DeMorgan $(\bigcup_{i \in \mathcal{I}} A_i)^c = \bigcap_{i \in \mathcal{I}} A_i^c$.
Let $x \in (\bigcup_{i \in \mathcal{I}} A_i)^c$. Then $x \notin \bigcup_{i \in \mathcal{I}} A_i$ and $x \notin A_i$ for $i \in \mathcal{I}$, and so $x \in A_i^c$ for all $i$. Hence $x \in \bigcap_{i \in \mathcal{I}} A_i^c$. We have shown that $(\bigcup_{i \in \mathcal{I}} A_i)^c \subset \bigcap_{i \in \mathcal{I}} A_i^c$.
We must now show that $\bigcap_{i \in \mathcal{I}} A_i^c \subset (\bigcup_{i \in \mathcal{I}} A_i)^c$. Now let $x \in \bigcap_{i \in \mathcal{I}} A_i^c.$ Then $x \in A_i^c$ for all $i \in \mathcal{I}$ and so $x \notin A_i$ for all $i \in \mathcal{I}$. Hence $x \in A_i^c$ for all $i \in \mathcal{I}$ and so $x \in (\bigcup_{i \in \mathcal{I}} A_i)^c$.
Then $\bigcap_{i \in \mathcal{I}} A_i^c \subset (\bigcup_{i \in \mathcal{I}} A_i)^c$, and since $(\bigcup_{i \in \mathcal{I}} A_i)^c \subset \bigcap_{i \in \mathcal{I}} A_i^c$ we have that $(\bigcup_{i \in \mathcal{I}} A_i)^c = \bigcap_{i \in \mathcal{I}} A_i^c$, which is what we set out to show.
I just want to make sure that my proof makes sense and I was hoping for some constructive criticism regarding proof style/format -- would really appreciate any feedback whatsoever.
You should not write
for $i= 1,2, \dots$
since you do not know that $\mathcal{I}$ is countable (or even consists of integers - perhaps $\mathcal{I}=\mathbb{Q}$ or $\mathcal{I}=\mathbb{R}^2$ or some set that doesn't even contain numbers).
In the reverse direction, you jump from $y \notin A_i$ for all $i \in \mathcal{I}$ to $y \in (\bigcup_{i \in \mathcal{I}} A_i)^c$. That jump is probably fine in most proofs, though you may want to stick closer to the definitions like you did in the forward direction if this is for an intro to proofs course. I would recommend adding an intermediate step there.
• Thanks a lot for the feedback. It's not for any class, I'm just trying to learn proof on my own or at least work on my proofs because I haven't gotten much feedback/critique before. I see what you mean re I being countable and adding another intermediate step in the forward direction. Is it okay to say instead "for all $i \in I$" then? – Hugo Nov 23 '15 at 9:07
• Is it necessary to switch from x to y when going from one direction to the next as I did? I assumed it made sense to do so because it hadn't been established yet the two sets were equal. – Hugo Nov 23 '15 at 9:14
• Yes, "for all $i \in I$" is probably the best way to phrase it. And no, it is not necessary to switch from $x$ to $y$. I think it's understood that when you switch directions and say "now let $x \in$..." this is a different $x$ from the $x$ you chose in the first direction. I hope that makes sense. – kccu Nov 24 '15 at 0:06
• The fact that $\mathcal{I}$ may not be a countable set prevents me from using a proof by induction here -- is that correct? I know the standard way to show equality between sets is to show that the two sets contain one another, but I was thinking to mess around with a proof by induction, but that seems to not be possible if $\mathcal{I}$ is uncountable. – Hugo Nov 24 '15 at 9:10
• It's not even possible if $I$ is countable. The only sort of result you could show by induction is $\left(\bigcup_{i=1}^n A_i\right)^c = \bigcap_{i=1}^n A_i^c$ for all $n \in \mathbb{N}$. But this only holds for finite (though arbitrarily large) $n$. A proof by induction would not tell you that $\left(\bigcup_{i=1}^\infty A_i \right) ^c = \bigcap_{i=1}^\infty A_i^c$. – kccu Nov 24 '15 at 17:37 | 2019-10-20T22:15:26 | {
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https://byjus.com/question-answer/in-the-given-figure-ab-6-and-bc-4-bc-is-extended-to-e-such-1/ | Question
# In the given figure, AB = 6 and BC = 4. BC is extended to E such that C is the midpoint of BE and join AE. It cuts CD in P. Now consider the two points: i) P is the midpoint of AE ii) P is the midpoint of CD
A
Both (i) and (ii) are true
B
Only (i) is true
C
Only (ii) is true
D
Neither (i) nor (ii) is true
Solution
## The correct option is B Both (i) and (ii) are true Consider △AEB and (\triangle PEC\) . Since PC is part of DC and DC is parallel to AB. ∠EAB=∠EPC [corresponding angles] ∠AEB=∠PEC [common angle] By AA similarity criterion,△AEB ~ △PEC So, APPE=BCCE --------------------------(I) Since C is the mid point of BE, BC=CE. -------------------------(II) From (II) and (I), APPE = BCCE = 1, So AP = PE and hence P is the mid point of AE. Now in right triangle △ABE , AB = 6, BE = 2BC = 8. So by Pythagoras theorem, ⇒AE2=AB2+BE2 ⇒AE2=36+64=100 ⇒AE=10 Hence PE = AE2 = 102 = 5. In right triangle PCE, CE = 4 = BC, PE = 5, by Pythagoras theorem, PE2 = PC2 + CE2 ⇒ PC2=25−16=9 ⇒PC=3 . Since AB = CD (rectangle), CD = 6 and hence DP=CD−PC=6−3=3. That is CD = PC. So P is the midpoint of BC also. Mathematics
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https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_19&oldid=33185 | # 2009 AMC 12B Problems/Problem 19
## Problem
For each positive integer $n$, let $f(n) = n^4 - 360n^2 + 400$. What is the sum of all values of $f(n)$ that are prime numbers?
$\textbf{(A)}\ 794\qquad \textbf{(B)}\ 796\qquad \textbf{(C)}\ 798\qquad \textbf{(D)}\ 800\qquad \textbf{(E)}\ 802$
## Solution
### Solution 1
To find the answer it was enough to play around with $f$. One can easily find that $f(1)=41$ is a prime, then $f$ becomes negative for $n$ between $2$ and $18$, and then $f(19)=761$ is again a prime number. And as $41 + 761 = 802$ is already the largest option, the answer must be $\boxed{802}$.
### Solution 2
We will now show a complete solution, with a proof that no other values are prime.
Consider the function $g(x) = x^2 - 360x + 400$, then obviously $f(x) = g(x^2)$.
The roots of $g$ are: $$x_{1,2} = \frac{ 360 \pm \sqrt{ 360^2 - 4\cdot 400 } }2 = 180 \pm 80 \sqrt 5$$
We can then write $g(x) = (x - 180 - 80\sqrt 5)(x - 180 - 80\sqrt 5)$, and thus $f(x) = (x^2 - 180 - 80\sqrt 5)(x^2 - 180 - 80\sqrt 5)$.
We would now like to factor the right hand side further, using the formula $(x^2 - y^2) = (x-y)(x+y)$. To do this, we need to express both constants as squares of some other constants. Luckily, we have a pretty good idea how they look like.
We are looking for rational $a$ and $b$ such that $(a+b\sqrt 5)^2 = 180 + 80\sqrt 5$. Expanding the left hand side and comparing coefficients, we get $ab=40$ and $a^2 + 5b^2 = 180$. We can easily guess (or compute) the solution $a=10$, $b=4$.
Hence $180 + 80\sqrt 5 = (10 + 4\sqrt 5)^2$, and we can easily verify that also $180 - 80\sqrt 5 = (10 - 4\sqrt 5)^2$.
We now know the complete factorization of $f(x)$:
$$f(x) = (x - 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5)(x - 10 + 4\sqrt 5)(x + 10 - 4\sqrt 5)$$
As the final step, we can now combine the factors in a different way, in order to get rid of the square roots.
We have $(x - 10 - 4\sqrt 5)(x - 10 + 4\sqrt 5) = (x-10)^2 - (4\sqrt 5)^2 = x^2 - 20x + 20$, and $(x + 10 - 4\sqrt 5)(x + 10 + 4\sqrt 5) = x^2 + 20x + 20$.
Hence we obtain the factorization $f(x) = (x^2 - 20x + 20)(x^2 + 20x + 20)$.
For $x\geq 20$ both terms are positive and larger than one, hence $f(x)$ is not prime. For $1 the second factor is positive and the first one is negative, hence $f(x)$ is not a prime. The remaining cases are $x=1$ and $x=19$. In both cases, $f(x)$ is indeed a prime, and their sum is $f(1) + f(19) = 41 + 761 = \boxed{802}$.
### Solution 3
Instead of doing the hard work, we can try to guess the factorization. One good approach:
We can make the observation that $f(x)$ looks similar to $(x^2 + 20)^2$ with the exception of the $x^2$ term. In fact, we have $(x^2 + 20)^2 = x^4 + 40x^2 + 400$. But then we notice that it differs from the desired expression by a square: $f(x) = (x^2 + 20)^2 - 400x^2 = (x^2 + 20)^2 - (20x)^2$.
Now we can use the formula $(x^2 - y^2) = (x-y)(x+y)$ to obtain the same factorization as in the previous solution, without all the work. | 2021-04-14T23:01:18 | {
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https://plainmath.net/discrete-math/81500-simple-question-about-number-theory-in-c | Frank Day
2022-07-09
Simple question about number theory in CLRS book
This is theorem 11.5 from CLRS book. Suppose $a\in {\mathbb{Z}}_{p}^{\ast },b\in {\mathbb{Z}}_{p}$.
Consider two distinct keys k and l from ${\mathbb{Z}}_{p}$, so that $k\ne l$. For a given hash function ${h}_{ab}$ we let
$r=ak+b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$
$s=al+b\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}p$
We first note that $r\ne s$. Why? Observe that
$r-s\equiv a\left(k-l\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}p\right).$
I'm not very well familiar with Number Theory, so my question is why $r-s\equiv a\left(k-l\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}p\right)$ is correct?
Dobermann82
Expert
Step 1
According to Knuth's definition for the mod operation, we have $\begin{array}{r}amodn=a-n⌊\frac{a}{n}⌋\end{array}$
And with the general definition of Congruence, $a\equiv b\phantom{\rule{0.444em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}n\right)$ can be rewritten as $\begin{array}{r}a-b=kn\end{array}$.
We then have $\begin{array}{rl}r& =ak+bmodp=ak+b-p⌊\frac{ak+b}{p}⌋\\ s& =al+bmodp=al+b-p⌊\frac{al+b}{p}⌋\end{array}$
Subtract r by s, we have
$\begin{array}{}\text{(A)}& r-s=a\left(k-l\right)-p\left(⌊\frac{ak+b}{p}⌋+⌊\frac{al+b}{p}⌋\right)\end{array}$
With the definition of congruence quoted above, we can then rewrite (A) as $\begin{array}{r}r-s\equiv a\left(k-l\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}p\right)\end{array}$.
daktielti
Expert
Explanation:
I’m turning my comment into an answer. If $r=ak+b$ mod p, and $s=al+b$ mod p, then $r-s=\left(ak+b\right)-\left(al+b\right)$ mod p, thus $r-s=ak-al$ mod p, hence $r-s=a\left(k-l\right)$ mod p.
Do you have a similar question? | 2023-02-06T16:45:58 | {
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https://math.stackexchange.com/questions/1728372/do-the-level-sets-of-a-constant-rank-map-give-a-foliation-of-the-domain | Do the level sets of a constant rank map give a foliation of the domain?
I'm working through Smooth Manifolds by Lee and came to problem 19-8, where you're asked to prove that the level sets of a submersion form a foliation of the domain. However when I try to solve this problem it seems like the assumption that the map is a submersion is not needed, ie. one need only assume the map is of constant rank. Lee's problems are usually carefully written to use minimal hypotheses, so I'm not sure if I'm wrong.
Let $$F: M \rightarrow N$$ be a map of constant rank $$r$$, with $$m = dim(M)$$ and $$n = dim(N)$$. The non-empty level sets are a collection $$\mathcal{F}$$ of disjoint immersed submanifolds of codimension $$r$$ whose union is $$M$$. To show that $$\mathcal{F}$$ is a foliation, I have to show that each $$p \in M$$ is contained in a flat chart for $$\mathcal{F}$$:
A smooth chart $$(U,\phi )$$ for $$M$$ is said to be flat for $$\mathcal{F}$$ if $$\phi (U)$$ is a cube in $$\mathbb{R}^n$$, and each submanifold in $$\mathcal{F}$$ intersects $$U$$ in either the empty set or a countable union of k-dimensional slices
The definition of a k-slice is
If $$U$$ is an open subset of $$\mathbb{R}^n$$ and $$k \in \{0,...,n\}$$, a k-dimensional slice of U (or simply a k-slice) is any subset of the form
$$S =\{ {(x^1,...,x^k,x^{k+1},...x^n) \in U: x^{k+1} = c^{k+1}, x^n = c^n } \}$$
for some constants $$c^{k+1},...,c^n$$.
By the rank theorem, for each $$p \in M$$ there is a coordinate representation of $$F$$ in a neighborhood $$U$$ of $$p$$ with the form $$F(x^1,...,x^r,x^{r+1},...,x^m) = (x^1,...,x^r,0,...,0)$$. We can assume $$U$$ is a cube, shrinking it if necessary.
Every submanifold in $$\mathcal{F}$$ is of the form $$F^{-1}(q), q \in N$$. There are two possibilities.
• If $$q \notin F(U)$$, then $$F^{-1}(q) \cap U$$ is empty.
• If $$q \in F(U)$$, then $$q = F(a^1,...,a^m)$$ for $$(a^1,...,a^m) \in U$$, and by the coordinate representation of $$F$$, we have that $$F^{-1}(q)$$ is the (m-r)-slice $$\{ {(x^1,...,x^r,x^{r+1},...x^n) \in U : x^1=a^1,...,x^r=a^r} \}$$
Thus I've shown that each submanifold in $$\mathcal{F}$$ intersects $$U$$ in a single (m-r)-slice, so $$U$$ is flat for $$\mathcal{F}$$. Is this correct?
• Yes, it's correct. As a challenge, come up with an interesting foliation that comes from a map of constant rank but not from a submersion. If you can do so I'll bounty you 200 points.
– user98602
Apr 5, 2016 at 2:35
• @MikeMiller, locally I can reduce to the case of a submersion. You have some global scenario in mind? Apr 5, 2016 at 3:13
• @Ted Yup, such a thing would make me care about constant rank foliations.
– user98602
Apr 5, 2016 at 3:17
• Mobius bandy thing?! Mar 20, 2018 at 3:49 | 2022-10-02T10:13:25 | {
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https://mathhelpboards.com/threads/integration-involving-substitutions.5571/ | # Integration involving substitutions
#### paulmdrdo
##### Active member
1. \begin{align*}\displaystyle \int\frac{dx}{(1+x){\sqrt{x}}}\end{align*}
2. \begin{align*}\displaystyle \int\frac{ds}{\sqrt{2s-s^2}}\end{align*}
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Re: Integration Inverse trig
1. \begin{align*}\displaystyle \int\frac{dx}{(1+x){\sqrt{x}}}\end{align*}
rewrite as
$$\displaystyle \int\frac{2dx}{(1+(\sqrt{x})^2){2\sqrt{x}}}$$
2. \begin{align*}\displaystyle \int\frac{ds}{\sqrt{2s-s^2}}\end{align*}
complete the square .
#### paulmdrdo
##### Active member
Re: Integration Inverse trig
\begin{align*}\displaystyle \int\frac{dx}{\sqrt{2s-s^2}} = \int\frac{ds}{\sqrt{-[(s^2-2s +1)-1]}} \\ = \int\frac{ds}{\sqrt{-[(s-1)^2-1]}} \\ = \int\frac{ds}{\sqrt{1-(s-1)^2}}\\ let \,u = s-1\\ a=1 \\\ = sin^{-1}\, (s-1) +C \end{align*}
is my answer correct? did i use correct algebra in completing the square?
i still don't know how did you get 2 to be in the numerator and denominator of prob one.please explain. thanks!
Last edited:
#### topsquark
##### Well-known member
MHB Math Helper
Re: Integration Inverse trig
$$\displaystyle \int\frac{2dx}{(1+(\sqrt{x})^2){2\sqrt{x}}}$$
Ignore the 2's and the extra $$\sqrt{x}$$ in the denominator and think strategically for a moment. Look at the rest of the denominator....The $$(1 + (\sqrt{x})^2)$$. What substitution do you think you are likely to use? Then make the substitution and solve for du. What terms arise?
-Dan
#### paulmdrdo
##### Active member
Re: Integration Inverse trig
\begin{align*}\displaystyle let\, u = x^{\frac{1}{2}}\\ du = \frac{1}{2}x^{-\frac{1}{2}}dx \\ dx = 2\sqrt{x}\\... my\,\, answer\,\, would\,\, be\, = 2tan^{-1}\,\sqrt{x}+C\end{align*}
but i still don't get the new form of the integrand. where the 2 came from and
\begin{align*}\displaystyle(\sqrt{x})^2 \end{align*} in the denominator.
#### topsquark
##### Well-known member
MHB Math Helper
Re: Integration Inverse trig
$$\displaystyle u = x^{\frac{1}{2}}$$
$$\displaystyle du = \frac{1}{2}x^{-\frac{1}{2}}dx$$
Look at the du equation. We need a 2 in the denominator to make a du. If we need one in the bottom, then we also need one in the top.
Perhaps the better idea right now is, now that you know the substitution you want, plug your u and du into the original integral and see what happens.
-Dan
#### Prove It
##### Well-known member
MHB Math Helper
Re: Integration Inverse trig
Do you understand how substitution works? You need to see if there is an "inner" function and if this inner function's derivative is a factor in your integrand. Surely you can see that \displaystyle \displaystyle \begin{align*} x = \left( \sqrt{x} \right) ^2 \end{align*}. Why did we choose to do that? Because if you know your derivatives, you will know that \displaystyle \displaystyle \begin{align*} \frac{d}{dx} \left( \sqrt{x} \right) = \frac{1}{2\sqrt{x}} \end{align*}. Notice that you ALREADY have \displaystyle \displaystyle \begin{align*} \frac{1}{\sqrt{x}} \end{align*} in your denominator, which means that if you can turn it into \displaystyle \displaystyle \begin{align*} \frac{1}{2\sqrt{x}} \end{align*}, then a substitution of the form \displaystyle \displaystyle \begin{align*} u = \sqrt{x} \end{align*} is appropriate.
#### soroban
##### Well-known member
Hello, paulmdrdo!
You can avoid that hassle . . .
