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https://math.stackexchange.com/questions/4013760/clarifying-the-notion-of-basic-free-variables-in-a-system-of-equations | # Clarifying the notion of basic/free variables in a system of equations
I have a fundamental confusion with regards to the notion of free/basic variables.
Consider the following linear system $$\begin{pmatrix} 1 & 2 & -1 \\ 2 & -4 & 0 \end{pmatrix}\begin{pmatrix} x_1\\ x_2 \\ x_3 \end{pmatrix}=\begin{pmatrix} 4\\ 5 \end{pmatrix}$$ This system has augmented matrix $$\begin{pmatrix} 1 & 2 & -1 & 4 \\ 2 & -4 & 0 & 5 \end{pmatrix}$$ which row reduces to $$\begin{pmatrix} 1 & 2 & -1 & 4 \\ 0 & -8 & 2 & -3 \end{pmatrix}$$ The last matrix is in echelon form. Columns 1 and 2 are pivot columns, so $$x_1,x_2$$ are basic variables. The third column is not a pivot column, so $$x_3$$ is a free variable. Finally, the last column is not a pivot column, so the system is consistent.
The above analysis tells me that $$x_3$$ is a free variable. I interpret this as saying "$$x_1,x_2$$ can be expressed as a function of $$x_3$$, while $$x_3$$ can be set equal to any number". For example, I can set $$x_3=200$$. In turn, $$\begin{cases} x_1+2x_2-200=4 \\ 2x_1-4x_2=5 \end{cases}\Rightarrow x_2=\frac{403}{8}, x_1=\frac{413}{4}$$
Nevertheless, just by doing naive computations, I realised that we could also "express $$x_1$$, $$x_3$$ as a function of $$x_2$$, while setting $$x_2$$ equal to any number". For example, I can set $$x_2=7$$. In turn, $$\begin{cases} x_1+14-x_3=4\\ 2x_1-28=5 \end{cases}\Rightarrow x_1=\frac{33}{2}, x_3=\frac{53}{2}$$ Alternatively, we could also "express $$x_2$$, $$x_3$$ as a function of $$x_1$$, while setting $$x_1$$ equal to any number". For example, I can set $$x_1=0$$. In turn, $$\begin{cases} 2x_2-x_3=4\\ -4x_2=5 \end{cases}\Rightarrow x_2=-\frac{13}{2}, x_3=-\frac{5}{4}$$
Hence, my question: what is the relation between
• the fact that $$x_3$$ is a free variable and $$x_1,x_2$$ are basic variables (from the row-reduced matrix)
• and the fact that in the system above I am free to fix the value of anyone among $$x_1,x_2,x_3$$ and I will always be able to solve for the other two unrestricted variables
• Note that the order of the columns is arbitrary in a way. Feb 5 at 13:22
You've got three column vectors, and any pair of them are independent and thus span $$\Bbb R^2$$. The solution set is one-dimensional (affine) (aka a line) so of the form $$v+tw$$, where $$v,w$$ are some vectors and $$t$$ is a free real variable. We can choose this to be $$x_3$$, or any of the other variables if you prefer.
You can then transform the equations to \begin{align}x_1+2x_2 &= 4+x_3\\ x_1 - 4x_2 &= 5\end{align}
and solve these in terms of $$x_3$$ uniquely. We get (add twice the first equation to the second and we get $$3x_1 = 13+2x_3$$ so $$x_1 = \frac{1}{3}(13+2x_3)$$, and subtraction the equations: $$6x_2 = x_3-1$$ and hence $$x_2 = \frac{1}{6}(x_3 - 1)$$ (we could also do another elimination step on this new system) so we can write the solution line as
$$\begin{pmatrix} \frac{13}{3}\\ -\frac16\end{pmatrix} + x_3 \begin{pmatrix} \frac{2}{3}\\ \frac16\end{pmatrix}$$
• Thanks. (1) Can we say that in this specific case "$x_3$ is a free variable, $x_1,x_2$ are basic variables" or, equivalently, "$x_1$ is a free variable, $x_3,x_2$ are basic variables" or, equivalently, "$x_2$ is a free variable, $x_1,x_3$ are basic variables"? (2) Is there any way to realise from the echelon form that we are "free" to permute the columns? This somehow relates to my other question math.stackexchange.com/questions/4012647/… if you can help.
– TEX
Feb 5 at 13:27
• @TEX They can all be free, that's the point. It doesn't matter. Feb 5 at 13:28
• Thanks. My doubt is: is there a formal way to understand from the echelon form (or any other form) when any variable can be set as the free variable? This is not always the case.
– TEX
Feb 5 at 13:29
• @TEX you just use $x_3$ and get the full solution set. Why do more? Feb 5 at 13:31
• Because this relates to a more general result that I need to prove. I want to understand when I'm free to set the free variables and how I can detect that.
– TEX
Feb 5 at 13:32 | 2021-10-20T19:16:06 | {
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https://math.stackexchange.com/questions/300393/group-tables-for-a-group-of-four-elements | # Group tables for a group of four elements.
I should consider group tables obtained by renaming elements as essentially the same and then show that there are only two essentially different groups of order 4.
There seems to be so many different possible group tables for all the different binary operations - which is why I'm confused. I was thinking about using Cayley's table to show their commutativity but I'm really not too sure. Any help please!
Edit: After all of your help, I completely understand how to show that there are only two different groups of order four. Thank you. The only thing which I am still unclear of is how to note down the 'other tables' before stating that they are essentially the same as one of the other tables - meaning that there are just two different ones. It's pointless work but I think it's what the question requires.
Answer: After a bit of playing around - I've realised that there are four different tables, however, 3 tables are the same as each other, just with different values (the Klein 4 Group with 3 different generators). Hence there are two tables.
• It might seem like there are many possibilities, but try to start writing out some of them and note that there are certain properties they must have in order to satisfy the group axioms. – Tobias Kildetoft Feb 11 '13 at 17:47
• Re: duplicate vote. It's the same question, but the discussion in the older one is left unfinished. – user53153 Feb 11 '13 at 19:12
Hints: we are talking about groups of order 4, which narrows the possible binary relations and elements available:
• each must contain the identity,
• each must be associative,
• each must be closed under inverses (if $a \in G,\;a^{-1} \in G$)
• (and of course, closed under the group operation).
There are essentially (up to isomorphism) only two groups of order 4:
• one of course will be the additive cyclic group $\mathbb{Z}_4$, and
• the other will be the Klein 4-group, which is indeed abelian.
If you follow the suggestions for solving the problem, you'll find, indeed, that any possible GROUP of order 4 can be shown isomorphic to one of the two groups mentioned simply by a renaming elements.
Yes: use of the Cayley table will be of great importance:
• for a group: no element can appear twice in any column,
• no element can appear twice in any row.
You'll find that the only ways of completing a table satisfying these criteria are limited, and then you show that a simple renaming of elements will reveal the group is isomorphic to $\mathbb{Z}_4$ or else the Klein-4 group.
• Yes, this has been very helpful, thank you. Your answer is so neatly written too! The only question I have is that I believe I should write down all the group tables... Is this just 2 group tables (of what binary operation???) OR do I list more but eliminate the rest. With Cayley's table, to show commutativity, is it sufficient to use the symmetry of the diagonal line? Thanks again. PS, I have encountered Lagrange - just struggling with Algebra! – user61854 Feb 11 '13 at 21:00
• If you have encountered Lagrange, than it suffices to argue that the only groups of order 4 have are the cyclic group $\mathbb{Z}$ (generated by one element), and a group whose non-identity elements have order 2 (since two and four divide 4). Completing any table that satisfies the properties of a group (one and only one element in each column and each row will limit the number of Cayley tables to construct. You can then show that of those, each is isomorphic to $Z_4$ or the Klein 4-group. And yes, you can show that those two non-isomorphic groups are indeed commutative by the Cayley table. – amWhy Feb 12 '13 at 1:47
There aren't that many. One of the elements will be the identity, let's call it $1$. (I am writing the operation as multiplication.) So the table looks like $$\begin{matrix} 1 & a & b & c\\ a\\ b\\ c\\ \end{matrix}$$ (I'm not writing the row and column labels, as they are the same as the first column and row). Now what could $ab$ be? Not $a$, not $b$... And $ba$? Proceed with $ac$, $ca$, $bc$, $cb$, and then you are left with the squares. They might all be $1$, or...
This simplifies it:
If a group has 4 elements, we separate two cases:
Case1: all the elements have order 2, then the group is abelian, and $a^2=1\Rightarrow a=a^{-1}$ for all $a\in G$: $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$ the table should be easy to make in this case: there is only one.
Case2: there is an element $a\not =1$ that has order different from 2, then by LAgrange's theorem, the order must be 4, and so $G=\langle a\rangle$, and so the group is cyclic, and therefore it's abelian. We have proved that a 4-element group is always abelian. The table shouldn't be difficult in this case either.
• You're assuming Lagrange, and chances are, at this stage, the OP has not yet encountered Lagrange. See my comment below Michael Hardy's answer. – amWhy Feb 11 '13 at 19:17
• @amWhy I would have never imagined, Lagrange always comes really soon. I guess that then he will have to do every possible table to get to the same conclusion. – MyUserIsThis Feb 11 '13 at 21:25
• so b^2 = a, c^2 = a in one table and in the other, b^2 = 1, c^2 = 1. – user61854 Feb 12 '13 at 16:45
• @user61854 Yes, in my case 1, you have that: $a^2=b^2=c^2=1$, so: $ab=c, bc=a$, it's abelian, so the whole table is there. In my case 2, $a$ is the generator, so $a=a,a^2=b,a^3=c$, and you can compose to get the rest of the combinations. – MyUserIsThis Feb 12 '13 at 17:00
The order of an element must divide $$4$$, so it must be either $$1$$, $$2$$, or $$4$$. Only the identity element can have order $$1$$. If one element has order $$4$$, then it generates the whole group and you have a cyclic (hence abelian) group.
Otherwise the three non-identity elements each have order $$2$$.
$$e=\text{the identity}$$.
$$a,b,c=\text{the other three}$$.
$$a^2=b^2=c^2=e$$, and each of these is its own inverse.
So what is $$ab$$? It can't be $$e$$ since if $$ab=e$$ then $$(ab)b^{-1} = b^{-1}=b$$, and hence $$a=b$$. It can't be $$a$$ or $$b$$ since if $$ab=b$$ then $$(ab)b^{-1}=bb^{-1}=e$$ and so $$a=e$$, and similar reasoning shows it can't be $$a$$ (but you have to multiply on the left be the inverse).
So $$ab=c$$.
Similar reasoning shows $$ab=ba=c$$, and $$ac=ca=b$$ and $$bc=cb=a$$.
So the group is abelian.
• Chances are, at this stage of the game, the OP hasn't encountered Lagrange. I know Fraleigh includes this exercise long before introducing Lagrange... – amWhy Feb 11 '13 at 19:15
• @amWhy Because this is such a toy example, you don't need Lagrange; it's trivial to see that only the identity can have order 1, or that if one element has order 4 then the group must be cyclic. The only other possibility aside from the all-order-2 case is that some element (and therefore two elements) has order 3, and it's easy to eliminate this possibility. (Suppose that $a$ and $b$ have order 3, with $a^2=b$. $ca$ can't be $a$ and can't be $c$ by cancellation; it can't be $b$ because $ca=aa\implies c=a$, and it can't be the identity because then $ca=1=ba$ and again by cancellation $c=b$.) – Steven Stadnicki Feb 12 '13 at 23:33 | 2021-04-21T02:34:19 | {
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https://www.physicsforums.com/threads/what-does-ceil-mean.116822/ | # What does ceil mean ?
1. Apr 7, 2006
### momentum
Its diifcult to express my question....so, i am posting this
ceil(4.5) =?
ceil(4.1)=?
ceil(4.6)=?
2. Apr 7, 2006
### AKG
ceil(x) is the smallest integer which is greater than or equal to x. In particular, if x is an integer, then ceil(x) = x, and if x is not an integer, then ceil(x) > x.
3. Apr 7, 2006
### momentum
ceil(4.5)=4 // is it ok ?
ceil(4.1)=4 //is it ok ?
ceil(4.6)=5 //is it ok ?
4. Apr 7, 2006
### Integral
Staff Emeritus
Nope, read the defintion given above, then try again.
5. Apr 7, 2006
### momentum
ah...i see, all of them should be 5 .....i had confusion on fractional part .5.
but i see ..it does not care for .5 which we use for round-off.
6. Apr 7, 2006
Yes, all 5.
7. Apr 7, 2006
### momentum
thank you for the clarifcation
8. Apr 7, 2006
### bomba923
Recall that
$$\begin{gathered} \forall x \in \left( {a,a + 1} \right)\;{\text{where }}a \in \mathbb{Z}, \hfill \\ {\text{floor}}\left( x \right) = \left\lfloor x \right\rfloor = a \hfill \\ {\text{ceil}}\left( x \right) = \left\lceil x \right\rceil = a + 1 \hfill \\ \end{gathered}$$
$$\forall x \in \mathbb{Z},\;\left\lfloor x \right\rfloor = \left\lceil x \right\rceil = x$$
Last edited: Apr 7, 2006
9. Apr 7, 2006
### Staff: Mentor
ceil -> "goes up" if it needs to, in order reach an integer
floor -> "goes down" as it needs to, in order to reach an integer
What happens with negative numbers:
$$floor( -1.1 ) = -2 \; ceil( -1.1 ) = -1$$
$$floor( -0.1 ) = -1 \; ceil( -0.1 ) = 0$$
$$floor( 0.9 ) = 0 \; ceil( 0.9 ) = 1$$ | 2017-10-19T18:53:57 | {
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https://math.stackexchange.com/questions/3690609/how-to-translate-this-statement-into-a-mathematical-oneusing-appropriate-quanti/3699299#3699299 | # How to translate this statement into a mathematical one(using appropriate quantifiers)?
The statement I'd like to translate into a mathematical one is
"Every American has a dream".
Let $$A$$ and $$D$$ denote the set of all Americans and the set of all dreams, respectively, and $$P(a,d)$$ denote the proposition "American $$a$$ has a dream $$d$$". The mathematically equivalent statement I've deduced is $$\forall a\in A.\exists d\in D.P(a, d)$$
However, I suspect the above statement implies that for every American there exists a common dream $$d$$ such that $$P(a,d)$$ holds true. I would like to know how to rectify this error(if there is one).
• Correct: "forall a there is a d..." does not mean that the d is the same for all a. To state that the d is the same, you have to write "there is a d for all a...". Compare $\forall n \exists m (n < m)$ and $\exists m \forall n (n < m)$. May 25, 2020 at 10:01
1. The verifier of a sentence of the form "$$∀a{∈}A\ ( P(a) )$$" must let the refuter first choose any arbitrary $$a∈A$$ and then verify $$P(a)$$ no matter what $$a∈A$$ was chosen.
2. The verifier of a sentence of the form "$$∃d{∈}D\ ( Q(d) )$$" must first choose some $$d∈D$$ and then verify $$Q(d)$$ for that chosen $$d∈D$$.
In your example, the verifier of "$$∀a{∈}A\ ∃d{∈}D\ ( P(a,d) )$$" must let the refuter make the first move in choosing an $$a∈A$$, and then verify "$$∃d{∈}D\ ( P(a,d) )$$" no matter what $$a$$ was chosen. But since the verifier makes the second move in choosing some $$d∈D$$, the verifier can choose this $$d$$ based on the refuter's first move (i.e. based on $$a$$). That is why "Every American has a dream." corresponds to this sentence.
In contrast, the verifier of "$$∃d{∈}D\ ∀a{∈}A\ ( P(a,d) )$$" must make the first move in choosing some $$d∈D$$, before the refuter makes the second move in choosing an $$a∈A$$. You can see easily that the verifier can win only if there is a single choice of $$d∈D$$ that defeats every possible choice of $$a∈A$$. That is why "All Americans have a common dream." corresponds to this sentence and not the other one. | 2023-03-29T03:00:09 | {
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http://simonedavis.com/h-xbu/10hec.php?page=acute-isosceles-triangle-421da2 | A right triangle may be isosceles … Input 3 triangle side lengths (A, B and C), then click "ENTER". Constructing an isosceles triangle Constructing an isosceles triangle also known as drawing an isosceles triangle using only a straightedge and a compass is what I will show you here. B - isosceles triangle. Yes, because it has an obtuse angle. Please update your bookmarks accordingly. Obtuse isosceles triangle with an irrational but algebraic ratio between the lengths of its sides and its base. This concept will teach students the properties of isosceles triangles and how to apply them to different types of problems. An isosceles right triangle therefore has angles of 45 degrees, 45 degrees, and 90 degrees. Apply the properties of isosceles triangles. An isosceles triangle is a special case of a triangle where 2 sides, a and c, are equal and 2 angles, A and C, are equal. Given below are the properties of isosceles and acute triangles. Parts of an Isosceles Triangle. The angles opposite the equal sides are also equal. Find information related to equilateral triangles, isosceles triangles, scalene triangles, obtuse triangles, acute triangles, right angle triangles, the hypotenuse, angles of a triangle and more. C - obtuse triangle. Let the coordinates of the right-angled isosceles triangle be O(0, 0), A(a, 0) and B(0, a). Area of the acute triangle is $$A = {1 \over 2} \times b \times h$$ The Perimeter of the Acute triangle is: $$P=a+b+c$$ The types of acute triangles are: a) Acute Equilateral Triangle b) Acute Isosceles Triangle c) Acute Scalene Triangle All three angles are different implies that all three sides of an acute scalene triangle are also different. The two shorter sides measure x cm and 3x cm. Without Using The Calculator When given 3 triangle sides, to determine if the triangle is acute… An acute angle triangle (or acute-angled triangle) is a triangle in which all the interior angles are acute angles. Step-by-step explanation: Correct statements and reasons are in bold. A - acute triangle. For an acute triangle, all three angles of the triangle are from $$0^o$$ to $$90^o$$. This calculator will determine whether those 3 sides will form an equilateral, isoceles, acute, right or obtuse triangle or no triangle at all. In a right triangle, one of the angles is a right angle—an angle of 90 degrees. For example, if we know a and b we know c since c = a. Addtionally, like all triangles, the three angles will sum to . All the three angles situated within the isosceles triangle are acute, which signifies that the angles are less than 90°. We have moved all content for this concept to for better organization. Step 1: One of the angles of the given triangle is a right angle. Add these two angles together and subtract the answer from 180° to find the remaining third angle. Answer : D Explanation. the isosceles triangle can be acute, right or obtuse, but it depends only on the vertex angle (base angles are always acute) The equilateral triangle is a special case of a isosceles triangle. PROPERTIES OF ISOSCELES TRIANGLE ABC Let ABC be an isosceles triangle with
Tourist Places In Coimbatore Near Gandhipuram, Sunrunner Pembroke Welsh Corgis, Where Is Serana In Fort Dawnguard, Hello Etch A Sketch Font, Monkey King Characters, Adam Bradley Amazon, | 2021-07-27T21:39:46 | {
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https://mathhelpboards.com/threads/c-f-gauss-on-how-to-add-all-numbers-of-1-to-100.1832/ | # C F Gauss, on how to add all numbers of 1 to 100?
#### CaptainBlack
##### Well-known member
Carl gauss 1777 1855 , on how to add all numbers of 1 to 100?
I can only add from 1 to 100 in order, apparently he could add lighteningly fast, and backwards, can anyone explain how in the confines of one message on here please
Allegedly his method was something like:
$(1+100)+(2+99)+ ... +(99+2)+(100+1)=2(1+2+...+100)$
But the left hand side is $101\times 100$, so $1+2+..100=101\times 100/2$
(Like Marlow's Dr Faustus he could sum them forwards and backwards - but had no need of anagramatisation)
CB
Last edited:
#### Bacterius
##### Well-known member
MHB Math Helper
This also works for any sum:
$\displaystyle \sum_{k = 1}^{n} k = 1 + 2 + \cdots + n$
This sum can be reorganized in this fashion:
$\displaystyle \left ( 1 + \left ( n \right ) \right ) + \left ( 2 + \left ( n - 1 \right ) \right ) + \cdots + \left (n + \left (1 \right ) \right )$
This produces $n$ sums of $n + 1$, so the total sum is $n(n + 1)$. But note that this goes through the list twice, from the left and from the right (for instance, $1$ is added to $n$ at the beginning and at the end). So this is actually twice the sum we need, we conclude:
$\displaystyle \sum_{k = 1}^{n} k = \frac{n(n + 1)}{2}$
$n = 100$ can then be easily calculated from the formula, just as can $n = 4885$ or $n = 6$
#### QuestForInsight
##### Member
I would like to share two more ways of doing this.
Proof #0 (Notice that what Gauss did is basically this):
\begin{aligned} \displaystyle & \sum_{0 \le k \le n}k = \sum_{0 \le k \le n}(n-k) = n\sum_{0 \le k \le n}-\sum_{0 \le k \le n}k \\& \implies 2\sum_{0 \le k \le n}k = n(n+1) \implies \sum_{0 \le k \le n}k = \frac{1}{2}n(n+1).\end{aligned}
Proof #1 (Now try that with $\sum_{0 \le k \le n}k^2$ and you will be pleasantly surprised):
\begin{aligned} \displaystyle & \sum_{0 \le k \le n}k^2 = \sum_{0 \le k \le n}(n-k)^2 = n^2\sum_{0 \le k \le n}-2n\sum_{0 \le k \le n}k+\sum_{0 \le k \le n}k^2 \\& \implies 2n\sum_{0 \le k \le n}k = n^2(n+1) \implies \sum_{0 \le k \le n}k = \frac{1}{2}n(n+1).\end{aligned}
Proof #2 (Writing it as a double sum and switching the order of summation):
\begin{aligned}\displaystyle & \begin{aligned}\sum_{1 \le k \le n}k & = \sum_{1 \le k \le n}~\sum_{1 \le r \le k} = \sum_{1 \le r \le n} ~\sum_{r \le k \le n} = \sum_{1 \le r \le n}\bigg(\sum_{1 \le k \le n}-\sum_{1 \le k \le r-1}\bigg) \\& =\sum_{1 \le r \le n}\bigg(n-r+1\bigg) = n\sum_{1 \le k \le n}-\sum_{1 \le k \le n}k+\sum_{1 \le k \le n}\end{aligned} \\& \implies 2\sum_{1 \le k \le n}k = n^2+n \implies \sum_{1 \le k \le n}k = \frac{1}{2}n(n+1), ~ \mathbb{Q. E. D.}\end{aligned}
Last edited:
#### MarkFL
Staff member
Here are two closely related methods to derive the formulas for summations of sequences of natural number powers of integers from 0 - n.
Method 1: A "bottom-up" approach...
We may state:
$\displaystyle\sum_{k=0}^n(k)=\sum_{k=0}^n(k+1)-(n+1)$
$\displaystyle\sum_{k=0}^n(k)=\sum_{k=0}^n(k)+\sum_{k=0}^n(1)-(n+1)$
$\displaystyle0=\sum_{k=0}^n(1)-(n+1)$
$\displaystyle\sum_{k=0}^n(1)=n+1$
Now we may compute:
$\displaystyle\sum_{k=0}^n(k)$
We may state:
$\displaystyle\sum_{k=0}^n\left(k^2 \right)=\sum_{k=0}^n\left((k+1)^2 \right)-(n+1)^2$
$\displaystyle\sum_{k=0}^n\left(k^2 \right)=\sum_{k=0}^n\left(k^2+2k+1 \right)-(n+1)^2$
$\displaystyle\sum_{k=0}^n\left(k^2 \right)=\sum_{k=0}^n\left(k^2 \right)+2\sum_{k=0}^n(k)+\sum_{k=0}^n(1)-(n+1)^2$
$\displaystyle2\sum_{k=0}^n(k)=-\sum_{k=0}^n(1)+(n+1)^2$
Now, using our previous result, we have:
$\displaystyle2\sum_{k=0}^n(k)=-(n+1)+(n+1)^2$
$\displaystyle2\sum_{k=0}^n(k)=(n+1)\left(-1+(n+1) \right)$
$\displaystyle2\sum_{k=0}^n(k)=(n+1)(n)$
$\displaystyle\sum_{k=0}^n(k)=\frac{n(n+1)}{2}$
Now we may continue and find:
$\displaystyle\sum_{k=0}^n\left(k^2 \right)$
$\displaystyle\sum_{k=0}^n\left(k^3 \right)=\sum_{k=0}^n\left((k+1)^3 \right)-(n+1)^3$
$\displaystyle\sum_{k=0}^n\left(k^3 \right)=\sum_{k=0}^n\left(k^3+3k^2+3k+1 \right)-(n+1)^3$
$\displaystyle\sum_{k=0}^n\left(k^3 \right)=\sum_{k=0}^n\left(k^3 \right)+3\sum_{k=0}^n\left(k^2 \right)+3\sum_{k=0}^n(k)+\sum_{k=0}^n(1)-(n+1)^3$
$\displaystyle3\sum_{k=0}^n\left(k^2 \right)=-3\sum_{k=0}^n(k)-\sum_{k=0}^n(1)+(n+1)^3$
Using our previous results, we have:
$\displaystyle3\sum_{k=0}^n\left(k^2 \right)=-3\left(\frac{n(n+1)}{2} \right)-(n+1)+(n+1)^3$
$\displaystyle3\sum_{k=0}^n\left(k^2 \right)=(n+1)\left(-\frac{3n}{2}-1+(n+1)^2 \right)$
$\displaystyle3\sum_{k=0}^n\left(k^2 \right)=(n+1)\left(-\frac{3n}{2}-1+n^2+2n+1 \right)$
$\displaystyle3\sum_{k=0}^n\left(k^2 \right)=(n+1)\left(n^2+\frac{n}{2} \right)$
$\displaystyle3\sum_{k=0}^n\left(k^2 \right)=\frac{n(n+1)(2n+1)}{2}$
$\displaystyle\sum_{k=0}^n\left(k^2 \right)=\frac{n(n+1)(2n+1)}{6}$
We may continue this process, to find further power summations.
Method 2: A recursion approach...
Suppose we want to find:
$\displaystyle S_n=\sum_{k=0}^n\left(k^3 \right)$
We may use the inhomogeneous recursion to state:
$\displaystyle S_n=S_{n-1}+n^3$
Now, we may use symbolic differencing to ultimately derive a homogeneous recursion.
We begin by replacing n with n + 1:
$\displaystyle S_{n+1}=S_{n}+(n+1)^3$
Subtracting the former from the latter, there results:
$\displaystyle S_{n+1}=2S_{n}-S_{n-1}+3n^2+3n+1$
$\displaystyle S_{n+2}=2S_{n+1}-S_{n}+3(n+1)^2+3(n+1)+1$
Subtracting again:
$\displaystyle S_{n+2}=3S_{n+1}-3S_{n}+S_{n-1}+6n+6$
$\displaystyle S_{n+3}=3S_{n+2}-3S_{n+1}+S_{n}+6(n+1)+6$
Subtracting again:
$\displaystyle S_{n+3}=4S_{n+2}-6S_{n+1}+4S_{n}-S_{n-1}+6$
$\displaystyle S_{n+4}=4S_{n+3}-6S_{n+2}+4S_{n+1}-S_{n}+6$
Subtracting again:
$\displaystyle S_{n+4}=5S_{n+3}-10S_{n+2}+10S_{n+1}-5S_{n}+S_{n-1}$
Now we have a homogeneous recursion whose associated characteristic equation is:
$\displaystyle (r-1)^5=0$
Since we have the root $\displaystyle r=1$ of multiplicity 5, we know the closed form for the sum will take the form:
$\displaystyle S_n=k_0+k_1n+k_2n^2+k_3n^3+k_4n^4$
We may use known initial values to determine the constants $\displaystyle k_i$.
$\displaystyle S_0=k_0=0$
This means, we are left with the 4X4 system:
$\displaystyle k_1+k_2+k_3+k_4=1$
$\displaystyle 2k_1+4k_2+8k_3+16k_4=9$
$\displaystyle 3k_1+9k_2+27k_3+81k_4=36$
$\displaystyle 4k_1+16k_2+64k_3+256k_4=100$
Did you notice we get the squares of successive triangular numbers? Anyway, solving this system, we find:
$\displaystyle k_1=0,k_2=\frac{1}{4},k_3=\frac{1}{2},k_4=\frac{1}{4}$
And so, we have:
$\displaystyle S_n=\frac{1}{4}n^2+\frac{1}{2}n^3+\frac{1}{4}n^4= \frac{n^4+2n^3+n^2}{4}=$
$\displaystyle\frac{n^2(n+1)^2}{4}=\left( \frac{n(n+1)}{2} \right)^2$
#### melese
##### Member
a nice combinatorial proof
To each yellow circle we can associate a pair of green circles in the way suggested by the drawing. This shows a bijection between (and hence the same number of) all the yellow circles and all the pairs of green circles.
Here, there are $1+2+3+4+5+6+7+8$ yellow circles, and the number of green-circle pairs is $\binom{9}{2}$; so $1+2+3+4+5+6+7+8=\binom{9}{2}$.
(I saw this 'proof without words' in a lecture by Gil Kalai)
#### HallsofIvy
##### Well-known member
MHB Math Helper
$(1+100)+(2+99)+ ... +(99+2)+(100+1)=2(1+2+...+100)$
But the left hand side is $101\times 100$, so $1+2+..100=101\times 100/2$ | 2020-11-30T23:10:27 | {
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https://www.jiskha.com/questions/1799344/13-exercise-convergence-in-probability-a-suppose-that-xn-is-an-exponential-random | math, probability
13. Exercise: Convergence in probability:
a) Suppose that Xn is an exponential random variable with parameter lambda = n. Does the sequence {Xn} converge in probability?
b) Suppose that Xn is an exponential random variable with parameter lambda = 1/n. Does the sequence {Xn} converge in probability?
c) Suppose that the random variable in the sequence {Xn} are independent, and that the sequence converges to some number a, in probability.
Let {Yn} be another sequence of random variables that are dependent, but where each Yn has the same distribution (CDF) as Xn. Is it necessarily true that the sequence {Yn} converges to a in probability?
1. 👍
2. 👎
3. 👁
1. a) yes
b) no
c) yes
1. 👍
2. 👎
2. y
n
n
1. 👍
2. 👎
3. 142n=47 find n
1. 👍
2. 👎
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http://www.mathworks.com/help/matlab/ref/istriu.html?s_tid=gn_loc_drop&nocookie=true | Accelerating the pace of engineering and science
# istriu
Determine if matrix is upper triangular
## Description
example
tf = istriu(A) returns logical 1 (true) if A is an upper triangular matrix; otherwise, it returns logical 0 (false).
## Examples
expand all
### Test Upper Triangular Matrix
Create a 5-by-5 matrix.
`A = triu(magic(5))`
```A =
17 24 1 8 15
0 5 7 14 16
0 0 13 20 22
0 0 0 21 3
0 0 0 0 9```
Test A to see if it is upper triangular.
`istriu(A)`
```ans =
1
```
The result is logical 1 (true) because all elements below the main diagonal are zero.
### Test Matrix of Zeros
Create a 5-by-5 matrix of zeros.
`Z = zeros(5);`
Test Z to see if it is upper triangular.
`istriu(Z)`
```ans =
1
```
The result is logical 1 (true) because an upper triangular matrix can have any number of zeros on the main diagonal.
## Input Arguments
expand all
### A — Input arraynumeric array
Input array, specified as a numeric array. istriu returns logical 0 (false) if A has more than two dimensions.
Data Types: single | double
Complex Number Support: Yes
expand all
### Upper Triangular Matrix
A matrix is upper triangular if all elements below the main diagonal are zero. Any number of the elements on the main diagonal can also be zero.
For example, the matrix
$A=\left(\begin{array}{cccc}1& -1& -1& -1\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -2& -2\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& -3\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\right)$
is upper triangular. A diagonal matrix is both upper and lower triangular.
### Tips
• Use the triu function to produce upper triangular matrices for which istriu returns logical 1 (true).
• The functions isdiag, istriu, and istril are special cases of the function isbanded, which can perform all of the same tests with suitably defined upper and lower bandwidths. For example, istriu(A) == isbanded(A,0,size(A,2)). | 2014-12-28T17:35:44 | {
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http://math.stackexchange.com/questions/92393/generalized-fibonacci-sequences | # Generalized Fibonacci sequences
Why Fibonacci sequence start at $0$, Tribonacci sequence with $0,0$, Tetranacci with $0,0,0$, etc. [ref OEIS] Has any good reasons for that?
These sequences arise in generalization of Pascal Triangle as diagonal sums and there they start at $1$.