$$1.\;\int\frac{dx}{(1+x)\sqrt{x}}$$
Let $$u \,=\,\sqrt{x} \quad\Rightarrow\quad x \,=\,u^2 \quad\Rightarrow\quad dx \,=\,2u\,du$$
Substitute: .$$\int \frac{2u\,du}{(1+u^2)u} \;=\;2\int\frac{du}{1+u^2} \;=\;2\arctan u + C$$
Back-substitute: .$$2\arctan(\sqrt{x}) + C$$ | 2020-09-20T23:46:55 | {
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http://tuningfly.it/saiv/qda-decision-boundary.html | The model at the right joins these smaller blobs together into a larger blob where the model classifies data as responders. Find books. The Bayes decision boundary is a hypersphere determined by QDA, the covariance structure is the same as in the high-curvature nonlinear model, the decision boundary has high curvature in comparison to the low-curvature nonlinear model, and this model is more difficult than the high-curvature nonlinear model. The decision boundary in QDA is non-linear. If the actual boundaries are indeed linear, QDA may have a higher model bias. The percentage of the data in the area where the two decision boundaries differ a lot is small. In order to use LDA or QDA, we need: An estimate of the class probabilities ˇ j. 0001, store_covariances=None) [source] Quadratic Discriminant Analysis. BAYESIAN DECISION THEORY assume that any incorrect classiflcation entails the same cost or consequence, and that the only information we are allowed to use is the value of the prior probabilities. Recall that the optimization problem has the following form: argmin w Xn i=1 (w>x i 2y i) + kwk2 2 2. Solution: D. However, I am applying the same technique for a 2 class, 2 feature QDA and am having trouble. To perfectly solve this problem, a very complicated decision boundary is required. If the Bayes decision boundary is linear, we expect QDA to perform better on the training set because it's higher flexiblity will yield a closer fit. Lecture9: Classification,LDA Reading: Chapter 4 STATS 202: Data mining and analysis Jonathan Taylor, 10/12 Slide credits: Sergio Bacallado 1/21. For quadratic discriminant analysis, there is nothing much that is different from the linear discriminant analysis in terms of code. Note: when the number of covariates grow, the number of things to estimate in the covariance matrix gets very large. The point of this example is to illustrate the nature of decision boundaries. In this article we will study another very important dimensionality reduction technique: linear discriminant analysis (or LDA). The optimal Bayes decision boundary for the simulation example of Figures 2. Right: Details are as given in the left-hand panel, except that Σ 1 6= Σ 2. (b) If the Bayes decision boundary is non-linear, do we expect LDA or QDA to perform better on the training set? On the test set? (c) In general, as the sample size n increases, do we expect the test prediction accuracy of QDA relative to LDA to improve, decline, or be unchanged?. QDA can perform better in the presence of a limited number of training observations because it does make some assumptions about the form of the decision boundary. Since the decision boundary is not known in advance an iterative procedure is required. As we increase dimensions, we exponentially reduce the number of observations near the point in consideration. BINARY RESPONSE AND LOGISTIC REGRESSION ANALYSIS ntur <- nmale+nfemale pmale <- nmale/ntur #-----# # fit logistic regression model using the proportion male as the # response and the number of turtles as the weights in glm. How do we get there with our sigmoid function? 2. Figure 3: Example sets that contain mass τ of the conditional p. What is important to keep in mind is that no one method will dominate the oth- ers in every situation. The question was already asked and answered for LDA, and the solution provided by amoeba to compute this using the "standard Gaussian way" worked well. We will prove this claim using binary (2-class) examples for simplicity (class Aand class B). MLP and SVM are classifiers that work directly on the decision boundary and fall under the discriminative paradigm. 1 Introduction. datasets import make_blobs from sklearn. in4085 pattern recognition written examination 3-02—20 17, 9:00—12:00 there are questions you have 45 minutes to answer the first question (answer sheets and. I decision trees I SVM I multilayer perceptron (MLP) Examples of generative classifiers: I naive Bayes (NB) I linear discriminant analysis (LDA) I quadratic discriminant analysis (QDA) We will study all of the above except MLP. As we increase dimensions, we exponentially reduce the number of observations near the point in consideration. Nonlinear SVM. Previous neuroimaging studies reveal that MCI is associated with aberrant sensory–perceptual processing in cortical brain regions subserving auditory and language function. 3 for the same data. Since the decision boundary of logistic regression is a linear (you know why right?) and the dimension of the feature space is 2 (Age and EstimatedSalary), the decision boundary in this 2-dimensional space is a line that separates the predicted classes "0" and "1" (values of the response Purchased). Quadratic Decision Boundary for QDA Helloi would like to plot a quadratic decision boundary for a Quadratic Discriminant Analysis method i implemented, but i can't undertand how. (가 들어간 로지스틱함수에서 반응변수를 뽑음) 이땐 예상대로 QDA가 제일 잘했고 그다음이 KNN-CV였다. Chapter 4 Solutions for Classification Text book: An Introduction to Statistical Learning with Applications in R b) If the Bayes decision boundary is non-linear, do we expect LDA or QDA to perform better on the training set?. Linear Discriminant Analysis LDA on Expanded Basis I Expand input space to include X 1X 2, X2 1, and X 2 2. fit with lda and qda from the MASS package. Classification [email protected] may have 1 or 2 points. As we can see from the figure, the K-nearest neighbors decision boundary is quite jagged. 92% is a pretty good number!. The actual decision rule doesn’t take in con-sideration X 3, as we know that X 1 and X 2 are truly independent given Y and they are enough to predict Y. In our previous article Implementing PCA in Python with Scikit-Learn, we studied how we can reduce dimensionality of the feature set using PCA. Plot the test data with radius as the $$x$$ axis, and symmetry as the $$y$$ axis, with the points colored according to their tumor. A function for plotting decision regions of classifiers in 1 or 2 dimensions. Python was created out of the slime and mud left after the great flood. If the Bayes decision boundary is non-linear, do we expect LDA or QDA to perform better on the training set? On the test set? In general, as the sample size n increases, do we expect the test prediction accuracy of QDA relative to LDA to improve, decline, or be unchanged? Why?. [QDA works very nicely with more than 2 classes. We saw the same spirit on the test we designed to assess people on Logistic Regression. Basically, what you see is a machine learning model in action, learning how to distinguish data of two classes, say cats and dogs, using some X and Y variables. 5Σ−1 (µT 1 µ1 −µT 2 µ2)+ln P(ω1) P(ω2) = 0 (15) The decision boundary xDB is xDB = µ1 + µ2 2 + Σ µ2 − µ1 ln P(ω1) P(ω2) (16) When the two classes are equiprobable, then the second term will be neglected and the decision boundary is the point in the middle of the class. Helwig (U of Minnesota) Discrimination and Classification Updated 14-Mar-2017 : Slide 1. In a certain way(too mathematically convoluted for this post), LDA does learn a linear boundary between the points belonging to different classes. Since QDA assumes a quadratic decision boundary, it can accurately model a wider range of problems than can the linear methods. The decision boundary is computed by setting the above discriminant functions equal to each other. On the left is the decision boundary from a linear model built using linear discriminant analysis (like LDA or the Fisher Discriminant) and on the right, a decision boundary built by a model using quadratic discriminant analysis (like the Bayes Rule). Mahalanobis distance term. Finally, the decision boundary obtained by LDA is equivalent to the binary SVM on the set of support vectors 31. I will be using the confusion martrix from the Scikit-Learn library (sklearn. 5 \] This is equivalent to point that satisfy. A QDA classifier might be sufficient (see scikit-learn. 0001, store_covariances=None) [source] Quadratic Discriminant Analysis. Plot the confidence ellipsoids of each class and decision boundary. Neural Quadratic Discriminant Analysis 2295 resulting in curved decision boundaries in the population response space (see Figure 1a). If the covariance matrices for two distributions are equal and proportional to the identity matrix, then the distributions are spherical in d dimensions. score - 10 examples found. (d) True or False: Even if the Bayes decision boundary for a given problem is linear, we will probably achieve a superior test er-ror rate using QDA rather than LDA because QDA is flexible enough to model a linear decision boundary. We have seen that in p = 2 dimensions, a linear decision boundary takes the form β 0 +β 1 X 1 +β 2 X 2 = 0. DATA11002 Introduction to Machine Learning Lecturer: Antti Ukkonen I Decision boundary is given by N(x j +; +) QDA I In QDA, decision regions may be non-connected. –On L4 we showed that the decision rule that minimized 𝑃[ 𝑟𝑟 𝑟] could be formulated in terms of a family of discriminant functions • For normally Gaussian classes, these DFs reduce to simple expressions –The multivariate Normal pdf is 𝑋 =2𝜋−𝑁/2Σ−1/2 − 1 2 𝑥−𝜇𝑇Σ−1𝑥−𝜇. Although the decision boundaries between classes can be derived analytically, plotting them for more than two classes gets a bit complicated. Since QDA is specifically applicable in cases with very different covariance structures, it will often be a feature of this plot that one group is spread out while the other is extremely concentrated, typically close to the decision boundary. For the line, we got pretty good results. Since the decision boundary of logistic regression is a linear (you know why right?) and the dimension of the feature space is 2 (Age and EstimatedSalary), the decision boundary in this 2-dimensional space is a line that separates the predicted classes "0" and "1" (values of the response Purchased). colors import ListedColormap from sklearn. The RBF SVM has very nice decision boundary. The question was already asked and answered for linear discriminant analysis (LDA), and the solution provided by amoeba to compute this using the "standard Gaussian way" worked well. When the population responses are gaussian distributed the maximum likelihood solution (known as quadratic discriminant analysis, QDA; Kendall 1966) corresponds to (2. And, because of this assumption, LDA and QDA can only be used when all explanotary variables are numeric. If the Bayes decision boundary is nonlinear, do we expect LDA or QDA to perform better on the training set? On the test set? The Attempt at a Solution 1. Decision boundaries are most easily visualized whenever we have continuous features, most especially when we have two continuous features, because then the decision boundary will exist in a plane. by Pierre Paquay. For quadratic discriminant analysis, there is nothing much that is different from the linear discriminant analysis in terms of code. QDA – sometimes may be usefull, k-NN can capure any non-linear decision boundary. 92% is a pretty good number!. Right: Details are as given in the left-hand panel, except thatΣ 1 ≠Σ 2. In contrast with this, quadratic discriminant analysis (QDA). exact decision boundary may not do as well in the test set. Quadratic Discriminant Analysis for Binary Classification In Quadratic Discriminant Analysis (QDA), we relax the assumption of equality of the covariance matrices: 1 6= 2; (24) which means the covariances are not necessarily equal (if they are actually equal, the decision boundary will be linear and QDA reduces to LDA). What is important to keep in mind is that no one method will dominate the oth- ers in every situation. We now examine the differences between LDA and QDA. The RBF SVM has very nice decision boundary. The shading indicates the QDA decision rule. Of course SVM bypasses it via kernel trick but still not as much complex decision boundary as nueral nets; Despite the risk of non linearity in data linear algorithms tends to work well in practice and are often used as starting point. ,σ 1 = σ 2 = σ),inwhichcaseweobtain x= µ 1 +µ 2 2 + σ2 log(π 1/π 2) µ 2 −µ 1. Since QDA assumes a quadratic decision boundary, it can accurately model a wider range of problems than can the linear methods. Mahalanobis distance term. The dashed line in the plot below is a decision boundary given by LDA. Assume that the input data in population (stochastic) version include a G-dimensional random vector X~ = (x 1; ;x G) as covariates (e. proven inconsistent in [8]. Discriminant function: a function fit directly to the feature values without estimating probability distributions; can be thresholded to obtain a decision boundary. If it is not, then repeat (a)-(e) until you come up with an example in which the predicted class labels are obviously non-linear. Quadratic Discriminant Analysis (QDA) relaxes the common covariance assumption of LDA through estimating a separate covariance matrix for each class. preprocessing. The quadratic discriminant analysis algorithm yields the best classification rate. Dy = {x ∈Rn, f(x) = y}= f−1(y). Maximum-likelihood and Bayesian parameter estimation techniques assume that the forms for the underlying probability densities were known, and that we will use the training samples to estimate the values of their parameters. Continue reading Classification from scratch, linear discrimination 8/8 → Eighth post of our series on classification from scratch. It is smooth, matches the pattern and is able to adjust to all three examles. class sklearn. decision boundary. Custom legend labels can be provided by returning the axis object (s) from the plot_decision_region function and then getting the handles and labels of the legend. Equation 3 shows that δ k is a linear function of x, which stems from the assumed linear decision boundaries between the classes. These visuals can be great to understand…. There is not a single algorithm for training such classifiers, but a family of algorithms based on a common principle: all naive Bayes classifiers assume that the. A non-linear Bayes decision boundary is indicative that a common covariance matrix is inaccurate, which would tell us that the QDA would be expected to perform better. We introduce a notion of decision boundary instability (DBI) to assess the stability (Breiman (1996)) of a classi cation procedure arising from the randomness of training sam-ples. When referring to, for example, a model developed from 1996 data with a 2 year default horizon, we mean a model developed from the data set D96, where a response variable is defined as 1 if the year of bankruptcy is 1997 or 1998 and 0 otherwise. The decision boundaries for the support vector machines and for logistic regression look much more smooth. In a certain way(too mathematically convoluted for this post), LDA does learn a linear boundary between the points belonging to different classes. The expression inside the exponent is the Mahalanobis distance. score extracted from open source projects. We now investigate a non-linear decision boundary - 1742820. One needs to be careful. seed(123) x1 = mvrnorm(50, mu = c(0, 0), Sigma = matrix(c(1, 0, 0, 3), 2)) x2 = mvrnorm(50, mu = c(3, 3), Sigma = matrix(c(1, 0, 0, 3), 2)) x3 = mvrnorm(50, mu = c(1, 6), Sigma. Decision boundaries are most easily visualized whenever we have continuous features, most especially when we have two continuous features, because then the decision boundary will exist in a plane. Heteroscedasticity. 169 ISLR) This question examines the differences between LDA and QDA. This is therefore called quadratic discriminant analysis (QDA). How do we get there with our sigmoid function? 2. Discriminant analysis¶. Trending AI Articles: 1. Is it just. Researchers, practitioners, and policymakers have nowadays access to huge datasets (the so-called “Big Data”) on people, companies and institutions, web and mobile devices, satellites, etc. QuadraticDiscriminantAnalysis. Nonlinear SVM. methods: (1) Quadratic discriminant analysis (QDA) assumes that the feature values for each class are normally distributed. Decision boundary is a function of a quadratic combination of known observations (Σk is covariance matrix) Assumptions: Normally distributed observations. An estimate of the mean vectors j. So far, we've only seen the case where the two classes occur about equally often. 17 *QuadraticDiscriminantAnalysis* Read more in the :ref:User Guide . The decision boundary is here But that can’t be the linear discriminant analysis, right? I mean, the frontier is not linear… Actually, in Fisher’s seminal paper, it was assumed that \mathbf{\Sigma}_0=\mathbf{\Sigma}_1. I am trying to find a solution to the decision boundary in QDA. discriminant_analysis. Neural Quadratic Discriminant Analysis 2295 resulting in curved decision boundaries in the population response space (see Figure 1a). observations = [rand(Bool) ?. predict_proba (X)) class QuadraticDiscriminantAnalysis (BaseEstimator, ClassifierMixin): """ Quadratic Discriminant Analysis A classifier with a quadratic decision boundary, generated by fitting class conditional densities to the data and. (2) Linear discriminant analysis (LDA) makes the additional assumption that the covariance of each of the. Custom handles (i. Visualize classifier decision boundaries in MATLAB W hen I needed to plot classifier decision boundaries for my thesis, I decided to do it as simply as possible. Python was created out of the slime and mud left after the great flood. QDA (priors=None, reg_param=0. A classifier with a quadratic decision boundary, generated by fitting class conditional densities to the data and using Bayes' rule. We have seen that in p = 2 dimensions, a linear decision boundary takes the form β 0 +β 1 X 1 +β 2 X 2 = 0. Since the Bayes decision boundary is non-linear, it is more accurately approximated by QDA than by LDA. Lecture9: Classification,LDA Reading: Chapter 4 STATS 202: Data mining and analysis Jonathan Taylor, 10/12 Slide credits: Sergio Bacallado 1/21. Python BaggingClassifier. Read more in the User Guide. When the true boundary is moderately non-linear, QDA may be better. 이번 포스팅에선 선형판별분석(Linear Discriminant Analysis : LDA)에 대해서 살펴보고자 합니다. 2）二次分类判别（Quadratic discriminant analysis, QDA） Plot the confidence ellipsoids of each class and decision boundary """ print(__doc__) from scipy. 2 Quadratic Discriminant Analysis (QDA) Quadratic Discriminant Analysis is a more general version of a linear classi er. Only for GradStudents Due: Friday November 23, at the start of class 4. Remark: In step 3, plotting the decision boundary manually in the case of LDA is relatively easy. BaggingClassifier. If the covariance matrices for two distributions are equal and proportional to the identity matrix, then the distributions are spherical in d dimensions. of X formed by the three different construction methods for A proposed Section 3. Unlike LDA, QDA assumes that each class has its own covariance matrix, and hence the decision boundary is quadratic. The le tumor. Right: Details are as given in the left-hand panel, except that Σ 1 6= Σ 2. Note: when the number of covariates grow, the number of things to estimate in the covariance matrix gets very large. Estimate covariance matrices from data, or fit d(d+2)/2 parameters to the data directly to obtain QDA – both ways are rather inaccurate. A linear decision boundary is easy to understand and visualize, even in many dimensions. In Fig 4 (a) to 4(d), the decision boundary curves reveal more misclassification with un-normalized data than with normalized data. Is the decision boundary ~linear? Are the observations normally distributed? Do you have limited training data? Start with LDA Start with Logistic Regression Start with QDA Start with K-Nearest Neighbors YES NO YES NO YES NO Linear methods Non-linear methods. As we will see, the term quadratic in QDA and linear in LDA actually signify the shape of the decision boundary. If the decision boundary can be visualised as a graph added to the scatter plot of the two variables. Quadratic Decision Boundary for QDA Helloi would like to plot a quadratic decision boundary for a Quadratic Discriminant Analysis method i implemented, but i can't undertand how. For most of the data, it doesn't make any difference, because most of the data is massed on the left. discriminant_analysis. For the line, we got pretty good results. Each record was generated. Quantitative Value: A value expressed or expressible as a numerical quantity. , j can be distinct. Right: Details are as given in the left-hand panel, except thatΣ 1 ≠Σ 2. Can anyone help me with that? Here is the data I have: set. In order to use LDA or QDA, we need: An estimate of the class probabilities ˇ j. discriminant analysis. The data itself is organized in two files, 1599 red wines and 4898 white wines, each containing 11 chemical attributes as well as one quality attribute. Linear discriminant function analysis (i. Is it just. This example plots the covariance ellipsoids of each class and decision boundary learned by LDA and QDA. What is important to keep in mind is that no one method will dominate the oth- ers in every situation. of X formed by the three different construction methods for A proposed Section 3. The double matrix meas consists of four types of measurements on the flowers, the length and width of sepals and petals in centimeters, respectively. Then r⇤(x) = (C if Q C(x) Q D(x) > 0, D otherwise. We want to find someclassification. 1) holds, then the random projection ensemble classifier performs nearly as well as the projected data base classifier with the oracle projection A∗. That was a visual intuition for a simple case of the Bayes classifier, also called: •Idiot Bayes •Naïve Bayes •Simple Bayes We are about to see some of the mathematical formalisms, and more examples, but keep in mind the basic idea. 2）二次分类判别（Quadratic discriminant analysis, QDA） Plot the confidence ellipsoids of each class and decision boundary """ print(__doc__) from scipy. For simple boundaries may give worst predictions. class QuadraticDiscriminantAnalysis (BaseEstimator, ClassifierMixin): """Quadratic Discriminant Analysis A classifier with a quadratic decision boundary, generated by fitting class conditional densities to the data and using Bayes' rule. Of K classes; they sum to one and remain in [0,1]: Linear Decision boundary: Logistic Regression Model fit: In max. Observation of each class are drawn from a normal distribution (same as LDA). Notice the slope though. Irisdata I TheIrisdata(Fisher,AnnalsofEugenics,1936)givesthe measurementsofsepalandpetallengthandwidthfor150 flowersusing3speciesofiris(50flowersperspecies). Johnson to illustrate the most classical supervised classification algorithms. An estimate of the mean vectors j. Like LDA, QDA models the conditional probability density functions as a Gaussian distribution, then uses the posterior distributions to estimate the class for a given test data. (c) If the Bayes decision boundary is linear, do we expect LDA or QDA to perform better on the training set? On the test set? (d) In general, as the sample size n increases, do we expect the test prediction accuracy of QDA relative to LDA to improve, decline, or be unchanged? Why? 2. Right: Details are as given in the left-hand panel, except that Σ 1 6= Σ 2. I decision trees I SVM I multilayer perceptron (MLP) Examples of generative classifiers: I naive Bayes (NB) I linear discriminant analysis (LDA) I quadratic discriminant analysis (QDA) We will study all of the above except MLP. It appears that the accuracy of both models is the same (let’s assume that it is), yet the behavior of the models is very different. class sklearn. Python source code: plot_lda_qda. quadratic discriminant analysis (QDA), Parzen window, mixture of Gaussian (MOG), and mixture of generalized Gaussian (MGG). observations = [rand(Bool) ?. QDA¶ class sklearn. Nonlinear SVM. YOUR DISCUSSION HERE: In the both datasets, the optimal decision boundary is quadratic (based on over all training accuracy and common sense). They are from open source Python projects. This results in a quadratic decision boundary. For example, quadratic discriminant analysis (QDA) uses a quadratic discriminant function to separate the two classes. One needs to be careful. plot the the resulting decision boundary. fit with lda and qda from the MASS package. This semester I'm teaching from Hastie, Tibshirani, and Friedman's book, The Elements of Statistical Learning, 2nd Edition. Visualize classifier decision boundaries in MATLAB W hen I needed to plot classifier decision boundaries for my thesis, I decided to do it as simply as possible. 334) Figure 2: A three class problem similar to that in figure 1, with the data in each class generated from a mixture of Gaussians. GS01 0163 Analysis of Microarray Data LDA, DLDA, QDA K Nearest Neighbors Validation divided by smooth curves defining a “decision boundary”. The following are code examples for showing how to use sklearn. [Solutions to a quadratic equation] – In d-D, B. Calculate the decision boundary for Quadratic Discriminant Analysis (QDA) I am trying to find a solution to the decision boundary in QDA. Remember: We use Bayes' theorem and only need p(xjy = k) to compute the posterior as: ˇ k(x) = P(y = k j x) = P(xjy = k)P(y = k)P(x)p(xjy = k)ˇ k Pg j=1 p(xjy = j)ˇ j NB is based on a simple conditional independence assumption: the features are conditionally independent given class y. In these one-, two-, and three-. discriminant_analysis. 1 Answer to This problem relates to the QDA model, in which the observations within each class are drawn from a normal distribution with a classspecific mean vector and a class specific covariance matrix. Since the discriminant yields a quadratic decision boundary, the method is known as quadratic discriminant analysis (QDA). My training data is stored in train which is a 145x2 matrix with height and weight as entries (males and females as classes). Since QDA assumes a quadratic decision boundary, it can accurately model a wider range of problems than can the linear methods. Continue reading Classification from scratch, linear discrimination 8/8 → Eighth post of our series on classification from scratch. However, I am applying the same technique for a 2 class, 2 feature QDA and am having trouble. Although the decision boundaries between classes can be derived analytically, plotting them for more than two classes gets a bit complicated. Nonlinear SVM. of X formed by the three different construction methods for A proposed Section 3. Classifier comparison¶ A comparison of a several classifiers in scikit-learn on synthetic datasets. decision boundary. Discriminant analysis¶. Quadratic Descriminant Analysis (QDA) Unlike LDA, QDA assumes that each class has its own covariance matrix. The latest one was on the SVM, and today, I want to get back on very old stuff, with here also a linear separation of the space, using Fisher's linear discriminent analysis. If there is new data to be classified that appears in the upper left of the plot, the LDA model will call the data point versicolor whereas the QDA model will call it virginica. Next we plot LDA and QDA decision boundaries for the same data. This is a little bit confusing (but on the other hand it increases the contrast of point on top of background). It can be classify according to the type of the decision tree that it is com-posed: orthogonal or oblique. About This Book Handle a variety of machine learning tasks effortlessly by leveraging the power of … - Selection from scikit-learn Cookbook - Second Edition [Book]. There are linear and quadratic discriminant analysis (QDA), depending on the assumptions we make. Python was created out of the slime and mud left after the great flood. 334) Figure 2: A three class problem similar to that in figure 1, with the data in each class generated from a mixture of Gaussians. By representing the variables on a calibrated axes [], sample values. Of course SVM bypasses it via kernel trick but still not as much complex decision boundary as nueral nets; Despite the risk of non linearity in data linear algorithms tends to work well in practice and are often used as starting point. • Consider the log-odds ratio (again, P(x)doesn’t matter for decision): This is a linear decision boundary! w 0 + xTw COMP-551: Applied Machine Learning 10 19 Joelle Pineau Linear discriminant analysis (LDA) • Return to Bayes rule: • Make explicit assumptions about P(x|y): – Multivariate Gaussian, with mean µ and covariance matrix Σ. These are the top rated real world Python examples of sklearnensemble. Johnson to illustrate the most classical supervised classification algorithms. When referring to, for example, a model developed from 1996 data with a 2 year default horizon, we mean a model developed from the data set D96, where a response variable is defined as 1 if the year of bankruptcy is 1997 or 1998 and 0 otherwise. An example of such a boundary is shown in Figure 11. Assume a linear classification boundary ˇ ˆ ˇ ˙ ˆ ˇ ˝ ˆ For the positive class the bigger the value of ˇ , the further the point is from the classification boundary, the higher our certainty for the membership to the positive class • Define ˛ ˚˜ as an increasing function of ˇ For the negative class the smaller the. A linear decision boundary is easy to understand and visualize, even in many dimensions. This might be due to the fact that the covariances matrices differ or because the true decision boundary is not linear. Quadratic Discriminant Analysis. QDA assumes that each class has its own covariance matrix (different from LDA). For a polyp candidate, the distance from the decision boundary, called polyp likelihood, provides the ranked ordering of the likelihood that the candidate is a polyp or a false-positive finding. Python source code: plot_lda_vs_qda. So suppose you have two classes, and 50% of your samples are in one class and the other 50% belong to another class then this simplifies the equation into this:. The decision boundary is now described with a quadratic function. We start with the optimization of decision boundary on which the posteriors are equal. Since QDA is more flexible, it can, in general, arrive at a better fit but if there is not a large enough sample size we will end up overfitting to the noise in the data. (가 들어간 로지스틱함수에서 반응변수를 뽑음) 이땐 예상대로 QDA가 제일 잘했고 그다음이 KNN-CV였다. So to handle this hierarchical setup, you probably need to do a series of binary classifiers manually, like group 1 vs. decision boundaries with those of linear discriminant analysis (LDA) and quadratic discriminant analysis (QDA). The decision boundary can be a bit jagged. The curved line is the decision boundary resulting from the QDA method. If the actual boundaries are indeed linear, QDA may have a higher model bias. discriminant_analysis. ISSN: 0167-9236. I This algorithm aims to find a separating hyperplane by minimizing the distance of misclassified points to the decision boundary: X. The Bayes decision boundary is a hypersphere determined by QDA, the covariance structure is the same as in the high-curvature nonlinear model, the decision boundary has high curvature in comparison to the low-curvature nonlinear model, and this model is more difficult than the high-curvature nonlinear model. Inaspecialcasewhereallclasses. In the absence of a. However, I am applying the same technique for a 2 class, 2 feature QDA and am having trouble. Maximum-likelihood and Bayesian parameter estimation techniques assume that the forms for the underlying probability densities were known, and that we will use the training samples to estimate the values of their parameters. Each record was generated. 