Pascal triangle:
$$\begin{array}{} \color{red}1& \color{blue}0& \color{green}0& \color{cyan}0&\dots\\ \color{blue}1& \color{green}1& \color{cyan}0& \color{magenta}0&\dots\\ \color{green}1& \color{cyan}2& \color{magenta}1& 0&\dots\\ \color{cyan}1& \color{magenta}3& 3& \color{red}1&\dots\\ \vdots&\vdots&\vdots&\vdots&\ddots \end{array}$$
diagonal sum gives $\color{red}1,\color{blue}1,\color{green}2,\color{cyan}3,\color{magenta}5,8,\color{red}{13}\dots$ Fibonacci sequence
First generalization:
$$\begin{array}{r} \color{red}1& \color{blue}0& \color{green}0& \color{cyan}0& \color{magenta}0& 0&\color{red}0&\dots\\ \color{blue}1& \color{green}1& \color{cyan}1& \color{magenta}0& 0& \color{red}0&\color{blue}0&\dots\\ \color{green}1& \color{cyan}2& \color{magenta}3& 2& \color{red}1& \color{blue}0&\color{green}0&\dots\\ \color{cyan}1& \color{magenta}3& 6& \color{red}7& \color{blue}6& \color{green}3&\color{cyan}1&\dots\\ \color{magenta}1&4&\color{red}{10}&\color{blue}{16}&\color{green}{19}&\color{cyan}{16}&\color{magenta}{10}&\dots\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\ddots \end{array}$$
gives sequence of diagonal sum $\color{red}1,\color{blue}1,\color{green}2,\color{cyan}4,\color{magenta}7,\dots$, Tribonacci sequence, etc.
-
Sorry, I'm slow: how are you computing your diagonal sums? – Guess who it is. Dec 18 '11 at 1:29
Also, if you look here, I'd say it's pretty convenient to have the generalized Fibonacci numbers take on positive values for positive index. – Guess who it is. Dec 18 '11 at 1:33
Thanks @Robert! I found this in the meantime. – Guess who it is. Dec 18 '11 at 1:37
@J.M.: I added some color to make it more obvious. – Brian M. Scott Dec 18 '11 at 2:19
@Peter: If $a(r,c)$ is the entry in row $r$, column $c$, $$a(r+1,c)=a(r,c)+a(r,c-1)+a(r,c-2)\;.$$ There’s an error in the table, which I’ve now fixed. – Brian M. Scott Dec 18 '11 at 10:04
The essential definition of a $k$-nacci sequence is that the earliest positive term is $1$ and all subsequent terms are the sum of the previous $k$ terms.
So to find the second earliest positive term (which is also $1$) you need at least $k-1$ zeros at the beginning of the sequence to do the sum. In your tables there are at least $k-1$ columns of implicit zeroes to the left of the columns you show.
Since the Fibonacci sequence conventionally starts with $F(1)=1$ and $F(0)=0$, it is a plausible but not necessary convention to start the $k$-nacci sequence with the initial leading zero also at a zero index, and this is what the OEIS has done.
You could do something different, but then you would need to translate between your indices and those used by others. That is the effect of conventions.
-
While it looks natural to start Fibonacci at $f_1=1$, the Fibonacci actually goes both ways. Indeed
$$f_{n+1}=f_n+f_{n-1} \,,$$ implies
$$f_{n-1}= f_{n+1}-f_n \,.$$
Thus you can calculate $f_0$ and $f_{-n}$.
Given a recurrence which goes in both directions, any "start" point is somewhat artificial.
I find the start $f_0=0$ useful when I work with the matrices
$$F:=\left( \begin{array}{rr} 1 & 1\\ 1 & 0 \end{array}\right)=\left(\begin{array}{rr} f_2 & f_1\\ f_1 & f_0 \end{array}\right) \,.$$
Note that without starting with $f_0=0$, the standard formula
$$F^n= \left(\begin{array}{rr} f_{n+1} & f_n\\ f_n & f_{n-1} \end{array}\right)$$
would not work for $n=1$. This would make some consequences weaker, since you'd need always the eliminate the case when the parameter(s) are 1. But this would be a not needed elimination.
It is somehow similar to the following situation: You have a function $f$ in the complex plane, and you can prove it is entire, but you only need the fact that it is Analytical in some region $R$. What do you do? Do you prove that it is Analytic in the region you need for the applications you are interested, or prove it is entire?
Who knows, at some point in the future you might need that the function it is entire. You also might need to work with $f_0$, and I explained above why $f_n$ can be extended in an unique way below 0. Same can be said about the other sequences.
-
As for "Why Fibonacci sequence start at $0$?" one compelling reason is that this indexing is the natural one that reveals their interesting divisibility properties, namely that they form a strong divisibility sequence, i.e.
$$\rm\ gcd(f_m,f_n)\ =\ f_{\:gcd(m,n)}$$
hence $\rm\ m\ |\ n\ \Rightarrow\ f_m\ |\ f_n\$ etc. Analogous results hold for the more general class of Lucas-Lehmer sequences, which prove convenient when studying elementary number theory of quadratic fields, e.g. generalizations of Fermat's little theorem, Euler's $\phi$ totient function, etc. If one changed the indexing then many of these results would be greatly obfuscated.
That said, it should be emphasized that such definitions are merely conventions that prove useful in the context at hand. In other contexts - where such divisibility properties play no role - another indexing might prove more convenient.
-
Can you be more specific about Lucas-Lehmer sequences as a general notion? The only somewhat authoritative mention I can find is in OEIS, but is is only one sequence, even though its title enigmatically uses the indefinite article. As far as I know the only thing really called after Lucas and Lehmer is the primality test (in which the above sequence occurs, and no other). Possibly you meant Lucas sequences? – Marc van Leeuwen Dec 18 '11 at 8:02
@Marc D.H. Lehmer developed an extended theory of Lucas sequences in his PhD thesis, which was published in Annals of Math., 1930 Due to such, many authors refer to the general class as Lucas-Lehmer sequences, e.g. see Ribenboim's The New Book of Prime Number Records, 1996. – Bill Dubuque Dec 18 '11 at 15:07
Once you have also an expression, which allows interpolating to noninteger n (thus putting the sequence into a greater framework), then this might provide an argument to select a specific n as beginning of the sequence. For the Fibonacci-numbers I think the Binet-formula is
$$\small fib(n) = {((1+\sqrt 5 )^n - (1-\sqrt 5)^n) \over 2^n \sqrt 5 }$$ and then $\small fib(0) = 0$. Surely one can insert an arbitrary offset for n into the defining expression, but the given definition may be taken as the most natural.
- | 2015-05-27T22:28:53 | {
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https://math.stackexchange.com/questions/816281/why-there-are-no-constant-functions-on-mathbbr-with-compact-support | # Why there are no constant functions on $\mathbb{R}$ with compact support?
For some the question might seem trivial but the concept is new to me and I have been wondering why there are no constant functions on $$\mathbb{R}$$ with compact support?
Following wiki:
Def.1) Functions with compact support on a topological space $$X$$ are those whose support is a compact subset of $$X$$.
Def.2) $$\operatorname{supp}f=:\overline{\{x\in X\;,f(x)\neq 0\}}$$
What if we take such function $$f:\mathbb{R}\to\mathbb{R}$$ defined as $$f(x)=0\;,\forall x\in\mathbb{R}$$ then $$\operatorname{supp}f=\overline{\{x\in\mathbb{R}\;,f(x)\neq 0\}}=\overline{\emptyset}=\emptyset$$ but $$\emptyset$$ is a compact subset of $$\mathbb{R}$$ I think....
Could someone enlighten me? Thank you.
• Where were you told that no constant functions on $\mathbb{R}$ have compact support? – Hayden May 31 '14 at 18:28
• well, anyway that's the only one there is, $f \equiv 0$ – mm-aops May 31 '14 at 18:30
• @Hayden: here for example www-personal.umich.edu/~wangzuoq/437W13/Notes/Lec%2030.pdf (12th line) – user124471 May 31 '14 at 18:32
• Well, I'd beg to differ with the claim, for precisely the example you gave. But as mm-aops pointed out, it's the only example. – Hayden May 31 '14 at 18:35
• the statement is very clear : ''there are NO functions'', (such functions don't exists). The real question here is if my counter-example is really correct (i.e that I followed all definitions correctly and so on). – user124471 May 31 '14 at 18:37
## 1 Answer
The author in the article you link to (www-personal.umich.edu/~wangzuoq/437W13/Notes/Lec%2030.pdf) is just being sloppy. He really should have said there are no non-zero constant functions with compact support on $\mathbb R$ (and thus $H_c^0(\mathbb R)$.
So yes, your example is correct. The empty set is (tautologically) compact.
• Hi Fredrik. Thank you. Unfortunately it is not the first time I have seen it....Exactly the same statement I found in Bott, Tu, ''Differential forms in algebraic topology''. Cheers – user124471 May 31 '14 at 18:44
• @user124471 Yep, I've seen it as well. In fact, I was just reading in the book by Bott & Tu a week ago, and had to think about the exact same question as you asked. – Fredrik Meyer May 31 '14 at 18:47
• @user124471 I was wondering the same thing! Now I know. Can you include Bott Tu in your question? – user636532 May 29 '19 at 11:27 | 2020-01-17T19:30:21 | {
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https://math.stackexchange.com/questions/2576298/x-1-cdot-dots-cdot-x-n-1-implies-x-1-dots-x-n-ge-n | # $x_1 \cdot \dots \cdot x_n=1 \implies x_1 + \dots + x_n \ge n$
Suppose $x_1, \dots, x_n$ are positive real numbers such that $x_1 \cdot \ldots \cdot x_n = 1$. Prove $x_1 + \dots + x_n \ge n$.
I don't want to use the AM-GM inequality to prove this (from which this statement would follow).
I suppose induction is the way to go. The base case is true. The case for $n=2$ is true by $$x_1 + x+2 - 2 \sqrt{x_1 x_2} = (\sqrt{x_1} - \sqrt{x_n})^2 \ge 0.$$
Now suppose the statement holds for $n$, then $$x_1 \cdot \ldots \cdot x_n = 1 \implies x_1 + \dots + x_n \ge n.$$
So if $x_1 \cdot \ldots \cdot (x_n x_{n+1}) = 1$, then $$x_1 + \dots + x_n x_{n+1} \ge n.$$
Then we also have $$x_1 + \dots + x_{n-1} + x_n + x_{n+1} \ge n - x_{n}x_{n+1} + x_n + x_{n+1}.$$
Now I'm not sure where else to go from here.
• Seem similar in proving AM-GM by induction I think – Azlif Dec 21 '17 at 23:25
• @MathematicianByMistake Induction certainly works, since this is just a special case of AM-GM which is proven by induction. – B. Mehta Dec 21 '17 at 23:29
Thus, it's enough to prove that $$n-x_nx_{n+1}+x_n+x_{n+1}\geq n+1$$ or $$(x_n-1)(x_{n+1}-1)\leq0,$$ which you can assume before.
Indeed, since $\prod\limits_{k=1}^{n+1}x_k=1$, there are $x_i$ and $x_j$ for which $(x_i-1)(x_j-1)\leq0.$
• It's not entirely clear to me why this can be assumed. – B. Mehta Dec 21 '17 at 23:31
• @B. Mehta Since $\prod\limits_{k=1}^{n+1}x_k=1$, there are $x_i$ and $x_j$ for which $(x_i-1)(x_j-1)\leq0.$ – Michael Rozenberg Dec 21 '17 at 23:33
• Ah, so there's a relabeling step added earlier. – B. Mehta Dec 21 '17 at 23:34
• @B.Mehta I believe the logic is: there has to be at least one value greater-than-or-equal-to 1 and at least one less-than-or-equal-to 1. Rearrange so that these two values come last. – Addem Dec 21 '17 at 23:34
• @MichaelRozenberg Thanks, very helpful. I would like to verify this last part with there needing to be a pair with one greater and one and one less. By contradiction, suppose for every $x_i$ and $x_j$, $(x_i -1)(x_j-1) > 0$, then either all pairs are greater than one or less than one which in either case would result in a product of the $x_k$'s being greater than 1 or less than one, respectively. – user330531 Dec 21 '17 at 23:42
The inequality follows from AM-GM, indeed: $\;x_1 + \dots + x_n \ge n \cdot \sqrt[n]{x_1 \cdot \ldots \cdot x_n}\,$.
Below is a proof of AM-GM by induction, along a different idea than the ones posted already.
Lemma. $\;(n-1)t^n+1 \ge n t^{n-1}\;$ for all $\;t \ge 1\,$.
Let $f\,(t)=(n-1)t^n - n t^{n-1} + 1\,$, then $\,f(1) = 0\,$ and $\,f'(t)=n(n-1)t^{n-2}(t-1) \ge 0\,$, so $\,f\,$ is increasing from $\,f(1)=0\,$ for $\,t \ge 1\,$ and therefore $\,f(t) \ge 0\,$ for $\,t \ge 1\,$.
(The stronger statement is true that $\,f(t) \ge 0\,$ for all $\,t \ge 0\,$. That follows from the identity $\,(n-1)t^n - n t^{n-1} + 1 = (t-1)^2 \cdot \sum_{k=1}^{n-1} k \cdot t^{k-1}\,$ which has an elementary no-calculus proof.)
Induction step. The AM-GM inequality is homogeneous, so it can be assumed WLOG that the smallest $x_j=1\,$. Assume again WLOG that $\,j=n\,$, then the inequality reduces to:
$$x_1+x_2+ \cdots + x_{n-1} + 1 \ge n \cdot \sqrt[n]{x_1 \cdot \ldots \cdot x_{n-1}} \quad\style{font-family:inherit}{\text{with}}\quad x_i \ge 1, \;i=1,2,\ldots,n-1$$
By the induction hypothesis:
$$\,x_1+\cdots+x_{n-1} \ge (n-1) \sqrt[n-1]{x_1\cdot \ldots \cdot x_{n-1}} \tag{1}$$
Also, it follows from the lemma with $\,t = \sqrt[n(n-1)]{x_1\cdots x_{n-1}}\ge 1\,$ that: $$\,(n-1)\cdot\sqrt[n-1]{x_1\cdot \ldots \cdot x_{n-1}} + 1 \ge n \cdot \sqrt[n]{x_1\cdot \ldots \cdot x_{n-1}} \tag{2}$$
Therefore:
$$x_1+ \ldots +x_{n-1}+1 \;\stackrel{(1)}{\ge}\; (n-1)\sqrt[n-1]{x_1\cdot \ldots \cdot x_{n-1}} + 1 \;\stackrel{(2)}{\ge}\; n \sqrt[n]{x_1 \cdot \ldots \cdot x_{n-1}}$$ | 2019-09-17T23:18:57 | {
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https://math.stackexchange.com/questions/434275/what-does-it-mean-to-divide-a-complex-number-by-another-complex-number | # What does it mean to divide a complex number by another complex number?
Suppose I have: $w=2+3i$ and $x=1+2i$. What does it really mean to divide $w$ by $x$?
EDIT: I am sorry that I did not tell my question precisely. (What you all told me turned out to be already known facts!) I was trying to ask the geometric intuition behind the division of complex numbers.
• Jul 2, 2013 at 6:23
• suitcaseofdreams.net/Geometric_division.htm Jul 2, 2013 at 6:25
• Perhaps a polar form might be a little more 'intuitive'? Jul 2, 2013 at 6:33
• While multiplication/division of complex numbers can be interpreted geometrically, I don't think it is meant to be interpreted that way. Jul 2, 2013 at 6:40
• @user1551 au contraire it is meant to be interpreted geometrically. Jul 2, 2013 at 16:40
Complex numbers' multiplication is better understood if you forget the "cartesian vector in the complex plane" analogy: $$z = a + b i \quad z \in \mathbb{C}, a, b \in \mathbb{R}$$ And in stead think in polar coordinates: $$z = r \angle \theta = r e^{i \theta} \quad z \in \mathbb{C}, r \in \mathbb{R}^+, \theta \in [0, 2 \pi)$$ Wherein $r$ is the magnitude, $\theta$ is the angle.
When multiplying it is easy: $$z w = (r_z r_w) \angle (\theta _z + \theta _w)$$ You add the angles and multiply the magnitudes.
When dividing you do what comes naturally: $$\frac z w = \left(\! \frac{r_z}{r_w} \!\right) \angle (\theta _z - \theta _w)$$ To divide means to find the difference in angles and the factor in magnitude.
It means to find another complex number $y$, such that $xy=w$. (Just as it does for real numbers.)
When you divide $a$ by $b$, you're asking "What do I multiply $b$ by in order to get $a$?".
Multiplying two complex numbers multiplies their magnitudes and adds their phases:
So when you divide a complex $a$ by a complex $b$ you are asking: "How much do I need to scale $b$ and rotate $b$ in order to get $a$? Please give me a complex number with a magnitude equivalent to how much I must scale and a phase equivalent to how much I must rotate.".
Example
Consider $\frac{1 + i}{1 - i}$. How much do we need to scale and rotate $1-i$ in order to make it the same as $1+i$?
Well, when graphed on the complex plane you can see that $1-i$ has a 45 degree clockwise rotation and a magnitude of $\sqrt{2}$. $1+i$, on the other hand, has a 45 degree counter-clockwise rotation and the same magnitude of $\sqrt{2}$.
Since the magnitudes are the same, we don't need any scaling. Our result's magnitude will be 1.
To rotate from 45 degrees clockwise to 45 degrees counter-clockwise, we must rotate 90 degrees counter-clockwise. Thus our result will have a phase of 90 degrees counter-clockwise (which is upwards along the imaginary Y axis).
Move a distance of 1 up the imaginary Y axis and you get the answer... $\frac{1 + i}{1 - i} = i$. We can confirm this by doing the multiplication: $(1-i) \cdot i = i+1$.
• wow for the gif Jul 2, 2013 at 16:39
• @Harsh I'm glad you like it. I made it (and a bunch of others) myself as part of an explanation of Grover's Quantum Search. Jul 3, 2013 at 21:32
• man... I was thinking you were very resourceful, and had taken it from somewhere... Now, to learn that you MADE IT! Salute Jul 3, 2013 at 22:01
Since multiplication can be nicely visualized as a rotation in the complex plane, you may find it helpful to think of division as a form of multiplication: $$\frac{2+3i}{1+2i} = \frac{2+3i}{1+2i}\cdot \frac{1-2i}{1-2i} = -\frac{1}{3}\cdot(2+3i)(1-2i)$$ So, instead of thinking about division, you can think of it as multiplying by the conjugate.
Geometrically, it means that the magnitudes get divided, and the angles get subtracted.
That is, imagine the complex numbers plotted in polar form.
ie: if
$w = r_1 e ^ {i \theta_1 }$
$x = r_2 e ^ {i \theta_2 }$
then
$w/x = (r_1 / r_2)e^{i(\theta_1 - \theta_2)}$
There are multiple ways by which you can describe this relation;
Since you have asked to give geometric representation so phasor form must be better to understand.
Dividing two complex numbers means to take the complex number that is of the magnitude equal to that of the division of amplitudes of the $$X$$ and $$Y$$ complex number and the phase of the new generated complex number is actually the difference of the phase between them.
Eg. $$X = 2+3i \ , \ Y = 9+3i$$
so the $$|Z|= \frac{ |X|}{|Y|} = \dfrac{\sqrt{2^2+3^2}}{\sqrt{9^2+3^2}}$$ and $$\arg(Z)= \tan^{-1}\left(\frac32\right) - \tan^{-1}\left(\frac 39 \right)$$
This $$Z$$ is also a complex number with amplitude and phase in geometric way.. but you to visualize the complex imaginary axis and real axis :D
Since $\frac wx = \frac{w\cdot x^*}{|x|^2}$, complex division is basically equivalent to multiplication of $w$ and the complex conjugate of $x$, but rescaled by the squared absolute value $|x|^2$. | 2022-05-23T03:08:06 | {
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# Rapid Math
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23 Oct 2017, 01:32
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Hi Everyone,
Although GMAT doesn't test your ability to perform complicated calculations, it would be beneficial to know some calculation tricks that would save you some time.
Trick 7: Rapidly Square Any Number Ending in 5
Multiply the tens digit by the next whole number then affix the number 25
$$65^2: 6*7=42 ==> 'affix' 25 ==> 4225$$
Trick 8: Rapidly Multiply Any Two-Digit Number by 11
Write the number leaving some space between the two digits, then insert the sum of the number's two digit in between.
24*11 ==> 2 ? 4 ==> 2+4=6 ==> 264
Trick 9: Rapidly Multiply by 25 (or 0.25, 2.5, 250, etc.)
Divide the number by 4 and affix or insert Zeroes or decimal points
Example: 28*25 ==> 28/4 = 7 ==> 700 (add 2 Zeroes)
Trick 10: Rapidly Divide by 25 (or 0.25, 2.5, 250, etc.)
Remove Zeroes or decimal points then multiply the number by 4
Example: 700/25 ==> 7/25 ==> 7*4=28
Trick 24: Rapidly Square Any Two-Digit Number ending in 1
$$21^2 ==> 20^2=400 ==> Add: 20+21=41 ==> Add: 400+41= 441 ==> 21^2 = 441$$
Trick 59: Rapidly Multiply Any Three-Digit Number by 11 (advanced variation of trick 8)
Write the number leaving some space between the two digits, then insert the summations of the numbers in between (summations of 2 digits everytime; as shown below).
Example:
342*11=??
342*11= 3--2 (write the first & the last number "3"& "2")
342*11=3-62 (the sum of 4+2)
342*11=3762 (the sum of 3+4)
Source: Rapid Math Tricks & Tips. EDWARD H. JULIUS
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Last edited by NDND on 12 Nov 2017, 04:48, edited 5 times in total.
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23 Oct 2017, 01:47
Thanx. It will help
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01 Nov 2017, 07:43
Updated with Tricks 9 & 10
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Re: Rapid Math [#permalink] 01 Nov 2017, 07:43
Display posts from previous: Sort by | 2018-03-21T11:05:45 | {
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https://math.stackexchange.com/questions/781365/find-the-parametric-equation-to-the-curve | # Find the parametric equation to the curve
Find the parametric equation for the curve.
$$x^{2}+y^{2}=10$$
I haven't learned parametric equations fully yet, so I wanted to check with you guys and see if you can confirm if I'm doing this correctly and possibly go more in depth on the problem if you can?
Because it's centered at (0,0) the origin, and it has a radius of sqrt(10) then the answer is this, right?
$$(x(t),y(t)) = (\sqrt{10}\cos t\,,\, \sqrt{10}\sin t)$$
• Yes, it is...and you should probably remark that $\;0\le t\le 2\pi\;$ – DonAntonio May 4 '14 at 21:15
taking $x=a\sin t,y=a\cos t$ from $x^2+y^2=10$ we get $$a^2\sin^2t+a^2\cos^2t=a^2(\sin^2t+\cos^2t)=a^2=10$$ from above $a=\sqrt{10}$
You are correct. Here's an easy check: note that the position vector at any point is given by $\langle\sqrt{10}\cos(t), \sqrt{10}\sin(t) \rangle$. If we take the magnitude of this vector, we get:
$$\sqrt{10\cos^2{t} + 10\sin^2{t}} = \sqrt{10}$$
And therefore the curve is a constant $\sqrt{10}$ distance from the origin, so it must be at least a part of a circle centered at the origin with radius $\sqrt{10}$. To make the argument that it's a complete circle, you can use a bit of calculus to verify that the curve never turns back on itself. To do this, take a derivative of the position function to yield the velocity function, and note that it is never $0$. | 2019-10-20T22:28:31 | {
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https://math.stackexchange.com/questions/853615/proving-an-expression-is-composite | # Proving an expression is composite
I am trying to prove that $n^4 + 4^n$ is composite if $n$ is an integer greater than 1. This is trivial for even $n$ since the expression will be even if $n$ is even.
This problem is given in a section where induction is introduced, but I am not quite sure how induction could be used to solve this problem. I have tried examining expansions of the expression at $n+2$ and $n$, but have found no success.
I would appreciate any hints on how to go about proving that the expression is not prime for odd integers greater than 1.
• Write it as a difference of squares. – Adam Hughes Jul 1 '14 at 19:54
• How? It's $n^4 + 4^n$ not $n^4 - 4^n$ – Mathmo123 Jul 1 '14 at 19:55
• @Mathmo123: see my answer below. – Adam Hughes Jul 1 '14 at 20:03
$(n^2)^2+(2^n)^2=(n^2+2^n)^2-2^{n+1}n^2$. Since $n$ is odd...
• Ah. Very neat!! – Mathmo123 Jul 1 '14 at 20:05
• Originally, I was not sure how I should go about writing it as a difference of squares. Thanks for your help. – pidude Jul 1 '14 at 20:12
Hint: calculate this value explicitly for $n=1,3$ (or predict what will happen). Can you see any common factors? Can you prove that there is a number $m$ such that if $n$ is odd, then $m|(n^4 + 4^n)$?
Let me know if you need further hints.
• I originally tried doing that but to no avail. Evaluated at 3 I obtain 145 which has factors of 5 and 29. At 5, the expression equals 1649, which has factors of 17 and 97. I will keep looking for a pattern and let you know if I need more hints. Thanks for your help. – pidude Jul 1 '14 at 19:59
• Ah... using this method, there will be a difference between multiples of 5 and other odd numbers. You will find that for odd numbers that are not a multiple of 5, 5 will be a divisor – Mathmo123 Jul 1 '14 at 20:01
• Ok, so I was able to prove that. Now I'm working numbers which are multiples of 5. – pidude Jul 1 '14 at 20:09
I am very impressive with Adam's solution. There is very neat. So, I beg for a chance to write the full description about the proof step-by-step.
• We can transform $n^4+4^n$ to $(n^2+2^n)^2-2^{n+1}n^2$ as Adam's suggestion by
1. $n^{(2^2)}+(2^2)^n = (n^2)^2+(2^n)^2$ associative law
2. Now, we mention $(a+b)^2 = (a+b)(a+b) = a^2+2ab+b^2$ algebraic multiplication
3. $(n^2)^2+(2^n)^2+2(n^2)(2^n)-2(n^2)(2^n)$ adding $+2ab-2ab$ to expression
4. $(n^2)^2+2(n^2)(2^n)+(2^n)^2-2(n^2)(2^n)$ re-arrange the expression
5. $(n^2+2^n)^2-2(n^2)(2^n)$ from step 2
6. $(n^2+2^n)^2-2^{n+1}n^2$ law of Exponential
• We try to get the $(n^2+2^n)^2-2^{n+1}n^2$ to conform to $a^2-b^2$ because $a^2-b^2=(a+b)(a-b)$ algebraic multiplication, again
1. Treat $n^2+2^n$ as $a$
2. Since $n$ is odd, n+1 is even. So, we can assume $2m=n+1$, where $m$ is integer
3. So, re-write the $2^{n+1}n^2$ to be $2^{2m}n^2$
4. $2^{2m}n^2=(n2^m)^2$ associative law
5. Treat $n2^m$ as $b$
• It implies that both $a$ and $b$ are both positive integer
• From $a^2-b^2=(a+b)(a-b)$ and the result of $n^4 + 2^4$, it implies that $a$ is greater than $b$
• Hence both $(a+b)$ and $(a-b)$ are positive integer, that causes the result of $n^4 + 2^4$ is combination of $(a+b)$ and $(a-b)$
Some interesting factorizations of a polynomial of type $x^4+\text{const}$: $$x^4+4=(x^2+2x+2)(x^2-2x-2) \tag{1}$$
$$x^4+1=(x^2+\sqrt[]{2}x+1)(x^2-\sqrt[]{2}x+1) \tag{2}$$
So one can ask, how to select the coefficients $a,b,c,d$ in
$$(x^2+ax+b)(x^2+cx+d) \tag{3}$$
such that all coefficients of the resulting polynomial are zero except the constant term and the coefficient of the 4th power. The latter is $1$.
If we expand $(3)$ we get
$$x^4+(c+a)x^3+(d+a c+b)x^2+(a d+b c)x+b d$$
And the coefficients disappear, if
$$\begin{eqnarray} c+a &=& 0 \\ d+ ac +b &=& 0 \\ ad+bc &=& 0 \end{eqnarray}$$
When solving for $b,c,d$ we get
$$\begin{eqnarray} c &=& -a \\ b &=& \frac{a^2}{2} \\ d &=& \frac{a^2}{2} \end{eqnarray}$$
and therefore
$$x^4+\frac{a^4}{4} = (x^2+a x+\frac{a^2}{2})(x^2-ax+\frac{a^2}{2})$$
For $a=2$ this gives $(1)$, $a=\sqrt[]{2}$ this gives $82)$ . Substituting $a=2^{t+1}$ we get
$$x^4+4^{2t+1} = (x^2+2\cdot 2^t x+2^{2t+1})(x^2-2\cdot 2^tx+2^{2t+1})$$
Substituting $x=n=2t+1$ gives the required result for odd $n$. | 2020-07-07T13:41:31 | {
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https://www.cnidaria.nat.uni-erlangen.de/shortcourse/S4/Assignment_Kendall_Tau_KEY.html | Write a function that computes Kendall rank coefficient Tau (see lecture notes for additional details on Kendall Tau).
Kendall Coefficient {as quoted from Wikipedia):
Let (x1, y1), (x2, y2), ., (xn, yn) be a set of observations of the joint random variables X and Y respectively, such that all the values of xi and yi are unique. Any pair of observations (xi and yi) are said to be concordant if the ranks for both elements (more precisely, the sort order by x and by y) agree: that is, if both xi > xj and yi > yi or if both xi < xj and yi < yj. They are said to be discordant, if xi > xj and yi < yj or if xi < xj and yi > yj. If xi = xj or yi = yj, the pair is neither concordant nor discordant.
Equation 1
TAU=(number of concordant pairs-number of discordant pairs)/(0.5 x n x (n - 1))
The above equations ignore ties (sets of pairs where either xi = xj or yi = yj). One way of correcting for ties is to consider how many ties (or non-ties) are present in the data. This variant known as Kendall’s tau_b is computed by modifying the denominator 0.5n(n-1) in Equation 1 to consider ties:
sqrt(mx) x sqrt(my)
Where mx is number of non-ties for x and my is number of non-ties for y.
Note: The R function {cor} returns tau_b.
YOUR TASKS: 1. Write a custom function that computes Kendall ignoring ties (Equation 1) and correcting for ties (as explained above).
1. Your function should return a single value (Kendall’s coefficient)
2. Your function should allow the user to choose to ignore ties or not and thus produce either tau or tau_b.
3. Apply the function to two vectors provided below and check the results provided by your function against those obtained using the R function {cor}.
4. Check how fast is your function relative to {cor} function using microbenchmark function from package ‘microbenchmark’.
5. Finally, generate large vectors x and y (n = 1000) from normal distribution using rnorm function. Check again how fast is your function relative to {cor} function using microbenchmark function from package ‘microbenchmark’.
To carry out this exercise, please start by uploading the library ‘microbenchmark’ and please use x and y variables as defined below. This will ensure that all results are the same for all of us. It will also help you to evaluate if your outcomes agree with those provided in solution functions or alternative R functions.
library(microbenchmark)
x <- c(-30, 0, 14, 5, 6, 7)
y <- c(-5, 0, 12, 16, 16, 8)
Once your function is ready, you can continue to the next step. The function solutions included in this tutorial are hidden here, but can be reviewed in the .rmd file or in the answer key file (.html). Please try to solve this on your own before checking the solution.
my.kend <- function(x, y, ties=T) {
n <- length(x)
nzero <- 0.5*n*(n-1)
s1 <- sign(x - t(matrix(x, n, n, byrow=F)))
s2 <- sign(y - t(matrix(y, n, n, byrow=F)))
t1 <- sum(s1[lower.tri(s1)]!=0)
t2 <- sum(s2[lower.tri(s2)]!=0)
n2 <- t1^0.5*t2^0.5
if (!ties) tau <- sum(s1[lower.tri(s1)]*s2[lower.tri(s2)])/nzero
if (ties) tau <- sum(s1[lower.tri(s1)]*s2[lower.tri(s2)])/n2
return(tau)
}
my.kend2 <- function(x, y, ties=T) {
n <- length(x)
nzero <- 0.5*n*(n-1)
k <- 0
s1 <- 0
s2 <- 0
out <- vector(mode='numeric', length=nzero)
for (i in 1:length(x)) {
for (j in 1:length(x)) {
if (j > i) {
k <- k + 1
out[k] <- sign(x[i] - x[j])*sign(y[i] - y[j])
if(x[i] - x[j]==0) s1 = s1 + 1
if(y[i] - y[j]==0) s2 = s2 + 1
}
}
}
if (!ties) tau <- sum(out)/nzero
if (ties) tau <- sum(out)/((nzero-s1)^0.5*sum(nzero-s2)^0.5)
tau
}
Now check if your function yields comparable results to those provided by standard R functions or custom functions developed by your instructor.
x <- c(-30, 0, 14, 5, 6, 7)
y <- c(-5, 0, 12, 16, 16, 8)
# compute kendall tau using generic function {cor}
cor(x, y, method='kendall')
## [1] 0.4140393
# check if your function is consistent with standard cor function
my.kend(x, y, ties=T)
## [1] 0.4140393
# if you developed a second solution check if that solution works as well.
my.kend2(x, y, ties=T) # check if your function works as well
## [1] 0.4140393
# now check if your function(s) also compute(s) uncorrected tau correctly
my.kend(x, y, ties=F)
## [1] 0.4
my.kend2(x, y, ties=F)
## [1] 0.4
Once you established that your function works, you should investigate if your function is faster or slower than cor function. Keep in mind, however, that the performance of the function may depend on sample size, so your function may work better than ‘cor’ function for small sample sizes, but perform poorly for large sample sizes (or vice versa).
microbenchmark(cor(x, y, method='kendall'), my.kend(x, y, ties=T),
my.kend2(x, y, ties=T)) # which is faster for a much larger dataset?