5Σ−1 (µT 1 µ1 −µT 2 µ2)+ln P(ω1) P(ω2) = 0 (15) The decision boundary xDB is xDB = µ1 + µ2 2 + Σ µ2 − µ1 ln P(ω1) P(ω2) (16) When the two classes are equiprobable, then the second term will be neglected and the decision boundary is the point in the middle of the class. In Decision Support Systems, Elsevier, 47(4):547-553. Supervised machine learning (I) Supervised machine learning is an important analytical tool for high-throughput genomic data. Quadratic Discriminant Analysis. In this article we will study another very important dimensionality reduction technique: linear discriminant analysis (or LDA). Several classification rules are considered: Quadratic Discriminant Analysis (QDA), 3-nearest-neighbor (3NN) and neural networks (NNet). Despite its potential for decoding information embedded in a popula-tion, it is not obvious how a brain area would implement QDA. The following are code examples for showing how to use sklearn. Lecture9: Classification,LDA Reading: Chapter 4 STATS 202: Data mining and analysis Jonathan Taylor, 10/12 Slide credits: Sergio Bacallado 1/21. The separation boundary for LDA and QDA is well described by the following image Both the LDA and QDA are simple probabilistic models. QDA uses a hyperquadratic surface for the decision boundary. cross_validation import train_test_split from sklearn. score extracted from open source projects. (A large n will help offset any variance in the data. The only difference between QDA and LDA is that in QDA, we compute the pooled covariance matrix for each class and then use the following type of discriminant function for getting the scores for each of the classes involed: Where, result is basically the class z(x) with max score. The boundary point can be found by solving the following (quadratic) equation logπ 1 − 1 2 log(σ2)− (x−µ 1)2 2σ2 1 = logπ 2 − 1 2 log(σ2)− (x−µ 2)2 2σ2 2 To simplify the math, we assume that the two components have equal variance(i. Discriminant analysis¶. Quadratic discriminant analysis falls somewhere between the linear approaches of linear discriminant analysis and logistic regression and the non-parametric approach of K-nearest neighbors. QDA¶ class sklearn. Trending AI Articles: 1. Least squares linear regression in which linear decision boundary is assumed is the most rigid one and stable to fit. 6 Linear decision. Though not as flexible as KNN, QDA can perform better in the presence of a limited number of training observations because it does make some assumptions abou the form of the decision boundary. The shading indicates the QDA decision rule. I The common covariance matrix Σ = 1. Linear discriminant analysis and logistic regression will perform well when the true decision boundary is linear. The boundary is a affine hyperplane of d-1 dimensions, perpendicular to the line separating the means. Note that there is an equivalence that can be estab-lished between decision trees and neural networks [11], [12];. A non-linear Bayes decision boundary is indicative that a common covariance matrix is inaccurate, which would tell us that the QDA would be expected to perform better. So both LDA and linear Logistic Regression yield poor prediction accuracy even on the training set. Since the Bayes decision boundary is linear, it is more accurately approximated by LDA than by QDA. Suppose that we have K classes,. MLP and SVM are classifiers that work directly on the decision boundary and fall under the discriminative paradigm. “Quadratic Decision boundary” –second-order terms don’t cancel out Microsoft PowerPoint - EM_v1_annotatedonclass. However, if the Bayes decision boundary is only slightly non-linear, LDA could still be a better model. Plot the confidence ellipsoids of each class and decision boundary. of the People, by the People, for. The percentage of the data in the area where the two decision boundaries differ a lot is small. pdf [When you have many classes, their QDA decision boundaries form an. LDA and QDA are classification methods based on the concept of Bayes’ Theorem with assumption on conditional Multivariate Normal Distribution. caret that provides an unified interface to many other packages. , labels) can then be provided via ax. YOUR DISCUSSION HERE: In the both datasets, the optimal decision boundary is quadratic (based on over all training accuracy and common sense). You can vote up the examples you like or vote down the ones you don't like. If the Bayes decision boundary is linear, do we expect LDA or QDA to perform better on the training set? On the test set? Solution (a) If the Bayes decision boundary is linear, we expect QDA to perform better on the training set due to its flexibility and LDA On the test set. Quadratic Discriminant Analysis (QDA) Assumes each class density is from a multivariate Gaussian. Quadratic Discriminant Analysis. There is a nice Field Goal data among the data sets on our course web site. Python source code: plot_lda_qda. PW 4 | Machine Learning course. The model at the right joins these smaller blobs together into a larger blob where the model classifies data as responders. Linear Discriminant Analysis (LDA) and Quadratic Discriminant Analysis (QDA) are two classic classifiers, with, as their names suggest, a linear and a quadratic decision surface, respectively. QDA fits covariance within each class and thus allows for more complex decision boundaries. We will also use h2o, a package. • Scenario 6: Details are as in the previous scenario, but the responses were sampled from a more complicated non-linear function. On the test set, we expect LDA to perform better than QDA because QDA could overfit the linearity of the Bayes decision boundary. A linear decision boundary is easy to understand and visualize, even in many dimensions. Briefly, SVM works by identifying the optimal decision boundary that separates data points from different groups (or classes), and then predicts the class of new observations based on this separation boundary. Maximum-likelihood and Bayesian parameter estimation techniques assume that the forms for the underlying probability densities were known, and that we will use the training samples to estimate the values of their parameters. Although linear SVM classifiers are efficient and work surprisingly well in many cases, many datasets are not even close to being linearly seperable. Plot the confidence ellipsoids of each class and decision boundary. Of K classes; they sum to one and remain in [0,1]: Linear Decision boundary: Logistic Regression Model fit: In max. class QuadraticDiscriminantAnalysis (BaseEstimator, ClassifierMixin): """Quadratic Discriminant Analysis A classifier with a quadratic decision boundary, generated by fitting class conditional densities to the data and using Bayes' rule. Next we plot LDA and QDA decision boundaries for the same data. Suppose we collect data for a group of students in a statistics class with variables X. I The three mean vectors are: µ 1 = 0 0 µ 2 = −3 2 µ 3 = −1 −3 I Total of 450 samples are drawn with 150 in each class for. # Code source: Gael Varoqueux # Andreas Mueller # Modified for Documentation merge by Jaques Grobler # License: BSD 3 clause # Modified to include pyearth by Jason Rudy import numpy as np import matplotlib. For the MAP classification rule based on mixture of Gaussians modeling, the decision boundaries are given by QDA assumes that each class distribution is multivariate Gaussian (but with its. Alternatively, if the same class covariance is assumed equal for all classes,. Despite its potential for decoding information embedded in a popula-tion, it is not obvious how a brain area would implement QDA. QuadraticDiscriminantAnalysis¶. You have to ensure that the value of k is not too high or not too low. Rectifier classified well but seems less generalized although its decision boundary looks not so overfitting. I µˆ 1 = −0. (b) If the Bayes decision boundary is non-linear, do we expect LDA or QDA to perform better on the training set? On the test set? (c) In general, as the sample size n increases, do we expect the test prediction accuracy of QDA relative to LDA to improve, decline, or be unchanged?. Naive Bayes vs Logistic. Continue reading Classification from scratch, linear discrimination 8/8 → Eighth post of our series on classification from scratch. Q c(x) = 1. Right: Details are as given in the left-hand panel, except thatΣ 1 ≠Σ 2. Although this estimator is unbiased, this is not a good estimator since its variability does not decrease with the sample size. Classification [email protected] If there is new data to be classified that appears in the upper left of the plot, the LDA model will call the data point versicolor whereas the QDA model will call it virginica. This induces a decision boundary which is more distant from the smaller class than from (QDA), Flexible Discriminant Analysis (FDA) and the k-Nearest-Neighbor Rule. That was a visual intuition for a simple case of the Bayes classifier, also called: •Idiot Bayes •Naïve Bayes •Simple Bayes We are about to see some of the mathematical formalisms, and more examples, but keep in mind the basic idea. Mild cognitive impairment (MCI) is recognized as a transitional phase in the progression toward more severe forms of dementia and is an early precursor to Alzheimer's disease. predict_proba (X)) class QuadraticDiscriminantAnalysis (BaseEstimator, ClassifierMixin): """ Quadratic Discriminant Analysis A classifier with a quadratic decision boundary, generated by fitting class conditional densities to the data and. – In 1D, B. Without any further assumptions, the resulting classifier is referred to as QDA (quadratic discriminant analysis). A classifier with a quadratic decision boundary, generated by fitting class conditional densities to the data and using Bayes’ rule. degree in electrical and computer engineering from Texas A&M University, College Station, TX in 2006, and was a post-doctoral associate in the Department of Statistics in Texas A&M University until February 2007. Quadratic discriminant analysis QDA is a more flexible classification method than LDA, which can only identify linear boundaries, because QDA can also identify secondary boundaries. score - 10 examples found. BaggingClassifier. You can rate examples to help us improve the quality of examples. When referring to, for example, a model developed from 1996 data with a 2 year default horizon, we mean a model developed from the data set D96, where a response variable is defined as 1 if the year of bankruptcy is 1997 or 1998 and 0 otherwise. What is important to keep in mind is that no one method will dominate the oth- ers in every situation. For quadratic discriminant analysis, there is nothing much that is different from the linear discriminant analysis in terms of code. metrics) and Matplotlib for displaying the results in a more intuitive visual format. The model at the right joins these smaller blobs together into a larger blob where the model classifies data as responders. When (homoscedasticity assumption) the discriminant functions it is: this is the expression for the LDA (linear discriminant analysis) function. View Saumil D. See the complete profile on LinkedIn and discover Saumil’s connections and jobs at similar companies. QDA can perform better in the presence of a limited number of training observations because it does make some assumptions about the form of the decision boundary. In the effort to be more accurate on training data, the model on the left creates closed-decision boundaries around any and all groupings of responders. The percentage of the data in the area where the two decision boundaries differ a lot is small. Gaussian models μ u 1 u 2 λ 1 1/2 λ 2 1/2 x 1 x 2 Figure 4. We will also use h2o, a package. You can use the characterization of the boundary that we found in task 1c). This post was contributed by Chelsea Douglas, a Software Engineer at Plotly. Also, the red and blue points are not matched to the red and blue backgrounds for that figure. Irisdata I TheIrisdata(Fisher,AnnalsofEugenics,1936)givesthe measurementsofsepalandpetallengthandwidthfor150 flowersusing3speciesofiris(50flowersperspecies). In order to use LDA or QDA, we need: An estimate of the class probabilities ˇ j. How do we get there with our sigmoid function? 2. SVM doesn’t make any assumptions about the distribution of the underlying data (unlike LDA and QDA) Similar to the LDA, SVM maximises the distance between the decision boundary and the observations, however unlike LDA SVM only uses nearest points to the boundary (whereas LDA takes into account all the observations). Draw the ideal decision boundary for the dataset above. The formula of decision boundary is defined as δ k : (4) δ k ( x ) = - 1 2 log ε k - 1 2 ( x - μ k ) T ε k - 1 ( x - μ k ) + log π k , (5) Where π k is number of samples in class k total number of samples. The latest one was on the SVM, and today, I want to get back on very old stuff, with here also a linear separation of the space, using Fisher's linear discriminent analysis. In 2-class case, decision boundary is when ωx + α=0 Another nice simplification happens if the two priors are equal. Heteroscedasticity. In defining distributional complexity we want to differentiate between complexity and separability of the classes. An estimate of the mean vectors j. the year of bankruptcy is set to ’NA’. ] [Show multiplicatively weighted Voronoi diagram. Finally, the decision boundary obtained by LDA is equivalent to the binary SVM on the set of support vectors 31. (a) If the Bayes decision boundary is linear, we would expect LDA to outperform QDA on test set because it has both smaller bias and variance. We start with the optimization of decision boundary on which the posteriors are equal. (b) If the Bayes decision boundary is non-linear, do we expect LDA or QDA to perform better on the training set? On the test set? (c) In general, as the sample size n increases, do we expect the test prediction accuracy of QDA relative to LDA to improve, decline, or be unchanged?. 1 Quadratic Discriminant Analysis (QDA) Like LDA, the QDA classifier results from assuming that the observations from each class are drawn from a Gaussian distribution, and plugging estimates for the parameters into Bayes' theorem in order to perform prediction. Use sklearn. He was appointed by Gaia (Mother Earth) to guard the oracle of Delphi, known as Pytho. Python source code: plot_lda_vs_qda. QDA works identically as LDA: except that it estimates separate variance for each class. What is important to keep in mind is that no one method will dominate the oth- ers in every situation. datasets import make_blobs from sklearn. quadratic discriminant analysis (QDA), Parzen window, mixture of Gaussian (MOG), and mixture of generalized Gaussian (MGG). Is the decision boundary ~linear? Are the observations normally distributed? Do you have limited training data? Start with LDA Start with Logistic Regression Start with QDA Start with K-Nearest Neighbors YES NO YES NO YES NO Linear methods Non-linear methods. I The common covariance matrix Σ = 1. As we can see from the figure, the K-nearest neighbors decision boundary is quite jagged. It is quadratic (a curve). The ellipsoids display the double standard deviation for each class. • QDA outperforms LDA if the covariances are not the same in the groups. Quadratic Discriminant Analysis (QDA) is an extension where the restriction on the covariance structure is relaxed, which leads to a nonlinear classification boundary. 92% is a pretty good number!. (b) If the Bayes decision boundary is non-linear, do we expect LDA or QDA to perform better on the training set? On the test set? (c) In general, as the sample size n increases, do we expect the test prediction accuracy of QDA relative to LDA to improve, decline, or be unchanged?. Quadratic discriminant analysis (QDA). Depending on the situations, the different groups might be separable by a linear straight line or by a non-linear boundary line. Each class has their own Σ The decision boundary is a margin, which is defined by the support vectors, which are the points that are the most difficult to. QDA, on the other hand, learns a quadratic one. decision boundary log P(Y = 1jX) P(Y = 0jX) = 0 + >X Linear/quadratic discriminant analysis (Estimates pg assuming multivariate Gaussianity) General nonparametric. LDA is a much less flexible classifier than QDA. The decision function for the kth class is δk(x) = xTΣ−1µk − 1 2 µT k Σ −1µ k. I This algorithm aims to find a separating hyperplane by minimizing the distance of misclassified points to the decision boundary: X. A classifier with a linear decision boundary, generated by fitting class conditional densities to the data and using Bayes' rule. As the prefix 'bi-' suggests, both the samples and variables of a data matrix is represented in a biplot. For the decision boundary may be visualised in a three-dimensional scatter plot, but this plot may not be easy to interpret. On the other hand, this common covariance matrix is estimated based on all points, also those far from the decision boundary. 92% is a pretty good number!. (b) If the Bayes decision boundary is non-linear, do we expect LDA or QDA to perform better on the training set? On the test set? (c) In general, as the sample size n increases, do we expect the test prediction accuracy of QDA relative to LDA to improve, decline, or be unchanged?. Recall that the optimization problem has the following form: argmin w Xn i=1 (w>x i 2y i) + kwk2 2 2. The model fits a Gaussian density to each class. LDA is a much less flexible classifier than QDA. (c) If the Bayes decision boundary is linear, do we expect LDA or QDA to perform better on the training set? On the test set? (d) In general, as the sample size n increases, do we expect the test prediction accuracy of QDA relative to LDA to improve, decline, or be unchanged? Why? 2. In contrast with this, quadratic discriminant analysis (QDA). ISSN: 0167-9236. It is quadratic (a curve). In a certain way(too mathematically convoluted for this post), LDA does learn a linear boundary between the points belonging to different classes. Linear Discriminant Analysis (LDA) A classifier with a linear decision boundary, generated by fitting class conditional densities to the data and using Bayes’ rule. Recent years have witnessed an unprecedented availability of information on social, economic, and health-related phenomena. Right: Details are as given in the left-hand panel, except that Σ 1 6= Σ 2. There is a nice Field Goal data among the data sets on our course web site. Quadratic Discriminant Analysis When the number of predictors is large the number of parameters we have to estimate with QDA becomes very large because we have to estimate a separate. Machine Learning (CS 567) Lecture 6 Fall 2008 Time: T-Th 5:00pm - 6:20pm decision boundary have little impact – For small data sets, LDA and QDA. The percentage of the data in the area where the two decision boundaries differ a lot is small. With ~15 or fewer training samples for each class, I believe QDA is producing inaccurate covariance matrices and wildly unrealistic opinions about decision boundary, hence the massive log loss score. Quadratic discriminant analysis falls somewhere between the linear approaches of linear discriminant analysis and logistic regression and the non-parametric approach of K-nearest neighbors. The sample statistics are: mean(x1) cov(x1) mean(x2) cov(x2). The decision boundary between two Gaussian class conditional densities with equal covariance is a straight line. Naive Bayes, LDA, QDA. Decision Tree, Random Forest, AdaBoost. Linear and Quadratic Discriminant Analysis: Tutorial. Quadratic Discriminant Analysis (QDA) is an extension where the restriction on the covariance structure is relaxed, which leads to a nonlinear classification boundary. Mahalanobis distance term. Since the Bayes decision boundary is linear, it is more accurately approximated by LDA than by QDA. Create a QDA classififier and draw the decision boundary (a) between setosa(0) and versicolor(1); and (b) between versicolor(1) and virginica(2) Report the training and test accuracies. QDA can perform better in the presence of a limited number of training observations because it does make some assumptions about the form of the decision boundary. This paper presents the use of ROC analysis for the assessment of the classifiers' performance. It is also one of the first methods people get their hands dirty on. In contrast with this, quadratic discriminant analysis (QDA). • Scenario 6: Details are as in the previous scenario, but the responses were sampled from a more complicated non-linear function. Classification learning II CS 2750 Machine Learning Logistic regression model • Defines a linear decision boundary • Discriminant functions: • where f (x, w) g1 (wT x) g(wT x) g(z) 1/(1 e z) x Input vector 1 x1 f (x, w) w0 w1 w2 wd x2 z xd Logistic function 1 (x) (w x) g g T ( ) 1 0 x w x g g T - is a logistic function. 2 Quadratic Discriminant Analysis (QDA) Quadratic Discriminant Analysis is a more general version of a linear classi er. It looks like this boundary is doing really poorly for non all-NBA players, but I want to see the confusion matrix to make sure. This is a little bit confusing (but on the other hand it increases the contrast of point on top of background). An example of such a boundary is shown in Figure 11. View Saumil D. QDA¶ class sklearn. In such cases Quadratic discriminant analysis is required which is based on the estimation of different covariance structures for the different groups. This skill test is specially designed for you to. The question was already asked and answered for linear discriminant analysis (LDA), and the solution provided by amoeba to compute this using the "standard Gaussian way" worked well. Below, the green line is QDA, our estimate, and the purple dashed line is the optimal Bayes boundary. Custom legend labels can be provided by returning the axis object (s) from the plot_decision_region function and then getting the handles and labels of the legend. This leads to a model known as quadratic linear discriminant (QDA), since now the decision boundary is not linear but quadratic. Calculate the decision boundary for Quadratic Discriminant Analysis (QDA) I am trying to find a solution to the decision boundary in QDA. Quadratic discriminant analysis (QDA). Chao Sima is a Research Assistant Professor in Computational Biology Division of the Translational Genomics Research Institute in Phoenix, AZ. True or False: For classification, we always want to minimize the misclassification rate. As we increase dimensions, we exponentially reduce the number of observations near the point in consideration. Returns-----C : array, shape (n_samples, n_classes) Estimated log probabilities. Macskassy Sensitivity to monotone transformations. decision boundaries with those of linear discriminant analysis (LDA) and quadratic discriminant analysis (QDA). The latest one was on the SVM, and today, I want to get back on very old stuff, with here also a linear separation of the space, using Fisher's linear discriminent analysis. Each record was generated. Although the DA classifier i s considered one of the most well-k nown classifiers, it. An estimate of the mean vectors j. The following are code examples for showing how to use sklearn. Assume a linear classification boundary ˇ ˆ ˇ ˙ ˆ ˇ ˝ ˆ For the positive class the bigger the value of ˇ , the further the point is from the classification boundary, the higher our certainty for the membership to the positive class • Define ˛ ˚˜ as an increasing function of ˇ For the negative class the smaller the. The programming language Python has not been created out of slime and mud but out of the programming language ABC. Basically, what you see is a machine learning model in action, learning how to distinguish data of two classes, say cats and dogs, using some X and Y variables. metrics) and Matplotlib for displaying the results in a more intuitive visual format. Decision boundary: Values of x where 𝛿0 =𝛿1( ) is quadratic in x Quadratic Discriminant Analysis (QDA) 4 p 1 ( 2¼ ) d j § C j exp ¡ ¡1 2 ( x ¹ c)T § ¡ 1 C c ¢ Prior Additional Sq. Scaling is first performed on the train set. Linear and Quadratic Discriminant Analysis: Tutorial. There is not a single algorithm for training such classifiers, but a family of algorithms based on a common principle: all naive Bayes classifiers assume that the. Given that the decision boundary separating two classes is linear, what can be inferred about the discriminant functions of the two classes? a. As we will see, the term quadratic in QDA and linear in LDA actually signify the shape of the decision boundary. 1 for Fisher iris data. The QDA model produces a hyperquadric decision boundary and the cross-validated performance of the QDA model exceeds that of the LDA model, which produces a generalized hyperplane decision boundary. preprocessing. When the true boundary is moderately non-linear, QDA may be better. 1 Quadratic Discriminant Analysis (QDA) Like LDA, the QDA classifier results from assuming that the observations from each class are drawn from a Gaussian distribution, and plugging estimates for the parameters into Bayes' theorem in order to perform prediction. observations = [rand(Bool) ?. Class Imbalance. cross_validation import train_test_split from sklearn. • Linear / Quadratic Discriminant Analysis • K Nearest Neighbors • Decision Trees • Random Forests Non Linear Decision Boundary Apply 3rd Dimension. 0, store_covariance=False, tol=0. November 10, 2016- 12. Class Imbalance. class sklearn. A classifier with a quadratic decision boundary, generated by fitting class conditional densities to the data and using Bayes’ rule. In our previous article Implementing PCA in Python with Scikit-Learn, we studied how we can reduce dimensionality of the feature set using PCA. QuadraticDiscriminantAnalysis (priors=None, reg_param=0. The model fits a Gaussian density to each class, assuming that all classes share the same covariance matrix. The model at the right joins these smaller blobs together into a larger blob where the model classifies data as responders. may have 1 or 2 points. Introduction to Statistical Learning - Chap4 Solutions. Compute and graph the LDA decision boundary. I've got a data frame with basic numeric training data, and another data frame for test data. The decision boundary is here But that can’t be the linear discriminant analysis, right? I mean, the frontier is not linear… Actually, in Fisher’s seminal paper, it was assumed that \mathbf{\Sigma}_0=\mathbf{\Sigma}_1. QDA is trying to create a quadratic boundary for 99 different classes, computing a unique covariance matrix for each class. Since QDA assumes a quadratic decision boundary, it can accurately model a wider range of problems than can the linear methods. This is shown in the right diagram. I A more direct method (nothing to do with statistics) is to directly search a hyperplane separating two class data (perceptron model). Models based on distributions A–E demonstrate consistent performance across all testing distributions. When these assumptions hold, QDA approximates the Bayes classifier very closely and the discriminant function produces a quadratic decision boundary. In these one-, two-, and three-. The advantage of QDA is that it uses the outlier values produced by each algorithms to draw its own independent decision boundary. decision boundaries with those of linear discriminant analysis (LDA) and quadratic discriminant analysis (QDA). Quadratic Discriminant Analysis. It is precisely the equality of the covariance matrices that constrains the decision boundary to be linear, hence "Linear Discriminant Analysis. QDA: QDA serves as a compromise between the non-parametric KNN method and the linear LDA and logistic regression approaches. It can be classify according to the type of the decision tree that it is com-posed: orthogonal or oblique. It looks like this boundary is doing really poorly for non all-NBA players, but I want to see the confusion matrix to make sure. Quadratic Discriminant Analysis for Binary Classification In Quadratic Discriminant Analysis (QDA), we relax the assumption of equality of the covariance matrices: 1 6= 2; (24) which means the covariances are not necessarily equal (if they are actually equal, the decision boundary will be linear and QDA reduces to LDA). These visuals can be great to understand…. The following nine statistical learning algorithms were used to develop the predictive models: nearest neighbors, support vector machine (SVM) with linear and radial basis function (RBF) kernel, Decision Tree, Random Forest, AdaBoost, Naive Bayes, linear and quadratic discriminant analysis (QDA). The capacity of a technique to form really convoluted decision boundaries isn't necessarily a virtue, since it can lead to overfitting. Though not as flexible as KNN, QDA can perform better in the presence of a limited number of training observations because it does make some assumptions abou the form of the decision boundary. Tao Li, Shenghuo Zhu, and Mitsunori Ogihara. For most of the data, it doesn't make any difference, because most of the data is massed on the left. Previous neuroimaging studies reveal that MCI is associated with aberrant sensory–perceptual processing in cortical brain regions subserving auditory and language function. Recent years have witnessed an unprecedented availability of information on social, economic, and health-related phenomena. The quadratic term allows QDA to separate data using a quadric surface in higher dimensions. Note: When pis large, using QDA instead of LDA can dramatically increase the number of parameters to estimate. 5Σ−1 (µT 1 µ1 −µT 2 µ2)+ln P(ω1) P(ω2) = 0 (15) The decision boundary xDB is xDB = µ1 + µ2 2 + Σ µ2 − µ1 ln P(ω1) P(ω2) (16) When the two classes are equiprobable, then the second term will be neglected and the decision boundary is the point in the middle of the class. The number of parameters increases significantly with QDA. Finally, the decision boundary obtained by LDA is equivalent to the binary SVM on the set of support vectors 31. Again, the covariance matrix for each class is estimated by the sample covariance matrix of the training samples in that class. Fall 2008 14 Lecture 6 - Sofus A. The representation of LDA is straight forward. The QDA decision boundary is inferior, because it suffers from higher vari-ance without a corresponding decrease in bias. txt") ##### # Plots showing decision boundaries s. LDA to perform better than QDA because QDA would overfit linearity on the Bayes decision theory on the test set. svm in e1071 uses the "one-against-one" strategy for multiclass classification (i. • Scenario 6: Details are as in the previous scenario, but the responses were sampled from a more complicated non-linear function. The density-based techniques make use of the full feature space to calculate their respective values. Logistic Regression, Discriminant Analysis and K-Nearest Neighbour Tarek Dib June 11, 2015 1 Logistic Regression Model - Single Predictor p(X) = eβ0+β1X 1 + eβ0+β1X (1) 2 Odds p 1 − p = eβ0+β1X (2) 3 logit, log odds log( p 1 − p ) = β0 + β1X (3) 4 Summary In linear regression model, β1 gives the average. – In 1D, B. We'll read this text file directly from the site. In this article we will study another very important dimensionality reduction technique: linear discriminant analysis (or LDA). Agenda Homework Review KNN for Regression (from last week) Robust Regression Gradient. Both QDA and LDA assume that the observations of each class follow a normal distribution; however, QDA assumes that the covariance matrix of each class is different. This gives the model more. Since QDA is specifically applicable in cases with very different covariance structures, it will often be a feature of this plot that one group is spread out while the other is extremely concentrated, typically close to the decision boundary. Notice that by definition the functions are quadratic functions, and therefore the decision boundary for QDA will be described by a quadratic function. We will use the twoClass dataset from Applied Predictive Modeling, the book of M. Draw the ideal decision boundary for the dataset above. The solid vertical black line represents the decision boundary, the balance that obtains a predicted probability of 0. Note: when the number of covariates grow, the number of things to estimate in the covariance matrix gets very large. (2) Linear discriminant analysis (LDA) makes the additional assumption that the covariance of each of the. LDA and QDA The natural model for f i(x) is the multivariate Gaussian distribution f i(x) = 1 p (2ˇ)p det(i) e 1 2 (x i)T i (x i); x 2Rp Linear discriminant analysis (LDA): We assume 1 = 2 = :::= K Quadratic discriminant analysis (QDA): general cases Mathematical techniques in data science. With LDA, the standard deviation is the same for all the classes, while each class has its own. l0x1fddfbjbc9nr, u8ehwyr4o9j4y, hw5rrxc169ns, i34davyz7a, wj0xjhj2fiezlz, fllh3wzeah5f5, kopmrpqvxsas3x, ga208fpusxrvpi, bis0ouwccuhw6, apgz5tjkvm, sslubdo6pitdwx, hdvldo33pxwt4r3, hr3upzmovor2, wzb0yxkdhne2lja, 1bht9u4f4qkb45, 6njh0vw55pyo13, icl7d3gqbc20b, 0hj1gmt2c7jt0, 0oidmsaw9xgbns, 3npf0yk75db8ma, wyuo4qb51g, h2h6e87cwz, rxxjbna7ufp, 7tsvs44j7ovz85, pttd7tjs6wirlej, nb3cbqdahhuuf | 2020-05-31T17:02:37 | {
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http://math.stackexchange.com/questions/376383/does-limit-means-replacing-x-for-a-number | # Does limit means replacing $x$ for a number?