## Unit: microseconds
## expr min lq mean median uq
## cor(x, y, method = "kendall") 65.700 80.0480 95.31707 95.1510 109.6875
## my.kend(x, y, ties = T) 23.033 26.2425 36.04066 35.6815 42.2900
## my.kend2(x, y, ties = T) 10.195 10.9500 15.45499 12.0830 24.5435
## max neval
## 215.976 100
## 70.986 100
## 40.780 100
You can see that my custom functions my.kend and my.kend2 are much faster than cor function when analyzed data are tiny.
Let’s now check how the functions perform for much larger datasets.
x <- rnorm(500,1,1)
y <- rnorm(500,1,3) + x
cor(x, y, method='kendall')
## [1] 0.19301
my.kend(x, y, ties=T) # check if your function agrees
## [1] 0.19301
my.kend2(x, y, ties=T) # check if your function agrees
## [1] 0.19301
microbenchmark(cor(x, y, method='kendall'), my.kend(x, y, ties=T),
my.kend2(x, y, ties=T)) # which is faster for a much larger dataset?
## Unit: milliseconds
## expr min lq mean median
## cor(x, y, method = "kendall") 3.233603 3.529627 4.097588 3.924766
## my.kend(x, y, ties = T) 13.880630 18.497122 21.550797 20.513782
## my.kend2(x, y, ties = T) 50.971912 58.680038 64.747797 62.303116
## uq max neval
## 4.657272 5.69241 100
## 23.238596 52.78090 100
## 69.767135 94.52662 100
As clear from the report, the custom-written solutions that I wrote (you may come up with better algorithms) perform well at small sample size, but underperform at larger sample sizes. That is, compiled, professionally written R functions are faster for large samples than custom written solutions written by amatuers. This means that when we iterate small datasets, custom functions can be faster. When huge datasets are processed, R functions are likley to be superior to custom solutions.
Let us evaluate this issue here by using ‘cor’ and our custom function to carry out the same analysis.
r1 <- as.data.frame(matrix(rnorm(20000), 10, 2000))
# NOTE r1 is a dataframe of many (s=2000) tiny samples (n=10 observations). Each column is a sample.
r2 <- as.data.frame(matrix(rnorm(20000), 2000, 10))
# NOTE r2 is a dataframe of s=10 huge samples (n=2000 observations). Each column is a sample.
system.time(sapply(r1, function(x) cor(x, r1[,1], method='kendall')))
## user system elapsed
## 0.16 0.03 0.19
system.time(sapply(r1, function(x) my.kend(x, r1[,1])))
## user system elapsed
## 0.09 0.00 0.09
system.time(sapply(r1, function(x) my.kend2(x, r1[,1])))
## user system elapsed
## 0.09 0.00 0.09
system.time(sapply(r2, function(x) cor(x, r2[,1], method='kendall')))
## user system elapsed
## 0.59 0.00 0.60
system.time(sapply(r2, function(x) my.kend(x, r2[,1])))
## user system elapsed
## 3.92 1.01 4.93
system.time(sapply(r2, function(x) my.kend2(x, r2[,1])))
## user system elapsed
## 9.39 0.03 9.42
Clearly, my functions worked faster when we processed many, many short vectors (data frame r1), but underperformed badly when a few very large vectors were evaluated (data frame r2).
Always cite packages.
citation('microbenchmark')
##
## To cite package 'microbenchmark' in publications use:
##
## Olaf Mersmann (2018). microbenchmark: Accurate Timing Functions.
## R package version 1.4-4.
## https://CRAN.R-project.org/package=microbenchmark
##
## A BibTeX entry for LaTeX users is
##
## @Manual{,
## title = {microbenchmark: Accurate Timing Functions},
## author = {Olaf Mersmann},
## year = {2018},
## note = {R package version 1.4-4},
## url = {https://CRAN.R-project.org/package=microbenchmark},
## } | 2019-10-16T01:49:45 | {
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http://mathhelpforum.com/advanced-statistics/103904-conditional-probability-brain-teaser.html | # Thread: Conditional Probability Brain Teaser
1. ## Conditional Probability Brain Teaser
A sock contains 2 red marbles and an unknown number of blue marbles. Tom places a new marble in the sock. Jason puts his hand in the sock and pulls out a red marble.
What is the probability that the marble Tom put in the sock was red?
I've been trying to approach this with Bayes' Theorem but not making any progress:
P(A l B) = P(B l A) * P(A) / P(B)
I'm confused as to what A and B would represent in the problem.
Appreciate the help, thanks!
2. Originally Posted by Penguins
A sock contains 2 red marbles and an unknown number of blue marbles. Tom places a new marble in the sock. Jason puts his hand in the sock and pulls out a red marble.
What is the probability that the marble Tom put in the sock was red?
I've been trying to approach this with Bayes' Theorem but not making any progress:
P(A l B) = P(B l A) * P(A) / P(B)
I'm confused as to what A and B would represent in the problem.
Appreciate the help, thanks!
Let the number of blue marbles in the sock be x.
Draw a tree diagram. The first two branches are the events Tom putting in a blue marble or a red marble. The next branches are the events of pulling out a red marble or a blue marble.
I get $\frac{\frac{3}{2(3 + x)}}{\frac{3}{2(3 + x)} + \frac{1}{3 + x}} = \frac{3}{5}$.
3. I don't understand how you set up the problem like you did. What do the terms in your numerator and denominator represent?
Here is what I get:
P(A l B) = (P(B l A) * P(A)) / P(B)
A is Tom putting a red marble in the sock.
B is Jason taking a red marble out of the sock.
P (B l A) = 3 / (x+3)
P (A) = unknown, let it be p
P (B) = p*(3/(x+3)) + (1-p)*(2/(x+2))
So
P(A l B) = (P(B l A) * P(A)) / P(B) = (3p/(x+3)) / (p*(3/(x+3)) + (1-p)*(2/(x+2)))
I end up with 3/2 p. I know I did something wrong, but I'm not sure what.
4. Hello, Penguins!
I agree with Mr. F
A sock contains 2 red marbles and an unknown number of blue marbles.
Tom places a new marble in the sock.
Jason puts his hand in the sock and pulls out a red marble.
What is the probability that the marble Tom put in the sock was red?
We want: . $P(\text{add Red }|\text{ draw Red}) \;=\;\frac{P(\text{add Red }\wedge\text{ draw Red})}{P(\text{draw Red})}$ .[1]
The sock contains: 2 Red and $b$ Blue marbles.
There are two cases to consider:
[1] A Red marble is added.
. . .The sock contains: 3 Red and $b$ Blue.
. . . . Then: . $P(\text{add Red }\wedge\text{ draw Red}) \:=\:\frac{3}{b+3}$ .[2]
[2] A Blue marble is added.
. . .The sock contains: 2 Red and $b+1$ Blue.
. . . . Then: . $P(\text{add Blue }\wedge\text{ draw Red}) \:=\:\frac{2}{b+3}$
Hence: . $P(\text{draw Red}) \;=\;\frac{3}{b+3} + \frac{2}{b+3} \;=\;\frac{5}{b+3}$ .[3]
Substitute [2] and [3] into [1]: . $P(\text{add Red }|\text{ draw Red}) \;\;=\;\; \frac{\dfrac{3}{b+3}} {\dfrac{5}{b+3}} \;\;=\;\;\frac{3}{5}$
5. Originally Posted by Soroban
Hello, Penguins!
I agree with Mr. F
We want: . $P(\text{add Red }|\text{ draw Red}) \;=\;\frac{P(\text{add Red }\wedge\text{ draw Red})}{P(\text{draw Red})}$ .[1]
The sock contains: 2 Red and $b$ Blue marbles.
There are two cases to consider:
[1] A Red marble is added.
. . .The sock contains: 3 Red and $b$ Blue.
. . . . Then: . $P(\text{add Red }\wedge\text{ draw Red}) \:=\:\frac{3}{b+3}$ .[2]
[2] A Blue marble is added.
. . .The sock contains: 2 Red and $b+1$ Blue.
. . . . Then: . $P(\text{add Blue }\wedge\text{ draw Red}) \:=\:\frac{2}{b+3}$
Hence: . $P(\text{draw Red}) \;=\;\frac{3}{b+3} + \frac{2}{b+3} \;=\;\frac{5}{b+3}$ .[3]
Substitute [2] and [3] into [1]: . $P(\text{add Red }|\text{ draw Red}) \;\;=\;\; \frac{\dfrac{3}{b+3}} {\dfrac{5}{b+3}} \;\;=\;\;\frac{3}{5}$
I see what you're doing, but isn't it a problem that you're assuming the probability of Tom putting a red marble in the sock to be 1/2? I set up the problem using p as this probability and unfortunately made a careless mistake in my previous post, but get the answer to the question as (3p)/(2+p). When p = 1/2, the answer is 3/5.
I'm curious as to whether there is a rule in a situation like this as to whether lacking more information it is assumed that all possibilities have equal likelihood?
Thanks for the replies!
6. Originally Posted by Penguins
I see what you're doing, but isn't it a problem that you're assuming the probability of Tom putting a red marble in the sock to be 1/2? I set up the problem using p as this probability and unfortunately made a careless mistake in my previous post, but get the answer to the question as (3p)/(2+p). When p = 1/2, the answer is 3/5.
I'm curious as to whether there is a rule in a situation like this as to whether lacking more information it is assumed that all possibilities have equal likelihood?
Thanks for the replies!
If you're expected to get a numerical answer then you have to make an assumption regarding a priori knowledge. The best one (for various reasons which I won't get into here) is a uniform prior distribution. | 2017-01-16T19:01:13 | {
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https://math.stackexchange.com/questions/1481433/1-biased-coin-and-1-fair-coin-probability-of-2-heads | # 1 Biased Coin and 1 Fair Coin, probability of 2 Heads?
You have 1 fair coin and 1 coin with 2 heads. Given that the first flip was a heads what is the probability of getting another heads?
My Answer: P(2H|F=H) = P(2H|F=H, Biased Coin)*P(Biased Coin) + P(2H|F=H, Unbiased Coin)*P(Unbiased Coin) = 0.5 + 0.25 = 0.75. In my equation, F refers to the First Throw. But the answer is supposed to be 5/6 and I can't seem to understand how.
Edit: From Arthurs comment I get the following, however, I dont know if this is the correct method, despite getting the correct answer:
P(Biased|F=H) = 2/3.
P(2H|F=H) = P(2H|(Biased|F=H))*P(Biased|F=H) + P(2H|(Unbiased|F=H))*P(Unbiased|F=H) = (1*2/3) + (1/2 * 2/3) = 5/6.
Thank You
• Is it random (and unknown) which coin was flipped first? – Arthur Oct 15 '15 at 12:41
• @Arthur Yes. There is no information about that. – Jojo Oct 15 '15 at 12:42
• I think this is the biggest thing you've missed: Given that the first toss was a head, what is the proability that the first coin was the biased one? – Arthur Oct 15 '15 at 12:46
• @Arthur Could you please check the edit to my question. – Jojo Oct 15 '15 at 13:08
• @Jojo, yes, although that should be $P(2H\mid\text{ Biased} \cap F=H)$ and so forth. – Graham Kemp Oct 15 '15 at 13:11
The first flip was a head.
The $3$ heads have equal probabilities to be the head that appeared at the first flip.
$2$ of the $3$ heads have another head as other side.
$1$ of the $3$ heads has tail as other side.
So there is a chance of $\frac23.1+\frac13.\frac12=\frac56$ of throwing a second head with that coin.
• Very intuitive answer. Thanks. – Jojo Oct 15 '15 at 13:22
• Glad to help, and indeed cherish your intuition in maths. – drhab Oct 15 '15 at 13:24
• Does this assume that the same coin is flipped twice, or that one coin is flipped and then the other? I'm getting a different answer when I assume that latter, but the question is ambiguous. – Bill the Lizard Oct 15 '15 at 14:56
• It assumes that the same coin was flipped twice. – drhab Oct 15 '15 at 15:00
You missed a conditioning in your formula. Let the event of choosing biased coin be B and fair be F. Getting heads on second toss be 2H and on first toss 1H. You want to calculate the probability of getting heads on second throw given that you it landed heads on first throw, which is
P(2H/1H) = P(2H/1H,F)P(F/1H) + P(2H/1H,B)P(B/1H)
To evaluate P(F/1H) and P(B/1H) use bayes rue.
P(F/1H) = P(1H/F)*P(F)/P(1H) = 0.5*0.5/(0.5*0.5 + 1*0.5) = 1/3
P(B/1H) = 2/3
P(1H) = P(1H/F)P(F) + P(1H/B)P(B)
P(2H/1H,F) = 0.5 and P(2H/1H,B) = 1 Therefore, P(2H/1H) = 0.5*(1/3) + 1*(2/3) = 5/6 | 2019-06-16T06:29:17 | {
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https://math.stackexchange.com/questions/1644943/f-has-a-zero-integral-on-every-measurable-set-prove-f-is-zero-almost-everyw | $f$ has a zero integral on every measurable set. Prove $f$ is zero almost everywhere
I am trying to solve the following exercise:
Let $f$ be integrable. Assume that $\int_A f d\mu = 0$ for every measurable set $A$. Prove that $f = 0$ a.e. [$\mu$].
I have the following proof but it seems to me too simple to be true. What is wrong with it?
For any $a>0$, define $W_a:=\{w|f(w)\geq a\}$. Now we have: $$\int_{W_a} fd\mu = 0 \geq \int_{W_a} ad\mu=a\mu(W_a)$$ and therefore $\mu(W_a)=0$. Same holds for the negative $a$'s. This completes my proof.
• You might want to add why it suffices to show $\mu(W_a) = 0$ for $a>0$ and $\mu(f \leq a)=0$ for $a<0$. – saz Feb 7 '16 at 19:58
Indeed, your proof shows that $\{f > 0\}$ has measure zero. Similarly, you show $\{f < 0\}$ has measure zero, and hence $f=0$ a.s. Yes it really is this easy.
It remains to show that the set $\{w\mid f(w)\ne0\}$ has measure $0$. Consider the nonzero rational $a$'s to complete the proof. | 2019-12-16T13:33:06 | {
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https://math.stackexchange.com/questions/1397679/surprising-summation-1-sum-i-1nn-i12i-1-sum-i-1n-i2 | # Surprising Summation (1) $\sum_{i=1}^n(n-i+1)(2i-1)=\sum_{i=1}^n i^2$
Show that
$$\sum_{i=1}^n(n-i+1)(2i-1)=\sum_{i=1}^n i^2$$
without expanding the summation to its closed-form solution, i.e. $\dfrac 16n(n+1)(2n+1)$ or equivalent.
E.g., if $n=5$, then
$$5(1) + 4(3)+ 3(5)+2(7)+1(9)=1^2+2^2+3^2+4^2+5^2$$
Background as requested:
The summands for both equations are not the same but the results are the same. The challenge here is to transform LHS into RHS without solving the summation. It seems like an interesting challenge.
Further edit
Thanks to those who voted to reopen the question!
• Hmm I don't see what is so slow about expanding and collecting the terms by powers of $r$. Summation by parts works too, but would take about the same amount of computation. – user21820 Aug 15 '15 at 3:03
• I think this is a duplicate. Looks a lot like this: math.stackexchange.com/questions/1408182/… – marty cohen Apr 4 '16 at 3:21
• @martycohen - It is a duplicate only if the question asks to show that it is equivalent to the sum of squares without expanding to the closed-form solution. – hypergeometric Apr 4 '16 at 3:25
• @martycohen - The summations may be related but it is definitely not a duplicate - the summands are different and the other question is a double summation. In any case this question was posted first so it cannot be a duplicate! – hypergeometric Apr 4 '16 at 3:50
Drawing pictures for this kind of problem helps. If you draw
There are
• $n$ sets of $1$ black squares,
• $n-1$ sets of $3$ blue squares,
• $n-2$ sets of $5$ purple squares,
• $n-3$ setes of $7$ red squares,
• $\vdots$
• $n - r + 1$ sets of $2r-1$ squares of any given color
but the shapes together form a sum of squares pyramid. Algebraically that is:
\begin{align} % \sum_{r=1}^{n} (n - r + 1)(2r - 1) % &= \sum_{r=1}^{n} \left(\sum_{s=r}^n 1\right)(2r - 1) % \\ &= \sum_{r=1}^{n} \sum_{s=r}^n (2r - 1) % \\ &= \sum_{s=1}^{n} \sum_{r=1}^s (2r - 1) % \\ &= \sum_{s=1}^{n} s^2 % \end{align}
where the last step uses the fact that the sum of odd numbers up to $2k-1$ is $k^2$.