I don't understand limit so much. For example I see $\lim_{x \to -3}$. And I always just put $-3$ everywhere I see $x$. I feel like I'm doing something wrong, but it seems correct all the time.
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You cannot always replace the number. Consider the example $\lim\limits_{x\rightarrow 0}\dfrac{\sin x}{x}=1$ but you cannot replace $0$ – Dimitrios Ntalampekos Apr 29 '13 at 17:10
Three more examples: $\lim_{x\rightarrow 0} \frac{x^2}{x}=0$, $\lim_{x\rightarrow 0} \frac{x^2}{x^2}=1$, $\lim_{x\rightarrow 0} \frac{x}{x^2}$ undefined. – vadim123 Apr 29 '13 at 17:11
Noooo, I'm doing all the time wrong ? Math is more difficult than I thought ... :P – user75045 Apr 29 '13 at 17:12
If you take the function $f(x) = \begin{cases}0, & x=0 \\ 1, & x>0 \end{cases}$, then $\lim_{x \to 0} f(x) = 1$, but $f(0) = 0$. If a function is continuous, then you can evaluate at the limit point. – copper.hat Apr 29 '13 at 17:12
So there is problem if you get $0$ under ? – user75045 Apr 29 '13 at 17:13
Substitution "works" many times; it works but not always: $$\lim_{x\to a} f(x) = f(a)\quad \text{{\bf only} when f(x) is defined and continuous at a}$$
and this is why understanding the limit of a function as the limiting value (or lack of one) when $x$ is approaching a particular value: getting very very near that value, is crucial. That is, $$\lim_{x \to a} f(x) \not\equiv f(a) \qquad\qquad\tag{"\not \equiv"\; here meaning "not identically"}$$
E.g., your "method" won't work for $\;\;\lim_{x\to -3} \dfrac{x^2 - 9}{x + 3}\;\;$ straight off.
Immediate substitution $f(-3)$ evaluates to $\dfrac 00$ which is indeterminate: More work is required. Other examples are given in the comments.
When we seek to find the limit of a function $f(x)$ as $x \to a$, we are seeking the "limiting value" of $f(x)$ as the distance between $x$ and $a$ grows increasingly small. That value is not necessarily the value $f(a)$.
And understanding the "limit" as the "limiting value" or lack there of, of a function is crucial to understanding, e.g. that $\lim_{x \to +\infty} f(x)$ requires examining the behavior of $f(x)$ as $x$ gets arbitrarily (increasingly) large, where evaluating $f(\infty)$ to find the limit makes no sense and has no meaning.
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so you don't replace -3 but, -2.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 – user75045 Apr 29 '13 at 17:14
This is so difficult ! – user75045 Apr 29 '13 at 17:15
So how I can do your problem ? You can make from $x^2-9$ to $(x-3)(x+3)$. Then you have only $(x-3)$. Can I replace that one ? – user75045 Apr 29 '13 at 17:17
Exactly, the example I give requires more work. When you cancel $(x + 3)$ you are left with only $(x - 3)$...and it is indeed safe then to substitute $x = -3$ to see that the limit is $-6$. My caution is that substitution won't always work. When you do substitute, and you are able to obtain a value, then you know that value is the limit. But it is best not to equate the concept of the limit of a function as x approaches a quantity, with it's value AT that quantity. The fact that it is often the case that the $\lim_{x\to a} f(x) = f(a)$ is handy to know, but will not always work. – amWhy Apr 29 '13 at 17:23
So yes, you can oftentimes replace $x$ with the value to which x is approaching to evaluate a limit. But keep in mind what a limit actually means. Go back to the definition of a limit...if necessary, to remind yourself about what is happening when taking a limit, versus merely evaluating a function at a given point. – amWhy Apr 29 '13 at 17:26
(I need to put this onto a T-shirt. :)
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Would the downvoter care to explain the vote? To me, the slogan seems perfectly apt, and the image nicely illustrates (one possible instance of) the notion that a function's behavior when $x$ equals $a$ has no bearing on its behavior as $x$ approaches $a$ ... Confusion about this distinction is the essence of OP's question. – Blue Apr 30 '13 at 8:00
I agree with your comment above, @Blue so +1 from me! (+1 with or without the downvote...I like your post! And it IS a good slogan for the question at hand.) – amWhy Apr 30 '13 at 15:32
If this were on a T-shirt, I would wear it while teaching. – treble Apr 30 '13 at 15:58
Let's go back to the $\epsilon,\delta$ definition of a limit.
Let a function $f$ be defined on an open interval containing the real number $c$. -This mean we aren't going to do silly things like take the square root of negative numbers etc. IN some interval around $c$.
Let $L$ be a real number -Self explanatory.
Then $\lim_{x\rightarrow c}f(x)=L$ if and only if -This is about to be an equivalent statement to what comes next.
For any real number $\epsilon>0$ -So we're allowing $\epsilon$ to be as absolutely small as we like, as long as it's still positive.
There exists a real number $\delta$ with the following property -Ready for this?
For all $x$ if $0<|x-c|<\delta$ then $|f(x)-L|<\epsilon$ -This means that if $|x-c|$ (think of this as the distance between $x$ and $c$) is less than our number $\delta$ then $f(x)$ is arbitrarily close to the real number $L$.
I'm not sure if this was very helpful but let me end by suggesting that you look up visual representations of the above as well as visuals of where a limit does not exist.
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One of the most important limits is the definition of the derivative $$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$ and you cannot substitute $h=0$ in that fraction.
- | 2014-09-23T06:24:14 | {
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https://math.stackexchange.com/questions/2518268/subsets-that-is-linear-independent | # Subsets that is Linear independent
$A$ is the set of linearly independent vectors and $B$ is a nonempty subset of $A$. Then is $B$ also linearly independent?
I know that this is true, since $A$ is LI and this means that no vector in $A$ is a linear combination of others. Then since $B$ is a subset, $B$ is also linear independent.
But I do not know how to show this.
• You just did show it. Now write it more formal. If $(v_1,…,v_n)$ linear independant, then $$\sumλ_iv_i =0 ⇒ λ_i=0, i=1,…,n.$$ Let $w_1,…,w_m$ be a subset ... – P. Siehr Nov 13 '17 at 13:54
• I have to disagree, you didn't already show it. You just wrote the definition of linear independence and then stated the result you want to proof. Admittedly, the result follows easily from the definition, but you still need to write out, why exactly $B$ being a subset of $A$ implies linear independence. – klirk Nov 18 '17 at 23:28
• Two ways to prove it would be either: 1) Suppose $v \in B$. As $B \subset A$ and $A$ is linearly independent, it follows that... 2)Proof by contradiction: Assume for contradiction, that $B$ is not linear independent, then ... – klirk Nov 18 '17 at 23:31
We know that $A$ is a linearly independent set of vectors and $B \subseteq A$. The set $A$ being linearly independent means that, if $x_1,x_2,...,x_n \in A$ and $\alpha_1, \alpha_2,...,\alpha_n$ are scalars such that $$\alpha_1 x_1 +\alpha_2 x_2 + ...+\alpha_nx_n = 0$$ then we have that $\alpha_1 = \alpha_2 = \ ...\ = \alpha_n = 0$.
Now let $B =\{b_1, b_2, ... , b_k \} \subseteq A$, and let's fix a linear combination of vectors from $B$ such that $$\alpha_1 b_1 +\alpha_2 b_2 + ...+\alpha_kb_k = 0$$ Since $B \subseteq A$, this is, $\{b_1, b_2, ... , b_k \} \subseteq A$ and since $A$ is linearly independent, then this means that $\alpha_1 = \alpha_2 = ... = \alpha_k = 0$, which makes $B$ linearly independent.
Assume that $A$ is linearly independent and $B \subseteq A$.
Take any $v_1, \ldots, v_n \in B$ and scalars $\lambda_1, \ldots, \lambda_n$ such that
$$\sum_{i=1}^n \lambda_iv_i = 0$$
Now, since $B\subseteq A$, we have that also $v_1, \ldots, v_n \in A$. So, $\sum_{i=1}^n \lambda_iv_i$ is a linear combination of vectors in $A$, equal to $0$. Because $A$ is linearly independent, we obtain that the scalars are $0$:
$$\lambda_1 = \ldots = \lambda_n = 0$$
Since $v_1, \ldots, v_n$ were arbitrary, this proves that $B$ is linearly independent.
A linearly independent set of vectors need not be a finite set. The definition of "$A$ is linearly independent" is that if $C$ is any non-empty finite subset of $A$ and $\sum_{v\in C}vk_v=0$ then $k_v=0$ for all $v\in C.$
Suppose $A$ is linearly independent and $B\subset A.$ Then any finite non-empty $C\subset B$ is a finite non-empty subset of $A$, so if $\sum_{v\in C}vk_v=0$ then $k_v=0$ for all $v\in C.$ | 2019-07-18T23:23:01 | {
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https://math.stackexchange.com/questions/1399008/using-right-hand-riemann-sum-to-evaluate-the-limit-of-fracnn21-cdots | Using right-hand Riemann sum to evaluate the limit of $\frac{n}{n^2+1}+ \cdots+\frac{n}{n^2+n^2}$
I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:
\begin{align*} \lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)&=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{n}{n^2+k^2}\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{n^2}{n^2+k^2})\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{1}{1+(k/n)^2})\\ &=\lim_{n \to \infty} \sum_{k=1}^{n}f(\frac{k}{n})(\frac{k-(k-1)}{n})\\ &=\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4} \end{align*}
where $f(x)=\frac{1}{1+x^2}$. Is this correct, are there any steps where I am not clear?
• Though the derivation is correct, you shouldn't use the $\lim$ symbol until you are sure that the limit exists. – yoann Aug 16 '15 at 14:59
• What steps must I go through in order to show that the limit exists? – lamyvista Aug 16 '15 at 18:49
• Usually you would perform your computations, and show that the last term has a limit. An easy way to avoid doing mistakes is to write $A_n = B_n = \dots = Z_n \longrightarrow_{n \to +\infty} \ell$. – yoann Aug 16 '15 at 19:52
Notice, we have $$\lim_{n\to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\dots +\frac{n}{n^2+n^2}\right)=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{n}{n^2+r^2}$$
$$\lim_{n\to \infty}\sum_{r=1}^{n}\frac{n}{n^2+r^2}=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\frac{1}{n}}{1+\left(\frac{r}{n}\right)^2}$$ Let, $\frac{r}{n}=x\implies \lim_{n\to \infty}\frac{1}{n}=dx\to 0$
$$\text{upper limit of x}=\lim_{n\to \infty }\frac{n}{n}=1$$ $$\text{lower limit of x}=\lim_{n\to \infty }\frac{1}{n}=0$$ Hence, using integration with proper limits, we get
$$\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\frac{1}{n}}{1+\left(\frac{r}{n}\right)^2}= \int_ {0}^{1}\frac{dx}{1+x^2}$$ $$=\left[\tan^{-1}(x)\right]_{0}^{1}$$ $$=\left[\tan^{-1}(1)-\tan^{-1}(0)\right]$$ $$=\left[\frac{\pi}{4}-0\right]=\frac{\pi}{4}$$
• Is the extra dx in your integrand a typo? Thanks! – lamyvista Aug 16 '15 at 8:27
• Thank you for observation! – Harish Chandra Rajpoot Aug 16 '15 at 8:29
• Forgive my ignorance, but could you explain how $lim_{n\to \infty}\frac{1}{n}=dx$ – mysatellite Aug 16 '15 at 15:44
• @Sky $\lim_{n\to\infty}\frac1n\ne\text{d}x$, that would be gibberish. What's happening is Harish is using the fact that $\lim_{n\to\infty}\frac1n\sum_{r=1}^nf\left(\frac{r}{n}\right)=\int_0^1f(x)\text{d}x$, which is the definition of Riemann integration. – user3002473 Aug 16 '15 at 15:58
Yes, what you've written is correct.
I think your write-up would be cleaner if you explicitly mentioned the antiderivative $$\int \frac{1}{1 + x^2} dx = \arctan x,$$ which makes it completely clear where $\frac{\pi}{4}$ comes from at the end. | 2019-12-08T03:05:15 | {
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http://linear.ups.edu/fcla/section-MR.html | We have seen that linear transformations whose domain and codomain are vector spaces of columns vectors have a close relationship with matrices (Theorem MBLT, Theorem MLTCV). In this section, we will extend the relationship between matrices and linear transformations to the setting of linear transformations between abstract vector spaces.
This is a fundamental definition.
##### DefinitionMRMatrix Representation
Suppose that $\ltdefn{T}{U}{V}$ is a linear transformation, $B=\set{\vectorlist{u}{n}}$ is a basis for $U$ of size $n\text{,}$ and $C$ is a basis for $V$ of size $m\text{.}$ Then the matrix representation of $T$ relative to $B$ and $C$ is the $m\times n$ matrix, \begin{equation*} \matrixrep{T}{B}{C}=\left[ \left.\vectrep{C}{\lteval{T}{\vect{u}_1}}\right| \left.\vectrep{C}{\lteval{T}{\vect{u}_2}}\right| \left.\vectrep{C}{\lteval{T}{\vect{u}_3}}\right| \ldots \left|\vectrep{C}{\lteval{T}{\vect{u}_n}}\right. \right]\text{.} \end{equation*}
We may choose to use whatever terms we want when we make a definition. Some are arbitrary, while others make sense, but only in light of subsequent theorems. Matrix representation is in the latter category. We begin with a linear transformation and produce a matrix. So what? Here is the theorem that justifies the term matrix representation.
##### Proof
This theorem says that we can apply $T$ to $\vect{u}$ and coordinatize the result relative to $C$ in $V\text{,}$ or we can first coordinatize $\vect{u}$ relative to $B$ in $U\text{,}$ then multiply by the matrix representation. Either way, the result is the same. So the effect of a linear transformation can always be accomplished by a matrix-vector product (Definition MVP). That is important enough to say again. The effect of a linear transformation is a matrix-vector product.
The alternative conclusion of this result might be even more striking. It says that to effect a linear transformation ($T$) of a vector ($\vect{u}$), coordinatize the input (with $\vectrepname{B}$), do a matrix-vector product (with $\matrixrep{T}{B}{C}$), and un-coordinatize the result (with $\vectrepinvname{C}$). So, absent some bookkeeping about vector representations, a linear transformation is a matrix. To adjust the diagram, we “reverse” the arrow on the right, which means inverting the vector representation $\vectrepname{C}$ on $V\text{.}$ Now we can go directly across the top of the diagram, computing the linear transformation between the abstract vector spaces. Or, we can around the other three sides, using vector representation, a matrix-vector product, followed by un-coordinatization.
Here is an example to illustrate how the “action” of a linear transformation can be effected by matrix multiplication.
We will use Theorem FTMR frequently in the next few sections. A typical application will feel like the linear transformation $T$ “commutes” with a vector representation, $\vectrepname{C}\text{,}$ and as it does the transformation morphs into a matrix, $\matrixrep{T}{B}{C}\text{,}$ while the vector representation changes to a new basis, $\vectrepname{B}\text{.}$ Or vice-versa.
# SubsectionNRFONew Representations from Old¶ permalink
In Subsection LT.NLTFO we built new linear transformations from other linear transformations. Sums, scalar multiples and compositions. These new linear transformations will have matrix representations as well. How do the new matrix representations relate to the old matrix representations? Here are the three theorems.
##### Proof
The vector space of all linear transformations from $U$ to $V$ is now isomorphic to the vector space of all $m\times n$ matrices.
##### Proof
This is the second great surprise of introductory linear algebra. Matrices are linear transformations (functions, really), and matrix multiplication is function composition! We can form the composition of two linear transformations, then form the matrix representation of the result. Or we can form the matrix representation of each linear transformation separately, then multiply the two representations together via Definition MM. In either case, we arrive at the same result.
A diagram, similar to ones we have seen earlier, might make the importance of this theorem clearer.
One of our goals in the first part of this book is to make the definition of matrix multiplication (Definition MVP, Definition MM) seem as natural as possible. However, many of us are brought up with an entry-by-entry description of matrix multiplication (Theorem EMP) as the definition of matrix multiplication, and then theorems about columns of matrices and linear combinations follow from that definition. With this unmotivated definition, the realization that matrix multiplication is function composition is quite remarkable. It is an interesting exercise to begin with the question, “What is the matrix representation of the composition of two linear transformations?” and then, without using any theorems about matrix multiplication, finally arrive at the entry-by-entry description of matrix multiplication. Try it yourself (Exercise MR.T80).
Click to open
# SubsectionPMRProperties of Matrix Representations¶ permalink
It will not be a surprise to discover that the kernel and range of a linear transformation are closely related to the null space and column space of the transformation's matrix representation. Perhaps this idea has been bouncing around in your head already, even before seeing the definition of a matrix representation. However, with a formal definition of a matrix representation (Definition MR), and a fundamental theorem to go with it (Theorem FTMR) we can be formal about the relationship, using the idea of isomorphic vector spaces (Definition IVS). Here are the twin theorems.
##### Proof
An entirely similar result applies to the range of a linear transformation and the column space of a matrix representation of the linear transformation.
##### Proof
Theorem KNSI and Theorem RCSI can be viewed as further formal evidence for the The Coordinatization Principle, though they are not direct consequences.
Figure 8.38 is meant to suggest Theorem KNSI and Theorem RCSI, in addition to their proofs (and so carry the same notation as the statements of these two theorems). The dashed lines indicate a subspace relationship, with the smaller vector space lower down in the diagram. The central square is highly reminiscent of Figure 8.24. Each of the four vector representations is an isomorphism, so the inverse linear transformation could be depicted with an arrow pointing in the other direction. The four vector spaces across the bottom are familiar from the earliest days of the course, while the four vector spaces across the top are completely abstract. The vector representations that are restrictions (far left and far right) are the functions shown to be invertible representations as the key technique in the proofs of Theorem KNSI and Theorem RCSI. So this diagram could be helpful as you study those two proofs.
##### SageLTRLinear Transformation Restrictions
Click to open
We have seen, both in theorems and in examples, that questions about linear transformations are often equivalent to questions about matrices. It is the matrix representation of a linear transformation that makes this idea precise. Here is our final theorem that solidifies this connection.
##### Proof
By now, the connections between matrices and linear transformations should be starting to become more transparent, and you may have already recognized the invertibility of a matrix as being tantamount to the invertibility of the associated matrix representation. The next example shows how to apply this theorem to the problem of actually building a formula for the inverse of an invertible linear transformation.
You might look back at Example AIVLT, where we first witnessed the inverse of a linear transformation and recognize that the inverse ($S$) was built from using the method of Example ILTVR with a matrix representation of $T\text{.}$
##### Proof
This theorem may seem gratuitous. Why state such a special case of Theorem IMR? Because it adds another condition to our NMEx series of theorems, and in some ways it is the most fundamental expression of what it means for a matrix to be nonsingular — the associated linear transformation is invertible. This is our final update.
Click to open
##### 1
Why does Theorem FTMR deserve the moniker “fundamental”?
##### 2
Find the matrix representation, $\matrixrep{T}{B}{C}$ of the linear transformation \begin{equation*} \ltdefn{T}{\complex{2}}{\complex{2}},\quad\lteval{T}{\colvector{x_1\\x_2}}=\colvector{2x_1-x_2\\3x_1+2x_2} \end{equation*} relative to the bases \begin{align*} B&=\set{\colvector{2\\3},\,\colvector{-1\\2}}& C&=\set{\colvector{1\\0},\,\colvector{1\\1}}\text{.} \end{align*}
##### 3
What is the second “surprise,” and why is it surprising?
# SubsectionExercises
##### C10
Example KVMR concludes with a basis for the kernel of the linear transformation $T\text{.}$ Compute the value of $T$ for each of these two basis vectors. Did you get what you expected?
##### C20
Compute the matrix representation of $T$ relative to the bases $B$ and $C\text{,}$ \begin{gather*} \ltdefn{T}{P_3}{\complex{3}},\quad \lteval{T}{a+bx+cx^2+dx^3}= \colvector{2a-3b+4c-2d\\a+b-c+d\\3a+2c-3d}\\ B=\set{1,\,x,\,x^2,\,x^3}\quad\quad C=\set{\colvector{1\\0\\0},\,\colvector{1\\1\\0},\,\colvector{1\\1\\1}}\text{.} \end{gather*}
Solution
##### C21
Find a matrix representation of the linear transformation $T$ relative to the bases $B$ and $C\text{,}$ \begin{align*} &\ltdefn{T}{P_2}{\complex{2}},\quad \lteval{T}{p(x)}=\colvector{p(1)\\p(3)}\\ &B=\set{2-5x+x^2,\,1+x-x^2,\,x^2}\\ &C=\set{\colvector{3\\4},\,\colvector{2\\3}}\text{.} \end{align*}
Solution
##### C22
Let $S_{22}$ be the vector space of $2\times 2$ symmetric matrices. Build the matrix representation of the linear transformation $\ltdefn{T}{P_2}{S_{22}}$ relative to the bases $B$ and $C$ and then use this matrix representation to compute $\lteval{T}{3+5x-2x^2}\text{,}$ \begin{align*} B&=\set{1,\,1+x,\,1+x+x^2} & C&=\set{ \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix},\, \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix},\, \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} }\\ \lteval{T}{a+bx+cx^2}&= \begin{bmatrix} 2a-b+c & a+3b-c \\ a+3b-c & a-c \end{bmatrix}\text{.} \end{align*}
Solution
##### C25
Use a matrix representation to determine if the linear transformation $\ltdefn{T}{P_3}{M_{22}}$ is surjective. \begin{equation*} \lteval{T}{a+bx+cx^2+dx^3}= \begin{bmatrix} -a+4b+c+2d & 4a-b+6c-d\\ a+5b-2c+2d & a+2c+5d \end{bmatrix} \end{equation*}
Solution
##### C30
Find bases for the kernel and range of the linear transformation $S$ below. \begin{equation*} \ltdefn{S}{M_{22}}{P_2},\quad\lteval{S}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}= (a+2b+5c-4d)+(3a-b+8c+2d)x+(a+b+4c-2d)x^2 \end{equation*}
Solution
##### C40
Let $S_{22}$ be the set of $2\times 2$ symmetric matrices. Verify that the linear transformation $R$ is invertible and find $\ltinverse{R}\text{.}$ \begin{equation*} \ltdefn{R}{S_{22}}{P_2},\quad\lteval{R}{\begin{bmatrix}a&b\\b&c\end{bmatrix}}= (a-b)+(2a-3b-2c)x+(a-b+c)x^2 \end{equation*}
Solution
##### C41
Prove that the linear transformation $S$ is invertible. Then find a formula for the inverse linear transformation, $\ltinverse{S}\text{,}$ by employing a matrix inverse. \begin{equation*} \ltdefn{S}{P_1}{M_{12}},\quad \lteval{S}{a+bx}= \begin{bmatrix} 3a+b & 2a+b \end{bmatrix} \end{equation*}
Solution
##### C42
The linear transformation $\ltdefn{R}{M_{12}}{M_{21}}$ is invertible. Use a matrix representation to determine a formula for the inverse linear transformation $\ltdefn{\ltinverse{R}}{M_{21}}{M_{12}}\text{.}$ \begin{equation*} \lteval{R}{\begin{bmatrix}a & b\end{bmatrix}} = \begin{bmatrix} a+3b\\ 4a+11b \end{bmatrix} \end{equation*}
Solution
##### C50
Use a matrix representation to find a basis for the range of the linear transformation $L\text{.}$ \begin{equation*} \ltdefn{L}{M_{22}}{P_2},\quad \lteval{T}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}= (a+2b+4c+d)+(3a+c-2d)x+(-a+b+3c+3d)x^2 \end{equation*}
Solution
##### C51
Use a matrix representation to find a basis for the kernel of the linear transformation $L\text{.}$ \begin{equation*} \ltdefn{L}{M_{22}}{P_2},\quad \lteval{T}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}= (a+2b+4c+d)+(3a+c-2d)x+(-a+b+3c+3d)x^2 \end{equation*}
##### C52
Find a basis for the kernel of the linear transformation $\ltdefn{T}{P_2}{M_{22}}\text{.}$ \begin{equation*} \lteval{T}{a+bx+cx^2}= \begin{bmatrix} a+2b-2c & 2a+2b \\ -a+b-4c & 3a+2b+2c \end{bmatrix} \end{equation*}
Solution
##### M20
The linear transformation $D$ performs differentiation on polynomials. Use a matrix representation of $D$ to find the rank and nullity of $D\text{.}$ \begin{equation*} \ltdefn{D}{P_n}{P_n},\quad \lteval{D}{p(x)}=p^\prime(x) \end{equation*}
Solution
##### M60
Suppose $U$ and $V$ are vector spaces and define a function $\ltdefn{Z}{U}{V}$ by $\lteval{Z}{\vect{u}}=\zerovector_{V}$ for every $\vect{u}\in U\text{.}$ Then Exercise IVLT.M60 asks you to formulate the theorem: $Z$ is invertible if and only if $U=\set{\zerovector_U}$ and $V=\set{\zerovector_V}\text{.}$ What would a matrix representation of $Z$ look like in this case? How does Theorem IMR read in this case?
##### T20
Construct a new solution to Exercise B.T50 along the following outline. From the $n\times n$ matrix $A\text{,}$ construct the linear transformation $\ltdefn{T}{\complex{n}}{\complex{n}}\text{,}$ $\lteval{T}{\vect{x}}=A\vect{x}\text{.}$ Use Theorem NI, Theorem IMILT and Theorem ILTIS to translate between the nonsingularity of $A$ and the surjectivity/injectivity of $T\text{.}$ Then apply Theorem ILTB and Theorem SLTB to connect these properties with bases.
Solution
##### T40
Theorem VSLT defines the vector space $\vslt{U}{V}$ containing all linear transformations with domain $U$ and codomain $V\text{.}$ Suppose $\dimension{U}=n$ and $\dimension{V}=m\text{.}$ Prove that $\vslt{U}{V}$ is isomorphic to $M_{mn}\text{,}$ the vector space of all $m\times n$ matrices (Example VSM). (Hint: we could have suggested this exercise in Chapter LT, but have postponed it to this section. Why?)
##### T41
Theorem VSLT defines the vector space $\vslt{U}{V}$ containing all linear transformations with domain $U$ and codomain $V\text{.}$ Determine a basis for $\vslt{U}{V}\text{.}$ (Hint: study Exercise MR.T40 first.)
##### T60
Create an entirely different proof of Theorem IMILT that relies on Definition IVLT to establish the invertibility of $T\text{,}$ and that relies on Definition MI to establish the invertibility of $A\text{.}$
##### T80
Suppose that $\ltdefn{T}{U}{V}$ and $\ltdefn{S}{V}{W}$ are linear transformations, and that $B\text{,}$ $C$ and $D$ are bases for $U\text{,}$ $V\text{,}$ and $W\text{.}$ Using only Definition MR define matrix representations for $T$ and $S\text{.}$ Using these two definitions, and Definition MR, derive a matrix representation for the composition $\compose{S}{T}$ in terms of the entries of the matrices $\matrixrep{T}{B}{C}$ and $\matrixrep{S}{C}{D}\text{.}$ Explain how you would use this result to motivate a definition for matrix multiplication that is strikingly similar to Theorem EMP.
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https://math.stackexchange.com/questions/2920366/c-p-mathbbz-p-times-mathbbz-p-is-a-field-for-some-prime-p | # $C_p = \mathbb{Z}_p \times \mathbb{Z}_p$ is a field for some prime $p$.
Let $p$ a prime number. $\mathbb{Z}_p \times \mathbb{Z}_p$ with sum given by $(a,b)+(c,d) = (a+c,b+d)$ and multiplication given by $(a,b)*(c,d)=(ac-bd,ad+bc)$ is a field for some prime greater than two and $C_2$ is not a field.
What I did: $C_2$ is not a field, because $(1,0)$ is the neutral, and $(1,1)*(c,d) = (1,0)$ gives $c-d=1, d+c=0$, which have no solution in $\mathbb{Z}_2$, then it is not a field. But I don't know how to do for $p>2$, because all I know is check case by case, and I think it us not supposed to be that much of work.
Thanks.
The idea is that this multiplication is the same multiplication for complex numbers:
$$(a + bi)(c + di) = (ac - bd) + (ad + bc)i.$$
Therefore, for $p > 2$, $C_p$ is a field if and only if $i \notin \mathbb Z_p$ meaning no element of $\mathbb Z_p$ has square $-1$. (Why is this true?)
• $i=(0,1)$ in Z_2, right? How can I see $i$ in the other Z_p? – User1618 Sep 17 '18 at 16:28
• @Alnitak $\mathbb Z_2$ is a special case. But for example in $\mathbb Z_5$ we have $2^2 = 4 = -1$ and accordingly $(2 + i)(2 - i) = 2^2 + 1^2 = 0$ which in $C_p$ implies that $(2,1)(2,-1) = (0,0)$. This creates an issue in $C_5$ because $2 = \sqrt{-1}$ so $(2,0) = i$ but we also want $(0,1) = i$. If $i \notin \mathbb Z_p$ then we can write $(a, b) = a + bi$ and perform multiplication as if we had adjoined a root of $-1$ to our ring. – Trevor Gunn Sep 17 '18 at 19:01
• Do you know which is the least prime that makes C_p field? I checked 3,7,11 and 13 and got not field. Maybe I'm not doing it right... – User1618 Sep 17 '18 at 23:00
• @Alnitak $p = 3$ should be a field. – Trevor Gunn Sep 18 '18 at 0:09
I find this question to be quite intrigueing because it addresses the issue of extending the underlying structure of the field of complex numbers $$\Bbb C$$ to arbitrary fields, in particular to arbitrary finite fields; in a sense we might say it is about the process of complexification, with which we are familiar from $$\Bbb R \subset \Bbb C$$, to general fields $$F$$.
In the case $$F = \Bbb R$$, which is somewhat prototypical, we may begin by considering the possibility of products on $$\Bbb R^2 = \Bbb R \times \Bbb R$$ which, given that addition is defined component-wise,
$$(a, b) + (c, d) = (a + c, b + d), \; a, b, c, d \in \Bbb R, \tag 1$$
are consistent with the field axioms. For example, we may immediately rule out component-wise multiplication,
$$(a, b)(c, d) = (ac, bd), \tag 2$$
the identity element of which is $$(1, 1)$$, because it has zero-divisors, e.g., assuming $$a \ne 0 \ne d$$,
$$(a, 0)(0, d) = (a \cdot 0, 0 \cdot d) = (0, 0), \tag 3$$
even though
$$(a, 0) \ne (0, 0) \ne (0, d); \tag 4$$
thus $$\Bbb R^2$$ with multiplication defined by (2) cannot be a field; if, however we take
$$(a, b)(c, d) = (ac - bd, ad + bc), \tag 5$$
to be our product, then if
$$(a, b)(c, d) = (0, 0), \tag 6$$
noting that
$$(a, -b)(a, b) = (a, b)(a, -b) = (a^2 + b^2, 0) \tag 7$$
we find
$$((a^2 + b^2)c, (a^2 + b^2)d) = (a^2 + b^2, 0)(c, d) = (a, -b)(a, b)(c, d) = (a, -b)(0, 0) = (0, 0), \tag 8$$
whence
$$(a^2 + b^2)c = (a^2 + b^2)d = 0, \tag 9$$
which, in the event that $$a^2 + b^2 \ne 0$$, forces
$$c = d = 0, \tag{10}$$
and thus we see there are no zero divisors in $$\Bbb R^2$$ with respect to the product rule (5); to put it another way, $$\Bbb R^2$$ is an integral domain with the operations (1), (5); in fact, we may show it to be a field under such circumstances, for then we have
$$(a(a^2 + b^2)^{-1}, -b(a^2 + b^2)^{-1})(a, b) = (a^2(a^2 + b^2)^{-1} + b^2(a^2 + b^2)^{-1}), 0)$$ $$= ((a^2 + b^2)(a^2 + b^2)^{-1}, 0) = (1, 0), \tag{11}$$
which shows that every $$(a, b) \ne (0, 0)$$ has an inverse in $$\Bbb R^2$$, and thus the field axioms are satisfied.
We wish to investigate when the choice of multiplication operation (5) makes $$F^2 = F \times F$$ into a field; i.e, when $$\Bbb R$$ is replaced by an arbitrary field $$F$$.
We accept that $$F \times F$$ with "addition" given by (1) and multiplication by (5) forms a commutative ring with unit given by $$(1, 0)$$; these assertions are most easily checked; our concern then is with the existence of multiplicative inverses.
If we trace through the calculations presented in (1) through (12) above we see that many of the essential points carry over directly from the case $$\Bbb R^2 = \Bbb R \times \Bbb R$$ to $$F^2 = F \times F$$; for instance, it is clearly shown above that
$$a^2 + b^2 \ne 0 \Longrightarrow \exists (a, b)^{-1}, \tag{12}$$
and we may simply prove the converse
$$\exists (a, b)^{-1} \Longrightarrow a^2 + b^2 \ne 0: \tag{13}$$
if $$(a, b)$$ has an inverse, then there is $$(c, d)$$ such that
$$(a, b)(c, d) = (1, 0); \tag{14}$$
then
$$(a, -b)(a, b)(c, d) = (a, -b), \tag{15}$$
or
$$(a^2 + b^2, 0)(c, d) = (a, -b), \tag{16}$$
or
$$(a^2 + b^2)c = a, \; (a^2 + b^2)d = -b; \tag{17}$$
now $$(a, b) \ne (0, 0)$$ lest (15) fail, and thus $$a^2 + b^2 \ne 0$$ as well. It follows then that $$F \times F$$ is a field with multiplication (5) provided
$$\forall (0, 0) \ne (a, b) \in F \times F, \; 0 \ne a^2 + b^2 \in F; \tag{18}$$
so then $$F \times F$$ will obey the field axioms if and only if the function $$a^2 + b^2$$ has no nontrivial zeroes. For finite $$F$$, this may be checked manually, one pair $$(a, b)$$ at a time; if $$\vert F \vert$$ is not too large, this is a feasible calculation. The amount of effort may be reduced a little bit by noting that
$$a^2 + b^2 = 0 \Longrightarrow [a, b \ne 0 ] \vee [a = b = 0]; \tag{19}$$
that is, either $$a$$ and $$b$$ are both zero, or neither is zero; this is easy to see: for example, if $$a = 0$$, then $$a^2 + b^2 = 0$$ reduces to
$$b^2 = 0 \Longrightarrow b = 0; \tag{20}$$
thus in fact we only need evaluate $$a^2 + b^2$$ on those pairs $$a \ne 0 \ne b$$.
Some examples: in $$\Bbb Z_2$$, have $$1^2 + 1^2 = 0$$, so $$\Bbb Z_2 \times \Bbb Z_2$$ is not a field with multiplication (5); for $$\Bbb Z_3$$, however, we have
$$1^2 + 1^2 = 1 + 1 = 2; \; 1^2 + 2^2 = 1 + 1 = 2; \; 2^2 + 2^2 = 1 + 1 = 2, \tag{21}$$
so $$\Bbb Z_3 \times \Bbb Z_3$$ is a field; in $$\Bbb Z_5$$, however,
$$1^2 + 2^2 = 1 + 4 = 0, \tag{22}$$
so we rule it out; next, in $$\Bbb Z_7$$,
$$1^2 + 1^2 = 2; \; 1^2 + 2^2 = 5; \; 1^2 + 3^2 = 3; \; 1^2 + 4^2 = 3; \; 1^2 + 5^2 = 5; \; 1^2 + 6^2 = 2; \tag{23}$$
$$2^2 + 2^2 = 1; \; 2^2 + 3^2 = 6; \; 2^2 + 4^2 = 6; \; 2^2 + 5^2 = 1; 2^2 + 6^2 = 5; \tag{24}$$
$$3^2 + 3^2 = 4; \; 3^2 + 4^2 = 4; \; 3^2 + 5^2 = 6; \; 3^2 + 6^2 = 3; \tag{25}$$
$$4^2 + 4^2 = 4; \; 4^2 + 5^2 = 6; \; 4^2 + 6^2 = 3; \tag{26}$$
$$5^2 + 5^2 = 1; \; 5^2 + 6^2 = 5; \tag{27}$$
$$6^2 + 6^2 = 2; \tag{28}$$
to check $$\Bbb Z_{11}$$ in this manner, that is, attempting to verify $$a^2 + b^2 \ne 0$$ for each pair $$a, b \in \Bbb Z_{11}$$, is really starting to look like a lot of work; however, there is an expedient which can save us hella lot of arithmetic base $$11$$ or in any other field, for that matter. We note that, for $$0 \ne a, b \in F$$, the equation
$$a^2 + b^2 = 0 \tag{29}$$
may be written
$$\left ( \dfrac{a}{b} \right )^2 + 1= 0; \tag{30}$$
setting
$$x = \dfrac{a}{b}, \tag{31}$$
we see that (29) has a legitimate solution only if there is $$x \in F$$ such that
$$x^2 + 1 = 0; \tag{32}$$
likewise, if (32) is satisfied for some $$x \in F$$, then we may set $$a = bx$$ for any $$0 \ne b \in F$$ and
$$a^2 + b^2 = x^2 b^2 + b^2 = (x^2 + 1) b^2 = 0; \tag{33}$$
so we see that $$F \times F$$ is a field with multiplication (5) whenever (32) has no solutions in $$F$$, for then every $$(a, b) \ne (0, 0)$$ has an inverse. In the light of (32), we may check $$\Bbb Z_{11}$$ with relative ease:
$$1^2 = 1, \; 2^2 = 4, \; 3^2 = 9, \; 4^2 = 5, \; 6^2 = 3, \; 7^2 = 5, \; 8^2 = 9, \; 9^2 = 4, \; 10^2 = 1; \tag{34}$$
we conclude that $$\Bbb Z_{11}$$ is a field with the product law (5).
To finish off our list, we check "Lucky" $$\Bbb Z_{13}$$: without presenting the list of all its squares, we simply note that
$$5^2 = 25 = -1 \mod{13}, \tag{35}$$
so $$\Bbb Z_{13}$$ may not be "complexified" according to our scheme.
The upshot of all this is that our "complexifcation" of $$F \times F$$ according to the multiplication (5) goes through precisely for those $$F$$ in which there is no $$x$$ for which $$x^2 + 1 = 0$$, that is, in which a so-called "imaginary unit" does not already exist; when such an $$x$$ does exist, there will be non-invertible elements with respect to (5) in $$F \times F$$.
One may in fact inquire further into finite fields $$F$$ which are not of the form $$\Bbb Z_p$$; for example, we may look into the field $$\Bbb Z_3[x] / (x^2 + x + 2)$$, which has $$3^2 = 9$$ elements all of which may be written as $$ay + b$$ with $$a,b \in \Bbb Z_3$$ and $$y^2 = -y - 2$$; can
$$a^2 y^2 + 2aby + b^2 + 1 = (ay + b)^2 + 1 = 0? \tag{36}$$
Well, (36) yields
$$(2ab - a^2)y + b^2 - 2a^2 + 1$$ $$= a^2 (-y - 2) + 2aby + b^2 + 1 = a^2 y^2 + 2ab y + b^2 + 1 = 0, \tag{37}$$
so (36) is equivalent to
$$2ab - a^2 = 0, \; 2a^2 - b^2 = -1; \tag{38}$$
if $$a \ne 0$$, then
$$a = 2b, \; 2a^2 - b^2 = 8b^2 - b^2 = b^2 = -1 \Longrightarrow b^2 + 1 = 0, \tag{39}$$
which we have already ruled out in $$\Bbb Z_3$$ (cf. ca. (21)); the case $$a = 0$$ has already been addressed since then $$ay + b \in \Bbb Z_3$$.
One could of course continue and attempt similar calculations for many other finite fields of the form $$\Bbb Z_p / (p(x))$$, or for other cases such as $$\Bbb Q(\omega)$$, where $$\omega$$ is a root of unity, but having illustrated these things, I'm ready to stop here.