• Yes very nice solution! (+1) That was how I started, hoping to find a shorter solution to the sum of squares by recognising the each square is the sum of odd numbers. – hypergeometric Aug 15 '15 at 6:39
• @hypergeometric I felt like the logic was backwards when I was writing it. Anyway, if you want a short solution to sum of squares, I suggest looking at $\sum (k^3 - (k - 1)^3) = \sum (3k^2 - 3k + 1)$ , use telescoping on the left and the known closed form for $\sum k$ on the right. – DanielV Aug 15 '15 at 7:32
$$(n-r+1)(2r-1)=(n+1)(-1)+r[2(n+1)+1]-2r^2$$
$$\implies\sum_{r=1}^n(n-r+1)(2r-1)=-(n+1)\sum_{r=1}^n1+\{2(n+1)+1\}\sum_{r=1}^nr-2\sum_{r=1}^nr^2$$
$$=-(n+1)\cdot n+(2n+3)\cdot\dfrac{n(n+1)}2-2\cdot\dfrac{n(n+1)(2n+1)}6$$
$\begin{array}\\ \sum_{r=1}^n(n-r+1)(2r-1) &=\sum_{r=1}^n(n+1)(2r-1)-\sum_{r=1}^nr(2r-1)\\ &=(n+1)n^2-2\sum_{r=1}^nr^2+\sum_{r=1}^nr\\ &=(n+1)n^2-2\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\\ &=\frac{6(n+1)n^2-2n(n+1)(2n+1)+3n(n+1)}{6}\\ &=n(n+1)\frac{6n-2(2n+1)+3}{6}\\ &=n(n+1)\frac{2n+1}{6}\\ &=\frac{n(n+1)(2n+1)}{6}\\ \end{array}$
Now that's quite a surprise!
I'll stop here a try to find a magic trick later.
• Well spotted! (+1) – hypergeometric Aug 15 '15 at 6:38
Make the substitution $i = n - r + 1$ (like change of variable in integration), then \begin{align} & \sum_{r = 1}^n (n - r + 1)(2r - 1) = \sum_{i = 1}^n i(2n - 2i + 1) \\ = & (2n + 1) \sum_{i = 1}^n i - 2\sum_{i = 1}^n i^2 \end{align} I think I didn't expand the term "brutally", but the last step is still slight expansion.
Here's the "magic" solution in more general form, followed by an continuous analog.
Let $f(n) =\sum_{i=1}^n (n+1-i)g(i)$. Note that $f(1) = g(1)$ and $f(2) =2g(1)+g(2) =g(1)+(g(1)+g(2))$. Then
$\begin{array}\\ f(n+1) &=\sum_{i=1}^{n+1} (n+2-i)g(i)\\ &=\sum_{i=1}^{n+1} (n+1+1-i)g(i)\\ &=\sum_{i=1}^{n+1} (n+1-i)g(i)+\sum_{i=1}^{n+1} g(i)\\ &=\sum_{i=1}^{n} (n+1-i)g(i)+\sum_{i=1}^{n+1} g(i) \quad\text{(since }n+1-i = 0 \text{ for } i=n+1)\\ &=f(n)+\sum_{i=1}^{n+1} g(i)\\ so\\ f(n+1)-f(n)&=\sum_{i=1}^{n+1} g(i)\\ so\ that\\ f(n)&=\sum_{j=1}^n\sum_{i=1}^{j} g(i)\\ \end{array}$
Setting $g(n) = 2n-1$, since $\sum_{i=1}^{n} g(i) =n^2$, $f(n) =\sum_{i=1}^{n} i^2 =\dfrac{n(n+1)(2n+1)}{6}$.
The continuous analog:
Let $f(x) =\int_0^x (x-y)g(y)dy$. Then
$\begin{array}\\ f(x) &=\int_0^x (x-y)g(y)dy\\ &=\int_0^x xg(y)dy-\int_0^x yg(y)dy\\ &=x\int_0^x g(y)dy-\int_0^x yg(y)dy\\ \text{so that}\\ f'(x)&=xg(x)+\int_0^x g(y)dy- xg(x)\\ &=\int_0^x g(y)dy\\ \text{Integrating,}\\ f(x)&=\int_0^x \int_0^y g(z)dz dy\\ \end{array}$
If $g(y) = 2y$, then $\int_0^x g(y)dy =x^2$ so $\int_0^x (x-y)g(y) dy =\int_0^x y^2 dy =\dfrac{x^3}{3}$
\begin{align} \sum_{i=1}^n(n-i+1)(2i-1) &=\sum_{i=1}^n\sum_{j=i}^n(2i-1)\\ &=\sum_{j=1}^n\sum_{i=1}^j(2i-1) &&(1\le i\le j \le n)\\ &=\sum_{j=1}^n\sum_{i=1}^ji^2-(i-1)^2\\ &=\sum_{j=1}^n j^2&&\text{(by telescoping)}\\ &=\sum_{i=1}^n i^2\qquad\blacksquare \end{align}
Posting another solution which has just been suggested by a friend:
\begin{align} \sum_{i=1}^n i^2 =&\sum_{i=1}^n \color{blue}{-}(\color{blue}{n-i})i^2+\sum_{i=1}^n(\color{blue}{n-i}+1)i^2\\ =&\sum_{i=2}^{n+1} -(n-i+1)(i-1)^2+\sum_{i=1}^n(n-i+1)i^2\\ =&\sum_{i=1}^{n} -(n-i+1)(i-1)^2+\sum_{i=1}^n(n-i+1)i^2\\ =&\sum_{i=1}^{n} (n-i+1)\big[i^2-(i-1)^2\big]\\ =&\sum_{i=1}^{n} (n-i+1)(2i-1)\qquad\blacksquare\\\ \end{align} | 2020-02-18T19:07:28 | {
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http://mathrefresher.blogspot.com/2005/05/coprime-numbers-more-results.html | ## Sunday, May 15, 2005
### Coprime Numbers: More Results
In this blog, I will review some basic results regarding coprime numbers.
Lemma 1: p,q coprime -> pq, p2 - q2 are coprime.
(1) Assume gcd(pq,p2 - q2) = d and d is greater than 1
(2) So, there is some prime P which divides d [Fundamental Theorem of Artihmetic]
(3) Since d divides pq. P either divides p or q. [Euclid's Lemma]
(4) Let's assume P divides p and the same argument will apply if it divides q. So that, there exists R such that p = PR.
(5) Since d divides p2 - q2, there exists a value Q such that p2 - q2 = PQ
(6) This means that q2 = p2 - PQ = (PR)2 - PQ = P(PR2 - Q).
(7) But this means that P divides q by Euclid's Lemma which is a contradiction since P also divides p. [Because gcd(p,q)=1]
(8) Therefore, we reject our assumption.
QED
Lemma 2: p,q are relatively prime and are of different parity (one is odd, one is even), then p + q, p - q are relatively prime.
(1) Assume that gcd(p+q,p-q) = d and d is greater than 1.
(2) Then there exists a prime x which divides d. [Fundamental Theorem of Arithmetic]
(3) We know that this x is odd since p+q and p-q are odd. [Since they are of different parity]
(4) Now x divideds 2p since 2p = (p + q) + (p - q) which means x divides p. [By Euclid's Lemma Since x is odd]
(5) Likewise x divides 2q since 2q = (p + q) - (p - q) = p + q - p + q which means x divides q. [Same reason as above]
(6) But this is a contradiction since p,q are relatively prime.
QED
Lemma 3: if S2 = P2 + Q2 and P2 = T2 + Q2, then Q,S,T are relatively prime.
(1) We can assume that S,P,Q are relatively prime and P,T,Q are relatively prime. [See Lemma 1, here for details]
(2) We can also assume that S,P are odd which means that Q is even and that T is odd. [See details above]
(3) From the two equations, we can derive that:
S2 = T2 + 2Q2
(4) Now we need only prove that any factor that divides 2 values, will divide the third.
Case I: f divides S,Q
(a) There exists s,q such that S = fs, Q = fq
(b) T2 = S2 - 2Q2 = f2 [ s2 - 2q2 ]
(c) Which means that f divides T [See here.]
Case II: f divides T,Q
(a) We can use the same argument as Case I.
Case III: f divides S,T
(a) There exists s,t such that S = fs, T = ft
(b) 2Q2 = f2[ s2 - t2 ]
(c) Since S,T are odd, f must be odd and s,t must be odd.
(d) Then, s2 - t2 must be even.
(e) Then, there exists u such that 2u = s2 - t2
(f) So, we have:
2Q2 = (2)u(f2)
(g) And canceling out the 2s gives us:
Q2 = f2u
QED
Lemma 4: S,T relatively prime, both odd, 2u = S - T, 2v = S + T, then u,v are relatively prime.
(1) Assume f divides u,v such that u = fU, v = fV.
(2) The f divides S
S - T + S + T = 2S = 2u + 2v = 2f(U + V)
(3) And f divides T
S + T - (S - T) = 2T = 2v - 2u = 2f(V - U)l.
(4) This contradicts our S,T beings coprime so we reject our assumption.
QED
Oscar Rojas said...
Hi Larry,
I think that in Lemma 3, the assumption 1 (about some of the numbers being relative primes) should be part of the hypothesis.
In fact the Lemma does not hold if that assumption is not satisfied.
The proper assumption must be made in the part of fermat's one proof that is using the Lemma 3.
Since Lemma 3 is only used in that proof, then one could say that it is irrelevant where do you place the assumption... but, for pure formalism...
Larry Freeman said...
Hi Oscar,
The assumption is justified by the proof in Lemma 1 in the link.
Since we can reduce any solution to x^n + y^n = z^n to a form where x,y,z are coprime (see Lemma 1, here), we can assume that S,P,Q and P,T,Q are relatively prime because if they weren't, then we could still find a form that was.
-Larry
Oscar Rojas said...
Thanks Larry.
I am sorry; I failed to state more clear my comment. I will try to do it better:
I agree that the assumption is justified... but it is justified in the context outside the lemma. If you take the lemma out of that context, it is not true as written.
The only fact in the hypothesis is that the numbers satisfy the equations, and that simple fact is not sufficient to derive the conclusion. If you take another set of numbers that are multiples of the originals, the new set still satisfy the unrestricted hypothesis but obviously they do not satisfy the conclusion.
So, I am not against the assumption, I am against using it as part of the proof of the lemma.
To save the consistency, the hypothesis must state that the original numbers are relative primes, precisely to bring the lemma in the restricted context.
As I said: pure formalism! | 2017-10-24T09:40:05 | {
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https://mathhelpboards.com/threads/conditional-probability.60/ | # Conditional probability
#### Romanka
##### New member
Please, help me to solve the problem
Details at a factory are tested randomly to check if they are faulty. It is known from previous experience that the probability of a
faulty detail is known to be 0.03. If a faulty detail is tested the probability of it testing faulty is 0.82. If a non-faulty detail is
tested the probability of it testing faulty is 0.06. Given that the detail was tested as not been faulty, calculate the
probability that it was faulty.
I understand that it's about conditional probability, but can't get it.
Thanks!
#### tkhunny
##### Well-known member
MHB Math Helper
Have you considered drawing a 2x2 grid and filling in the boxes?
#### Plato
##### Well-known member
MHB Math Helper
Details at a factory are tested randomly to check if they are faulty. It is known from previous experience that the probability of a faulty detail is known to be 0.03. If a faulty detail is tested the probability of it testing faulty is 0.82. If a non-faulty detail is tested the probability of it testing faulty is 0.06. Given that the detail was tested as not been faulty, calculate the probability that it was faulty.
Use $F$ for faulty and $T$ for a positive test.
From the given: $\mathcal{P}(F)=0.03,~\mathcal{P}(T|F)=0.82,~\&~ \mathcal{P}(T|F^c)=0.06$
Now you want $\mathcal{P}{(F|T^c)}$$=\dfrac{\mathcal{P}(F\cap T^c)}{\mathcal{P}(T^c)}. #### Romanka ##### New member thanks, I'll try ---------- Post added at 03:27 PM ---------- Previous post was at 03:26 PM ---------- Use F for faulty and T for a positive test. From the given: \mathcal{P}(F)=0.03,~\mathcal{P}(T|F)=0.82,~\&~ \mathcal{P}(T|F^c)=0.06 Now you want \mathcal{P}{(F|T^c)}$$=\dfrac{\mathcal{P}(F\cap T^c)}{\mathcal{P}(T^c)}$.
that's exactly what I need! thank you so much!
#### HallsofIvy
##### Well-known member
MHB Math Helper
Here's how I like to do problems like this: Imagine there are 10000 details (chosen to avoid fractions).
"It is known from previous experience that the probability of a faulty detail is known to be 0.03."
Okay, so our 10000 sample includes (0.03)(10000)= 300 faulty details and 10000- 300= 9700 that are not faulty.
"If a faulty detail is tested the probability of it testing faulty is 0.82."
Of the 300 faulty details, (0.82)(300)= 246 will test faulty, the other 54 will test not-faulty.
"If a non-faulty detail is tested the probability of it testing faulty is 0.06."
Of the 9700 non-faulty details, (.06)(9700)= 582 will test faulty. 9700- 582= 9118 will test not-faulty.
"Given that the detail was tested as not been faulty, calculate the probability that it was faulty."
There are a total of 9118+ 54= 9172 details that test non-faulty of which 54 are faulty.
thanks!
#### HallsofIvy
##### Well-known member
MHB Math Helper
By the way, "detail" doesn't seem like quite the word you want. "Detail" means a small part of something larger. I suspect this was translated from another language and you really wanted "item".
#### Romanka
##### New member
Yes, maybe. But I got the problem about "details"
#### Plato
##### Well-known member
MHB Math Helper
Yes, maybe. But I got the problem about "details"
YES. But what was language of the question?
Did you use a translation program to post it here? | 2021-10-24T10:21:10 | {
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https://math.stackexchange.com/questions/1860168/uncountability-of-increasing-functions-on-n/1860193 | # Uncountability of increasing functions on N
I believe I have made a reasonable attempt to answer the following question. I would like a confirmation of my proof to be correct, or help as to why it is incorrect.
Question: Let $f : \mathbb{N} \to \mathbb{N}$ be increasing if $f(n+1) \ge f(n)$ for all $n$. Is the set $A$ of functions $f$ countable or uncountable?
For each $f$ in $A$, there is a function $g: \mathbb{N}\cup \{0\} \to \mathbb{N} \cup\{0\}$ defined by $g(0)=f(1)$ and $g(n)=f(n+1)-f(n)$ for $n>0$. Conversely, for each $g: \mathbb{N}\cup \{0\} \to \mathbb{N}\cup\{0\}$ with $g(0)>0$, there is an increasing function $f$ defined recursively by $f(1)=g(0)$ and $f(n+1)=f(n)+g(n)$ for $n>0$. Consequently, there is a bijection between $A$ and the set $B$, of $g: \mathbb{N}\cup\{0\} \to \mathbb{N}\cup\{0\}$, $g(0)>0$. If $B$ is uncountable, then $A$ must be uncountable.
The set $C$ of $h: \mathbb{N}\cup\{0\} \to \{0,1,2,3,4,5,6,7,8,9\}$ with $h(0)=1$is a subset of $B$. For each real $x \in [1,2)$, there is a unique decimal expansion with 1 before the decimal point, ending NOT in trailing 9's. By defining for $n>0$, $h(n)=$ (n'th digit of x after decimal point), we have an injection from $[1,2)$ to the set $C$ (different $x$ => different decimal expansions => x maps to different h). Since the set of reals on $[1,2]$ is uncountable, the set C is uncountable, => the set B is uncountable, => the set A is uncountable.
This proof works but seems needlessly verbose. It is easier to prove that strictly increasing functions form an uncountable set (which implies that the monotone increasing are uncountable as well). This is because there is an obvious 1-1 correspondence between strictly increasing functions and infinite subsets of N (each such function is uniquely determined by its range). It is a standard result that there are uncountably many subsets of N, of which only countably many are finite. Hence, there are uncountably many infinite subsets.