Nota Bene: In writing up this answer, I did a little research and discovered that in the general theory of quadratic residues it is proved that $$-1$$ is a square in those $$\Bbb Z_p$$ with $$p \equiv 1 \mod 4$$ and is not a square if $$p \equiv 3 \mod 4$$, a fact corroborated by the cases considered here. End of Note.
Hint: $(2,1)\times(2,-1)=(0,0)$ in $C_5$
• You ought to show this is a field for some $p$, not that it isn't. – Wojowu Sep 17 '18 at 16:28
• Well, it dealt with half the possible $p$. – Empy2 Sep 17 '18 at 16:40 | 2020-08-09T23:29:42 | {
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http://techiemathteacher.com/2014/09/15/mtap-saturday-class-session-6/ | # Solving Problems in MTAP Saturday Class Session 6 for Grade 8
One of the Mathletes named Patrick John Uy, a grade 8 student from F Torres High School sent problems of their session 6 Saturday MTAP program. As a medium of help to those students like him who are eager to earn. We will demonstrate the solutions to problems that bothers them.
Problem 1:
Express the following products as reduced fractions.
$(1-\displaystyle\frac{1}{2})(1-\displaystyle\frac{1}{3})(1-\displaystyle\frac{1}{4})\ldots(1-\displaystyle\frac{1}{2013})$
Solution:
Observe the following,
$1-\displaystyle\frac{1}{2}=\displaystyle\frac{1}{2}$
$1-\displaystyle\frac{1}{3}=\displaystyle\frac{2}{3}$
$1-\displaystyle\frac{1}{1}=\displaystyle\frac{3}{4}$
.
.
.
$1-\displaystyle\frac{1}{2013}=\displaystyle\frac{2012}{2013}$
By expressing them to product of their differences we have,
$(\displaystyle\frac{1}{2})(\displaystyle\frac{2}{3})(\displaystyle\frac{3}{4})\ldots(\displaystyle\frac{2012}{2013})$
Cancel out terms from 2 to 2012 we only left with
$\displaystyle\frac{1}{2013}$
Problem 2:
One angle of the triangle has a measure of 40º. Find the measures of the other two angles if the difference between their measures is 10º.
Solution:
Use the fact that the sum of the interior of a triangle is 180º.
Let A, B, and C are the measures of the three angles,
$A+B+C=180$
But one of the angles is 40º
$40+B+C=180$
$B+C=140$
It is given in the remaining angles that the difference of their measures is 10º, hence
$C-10+C=140$
$2C=150$
$C=75^\circ$
The other angle (B) is 10 degrees less than the angle C. Thus B=75-10=65º.
Worked Problem 3:
The height of a cylindrical container of radius r is 30 cm. what will be the height of the water if it is poured in another cylindrical container which is four times the radius of the other.
Solution:
Let $r_1$ be the radius of the first container
Let $h_1$ be the height of the first container
Let $r_2$ be the radius of the second container
Let $h_2$ be the height of the water in the second container
Given: $h_1=30 cm$, $r_2=4r_1$
Using the formula for the volume of the 1st Cylinder we have,
$V_1=\pi r_1^2h_1$
$V_1=\pi r_1^2(30)$
$V_1=30\pi r_1^2$
Using the formula for the volume of the second Cylinder we have,
$V_2=\pi r_2^2h_2$
But, $r_2=4r_1$
$V_2=\pi (4r_1)^2h_2$
$V_2=16\pi r_1^2h_2$
Using a little common sense, the volume of the water in both container must be the same. ( Math approach)
$V_1=V_2$
$30\pi r_1^2=16\pi r_1^2h_2$
Cancel out like terms in both sides of equation and solving for $h_2$ we have,
$h_2=\displaystyle\frac{30}{16}=\displaystyle\frac{15}{8} cm$
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### Dan
Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.
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http://mathhelpforum.com/geometry/134179-circle-problem.html | # Math Help - Circle Problem
1. ## Circle Problem
Im currently studying for an exam which frankly is way over my head, so I want to see who can help me, right now with this particular problem. I need not only the solution but how to solve and how to apply it to other situations. Thank you and have a good day.
Denoted by D, the domain {(x,y)}| x>=0, y>=0}
Assume a circle C contained in D touches parabola y=(1/2)x^2 at the point (2,2) and also touches the x-axis. Find the radius of C.
Please show all steps and explain them to me as I might need to apply this to other problems.
2. Just a quick note, I will not have a calculator or any reference available so rather than the answer I must have the method you use to get to the answer, even if you dont solve it but can tell me how to solve it then I would be a happy man because right now Im failing in getting any real answers.
3. i)By differentiating the equation of parabola, find the slope m at the point P.(2,2).
ii) Find the equation tangent at (2,2) using the formula (y-y1) = m(x-x1)
iii) Find the point of intersection of the tangent and the x-axis. (x,0)
iv) Find the distance between the point of intersection X and P(2,2)
v) From the slope find the angle between the tangent and the x-axis.
vi) If C is the center of the circle, then CPX is a right angled triangle.
vii) The angle PXC is half the angle made by the tangent with x-axis.
viii) tan(PXC) gives you the required result.
4. Just making sure, this would work with any circle that touches only one point of a parabola or it just work in this case because it touches the x-axis?
PS: I have tried you method and all seems to be going well till step v). If you could show me with this problem how it done it would be great.
5. Originally Posted by Solaris123
Just making sure, this would work with any circle that touches only one point of a parabola or it just work in this case because it touches the x-axis?
PS: I have tried you method and all seems to be going well till step v). If you could show me with this problem how it done it would be great.
This method works only when the circle touches the x-axis.
Slope of the tangent is 2. So tanθ = 2
The angle made by the tangent with x-axis is 63.4 degrees. Half of that angle = angle PXC = 31.7 degrees.
PC/PX = tan(31.7).
So the radius PC = PX*tan(31.7)
6. I wanted to ask to see if anyone else had a method that didnt involve tangents? I will lack a calculator and finding those angles to do this seems a little unwidely seeing how this arent special right triangles, thus if someone else has another method which you dont need a calculator to solve it would be great.
7. ## Teacher confirmed it
My teacher confirmed I had to find the solution by another method other than this because he wont let us use calculators. Is there anyway to use the definition the formula of a circle to solve this?
8. I know you have a sketch ...
circle center is $(r,y)$
slope from $(r,y)$ to $(2,2)$ has magnitude $|m| = 1/2$
let $a$ = vertical distance from $(2,2)$ to $(r,y)$
$2+a = y$
then $2a$ = horizontal distance from $(r,y)$ to $(2,2)$
$r + 2a = 2$
also, using Pythagoras ... $r = \sqrt{a^2 + (2a)^2}$
$r = \sqrt{5} \cdot a$
$\sqrt{5} \cdot a + 2a = 2$
$a(\sqrt{5} + 2) = 2$
$a = \frac{2}{\sqrt{5} + 2}$
$r = \sqrt{5} \cdot \frac{2}{\sqrt{5} + 2} = \frac{2\sqrt{5}}{\sqrt{5} + 2}$ | 2015-05-06T13:43:12 | {
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https://dsp.stackexchange.com/questions/46648/solving-for-impulse-response-hn-given-input-output-pairs | # Solving for impulse response h[n] given input-output pairs
This is a homework problem that I've worked on and I want to confirm some of my reasoning.
I'm given two sets of input-output pairs of a particular system, $S$, that we know is linear time-invariant (LTI). Given those two pairs, find the impulse response h[n] of the system:
$$x_1[n] = [1, 0, 1] * S \longrightarrow y_1[n] = [0,1,0,2,0,1]$$
$$x_2[n] = [0, 1, 1] * S \longrightarrow y_2[n] = [0,0,1,1,1,1]$$
I found a linear combination of both inputs to yield the delta:
$$\delta[n] = x_1[n] - x_2[n-1] = [1, 0, 0]$$
Due to the system being LTI, the output to the delta is:
$$y[n] = h[n] = y_1[n] - y_2[n-1] = [0,1,0,1,-1,0]$$
Since the input is 3 samples long and the output is 6 samples long, then my impulse response $h[n]$ has to be 4 samples long. So I discard the last two to finally yield:
$$h[n] = [0,1,0,1]$$
Convolving the input signals with the newly found $h[n]$ yields the given output signals. I am however uncomfortable with having to invoke the required length of $h[n]$ to yield its final form by deleting the last two samples.
Did I go wrong somewhere?
Your instinct about the truncation being fishy is correct. Your solution is indeed wrong: you just got lucky to get the correct result. As @Fat32 has pointed out, $x_1[n]-x_2[n] = {1,0,0-1}$, which is not a delta impulse and is longer than the original sequence. The result is $h[n]-h[n-3]$, and not $h[n]$.
You actually only need one set of input and output to solve this. Given the impulse response length, we know that $$y[n]=\sum_{k=0}^{3}h[k] \cdot x[n-k]$$
You can work through the sum one element at a time and solve for the new coefficient of the impulse response. Let's use $x_1$ and $y_1$ $$0 = y[0]=h[0] \cdot x[0] = h[0] \cdot 1 \rightarrow h[0] = 0$$ $$1 = y[1]=h[0] \cdot x[1] + h[1] \cdot x[0] = 0 + h[1] \cdot 1 \rightarrow h[1] = 1$$ $$0 = y[2]=h[0] \cdot x[2] + h[1] \cdot x[1] + h[2] \cdot x[0]= 0 + 0 + h[2] \cdot 1 \rightarrow h[2] = 0$$
and so forth. It's a bit tedious but will give you the right answer with a single set of input and output.
Please note that you only need the first four outputs to calculate the entire impulse response. The last two outputs are "dependent variables". If they don't come out correctly, it would mean that your system is not LTI. The second set of input and output is redundant. You can try the method on either one and should get the same result
• Thanks for the response. I later did go through the convolution such as you suggest and got the same correct answer. I initially tackled this problem graphically by trying to invoke the linearity property of the LTI system. I realize now that my mistake was in decimating values that were shifted out of the window in $x_2[n-1]$ when doing the graphical approach. Jan 24 '18 at 16:15
From the lengths of the inputs and outputs, we conclude that the length of $h$ is $4$. Hence,
$$\begin{array}{ccccccc} & h_0 & h_1 & h_2 & h_3 & 0 & 0\\ + & 0 & 0 & h_0 & h_1 & h_2 & h_3\\ \hline & 0 & 1 & 0 & 2 & 0 & 1\end{array}$$
and, thus, we obtain a system of $6$ linear equations in $4$ unknowns. The solution is $$(h_0, h_1, h_2, h_3) = \color{blue}{(0,1,0,1)}$$
There is no need to consider the 2nd input-output pair.
Without any generalizations, this particular problem could be solved with the following: Let the causal FIR impulse response of the LTI system be $h[n]$ of length $L$, which can be inferred from the first set that $L = 4$, hence let the impulse response be $$h[n] = \{a,b,c,d\}$$
Then from the first set, it can be seen that $$y[n] = h[n]+h[n-2] = \{a,b,c+a,d+b,c,d\} = \{0,1,0,2,0,1\}$$
from which you can deduce that $$a=0 ~,~ b=1 ~,~ c=0 ~,~ d=1$$ which confirms that your result was right. For a longer impulse response you would need to solve a more involved set of (linear) equations...
• Thanks for the reply! My discomfort came from the mechanics of solving the problem which initially yielded an impulse response of length 6. I knew that the impulse response had to be length 4, but felt odd about deleting the last two samples to get the real answer. Jan 24 '18 at 1:14
• indeed $\delta[n] \neq x_1[n]-x_2[n-1] = \{1,0,0,-1\}$ But your restriction on the length of $h[n]$ helps to solve the problem... is this a coincidence? elaborate more to see. Jan 24 '18 at 1:20
• I made the dumb mistake of decimating the last 1 of $x_2[n]$ when shifting it to become $x_2[n-1]$. Jan 24 '18 at 16:28
• yes I saw that. you were lucky there, but now you can compute it firmly. Jan 24 '18 at 16:38 | 2021-10-24T22:08:13 | {
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https://engineering.stackexchange.com/questions/34871/resultant-force-coordinate-for-parallel-3d-forces-acting-on-a-slab | # Resultant force coordinate for parallel 3d forces acting on a slab
Apologies if I have done something wrong, first time posting on a stack exchange website.
Four forces act perpendicular to the surface of a circular slab (assume mass of slab is 0 and middle of slab is at 0,0,0)(all forces in -y direction)
Force 1 is located at 5,0,0 and has a magnitude of 77.4kN
Force 2 is located at 0,0,5 and has a magnitude of 48.6kN
Force 3 is located at 0,0,-5 and has a magnitude of 24.4kN
Force 4 is located at -5,0,0 and has a magnitude of 97.6kN
What is the coordinates of the resulting force (x,y,z) to two decimal places?
My working currently and my initial answer
The given answer is (0.49,0,-0.41) and I don't know where I went wrong.
After talking to my lecturer, she said the question was supposed to be determining the z coordinate only, which is why it would say the answer was correct when 0.49 was put first, meaning that the answers I calculated were correct.
• If you want help, show us what you've done so far. This isn't a "do my homework for me" site. – Fred Mar 29 at 9:37
• Sorry about that, thanks for letting me know, fixed. – Max Mar 29 at 10:16
• The book answer looks wrong IMO. Without doing any calculations, Force D is bigger than force C. Imagine the slab is balanced like a seesaw at x = 0. It would tip towards D, therefore the x position of the resultant will be negative. Also force A is bigger than force C so the z position of the resultant will be positive. – alephzero Mar 29 at 18:47
We start by the forces along the x-axis about D and call the resultant distance from D, X.
$$\Sigma M_{about\ D}=0 \rightarrow X*(97.6+24.4+77.4+48.6 ) \\= 73*5+77.4*10 \quad \rightarrow X=\frac{365+774}{248}=4.59$$
$$5-4.59=0.41$$ X is 0.41 meters to the left of the origin $$\ X=-0.41m \quad$$ as per the book.
The same steps apply to the Y-axis. Please check my numbers. also, I assumed the radius 5m not 6m as shown, because you asked 5m.
• The reason the distance was 5 instead of 6 is because the question is worded weirdly, the radius of the slab is 6m but the forces are placed 5m away from origin. Thanks for the response. – Max Mar 30 at 4:11
I haven't looked at your specific numbers but on quick review it appears that your approach is correct by summing the moments (force x distance) in each of the two directions, and that both directions are independent from each other.
Two things
1. To be an engineer, you have to learn to trust YOURSELF. The option that you are right is always an option. The book could be wrong. Find another way to do the math to double check your answer or confirm the books answer
2. An engineer solves problems, yes? Or how about this: engineers create solutions. The former is passive, the later is assertive. You are doing this problem passively 'by the book.' However, you can be assertive and find your own approach and use it if it is valid. I'll use your question as an example. Note that the distances are all the same... figured out where I'm going yet? The distance is not a variable in this problem, it's a fixed number. Therefore, the problem is a simple algebra problem. All you need is the ratios of the loads and then apply that to the distance. If the two loads equal themselves, the ratio is 1:1 and the coordinate is directly between them, since the distance is the same for both from the origin, then the resultant is at 0. What happens when the ratio is not 1:1, on what side is the resultant and how far over? If nothing else, with this, you can come up with a very good approximate answer in seconds, in your head!
• Thanks for your answer, I guess I automatically assumed the given answer was correct because it came from my lecturer (I realise that they can make mistakes too but that was why I made the assumption). I sort of did what you said in the second point a couple hours after I posted the question and looked at the opposing forces along the x and z axis and sort of realised that my answer must be right. But even then, I probably don't trust myself maybe as much as I should. Sorry for the long response, thanks again! – Max Mar 30 at 4:07 | 2020-10-26T07:48:16 | {
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https://mathoverflow.net/questions/297896/distance-from-nonnegativity-of-some-orthonormal-vectors | # Distance from nonnegativity of some orthonormal vectors
Suppose that $1 < k < n$. Does there exist a constant $\beta > 0$, such that for every $k$ orthonormal vectors $f_1,\ldots,f_k \in \mathbb R^n$, there exist $k$ orthonormal vectors with nonnegative elements, $x_1,\ldots,x_k\in \mathbb R_+^n$, such that
$$\sum_{i=1}^k \|x_i - f_i\|^2_2 \leq \beta \sum_{i=1}^k\|f_i^-\|_2^2$$
where $f_i^- := \max\{-f_i,0\}$ is the negative part of the vectors $f_i$?
In another way, I am interested in the estimation of the distance of a $n\times k$ dimensional matrix $F$ whose columns are orthonormal from the set of $n\times k$ matrices whose columns are nonnegative and orthonormal, i.e.,
$$\mathrm{dist}(F;St_+(n,k))$$
Where $St_+(n,k)$ is the set of $n\times k$ matrices whose columns are nonnegative and orthonormal. If we drop orthonormal condition and compute $\mathrm{dist}(F;\mathbb{R}^{n\times k }_+)$, we obtain $\|F^-\|$ as a lower bound for the above distance. In this term, my question is as follows: Is a multiple of $\|F^-\|$ an upper bound for the above distance, i.e.,
Is there a constant $c>0$, such that $$\|F^-\| \leq \mathrm{dist}(F;St_+(n,k)) \leq c \|F^-\|$$ for every $n\times k$ dimentional matrix $F$ whose columns are orthonormal?
In the special case $k=1$, the above statement is true, with $c = 2$. I'm interested in the special case of small values of $k$, such as $k=2$. Experimentally, for $k>1$ and random matrices $F$ and by using Frobenius norm, I get an upper bound for $\mathrm{dist}(F; St_+(n,k))$ by alternating projection to nonnegative matrices and orthonormal matrices. I guess that the above statement is true for $c \approx 2$, $(\beta \approx 4)$.
• This looks like a combinatorial problem. Indeed, for any $k>1$, there are only finitely many possible choices for $x_1,\ldots,x_k$.
– Surb
Apr 15, 2018 at 10:51
• @Mahdi What do you know? What have you tried? Also, it looks related your Steifel manifold question from earlier. Some background would be welcome. Apr 15, 2018 at 11:49
• What are you going to do with $(1,\sqrt{2\varepsilon},-\varepsilon)$ and $(-\varepsilon,\sqrt{2\varepsilon},1)$? Apr 15, 2018 at 19:19
• I think that it may work if you drop the squares on the right hand side. It definitely gives you the right scaling and if you don't care too much about the dependence of $\beta$ on $k$, it gets trivial with $\beta=k$ then. However, I still want to see if we can get a neat inequality before posting anything. Also, will such an estimate (if true) suffice for your purposes? Apr 15, 2018 at 21:26
• No, that one is fine: approximate by $(1,0,\varepsilon^{2/3},0)$ and $(0,1,0,\varepsilon^{1/4})$. Then $LHS=\varepsilon^{3/2}$ and $RHS=\varepsilon^{2/3}$ (without squares). I'll post the bound a bit later Apr 16, 2018 at 17:01
The following two simple lemmas are crucial.
Lemma 1. For any nonnegative numbers $a_1,\dots,a_k$, \begin{equation*} \sum_1^k a_i^2-\max_1^ka_i^2\le\sum_{1\le i<i'\le k}a_i a_{i'}. \end{equation*}
Proof. Without loss of generality, $a_1=\max_1^ka_i$. Then $a_i^2\le a_1a_i$ for all $i=2,\dots,k$. So, Lemma 1 follows.
For any $u\in\R^n$, let $u^+ :=\max\{u,0\}$ and $u^- :=\max\{-u,0\}$, so that $u=u^+-u^-$.
Lemma 2. For any orthonormal vectors $u$ and $v$, \begin{equation*} u^+\cdot v^+\le \|u^-\|+\|v^-\|, \end{equation*} where $\cdot$ denotes the dot product.
Proof. We have $0=u\cdot v=u^+\cdot v^+ - u^+\cdot v^- -u^-\cdot v^+ + u^-\cdot v^- \ge u^+\cdot v^+ - \|u^+\|\,\|v^-\| -\|u^-\|\,\| v^+\|$, whence Lemma 2 follows.
As in the question, let now $f_1,\ldots,f_k$ be any orthonormal vectors in $\R^n$. Write $f_i=(f_{ij})_{j=1}^n$ and $f^+_i=(f^+_{ij})_{j=1}^n$. Let $(J_1,\dots,J_k)$ be any partition of the set $[n]:=\{1,\dots,n\}$ such that for all $i\in[k]$ and $j\in[n]$ we have the implication \begin{equation*} j\in J_i\implies f^+_{ij}=\max_{q\in[k]}f^+_{qj}. \end{equation*} Define $y_i=(y_{ij})_{j=1}^n$ by \begin{equation*} y_{ij}:=f^+_{ij}\,\ii{j\in J_i}, \end{equation*} where $\ii{}$ denotes the indicator; so, $y_{ij}=\max_{q\in[k]}f^+_{qj}$ for $j\in J_i$ and $y_{ij}=0$ for $j\in[n]\setminus J_i$. Hence, in view of Lemmas 1 and 2, \begin{multline*} \sum_1^k\|y_i-f^+_i\|^2 =\sum_{j\in[n]}\Big(\sum_{i\in[k]}(f^+_{ij})^2-\max_{i\in[k]}(f^+_{ij})^2\Big) \le\sum_j\sum_{i<i'}f^+_{ij}f^+_{i'j} \\ =\sum_{i<i'}f^+_i\cdot f^+_{i'} \le\sum_{i<i'}(\|f^-_i\|+\|f^-_{i'}\|) =2(k-1)\sum_{i\in[k]}\|f^-_i\|. \tag{1} \end{multline*}
Also, $\sum_1^k\|f_i-f^+_i\|^2=\sum_1^k\|f^-_i\|^2\le\sum_1^k\|f^-_i\|$. So, by (1) and Minkowski's inequality,
\begin{equation*} \sum_{i\in[k]}\|y_i-f_i\|^2 \le(\sqrt{2(k-1)}+1)^2\sum_{i\in[k]}\|f^-_i\|\le3k\sum_{i\in[k]}\|f^-_i\|=:\ep. \tag{2} \end{equation*}
Next, $$0\le1-\|y_i\|=\|f_i\|-\|y_i\|\le\|y_i-f_i\|, \tag{3}$$ by the triangle inequality.
Consider now two possible cases:
Case 1: $\ep<1$. (This is hopefully the main case.) Then, by (2), $\|y_i-f_i\|<1$ for all $i$, whence, by (3), $y_i\ne0$ for all $i$, so that we can let
\begin{equation*} x_i:=y_i/\|y_i\|. \end{equation*}
Then $x_1,\dots,x_k$ are orthonormal vectors in $\R_+^n$, and
\begin{equation*} \sum_{i\in[k]}\|x_i-y_i\|^2=\sum_{i\in[k]}(1-\|y_i\|)^2\le\sum_{i\in[k]}\|y_i-f_i\|^2\le\ep \end{equation*} by (3) and (2), which yields \begin{equation*} \sum_{i\in[k]}\|x_i-f_i\|^2 \le4\ep=12k\sum_{i\in[k]}\|f^-_i\|. \end{equation*}
Case 2: $\ep\ge1$. Here for any orthonormal $x_1,\dots,x_k$ we have \begin{equation*} \sum_{i\in[k]}\|x_i-f_i\|^2\le 2\sum_{i\in[k]}(\|x_i\|^2+\|f_i\|^2)=4k\le4k\ep =12k^2\sum_{i\in[k]}\|f^-_i\|. \end{equation*}
Thus, \begin{equation*} \sum_{i\in[k]}\|x_i-f_i\|^2\le \left\{ \begin{aligned} 12k\sum_{i\in[k]}\|f^-_i\|&\text{ if }\ep<1,\\ 12k^2\sum_{i\in[k]}\|f^-_i\|&\text{ if }\ep\ge1. \end{aligned} \right. \end{equation*}
(As follows from the comment by user fedja, here $\|f^-_i\|$ cannot be replaced by $\|f^-_i\|^{1+\ep}$, for any real $\ep>0$.)
• It is possible that $y_i=0$, for example in the case that one of the $f_i$ has no positive elements. Also, why did you suppose that $\epsilon < 1$? Apr 17, 2018 at 15:32
• Thank you so much if you explain more why $\sum_{i\in[k]}(1-\|y_i\|^2)\le\sum_{i\in[k]}\|y_i-f_i\|^2$ in the 4th line from the end of your answer. Apr 17, 2018 at 16:56
• Thank you for your comments. I have now considered the case $\varepsilon\ge1$ as well. Also, I have added details showing that $y_i\ne0$ if $\varepsilon<1$. Concerning $(1-\|y_i\|^2)$, that was a typo: there should be (and now is) $(1-\|y_i\|)^2$ there instead. Apr 17, 2018 at 17:56 | 2023-01-31T13:53:27 | {
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http://mathhelpforum.com/calculus/87525-how-do-you-determine-number-inflection-points.html | Thread: How do you determine the number of inflection points
1. How do you determine the number of inflection points
for the function $y=2x+cos(x^2)$ on the interval of $0\leq x \leq 5$
$f\prime\prime = -2x(2xcosx^2)- 2sinx^2$
I don't need to find the inflection values. Just the number of inflections points for the graph over the interval.
2. Originally Posted by swatpup32
for the function $y=2x+cos(x^2)$ on the interval of $0\leq x \leq 5$
$f\prime\prime = -2x(2xcosx^2)- 2sinx^2$
I don't need to find the inflection values. Just the number of inflections points for the graph over the interval.
Solve $-4x^2\cos(x^2)-2\sin(x^2)=0$ on $[0,5]$
Let $u=x^2$ (this extends the domain to $[0,5^2]$)
$-4u\cos(u)-2\sin(u)=0 \implies 2u\cos(u)+\sin(u)=0$.
Clearly, a solution is $u=0 \implies x=0$. There are eight others, but I only know that because I graphed the equation and counted them. I honestly don't know how to solve for the other roots analytically.
But there are 9 zeroes and thus 9 inflection points.
3. Originally Posted by redsoxfan325
But there are 9 zeroes and thus 9 inflection points.
Not neccessarily true. An inflection point is a point in which (1.) the slope of f' is zero or undefined vertically and (2.) there is a change of the sign of the slope of f' from - to + or from + to - . A zero can exist without the slope sign changing, therefore it'd not be an inflection point.
4. Ok, but this is a continuous function defined on a compact set, so it can't be unbounded, and whenever it changes from positive to negative there will be a corresponding zero by the IVT.
5. I have an indeterminate form ie $0\cdot \infty$ this is not well defined, and I checked if the limits go to zero and they do not.
The zero factor principle does not apply becuase $0\cdot \infty \ne =$
Thanks for the graph skeeter.
So the 2nd derivative is
$f''(x)=-4x^2\cos(x^2)-2\sin(x^2)$
This can be factored as follows
$0=-2\cos(x^2)[2x^2+\tan(x^2)]$
The 2nd factor only has x=0 as a solution so the solutions are
$\cos(x^2)=0 \iff x^2=\frac{\pi}{2}+n \pi= \frac{(2n+1) \pi}{2}$
$x=\pm \sqrt{\frac{(2n+1)\pi}{2}}$
Note that in the factorization that $(2x^2+\tan(x^2)) > 0$ for all $x> 0$
Since $\cos(x^2)$ changes sign at each of it zero all of the above zero's are critical points
6. graph of the 2nd derivative ... | 2017-05-24T14:05:11 | {
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https://ledatascifi.github.io/ledatascifi-2023/content/05/02c_goodnessOfFit.html | # 6.3. Goodness of Fit¶
1. You can measure the goodness of fit on a regression
R$$^2$$ and Adjusted R$$^2$$
• $$R^2 = 1-SSE/TSS$$ = The fraction of variation in y that a model can explain, and always between 0 and 1.
• Adding extra variables to a model can not reduce R$$^2$$ even if the variables are random
• Adjusted R$$^2$$ adjusts R$$^2$$ by penalizing based on the number of variables in X. It can actually be negative.
• sm_ols(<model and data>).fit().summary() automatically reports R$$^2$$ and Adjusted R$$^2$$
A key goal of just about any model is describing the data. In other words, how well does the model explain the data?
Intuitively, you look at these graphs and know that Model 1 describes the data better.
# load some data to practice regressions
import seaborn as sns
import numpy as np
import matplotlib.pyplot as plt
from statsmodels.formula.api import ols as sm_ols
# this alteration is not strictly necessary to practice a regression
# but we use this in livecoding
np.random.seed(400)
diamonds2 = (diamonds
.assign(lprice = np.log(diamonds['price']),
lcarat = np.log(diamonds['carat']),
ideal = diamonds['cut'] == 'Ideal',
lprice2 = lambda x: x.lprice + 2*np.random.randn(len(diamonds))
)
.query('carat < 2.5') # censor/remove outliers
# some regression packages want you to explicitly provide
# a variable for the constant
.assign(const = 1)
)
fig, axes = plt.subplots(2,2,figsize = (10,10))
sns.regplot(data=diamonds2,ax=axes[0,0],y='lprice',x='lcarat',
scatter_kws = {'color': 'blue', 'alpha': 0.01},
line_kws = {'color':'red'},
).set(xlabel='',ylabel='',title='Model 1 - data and reg line')
sns.regplot(data=diamonds2,ax=axes[0,1], y='price',x='carat',
scatter_kws = {'color': 'blue', 'alpha': 0.01},
line_kws = {'color':'red'},
).set(xlabel='',ylabel='',title='Model 2 - data and reg line')
sns.residplot(data=diamonds2,ax=axes[1,0],y='lprice',x='lcarat',
scatter_kws = {'color': 'blue', 'alpha': 0.01},
line_kws = {'color':'red'},
).set(xlabel='',ylabel='',title='Model 1 - residuals ')
sns.residplot(data=diamonds2,ax=axes[1,1], y='price',x='carat',
scatter_kws = {'color': 'blue', 'alpha': 0.01},
line_kws = {'color':'red'},
).set(xlabel='',ylabel='',title='Model 2 - residuals')
plt.show()
The next graph shows two different y variables (shown in blue and orange). Again, it’s fairly obvious that the blue points are better explained by the blue line than the orange is by the orange line.
fig, ax = plt.subplots(figsize = (10,5))
sns.regplot(x='lcarat',y='lprice',ax=ax,
data=diamonds2.sample(75,random_state=20),ci=None)
sns.regplot(x='lcarat',y='lprice2',ax=ax,
data=diamonds2.sample(75,random_state=20),ci=None)
plt.xlabel("")
plt.ylabel("")
plt.show()
The way we can capture that statistically is by computing R$$^2$$.
• Step 1: The total sum of squares: TSS $$= \sum_i (y_i - \bar{y}_i)^2$$ captures the total amount of variation in $$y$$ around its mean $$\bar{y}$$. We’d like to explain as much of this as possible.
• Step 2: The sum of squared errors: SSE $$= \sum_i (y_i - \hat{y}_i)^2$$ captures the total amount of errors in the predictions. We’d like this to be as close to zero.
• Step 3: $$SSE/TSS$$ = The fraction of variation in y that a model can’t explain
Note
$$R^2 = 1-SSE/TSS$$ = The fraction of variation in y that a model can explain
Statsmodels reports R$$^2$$ and Adjusted R$$^2$$ by default with .summary().
Let’s confirm that Model 1 > Model 2, and that Blue > Orange:
from statsmodels.iolib.summary2 import summary_col # nicer tables
# fit the two regressions, and save these as objects (containing the results)
res1 = sm_ols('lprice ~ lcarat',data=diamonds2).fit()
res2 = sm_ols('price ~ carat',data=diamonds2).fit()
res3 = sm_ols('lprice ~ lcarat',data=diamonds2).fit()
res4 = sm_ols('lprice2 ~ lcarat',data=diamonds2).fit()
# now I'll format an output table
# I'd like to include extra info in the table (not just coefficients)
info_dict={'R-squared' : lambda x: f"{x.rsquared:.2f}",
'No. observations' : lambda x: f"{int(x.nobs):d}"}
# This summary col function combines a bunch of regressions into one nice table
print(summary_col(results=[res1,res2,res3,res4],
float_format='%0.2f',
stars = True,
model_names=['Model 1','Model 2','Blue','Orange'],
info_dict=info_dict,
#regressor_order=[ # you can specify which variables you
# want at the top of the table here]
)
)
====================================================
Model 1 Model 2 Blue Orange
----------------------------------------------------
Intercept 8.45*** -2335.81*** 8.45*** 8.45***
(0.00) (12.98) (0.00) (0.01)
R-squared 0.93 0.85 0.93 0.19
R-squared Adj. 0.93 0.85 0.93 0.19
carat 7869.89***
(14.14)
lcarat 1.68*** 1.68*** 1.69***
(0.00) (0.00) (0.01)
R-squared 0.93 0.85 0.93 0.19
Adj R-squared 0.93 0.85 0.93 0.19
No. observations 53797 53797 53797 53797
====================================================
Standard errors in parentheses.
* p<.1, ** p<.05, ***p<.01 | 2022-08-12T19:09:24 | {
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https://aspireinmath.org/Wiki-Sample | # Wiki-Sample
## Task 1: What degree could this polynomial be?
Why do you think what you think?
## Task 1A: Why does the degree have to be even?
We all seem to agree that the polynomial in the picture would have to be an even degree polynomial. Why is this the case?
## Task 2: Define “root” or “zero” of a polynomial
Let’s agree on a definition of “root” which is a key concept when you are talking about real valued functions.
Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is a function. A (real) root of $f$ is a number $x \in \mathbb{R}$ such that $f(x) = 0$.
## Task 3: Does every 5th degree polynomial have a real root?
Consider the conjecture that every 5th degree polynomial has at least one (real) root.
Is this true or false? Why?
We decided yes! An odd degree polynomial will have different end behavior in the two directions (because of how negative numbers work with even and odd powers) AND polynomials are continuous, so a fifth degree polynomial will have to cross the $x$-axis.
## Task 4: Formulating our BIG CONJECTURE for the term
Our justification that a 5th degree polynomial must have at least one real root: such a polynomial will be continuous and will go to infinity in one direction and negative infinity in the other. So it must cross the x-axis.
We all seem convinced by this argument. Maybe we can get a general conjecture/theorem out of this idea!
1. Does this argument only work for polynomials?
2. Does it only work for functions that go to $+ \infty$ in one direction and $- \infty$ in the other?
3. Can we generalize this idea to make a conjecture about conditions that would guarantee a function had a root?
ANSWER: We decided that it could work for any continuous function. And we don't need it to go to positive and negative infinity, we just need it to take on positive and negative values.
We came up with an informal version of a conjecture: Any function that is continuous and takes on positive and negative values will have at least one real root.
Since we missed two days of class due to winter storm shenanigans, I took the liberty of formalizing the conjecture myself:
Conjecture (IVT): Suppose $f: \mathbb{R} \to \mathbb{R}$ is a continuous function. If there exists inputs $a, b$ where $a < b$ and $f(a)$ and $f(b)$ have opposite signs, then there exits $r \in (a, b)$ such that $f(r) = 0$
Our BIG GOAL for the rest of the term will be to prove this conjecture. We will start the same way that Cauchy did back in the day - by becoming experts on approximation methods.
## Task 5: Approximating Roots
Find an approximation of a root of $f(x) = x^5 – 5x -5$ that is correct to four decimal places.
## Task 6: Approximation Rules of Engagement
What can we use a calculator for given that we need to have complete control and understanding of our approximation method?
Answer: Pretty much the only thing we can trust is plugging in nice values and getting outputs (that might be approximation after some number of decimals).
What exactly does it mean to be “within four decimals”?
There are a number of reasonable ways to interpret this! The one we are going to choose is that: We are correct to n decimals when we know that the first n digits after the decimal are the exact same as those of the exact value.
Given that are ultimately interested in the existence of an EXACT root, the picture below indicates why don't want to focus on how close the outputs are to zero:
## Task 7: Approximation Instruction Manual
Write an instruction manual that anyone could use to approximate a root of any function that satisfies the hypotheses of our IVT conjecture.
Explain the steps of the method as clearly as you can.
Use your work on Task 5 give give an example of how to use your method
Make sure your manual explains how you would get an approximation accurate to any number of decimal places.
Here is a fun video showing the method in action!
## Task 8: What is a Sequence?
These letters with subscripts remind me of sequences from calculus. What exactly is a sequence?
Definition (sequence): A sequence of real numbers is a function with domain $\mathbb{N}$ and codomain $\mathbb{R}$
## Task 9: Sequences Generated by our Approximation Method
Identify at least five sequences that are generated if you iterate the approximation method forever.
We came up with three input sequence, $a_n$, $b_n$,$c_n$, three output sequences, $f(a_n)$, $f(b_n)$, $f(c_n)$ and the sequence of interval lengths, $e_n = b_n - a_n = \frac{b_1 - a_1}{2^{n-1}}$
## Task 10: Conjectures about the Sequences Generated by our Approximation Method
Make some conjectures about the sequences we have identified:
1. What properties do they have?
2. Do the sequences have limits? If so, what are they?
3. Are there any interesting relationships between any of the sequences? Any other thoughts?
To make things simpler, let's name those sequences:
$a_n$ is the sequence of left endpoints $b_n$ is the sequence of right endpoints $c_n$ is the sequence of midpoints $e_n = b_n - a_n$ is the sequence of error bounds (AKA the sequence of interval lengths) $f(a_n)$ is the sequence of outputs of the left endpoints $f(b_n)$ is the sequence of outputs of the right endpoints $f(c_n)$ is the sequence of outputs of the midpoints
Here are some of the conjectures you came up with. Let's discuss them on the way to making a new list that everyone is happy with.
Conjectures NOT about limits:
1. $e_n = b_n - a_n = \frac{b_1-a_1}{2^{n-1}}$ for every $n \in \mathbb{N}$
2. $a_{n+1} \geq a_n$ for every $n \in \mathbb{N}$. In other words, $a_n$ is an increasing sequence
3. $b_{n}$ is decreasing
4. $e_n$ is strictly decreasing
5. $a_n$ is bounded above by $b_1$
6. $b_n$ is bounded below by $a_1$
7. $a_n < b_n$ for every $n \in \mathbb{N}$
8. $f(a_n) < 0$ for every $n \in \mathbb{N}$ (NOTE: Assumes $f(a_1) < 0$ and $f(b_1) > 0$)
9. $f(b_n) > 0$ for every $n \in \mathbb{N}$ (NOTE: Assumes $f(a_1) < 0$ and $f(b_1) > 0$)
1. $e_n$ converge to 0. Using limit notation: $\displaystyle{\lim_{n \to \infty}(b_n - a_n)} = 0$
2. $a_n$ converges to a root (from the left)
3. $\displaystyle{\lim_{n \to \infty}(a_n)} < b_k$ for every $k \in \mathbb{N}$
4. $a_n$ does converge
5. $\displaystyle{\lim_{n \to \infty}(a_n)} = \displaystyle{\lim_{n \to \infty}(b_n)} = \displaystyle{\lim_{n \to \infty}(c_n)}$.
6. $f(a_n)$ converges to 0
7. $f(b_n)$ converges to 0
## Task 11: What happens if we try our method on a function with no sign change?
What happens if you apply your approximation method to a function that does not have a sign change? Try it on this function:
## Task 12: What if you try it with a discontinuous function?
It is impossible to even start the approximation method when your function does not have a sign change. Now let’s consider the other part of our IVT Conjecture hypothesis!
What happens if you apply your method to a discontinuous function? Try it with these two functions:
NOTE: You aren’t trying to find numbers just trying to figure out what will happen if you apply the method.
Answer: We can definitely apply the method for these functions! The only difference is that the left and right endpoints appear to converge to the point where the function is discontinuous instead of to a root!
## Task 13: Sorting Conjectures
Sort our list of conjectures into two groups. Group 1 is the conjectures that will be true for any function that has a sign change. Group 2 is the conjectures that will be true for any continuous function that has a sign change.
KEEP IN MIND: We are ASSUMING that 1) the sign change goes from negative to positive and 2) the approximation process goes on forever (our goal is to prove there is a root, and if we happen to hit it on the number with one of our approximations - great! - so we don't need to worry about that).
(Group 1) True as long as there is a sign change:
Everything except the three below
(Group 2) True ONLY if the function has a sign change AND is continuous:
1. $a_n$ converges to a root (from the left)
2. $f(a_n)$ converges to 0
3. $f(b_n)$ converges to 0
One of our big goals the next 5 class sessions will be to PROVE that $a_n$ does converges. That turns out to be a big first step in proving IVT and in figuring out some of the foundational ideas of real analysis. But first we are going to build up some expertise about sequences. | 2023-04-02T08:28:56 | {
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https://math.stackexchange.com/questions/2057148/three-shooters-shoot-a-target | # Three shooters shoot a target
Three shooters shoot a target. The known probability to hit the target of the first, second and third shooter are 0.3, 0.4 and 0.5 respectively. What is the probability that the target is hit from only one shooter?
I try this.
The probabilities that two will not heat the target:
0.7 x 0.6 = 0.42 0.7 x 0.5 = 0.35 0.6 x 0.5 = 0.30
and finally I'm not sure what should i do with those results? suppose have to multiply them?
• Can you calculate the probability that the first shooter hit the target and the other don't? You should easily arrive to the result. – nicola Dec 13 '16 at 15:44
• How many ways there are that only 1 of them hit it? What are the probabilities for each of those outcomes? Multiply the probability of one of them hitting and the other 2 missing for each case. – Jesús Ros Dec 13 '16 at 15:45
Split it into disjoint events, and then add up their probabilities:
• The probability that the target is hit by the 1st shooter only is $\small(0.3)\times(1-0.4)\times(1-0.5)$
• The probability that the target is hit by the 2nd shooter only is $\small(1-0.3)\times(0.4)\times(1-0.5)$
• The probability that the target is hit by the 3rd shooter only is $\small(1-0.3)\times(1-0.4)\times(0.5)$
The overall probability is therefore:
$$\small(0.3)\times(1-0.4)\times(1-0.5)+(1-0.3)\times(0.4)\times(1-0.5)+(1-0.3)\times(1-0.4)\times(0.5)=0.44$$
Let 3 shoots A, B and C.
$P(A) = 0.3, P(A') = 1 - 0.3 = 0.7$
$P(B) = 0.4, P(B') = 1 - 0.4 = 0.6$
$P(C) = 0.5, P(C') = 1 - 0.5 = 0.5$
Probability of hitting by only one = $P(A) \times P(B') \times P(C') + P(B) \times P(C') \times P(A') + P(C) \times P(A') \times P(B')$
$= 0.3 \times 0.6 \times 0.5 + 0.4 \times 0.5 \times 0.7 + 0.5 \times 0.7 \times 0.6$
$= 0.09 + 0.14 + 0.21 = 0.44$
• thank you for the detailed answer – vachoh Dec 13 '16 at 20:32
• Mine pleasure. And I saw your questions. You not accept any answer in your all questions. I think you have to accept the answers (use tick). Because in future if anyone have similar problem. They get help from your questions easily. – Kanwaljit Singh Dec 14 '16 at 2:38 | 2020-02-28T07:24:32 | {
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https://www.physicsforums.com/threads/system-of-linear-second-order-differential-equations.344399/ | # System of linear second order differential equations
1. Oct 10, 2009
### matematikawan
I think I know how to solve
$$\frac{d\vec{x}}{dt}= A \vec{x}$$
where A is a constant nXn matrix. We just compute the eigenvalues and the corresponding eigenvectors.
But how do we solve
$$\frac{d^2\vec{x}}{dt^2}= A \vec{x}$$
Can we say straight away that the solution is (following that of one dependent variable)
$$\vec{x}(t) = exp(-Mt) \vec{x}_1+ exp(Mt) \vec{x}_2$$
where M2=A and x1 and x2 are constant vectors.
2. Oct 10, 2009
### tiny-tim
Hi matematikawan!
Try putting y = dx/dt + Mx
3. Oct 10, 2009
### matematikawan
This is great. How did you came out with the substitution y = dx/dt + Mx tiny-tim ?. I have try searching the internet for system of DE. Most of them are about first order DE.
$$\frac{d\vec{y}}{dt}=M\vec{y}$$
Using http://en.wikipedia.org/wiki/Matrix_exponential" [Broken]
$$\vec{y} = exp(Mt) \vec{c_1}$$
$$\frac{d\vec{x}}{dt} = -M\vec{x} + exp(Mt) \vec{c_1}$$
$$\vec{x}=exp(-Mt)\vec{c_2} + exp(-Mt)\int exp(2Mt) \vec{c_1}$$
$$\vec{x}=exp(-Mt)\vec{c_2} + exp(-Mt) (2M)^{-1} exp(2Mt) \vec{c_1}$$
So not exactly
$$\vec{x}(t) = exp(-Mt) \vec{x}_1+ exp(Mt) \vec{x}_2$$
as I predicted.
Last edited by a moderator: May 4, 2017
4. Oct 10, 2009
### matematikawan
Or it could possible that
$$\vec{x}=exp(-Mt)\vec{c_2} + exp(-Mt) exp(2Mt) (2M)^{-1}\vec{c_1}$$
Then it is of the form
$$\vec{x}(t) = exp(-Mt) \vec{x}_1+ exp(Mt) \vec{x}_2$$
5. Oct 10, 2009
### tiny-tim
Hi matematikawan!
(btw, we usually write the constant before the exponential, and we usually use A and B if we're expecting more than one )
That's right …
now write it dx/dt + Mx = BeMt,
and use the particular solution and general solution method (do you know what that is?) …
the g s gives you Ce-Mt, and the p s of course is (B/M - 1)eMt, or just BeMt.
And I got it from the "D" method …
we write D instead of d/dt, so the original equation is (D2 - A)x = 0,
then (D + A)(D - A)x = 0,
so either (D + A)x = 0 or (D - A)x = 0 …
this works for any polynomial in D (it's called the characteristic polynomial of the original equation) …
google "characteristic polynomial", or see the recurrence relation page in the PF Library for some details.
6. Oct 11, 2009
### matematikawan
That is a clever trick. I have seen in some old DE textbooks where they use similar 'D operator' technique to solve ODE. But that works only for DE with one dependent variable.