It is rather easy to apply the diagonalization argument directly. Let $\langle f_n\rangle$ be a sequence of increasing functions from the naturals to itself. Define a function $f$ recursively by $f(1)=1$ and $$f(n+1)=f(n)+f_n(n+1)-f_n(n)+1.$$ The resulting function $f$ is clearly different from every function in the sequence $\langle f_n\rangle$.
• Oh I really like this. +1 – Pete Caradonna Jul 15 '16 at 11:30
• The resulting function $f$ need not be increasing, so doesn't belong to the set being considered. – hunter Jul 15 '16 at 13:21
• (More precisely, $f(n+1) - f(n) = (f_n(n+1) - f_n(n)) - (f_{n-1}(n) - f_{n-1}(n-1))$; since there's no relation between the rates of growth of $f_n$ and $f_{n-1}$ you can't determine if this difference is positive or negative. – hunter Jul 15 '16 at 13:23
• @hunter Thank you, you are right, what I wrote was different from what I actually intended. I have fixed it now. That is exactly where the recursive definition comes in. – Michael Greinecker Jul 15 '16 at 13:38
• Note that this is exactly the diagonalisation argument for general functions composed with the OP's transformation. – filipos Jul 15 '16 at 20:49
Your proof looks OK to me. It's a little longer than needed though.
Right after you define your set $A$, what if we just restrict ourselves to a subset of $A$ given by the subset of monotone functions $t: \mathbb{N} \to \mathbb{N}$ such that $t(n+1)-t(n) \in \{0,1\}$ and $t(0)=k$.
Then in your spirit, we can identify each $t$ with the infinite binary sequence of its step-wise differences. Moreover, clearly for any infinite binary sequence we can construct the unique $t$ that gets related to it. Then by Cantor's diagonalization argument we know $\{0,1\}^\mathbb{N}$ is uncountable and we're done.
Another way to show that $A$ is uncountable : Show that if $B=\{f_n: n\in N\}\subset A$ then $B\ne A$ as follows:
Let $g(n)=\max \{1+f_j(n): j\leq n\}.$
We have $g(n)> f_n(n)$ for every $n.$ So $g\ne f_n$ for every $n.$ So $g \not \in B.$
For each $n$ we have $g(n)=1+f_{j_n}(n)$ for some $j_n\leq n .$ So $g(n+1)=\max \{1+f_j(n+1):j\leq n+1\}\geq 1+f_{j_n}(n+1)\geq 1+f_{j_n}(n)=g(n).$ So $g\in A.$ | 2019-10-15T02:46:09 | {
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https://math.stackexchange.com/questions/2752037/probability-involving-exponentially-distributed-random-variabl | # Probability involving exponentially distributed random variabl
Two buses A and B arrive independently at a bus station at random rate of 5/hour and 4/hour respectively. A passenger comes to the bus station at 10 am. What is the probability that it takes at most 5 minutes before the first bus arrives at the station?
Attempt
Let X and Y be random variables indicating time of arrivals of Bus A and B respectively. Then, X and Y are exponentially distributed random variables with rate 5 and 4 respectively. So the probability density functions of X and Y are 5*exp(-5*t) and 4*exp(-4*t).
Now, let Z = min(X,Y) be a new random variable. Then, I need to calculate P(Z<5/60).
Question: My solution sheet goes as follows: P(Z<5/60) = 1-P(Z>5/60) = 1-P(X>5/60, Y>5/60) and goes on like this. But why cannot I just calculate P(Z<5/60) = P(X<5/60, Y<5/60) directly? These two give me different answers, meaning my attempt might be wrong. Why do we need to calculate the complementary probability?
• It does not matter if the second bus arrives before 5 minutes if the first one does arrive in 5 minutes. You only care about the complement of the event that no busses arrive within 5 minutes. Apr 24, 2018 at 18:22
• I did not get it. @TimDikland Apr 24, 2018 at 18:24
• There are four possible interesting events. P(X<5, Y<5), P(X<5, Y>5), P(X>5, Y<5), P(X>5, Y<5). Now ask yourself which of those events are desirable and which are not. (Here X<5 means bus X arrives within 5 minutes) Apr 24, 2018 at 18:29
• I am still confused. So, if the question was: What is the probability that she waits at least 5 mins before the first bus arrives? Would I directly calculate P(Z>5/60) = P(X>5/60,Y>5/60)? Apr 24, 2018 at 18:34
• Exactly. Your mistake in your first calculation is that $P(Z < 5/60) \neq P(X<5/60,Y<5/60)$. If you would do it correctly then you would find that $P(Z<5/60) = P(X<5/60, Y<5/60) + P(X<5/60, Y>5/60) + P(X>5/60, Y<5/60)$, which is obviously more work to calculate Apr 24, 2018 at 18:38
Here's another interesting way to approach this problem. I'll start out with a more general version, and then we'll look at the case you need.
Let $X_1, X_2, \cdots X_n$ be independent and let $X_{(1)}$ be the minimum observation. Let's find the CDF of $X_{(1)}$.
\begin{align*} F_1(x) &= P(X_{(1)} \leq x) & \text{(definition)} \\ &= 1 - P(X_{(1)} > x) & \text{(complement)} \\ &= 1 - P(X_1 > x, \ X_2 > x, \ \cdots, \ X_n > x) & \text{(min > x $\Leftrightarrow$ all > x)} \\ &= 1 - P(X_1 > x)P(X_2 > x)\cdots P(X_n > x) & \text{(independence)} \\ \end{align*}
## Back to the problem
Let $$X \sim Exp(4) \quad Y \sim Exp(5) \quad Z = min(X, Y)$$ $$P(X > x) = e^{-4x}$$ $$P(Y > x) = e^{-5x}$$ Hence, by the above result we can write the CDF of the minimum as: $$P(Z \leq x) = 1 - e^{-4x}e^{-5x} = 1 - e^{-9x}$$
Now you can calculate the desired probability easily.
Note that this is equivalent mathematically to the approach that Davis gives, but with this approach we see that the the minimum of independent exponential random variables is itself Exponentially distributed! Specifically, $Z \sim Exp(4+5)$.
$Z < \frac{5}{60}$ might occur without $X < \frac{5}{60}$ occurring, specifically, if $Y < \frac{5}{60}$.
Multiplying probabilities of independent events gives you the probability of their intersection (both events happen), but what you want is the probability of their union. Call $E$ the event that bus A arrives before the given time, and $F$ the event that bus B arrives before the given time. Then, the event you are interested in is:
$E \cup F = (E^c \cap F^c)^c$
The equality is by DeMorgan's law. Now you can figure out the probability by multiplying the probabilities of $E^c$ and $F^c$. | 2022-05-22T13:57:56 | {
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https://logivan.com/ixi3dmsm/if-two-triangles-are-equal-in-area%2C-they-are-congruent-bbe282 | An included angleis an angle formed by two given sides. For example, both of these triangles are isosceles, since they have two equal sides and angles. (iii) If two figures have equal areas, they are congruent. You can replicate the SSS Postulate using two straight objects -- uncooked spaghetti or plastic stirrers work great. Congruence is our first way of knowing that magnitudes of the same kind are equal. Triangles are congruent when all corresponding sides & interior angles are congruent. = False (v) If two sides and one angle of a triangle are equal to the corresponding two sides and angle of another triangle, the triangles are congruent. Hence all squares are not congruent. You have one triangle. TRUE. find the price at which it was sold. Triangle Congruence Theorems (SSS, SAS, ASA), Congruency of Right Triangles (LA & LL Theorems), Perpendicular Bisector (Definition & Construction), How to Find the Area of a Regular Polygon, Do not worry if some texts call them postulates and some mathematicians call the theorems. asked by priyanka. Geometricians prefer more elegant ways to prove congruence. Since b/e = 1, we have a/d = 1. = False (iv) If two triangles are equal in area, they are congruent. - the answers to estudyassistant.com Asked on January 17, 2020 by Premlata Pandagre In a squared sheet, draw two triangles of equal areas such that (i) the triangles are congruent. It can be shown that two triangles having congruent angles (equiangular triangles) are similar, that is, the corresponding sides can be proved to be proportional. -If two squares have equal perimeters, then they have equal ...” in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. prove that they are congruent. Think: Two congruent triangles have the same area. Two triangles are congruent if their corresponding sides are equal in length and their corresponding interior angles are equal in measure. Yes, they are similar. Triangles can be similar or congruent. one could look upside-down compared to the other but if they have the same dimensions (i.e. (if you don't know the a False. Two triangles are congruent if they have the same shape and size, but their position or orientation may vary. they have to be congruent to have same perimeters AND area con. FALSE. All three triangle congruence statements are generally regarded in the mathematics world as postulates, but some authorities identify them as theorems (able to be proved). After you look over this lesson, read the instructions, and take in the video, you will be able to: Get better grades with tutoring from top-rated private tutors. You can think you are clever and switch two sides around, but then all you have is a reflection (a mirror image) of the original. Congruent triangles are triangles which are identical, aside from orientation. An included side is the side between two angles. It is congruent to ∠WSA because they are alternate interior angles of the parallel line segments SW and NA (because of the Alternate Interior Angles Theorem). it the shopkeeper sells it at a loss of 10% . Converse of the above statement : If the areas of the two triangles are equal, then the triangles are congruent. If the areas of two similar triangles are equal, prove that they are congruent. But just to be overly careful, let's compute a/d. So once you realize that three lengths can only make one triangle, you can see that two triangles with their three sides corresponding to each other are identical, or congruent. To be congruent two triangles must be the same shape and size. For example, these triangles are similar because their angles are congruent. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent. A, B and C are three sets, n(A) = 14, n(C) = 13, n(A U BUC) = 28, Move to the next side (in whichever direction you want to move), which will sweep up an included angle. Congruent Triangles. b=e. Those are the three magnitudes of plane geometry: length (the sides), angle, and area. If two triangles have the same area, they must be congruent. The given statement - "If two triangles are congruent, then their areas are equal." (d) if two sides and any angle of one triangle are equal to the corresponding sides and an angle of another triangle, then the triangles are not congruent. We use the symbol ≅ to show congruence. Log in. In the sketch below, we have △CAT and △BUG. (iv) If two triangles are equal in area, they are congruent. Their interior angles and sides will be congruent. In the simple case below, the two triangles PQR and LMN are congruent because every corresponding side has the same length, and every corresponding angle has the same measure. 3) They are equal areas. Local and online. …, how much loss and profit computed on his original capital. Answer: 3 question What would prove that the two triangles are congruent? Now you have three sides of a triangle. Similar triangles will have congruent angles but sides of different lengths. Why should two congruent squares have the same area? Note that if two angles of one are equal to two angles of the other triangle, the tird angles of the two triangles too will be equal. The SAS Postulate says that triangles are congruent if any pair of corresponding sides and their included angle are congruent. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. triangles have same shape/size but just placed/viewed differently cidyah for that to … Click here to get an answer to your question ️ if the area of two triangles are equal. the triangle which are equal in area also congruent, Here ur answer ....pls mark it brainlist..pls, This site is using cookies under cookie policy. It is equal in length to the included side between ∠B and ∠U on △BUG. Congruent triangles will have completely matching angles and sides. Join now. You could cut up your textbook with scissors to check two triangles. So let's just start That triangle BCD is congruent Let’s use congruent triangles first because it requires less additional lines. This forces the remaining angle on our △CAT to be: This is because interior angles of triangles add to 180°. Testing to see if triangles are congruent involves three postulates. Theorem 1 : Hypotenuse-Leg (HL) Theorem. If the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, then the two right triangles are congruent. Get help fast. You will see that all the angles and all the sides are congruent in the two triangles, no matter which ones you pick to compare. Congruence is denoted by the symbol ≅. AAS is equivalent to an ASA condition, by the fact that if any two angles are … So now you have a side SA, an included angle ∠WSA, and a side SW of △SWA. Similar triangles. You can't do it. 0 votes . We could also think this angle right over here. 1. (ii) If two squares have equal areas, they are congruent. A postulate is a statement presented mathematically that is assumed to be true. By applying the Side Angle Side Postulate (SAS), you can also be sure your two triangles are congruent. Triangles can be considered congruent if following conditions are satisfied. https://www.mathsisfun.com › geometry › triangles-congruent.html ... maths. Answer In two similar triangles, the ratio of their areas is the square of the ratio of their sides. Compare them to the corresponding angles on △BUG. Prove that if in two triangles,two angles and the included side of one triangle are equal to two angles and the included side of the other triangle,then two triangles are congruent. 1. For two triangles to be similar, it is sufficient if two angles of one triangle are equal to two angles of the other triangle. The legs of each of these isosceles triangles could have any lengths as long as they are equal, but the legs of these two triangles need not be the same. So the fact that two triangles have the same area does not necessarily tell us that they are congruent (they could be congruent but that is often not the case) since they can have different measures on their sides or a different size and still have the same area. monojmahanta4 monojmahanta4 09.04.2020 Math Secondary School If the area of two triangles are equal. Conditional Statements and Their Converse. You can check polygons like parallelograms, squares and rectangles using these postulates. There are exactly two isosceles triangles with given perimeter and area. https://www.mathsisfun.com/geometry/triangles-congruent.html Side-Angle-Sideis a rule used to prove whether a given set of triangles are congruent. Note that if two angles of one are equal to two angles of the other triangle, the tird angles of the two triangles too will be equal. Those are the three magnitudes of plane geometry: length (the sides), angle, and area. Thus, a=d. are both 90 degrees. If the hypotenuse and one leg of a right triangle are equal to the hypotenuse and one leg of another right triangle, then the two right triangles are congruent. If triangle RST is congruent to triangle WXY and the area of triangle WXY is 20 square inches, then the area of triangle RST is 20 in.². e.g. 2) Their corresponding angles are equal. Two triangles, ABC and A′B′C′, are similar if and only if corresponding angles have the same measure: this implies that they are similar if and only if the lengths of corresponding sides are proportional. And for sensible cases. If you have two similar triangles, and one pair of corresponding sides are equal, then your two triangles are congruent. (g) If two triangles are equal in area, they are congruent. Write the following statements in words, farhan sarted his bussiness by investing 75000 in the first year he made a profit of 20% he invested the capital in new business and made loss of 12% The SAS rule states that If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent. Now you will be able to easily solve problems on congruent triangles definition, congruent triangles symbol, congruent triangles Class 8, congruent triangles geometry, congruent t The SAS rule states that If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent. Two triangles are congruent if their corresponding sides are equal in length and their corresponding interior angles are equal in measure. A rectangle and a triangle with the same base and height have the same area. = False (vi) If two angles and any side of a triangle are equal to the corresponding angles of another triangles then the triangles are congruent. So go ahead; look at either ∠C and ∠T or ∠A and ∠T on △CAT. "If two triangles are congruent, then their areas are equal." You can compare those three triangle parts to the corresponding parts of △SAN: After working your way through this lesson and giving it some thought, you now are able to recall and apply three triangle congruence postulates, the Side Angle Side Congruence Postulate, Angle Side Angle Congruence Postulate, and the Side Side Side Congruence Postulate. Therefore, triangles are congruent. If the area of two similar triangles are equal, prove that they are congruent. Corresponding sides and angles mean that the side on one triangle and the side on the other triangle, in the same position, match. The two are not the same. You now have two triangles, △SAN and △SWA. Suppose you have parallelogram SWAN and add diagonal SA. Using any postulate, you will find that the two created triangles are always congruent. Theorem 1 : Hypotenuse-Leg (HL) Theorem. In most systems of axioms, the three criteria – SAS, SSS and ASA – are established as theorems. I could have one triangle with sides measuring 7 in + 5 in + 8 in (perimeter = 20 in) and another triangle with sides of 6 in + 4 in + 10 in (perimeter = 20 in). Congruent triangles sharing a common side. …, nswer then don't give) give me with full steps.don't spam ❌❌❌❌❌. This is the only postulate that does not deal with angles. That is not very helpful, and it ruins your textbook. You can now determine if any two triangles are congruent! For the two triangles to be congruent, those three parts -- a side, included angle, and adjacent side -- must be congruent to the same three parts -- the corresponding side, angle and side -- on the other triangle, △YAK. Ask your question. All the sides of a square are of equal length. Side-Angle-Sideis a rule used to prove whether a given set of triangles are congruent. You can only make one triangle (or its reflection) with given sides and angles. If we can show, then, that two triangles are congruent, we will know