In this case of system of DE
$$\frac{d^2\vec{x}}{dt^2}= A \vec{x}$$
I'm quite lucky to be able to find M such that M^2=A. To find M in matlab we just use the command sqrtm(A).
How about these system of DE
$$\frac{d^3\vec{x}}{dt^3}= A \vec{x}$$
or
$$\frac{d^2\vec{x}}{dt^2} + \frac{d\vec{x}}{dt}= A \vec{x}$$
How are we going to 'factorise' the so called characteristic polynomial ?
7. Oct 12, 2009
### tiny-tim
(erm … it's not "so-called" … that really is its name!)
For the first one, use the cube root of A (I don't know whether there's a matlab command for that: if not, is there one for lnA or expA?).
The second is easy: the roots are -(1/2)I ±√((1/4)I + A2).
8. Oct 12, 2009
### matematikawan
Oh I'm sorry. :shy: I didn't mean to offend anyone. May be I'm not using proper words there. English is not my first language. So it is characteristic polynomial. Really sorry.
But by the way when I Google 'characteristic polynomial'. This term gives the meaning as the polynomial whose roots are the eigenvalues of a given matrix A, i.e. det(A-aI)=0.
In matlab we use the command expm(A) for matrix exponential. I'm not aware of commands for cube root or log of A. I'm trying to solve second order pde numerically. I discretise the variables x and t. Solution for the third order system of DE is not that important for me at the moment compare to that of second order.
It looks like that we can factorise the characteristic quadratic by completing the square.
$$D^2+D-A =\left(D + \frac{1}{2}I\right)^2 - \frac{1}{4} I - A$$
$$= \left(D + \frac{1}{2}I - \sqrt{\frac{1}{4}I+A} \right) \left(D + \frac{1}{2}I + \sqrt{\frac{1}{4}I+A} \right)$$
9. Oct 12, 2009
### D H
Staff Emeritus
Alternatively, form a state vector comprising $\vec x$ and $d\vec x/dt$.
$$\vec u = \left [\vec x \\ d\vec x/dt \right]$$
Then
$$\dot{\vec u} = \bmatrix \boldsymbol 0_{N\times N} & \boldsymbol 1_{N\times N} \\ \boldsymbol A & \boldsymbol 0_{N\times N} \endbmatrix \vec u$$
where $\boldsymbol 1$ is the NxN identity matrix and $\boldsymbol 0$ is the NxN zero matrix. Denoting
$$\boldsymbol B \equiv \bmatrix \boldsymbol 0_{N\times N} & \boldsymbol 1_{N\times N} \\ \boldsymbol A & \boldsymbol 0_{N\times N} \endbmatrix$$
This is of course a constant matrix. Then
$$\dot{\vec u} = \boldsymbol B \vec u$$
and tada! A linear first order equation with a constant matrix.
10. Oct 13, 2009
### matematikawan
Of course, we could then solve the equation with Runge-Kutta method, thus avoiding the need for matrix exponential. Although we have double the matrix size but their entries are sparse. Anyway I'm looking for numerical answer. Analytical answer may be suitable for theoretical purposes.
Thanks D H, especially for formulating the equation with nice block matrices. Is there any chance of having B as
$$\boldsymbol B \equiv \bmatrix \boldsymbol A & \boldsymbol 0_{N\times N} \\ \boldsymbol 0_{N\times N} & \boldsymbol 1_{N\times N} \endbmatrix$$
I have try a few permutation but without success. | 2017-08-20T12:05:34 | {
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https://math.stackexchange.com/questions/3127980/vector-valued-function-derivative-with-matrix | # Vector valued function derivative with matrix
Given a function $$f: \mathbb{R}^n \rightarrow \mathbb{R}^n$$ and a matrix $$A \in \mathbb{R}^{n \times n}$$. Is there a general formula for calculating the following derivative:
$$\frac{\partial}{\partial x} f(x)^T A f(x) \tag{1} = ?$$
I know that
$$\frac{\partial}{\partial x} x^T A x = x^T(A + A^T) \overset{A = A^T}{=} 2 x^T A \tag{2}$$
and the solution to $$(1)$$ will probably look similar to $$(2)$$, but I am stuck here since I am not sure how to apply the chain rule in the matrix case.
Edit: Regarding notation, we have
$$\frac{\partial }{\partial x}f(x) = \begin{bmatrix} \frac{\partial}{\partial x_1} f_1(x) & \frac{\partial}{\partial x_2} f_1(x) & \cdots & \frac{\partial}{\partial x_n} f_1(x) \\ \frac{\partial}{\partial x_1} f_2(x) & \frac{\partial}{\partial x_2} f_2(x) & \cdots & \frac{\partial}{\partial x_n} f_2(x) \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial}{\partial x_1} f_n(x) & \frac{\partial}{\partial x_2} f_n(x) & \cdots & \frac{\partial}{\partial x_n} f_n(x) \end{bmatrix}$$
and
$$x = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} , f(x) = \begin{bmatrix} f_1(x) \\ f_2(x) \\ \vdots \\ f_n(x) \end{bmatrix}$$
• Is $\partial/\partial x$ the total differential (or Jacobian, however you want to call it)? The expression $f(x)^TAf(x)$ is a product of things, so you can appeal to the product rule (either a general product rule, or do it entry wise). Feb 26, 2019 at 20:56
• @Reveillark I updated the question. I know I could do everything elementwise using the product rule, but I am rather looking for a compact formula in matrix notation, similar to $(2)$. Feb 26, 2019 at 21:25
Given a differentiable vector field $$\mathrm v : \mathbb R^n \to \mathbb R^n$$ and a matrix $$\mathrm A \in \mathbb R^{n \times n}$$, let function $$f : \mathbb R^n \to \mathbb R$$ be defined by
$$f (\mathrm x) := \langle \mathrm v (\mathrm x), \mathrm A \mathrm v (\mathrm x) \rangle$$
whose directional derivative in the direction of $$\mathrm y \in \mathbb R^n$$ at $$\mathrm x \in \mathbb R^n$$ is
$$D_{\mathrm y} f (\mathrm x) := \lim_{h \to 0} \frac{f (\mathrm x + h \mathrm y) - f (\mathrm x)}{h} = \cdots = \langle \mathrm y, \mathrm J_{\mathrm v}^\top (\mathrm x) \, \mathrm A \, \mathrm v (\mathrm x) \rangle + \langle \mathrm J_{\mathrm v}^\top (\mathrm x) \, \mathrm A^\top \mathrm v (\mathrm x) , \mathrm y \rangle$$
where matrix $$\mathrm J_{\mathrm v} (\mathrm x)$$ is the Jacobian of vector field $$\rm v$$ at $$\mathrm x \in \mathbb R^n$$. Thus, the gradient of $$f$$ is
$$\nabla_{\mathrm x} f (\mathrm x) = \mathrm J_{\mathrm v}^\top (\mathrm x) \left( \mathrm A + \mathrm A^\top \right) \mathrm v (\mathrm x)$$
I find differential notation helpful here in organizing things. The total derivative is a linear operator, so we introduce its argument and apply the product rule: $$d(f(x)^TAf(x))=d(f(x)^T)\cdot Af(x)+f(x)^TA\cdot d(f(x)$$ $$d(f(x)^TAf(x))=\left(\frac{df}{dx}\cdot dx\right)^T\cdot Af(x)+f(x)^TA\cdot \left(\frac{df}{dx}\cdot dx\right)$$ Now, the transpose of a $$1\times 1$$ matrix is itself, so we transpose that first term: $$d(f(x)^TAf(x)) = f(x)^TA^T\cdot \left(\frac{df}{dx}\cdot dx\right)+f(x)^TA\cdot \left(\frac{df}{dx}\cdot dx\right)$$ $$d(f(x)^TAf(x)) = f(x)^T(A+A^T)\cdot \left(\frac{df}{dx}\cdot dx\right)$$ Now that's in the form we want for the derivative. The total derivative of $$f(x)^TAf(x)$$ is $$f(x)^T(A+A^T)\frac{df}{dx}$$ where $$\frac{df}{dx}$$ is the matrix of partial derivatives of $$f$$, written in your question as $$\frac{\partial}{\partial x}f(x)$$.
Well we want to differentiate $$f(x)^TAf(x)$$ then it is useful to break into pieces.
First we see how to differentiate $$g(x,y) = x^TAy$$ with $$A$$ constant. $$g(x+h,y+k) = (x+h)^TA(y+k) =(x^T+h^T)A(y+k) = x^TAy + h^TAy + x^TAk + h^TAk$$ From this we see that $$Dg_{(x,y)}(h,k) = h^TAy + x^TAk$$.
Now we use the chain rule $$D(f(x)^TAf(x))_x(v) = Dg_{(f(x),f(x))}(Df_x(v),Df_x(v)) = Df_x(v)^TAf(x) + f(x)^TADf_x(v)$$ | 2022-08-12T21:32:47 | {
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https://www.physicsforums.com/threads/trigonometry-sinusoid.878213/ | # Trigonometry - sinusoid
Tags:
1. Jul 8, 2016
### Rectifier
The problem
I am trying to write $-sin 2x + \sqrt{3} \cos 2x$ as $A \sin(2x+\phi)$ without using this formula below $a sin (x) + b cos (x) = c sin (x+v), \ \ c=\sqrt{a^2+b^2} \ ,\ \ \tan v = \frac{b}{a}$
The attempt
$A \sin(2x+\phi) = A \cos(\phi) \sin(2x) + A \sin(\phi) \cos(2x)$ comparison with the starting expression.
$$\begin{cases} A \cos(\phi) = -1 \\ A \sin(\phi) = \sqrt{3} \end{cases}$$
$\phi :$
$$\frac{A \sin(\phi)}{A \cos(\phi)} = \tan \phi = \frac{\sqrt{3}}{-1} = - \sqrt{3} \\ \tan -\phi = \sqrt{3} \\ \phi = -\frac{\pi}{3} + \pi n, \ \ n \in \mathbb{Z}$$
For $n=0$ -> $-\frac{\pi}{3}$ gives $$A \sin(-\frac{\pi}{3}) = \sqrt{3} \\ -A \sin(\frac{\pi}{3}) = \sqrt{3} \\ -A \frac{\sqrt{3}}{2}=\sqrt{3} \\ A = -2$$
$n = 1$ from above gives $\phi = \frac{2 \pi}{3}$
$$A \sin(\phi) = \sqrt{3} \\ A \sin(\frac{2 \pi}{3}) = \sqrt{3}$$
I can rewrite $\sin(\frac{2 \pi}{3})$ as $2 \sin(\frac{\pi}{3}) \cos(\frac{\pi}{3}) = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}$
$$A \sin(\frac{2 \pi}{3}) = \sqrt{3} \\ A \frac{\sqrt{3}}{2} = \sqrt{3} \\ A = 2$$
Which A should I use in my answer? And why?
2. Jul 8, 2016
### Irene Kaminkowa
An easier approach
$$-sin 2x + \sqrt{3}cos 2x = 2\left ( -\frac{1}{2}sin 2x+\frac{\sqrt{3}}{2} cos 2x\right )=2\left ( cos (2\pi/3) sin 2x+sin(2\pi/3)cos 2x\right )=2sin(2x + 2 \pi/3)$$
3. Jul 8, 2016
### Rectifier
Thank you for that elegant solution. I solved the problem myself just now, it does not really matter which A I use as long as I use it with the corresponding angle, since $-2 \sin(2x- \frac{\pi}{3}) = 2 \sin(2x + \frac{2 \pi}{3})$ . | 2018-02-25T08:23:55 | {
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https://ohhaskme.com/327973/inequalities-and-multiplying-dividing-by-a-variable | The book says that when you multiply/divide both sides of an inequality, the relation switches if what you're multiplying/dividing by is negative. That's fine if I'm multiplying/dividing by a constant, but what happens when I'm multiplying/dividing by a variable or an expression containing variables? Does the inequality then branch into two forms?
For example, let's say I have the inequality 5 < 10 and I want to multiply both sides by x-1. Does the inequality then branch out to multiple inequalities?
* 5\*(x-1) < 10\*(x-1), if x-1 > 0
* 5\*(x-1) > 10\*(x-1), if x-1 < 0
* ????, if x-1 = 0
Then what happens when you keep taking on stuff to the sides? Now lets say I want to divide both sides by y. Do the branches keep exploding out?
* (5\*(x-1))/y < (10\*(x-1))/y, if x-1 > 0 and y > 0
* (5\*(x-1))/y > (10\*(x-1))/y, if x-1 > 0 and y < 0
* (5\*(x-1))/y > (10\*(x-1))/y, if x-1 < 0 and y > 0
* (5\*(x-1))/y < (10\*(x-1))/y, if x-1 < 0 and y < 0
* ... cases where x-1 and y are 0 ...
Yes, though the case where the multiplicand is 0 is trivial: you just get 0 on both sides and 0 = 0. And you can't divide by 0 so that case doesn't apply.
Yes. Inequalities compare real numbers (for most purposes anyway) instead of general expressions/functions. So if you see an inequality in the real variable x,
f(x) < g(x)
we are actually evaluating the expressions one value of x at a time. It's easy to come up with an example wherein both
f(x_0) < g(x_0)
and
f(x_1) ≥ g(x_1)
with x_0 distinct from x_1, are true.
So an inequality involving a real variable x has an accompanying range of values for x for the inequality to be true.
by | 2023-03-28T11:32:28 | {
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https://math.stackexchange.com/questions/3135697/is-the-set-of-paths-between-any-two-points-moving-only-in-units-on-the-plane-cou/3135746 | # Is the set of paths between any two points moving only in units on the plane countable or uncountable?
Consider 2 arbitrary, fixed points A and B on the plane. Suppose you can move from point A in unit distance at any angle to another point and from this point you can again travel a unit distance at any angle to another point and so on. Ultimately your goal is to travel from point A to point B along a path of unit-distance line segments without repeating a point during your journey. Is the set of such paths countable or uncountable?
I believe it is uncountable and here is my thought process, but I'm not sure of my logic. Consider the line between the two points equidistant from them; call this line L. A path of only unit distances can be made between A and any point, P, which lies on L without crossing L(I don't know how to prove this statement but it seems true). This path can be mirrored on the other side of L to connect B to P. Thus for any point, P, which lies on L a path can be made from A to B crossing through P halfway through the path. Since the points on L are uncountable the set of paths between A and B are also uncountable.
## 7 Answers
Yes, this looks convincing. For the missing step, go directly from A towards P in unit steps until the distance left is less than 2. Then use the remaining distance as the base of an isosceles triangle with unit legs, which you make point away from L.
• It's not only convincing, it's simply correct. OP has shown a mapping of domain subset (domain here is all possible paths with the given constrains) to the straight line so the cardinality of domain has to be no less than the cardinality of a line (which of course is uncountable). – Ister Mar 6 at 12:28
Go one unit from $$a$$ at an angle of $$t$$ to $$c$$.
Go in unit steps along ca until one is less than a unit away from $$b$$ to a point $$p$$.
If $$p \neq b$$, then draw a triangle with base $$pb$$ and sides of unit length adding the sides as the final steps.
As for each $$t$$ in $$[0,2\pi)$$, I've constructed a different accepted zigzaging from $$a$$ to $$b$$, there are uncountably many ways of so staggering from $$a$$ to $$b$$.
Let $$C$$ be on the line through $$B$$ that is perpendicular to the segment $$AB$$ with the distance $$BC$$ equal to $$1/2.$$ Take any half-line $$L$$, not through $$B$$, that originates at $$A$$ and intersects the segment $$BC.$$ For some $$n\in \Bbb N$$ there is a path along $$L,$$ starting at $$A,$$ determined by $$n$$ points $$A=A_1,...,A_n$$ where $$A_j,A_{j+1}$$ are distance $$1$$ apart for each $$j and such that the distance from $$A_n$$ to $$B$$ is less than $$1.$$
Let the point $$D$$ be such that $$A_nD=BD=1$$ nd $$D\not \in \{A_1,...,A_n\}.$$ Then the path determined by $$\{A_1,...,A_n\}\cup \{D\}$$ is a path of the desired type.
The cardinal of the set all such $$L$$ is $$2^{\aleph_0}$$ so there at least this many paths of the desired type, joining pairs of points .
And each path is determined by a function from some $$\{1,2,...,m\}\subset \Bbb N$$ into $$\Bbb R^2.$$ The set of all such functions has cardinal $$2^{\aleph_0}$$ so there are at most $$2^{\aleph_0}$$ paths of the desired type.
Suppose first that the distance $$AB$$ between $$A$$ and $$B$$ is strictly between $$2$$ and $$3$$. Then for every point $$C$$ such that $$AC=1$$ and $$BC<2$$, there is a second point $$D$$ such that $$CD=1=BC$$ (draw an isosceles triangle). The set of such points $$C$$ forms an open arc on the unit circle centered at $$A$$, and so there are uncountably many such $$C$$, hence uncountably many such "unit paths".
For general points $$A$$ and $$B$$, just find one unit path between $$A$$ and $$B$$ that contains two intermediate points $$A'$$ and $$B'$$ with $$2 (for example, points on a giant circle eventually connecting $$A$$ to $$B$$), and then vary the path from $$A'$$ to $$B'$$ in the way described above.
This proof even shows that one can get uncountably many unit paths with the same number of unit-steps. A slight sharpening shows that for any integer $$k>AB$$, there are uncountably many unit paths from $$A$$ to $$B$$ with exactly $$k$$ unit-steps.
I believe it is uncountable also, and here is my thought process:
For each direction vector $$v\in\Bbb R^2=T_a\Bbb R^2$$, take a curve $$\alpha_v$$ from $$a$$ to $$b$$ with $$\alpha_v'(0)=v$$.
Then if $$v\neq w$$, we have $$\alpha_v\neq\alpha_w$$.
But clearly there are uncountably many direction vectors in $$\Bbb R^2=T_a\Bbb R^2$$.
• It seems OP wants movements made of straight line segments all of length $1,$ to go from fixed $A$ to fixed $b$ points. [rather than curves as you use] – coffeemath Mar 5 at 3:28
• I did notice that. But I thought, well, straight lines have derivatives. @coffeemath – Chris Custer Mar 5 at 4:39
• Cris-- Then to finish one has to show that after the first initial unit straight line step from $A,$ it is possible to continue such steps and somehow arrive at $B.$ – coffeemath Mar 5 at 5:17
• @coffeemath yes. Well there certainly appear to be some details missing. But this seems fairly reasonable. Thanks. – Chris Custer Mar 5 at 5:21
• William Elliot's answer does this. – coffeemath Mar 5 at 5:52
Your proof is correct. We can fill in the details on the first part: For any two distinct points $$A$$ and $$B$$ in the plane, we can construct a path consisting of unit distances from $$A$$ to $$B$$.
If $$A$$ is more than one unit away from $$B$$, start at $$A$$ and take unit steps toward $$B$$ until we are less than one unit from $$B$$. If $$B$$ is a whole number of steps away, we will land on $$B$$. If not, we will land on a point $$P$$ within the unit circle centered at $$B$$. Since $$P$$ is less than a distance of $$1$$ away from $$B$$, the unit circle centered at $$P$$ will intersect the unit circle centered at $$B$$ in two places. Take one step from $$P$$ to an intersection point and then one step from the intersection point to $$B$$ to finish the path.
Yes, and here is a very simple and intuitive argument without mirrors ;-)
Observation:
• Let C be any point on AB such that 0< |AC| < min( |AB|, 2)
• We can always travel from A to C in 2 x 1-unit steps. (Construct an appropriate unit side isosceles triangle to a point X on the perpendicular bisector of AC such that |AX| = |XC| = 1).
• We can then travel from C to B in < |AB| unit steps.
• So all such paths (whatever value of C we chose) will traverse AXCB in < (|AB| + 2) steps, and all such paths are distinct if their respective C' and C are distinct.
But as C can be any real number in that range of real numbers, the number of distinct C we could have chosen is uncountable. Each C gives a unique valid path, so the number of paths in uncountable.
The followup question would be the cardinality involved. There is a chance it might be more than $$\mathfrak c=2^{\aleph_0}$$ [cardinality of reals, corrected per comments below] - but that's far beyond my skill to figure out.
• The cardinality is 'only' aleph_0. You can unique specify any such path by listing the coordinates of all the intermediate points. Hence the set of all paths is contained in the set of finite lists of real numbers. This set has cardinality aleph_0. – quarague Mar 5 at 8:02
• The cardinality of the real numbers is not $\aleph_0$. It is $\mathfrak c=2^{\aleph_0}$. ($\aleph_0$ is the cardinality of the integers.) – TonyK Mar 5 at 10:32
• Thanks, both. Especially @quarague - a simple clear answer to that one – Stilez Mar 5 at 11:42
• The correction of TonyK is true. We both meant the cardinality of the reals, but aleph_0 is just the cardinality of the integers. – quarague Mar 5 at 11:47
• Corrected, thank you. But is "set of finite lists" actually the right descriptor? Eg if A=(0,0) B=(0,1) then any specific path of the form [[ ... (-1,0), (-1,0) {n times} ...(0,1), ...(1,0), (1,0) {n times} ... ]] is a valid path. After all, a list of "arbitrary long but finite" paths sounds indistinguishable from an infinite list? – Stilez Mar 5 at 15:34 | 2019-04-25T02:10:56 | {
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https://byjus.com/question-answer/a-and-b-play-a-match-to-be-decided-as-soon-as-either-has-won/ | Question
$$A$$ and $$B$$ play a match to be decided as soon as either has won two games. The chance of either winning a game is $$\displaystyle \dfrac{1}{20}$$ and of its being drawn $$\displaystyle \dfrac{9}{10}.$$ What is the the chance that the match is finished in $$10$$ or less games?
A
0.18 approx.
B
0.16 approx.
C
0.17 approx.
D
0.15 approx.
Solution
The correct option is B $$\displaystyle 0.17$$ approx.If the match is not finished in $$10$$ games, either of the following events. will possibly occur : (i) All the games are drawn (ii) $$A$$ and $$B$$ each win one game and the rest $$8$$ games are drawn. (iii) $$A$$ or $$B$$ wins one game and the rest $$9$$ games are drawn. The corresponding probabilities of the above events are (i) $$\displaystyle \left ( \frac{9}{10} \right )^{10},$$ (ii) 2.$$\displaystyle ^{10}C_{2}\left ( \frac{1}{20} \right )^{2}\left ( \frac{9}{10} \right )^{8},$$ (iii) 2.$$\displaystyle ^{10}C_{1}\left ( \frac{1}{20} \right )\left ( \frac{9}{10} \right )^{9},$$ Since these events are mutually exclusive, the probability of the game not being finished in 10 games $$\displaystyle =\left ( \frac{9}{10} \right )^{10}+2^{10}C_{2}\left ( \frac{1}{20} \right )^{2}\left ( \frac{9}{10} \right )^{8}+2^{10}C_{1}\left ( \frac{1}{20} \right )\left ( \frac{9}{10} \right )^{9}$$ $$\displaystyle =\frac{9^{8}\times 387}{2\times 10^{10}}$$ Hence the probability that the match is finished in $$10$$ or less number of games $$\displaystyle =1-\frac{9^{8}\times 387}{2x\times 10^{10}}=0.17$$ approx.Maths
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