the following: 1) Their corresponding sides are equal. Now shuffle the sides around and try to put them together in a different way, to make a different triangle. Definition: Triangles are congruent when all corresponding sides and interior angles are congruent.The triangles will have the same shape and size, but one may be a mirror image of the other. (iv) If two triangles are equal in area, they are congruent. So I guess the answer is not always. Perhaps the easiest of the three postulates, Side Side Side Postulate (SSS) says triangles are congruent if three sides of one triangle are congruent to the corresponding sides of the other triangle. An included angleis an angle formed by two given sides. This triangle has two angles with equal measures of 75^@. However, they do not have to be congruent in order to be similar. ALL of this is based on a single concept: That the quality that we call "area" is an aspect of dimensional lengths and angles. Both the areas would be 9, but the figures would not be congruent. Illustration of SAS rule: Two right triangles can be considered to be congruent, if they satisfy one of the following theorems. Light1729 Light1729 If two triangles are congruent then they must overlap each other completely, so, they have exactly equal area. Furthermore, your question is about congruent triangles and not similar triangles.l. prove that they are congruent. Notice we are not forcing you to pick a particular side, because we know this works no matter where you start. Here, instead of picking two angles, we pick a side and its corresponding side on two triangles. , the radic of two circles are 8 cm and 6 cm representively find the radius of the circle having area equal to the sum of the areas of the two circles, bnRlinP) = 16 and n(Q).10, what are the greatest and the least values of n(PU QI?8. The congruence of two objects is often represented using the symbol "≅". Only if the two triangles are congruent will they have equal areas. Comparing one triangle with another for congruence, they use three postulates. Isosceles triangles are triangles with two equal sides, and thus two equal angle measures. In this case, two triangles are congruent if two sides and one included angle in a given triangle are equal to the corresponding two sides and one included angle in another triangle. SSSstands for "side, side, side" and means that we have two triangles with all three sides equal. Since two triangles are similar therefore the ratio of the area is equal to the square of the ratio of its corresponding side a r e a ∆ A B C a r e a ∆ D E F = B C E F 2 = A B D E 2 = A C D F 2 B C E F 2 = A B D E 2 = A C D F 2 = 1 N o w , t a k i n g a n y o n e c a s e 1 = B C E F 2 1 = B C E F E F = B C However they can share a side, and as long as they are otherwise identical, the triangles are still congruent. …, n(A) = n(B), nn Byn(B, C) = 4 and n(An C) = 3. However, triangles … Two right triangles can be considered to be congruent, if they satisfy one of the following theorems. The ASA Postulate was contributed by Thales of Miletus (Greek). Testing to see if triangles are congruent involves three postulates, abbreviated SAS, ASA, and SSS. Pairs - The classic pairs game with simple congruent shapes. You already know line SA, used in both triangles, is congruent to itself. For two triangles to be congruent, the corresponding angles and the sides are to be equal. The four triangles are congruent with each other regardless whether they are rotated or flipped. Divide a square sheet in two triangles of equal area so that they are congruent. The Angle Side Angle Postulate (ASA) says triangles are congruent if any two angles and their included side are equal in the triangles. (f) Two circles having same circumference are congruent. Congruent triangles. But, it is not necessary that triangles having equal area will be congruent to one another they may or may not be congruent. Contrapositive of the given statement : If the areas of two traingles are not equal then the triangles are not congruent. So also, sides d and e must be equal because DEF is isosceles. They have the same area and the same perimeter. (ii) If two squares have equal areas, they are congruent. If we can show, then, that two triangles are congruent, we will know the following: 1) Their corresponding sides are equal. $\endgroup$ – Moti Feb 7 '15 at 16:42 add a comment | Your Answer In the figure below, the triangle LQR is congruent to PQR even though they share the side QR. Congruent triangles are triangles having corresponding sides and angles to be equal. If the areas of two similar triangles are equal, prove that they are congruent. But that does not mean that they have to be congruent. If the corresponding angles of two triangles are equal, then they are always congruent. Guess what? If the areas of two similar triangles are equal, prove that they are congruent. If 2 sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent. Cut a tiny bit off one, so it is not quite as long as it started out. So, for equal area, all sides are equal. FALSE. For example: (See Solving SSS Trianglesto find out more) Put them together. Cut the other length into two distinctly unequal parts. It is true that all congruent triangles have equal area. #2 That is true, they must both have the same side length to have the same perimeter, therefore they will also have the same area. Congruence in triangles is important in the study of plane geometry. Yes, if two triangles have two congruent angles and two congruent sides then the triangles are guaranteed to be congruent. What is the solution set of y=3x+16 and 2y=x+72? Join now. Only if the two triangles are congruent will they have equal areas. And why does a $1 \times 1$ square have an area of $1$ unit?) the cost price of a flower vase is rs 120 . Remember that the included angle must be formed by the given two sides for the triangles to be congruent. This will happen when the area is half the multiplication of the sides, or maximum area for such two sides. 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Presented mathematically that is assumed to be congruent their three sides and their included angle does! Side ) if all their three sides and angles be 9 but! Top-Rated professional tutors or its reflection ) with given sides ) congruent triangles then. The Postulate says you can not even do that, angle, and SSS 2019! Think this angle right over here in area, they use three postulates, abbreviated SAS, and..., if two triangles are congruent other based on their shape and size, or to put it another,... | 2021-04-23T09:16:52 | {
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https://www.physicsforums.com/threads/integrating-for-approximation-of-a-sum.753870/ | # Integrating for approximation of a sum
1. May 15, 2014
### goraemon
1. The problem statement, all variables and given/known data
Find an N so that $∑^{\infty}_{n=1}\frac{log(n)}{n^2}$ is between $∑^{N}_{n=1}\frac{log (n)}{n^2}$ and $∑^{N}_{n=1}\frac{log(n)}{n^2}+0.005.$
2. Relevant equations
Definite integration
3. The attempt at a solution
I began by taking a definite integral: $\int^{\infty}_{N}\frac{log(n)}{n^2}dn$ and, using integration by parts, arrived at the following answer: $\frac{log(N)+1}{N}$. (Is this right? If not, I could post the steps I used to try to see where I made an error)
Next we need $\frac{log(N)+1}{N}$ to be within 0.005 as given by the problem, so:
$\frac{log(N)+1}{N}=0.005=\frac{1}{200}$
But I'm having trouble how to solve for N algebraically. Would appreciate any help.
2. May 15, 2014
### BiGyElLoWhAt
I actually got -1 times what you have. remember: $\int \frac{1}{n^2} = -\frac{1}{n}$
3. May 15, 2014
### goraemon
Right but since we're taking a definite integral on the interval from N to ∞, and since $\frac{-log(N)-1}{N}$ as N approaches infinity equals zero, shouldn't the definite integral work out to:
$0-\frac{-log(N)-1}{N}=\frac{log(N)+1}{N}$?
4. May 15, 2014
### pasmith
The graph of $x^{-2}\log(x)$ has a maximum in $[1,2]$, so it's difficult to say whether the integral is an over-estimate or under-estimate of the sum. To be safe I'd require that $$\frac{\log(N) + 1}{N} < \frac{1}{400}$$.
Equations of that type have to be solved numerically. You're looking for zeroes in $x > 0$ of $$f(x) = \frac{x}{200} - 1 - \log x$$. The derivative of this function is $$f'(x) = \frac{1}{200} - \frac{1}{x}$$ so the second derivative ($x^{-2}$) is everywhere positive. There is a minimum at $x = 200$ where $f(200) < 0$. Since $f(x) \to +\infty$ as $x \to 0$ or $x \to \infty$ there are exactly two solutions. Logic dictates you want the larger, since if $N$ works then any larger $M$ should also work.
5. May 15, 2014
### goraemon
Thanks, I tried solving it numerically and came up with N = 1687, which the textbook confirms is correct. | 2017-11-18T17:01:54 | {
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https://math.stackexchange.com/questions/434967/combinations-with-exclusions/435217#435217 | # Combinations with exclusions
Given five objects $A, B, C , D , E$. I'd like to calculate all possible sets of these objects such that
1. Two pairs of objects, $B$ and $C$, and $D$ and $E$, cannot coexist. For example the set $\{ABD\}$ would be valid but the set $\{CDE\}$ would not.
2. Each object appears no more than once in a set, e.g. $\{ACD\}$ is valid but $\{ADD\}$ is not.
3. There can be no empty set.
4. Apart from the above, there is no other limit on the cardinality of a set, e.g. $\{D\}$, $\{AB\}$, $\{ACE\}$. I know there can be no set of cardinality greater than 3, but it would be nice to show this.
5. The objects can be listed in any order, i.e. $\{AB\} = \{BA\}$
Edit My intention is to generalize the process of generating these combinations given any number of elements and restrictions.
• "Sets of these objects" is somewhat incompatible with the ambiguous "objects can be listed in any order." So is AB different from BA? Jul 3 '13 at 4:13
• I apologize for the confusion. {A,B} = {B,A} Jul 3 '13 at 9:44
Since your options are $A$, one of $\{B, C\}$, and one of $\{D, E\}$, it's easy to see why the cardinality can only be at most 3.
I'm going to tackle this case by case:
# No objects
There is only one possibilitiy for the empty set.
# One object
We can choose any of the 5 objects, so there are 5 possibilities.
# Two objects
If we pick A first, we can choose any of the other four objects to go with it. If we pick any of the other objects first, then there are only three objects that can follow, due to the first restriction (e.g. $BA, BD, BE$ for $B$)
Thus there are $4 + 3 \times 4 = 16$ permutations for two objects.
# Three objects
Pick one item from each of the sets $\{A\}, \{B, C\}, \{D, E\}$. There are $1 \times 2 \times 2 = 4$ possible selections that can be made.
For each selection, there are $3! = 6$ ways to permute the objects, giving a total of $4 \times 6 = 24$ permutations for three objects.
Altogether, this is a grand total of $1 + 5 + 16 + 24 = 46$ permutations.
From a response to a question, it appears that order does not matter.
It is not clear whether the empty set ("none") is to be considered a valid choice. We will assume that it is allowed. If it is not, subtract $1$ from the answer we will obtain.
With such a small number of objects, a carefully drawn up list may be the best approach.
Another good option is the approach taken by Sp3000. There, permutations were being counted, but the ideas can be readily modified to disregard order. For choices of $0, 1, 2, 3$ objects, the answers become $1$, $5$, $8$, and $4$ for a total of $18$ (or $17$ if we don't allow the empty set).
But you may like the following idea. Line up the $5$ objects in front of us, with $B$ and $C$ close to each other, and also $D$ and $E$, like this: $$A\qquad BC \qquad DE.$$
Stand in front of $A$ and decide whether to include her in the set we are building. There are $2$ choices, yes or no.
Now walk over to the $BC$ group. For every choice we made about $A$, we now have $3$ choices, $B$, $C$, or neither. So far, we have $2\times 3$ distinct possibilities.
Finally, walk over and stand in front of the $DE$ group. For every choice we made earlier, there are again $3$ choices, this time $D$, $E$, or neither.
Thus the total number of possible choices is $2\times 3\times 3$, that is, $18$. Again, if the empty set is not allowed, we have $17$ possibilities only.
• You hit the nail on the head. My intention is to generalize the process of generating these combinations, for any number of elements and restrictions. I've updated my updated question accordingly. Jul 3 '13 at 11:17
• For setups precisely of this form (the people belong to disjoint groups of $1$ or more, possibly of different sizes, and at most one member of a group can be selected) the above idea generalizes immediately, and gives implicitly a mechanical way to generate and list. Jul 3 '13 at 12:28
Your question asks for the number of all subsets $S$ of $\{A,B,C,D,E\}$ which are not a superset of $\{B,C\}$ or $\{D,E\}$.
This is a typical application of the principle of inclusion-exclusion.
The total number of subsets of $\{A,B,C,D,E\}$ is $2^5 = 32$. The number of supersets of $\{B,C\}$ in $\{A,B,C,D,E\}$ is $2^3 = 8$ (each of the elements $A,B,C$ may be selected or not). In the same way, the number of supersets of $\{D,E\}$ in $\{A,B,C,D,E\}$ is $8$. The number of common supersets of $\{B,C\}$ and $\{D,E\}$ is 2.
Now the principle of inclusion-exclusion yields the result $$32 - 8 - 8 + 2 = 18.$$ | 2021-10-21T14:26:10 | {
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https://community.wolfram.com/groups/-/m/t/1620190 | # Statistical Distributions of Areas of Voronoi Cells
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Open code in Cloud | Download code to Desktop via Attachments BelowOne of the brightest characteristics of the Wolfram Language (WL) is that it works easily across many different scientific fields (thanks to numerous built-in functions and data) and that due to the coherence of WL design, combining such different sciences via computation comes as natural as speaking to a human. Here WL will enable us to move quickly from computational geometry to machine learning to statistics on a quest of exploring an interesting subject. This seemingly simple question is not easy to answer without right framework and tools: What is statistical distributions of areas of cells in Voronoi tessellation? From a brief web search it appears to follow closely GammaDistribution. But in the absence of analytical proof, how can one quickly deduce this fact or verify it is good model? We will do exactly that below. For reference see also a relevant paper by TANEMURA and a real-interest question by actual user which prompted this exploration.To start let's build a "large" VoronoiMesh on 5000 uniformly distributed points within a unit Disk: pts = RandomPoint[Disk[], 5000]; mesh = VoronoiMesh[pts, Axes -> True] We see the statistics we can gather from this will be obviously offset by the presence of "border" cells of much larger area than regular inner cells have: Let's exclude large cells by selecting only those within original region of random points distribution - unit Disk: vor=MeshPrimitives[mesh,2]; vor//Length 5000 vorInner=Select[vor,RegionWithin[Disk[],#]&]; vorInner//Length 4782 We got fewer elements of course and they all are nice regular cells without any large outliers: Graphics[{FaceForm[None], EdgeForm[Gray], vorInner}] Let's take a look at the statistics of areas' distribution: areas = Area /@ vorInner; hist = Histogram[areas, Automatic, "PDF", PlotTheme -> "Detailed"] We can see that there is a minimum of distribution at zero area, which is intuitively correct. Close to zero-area cells of course are not present much with the finite number of points per region (or finite density). So there is some prevalent finite mean area there. We can find a very close simple analytic distribution using WL machine learning tools, which turns out to be GammaDistribution as discussed in some publications: dis=FindDistribution[areas] GammaDistribution[3.3458431368450587,0.00018456010158014188] FindDistribution acted automatically without any additional options from our side. GammaDistribution matches very nicely the empirical histogram: Show[hist, Plot[PDF[dis, x], {x, 0, .0015}]] And now it is easy to find thing like: {Mean[dis], Variance[dis], Kurtosis[dis]} {0.0006175091492073445, 1.1396755130437449*^-7, 4.793269963533813} Probability[x > .0005, x \[Distributed] dis] 0.5740480899719699 Attachments:
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Posted 11 months ago
Very nice. Thanks for sharing. Some years ago I did this with 10^9 cells. If I recall correctly the left side is a bit higher (more probably) than a gamma distribution.To remove any edge-effects, what people do is: take random points in a square, and then copy the square 9 times in a 3*3 grid, such as to emulate period boundary conditions. Then take only the polygons which center is in the center unit square…Great stuff!–SH
Posted 11 months ago
Thanks! Awesome idea about edge-effects! Did you estimate any analytical models beyond Gamma or it was just pure empirical thing?
Posted 11 months ago
I looked at different things: how many sides, angles, area, perimeter, and many conditionally averaged quantities. But area of the PDF as well. But no analytical shape is known for it. I believe that there is something known for perimeter.
Posted 11 months ago
I don't find it all that surprising that the result is a Gamma distribution. The 1-d case is well known as it is the distribution of interarrival times of a Poisson process and is exponentially distributed. The exponential distribution is a special case of the Gamma so maybe you're onto something.
Posted 11 months ago
Dear All,that is indeed interesting. I also believe that the exact distribution is unknown but it can be approximated by a GammaDistribution, see:https://arxiv.org/pdf/1207.0608.pdfhttps://arxiv.org/pdf/1612.02375.pdfhttps://www.sciencedirect.com/science/article/pii/002608008990030Xhttp://www.scipress.org/journals/forma/pdf/1804/18040221.pdfhttp://www.eurecom.fr/~arvanita/PVT.pdfInterestingly, if you run the code in @Vitaliy Kaurov's cod for 50000 points you obtain: dis = FindDistribution[areas] (*GammaDistribution[3.2718, 0.0000197063]*) I managed to get it done for 500000 points. I think that the edge effects should be getting smaller; as the area increases faster than the circumference. So I obtain: dis = FindDistribution[areas, 5] (* {MixtureDistribution[{0.658342, 0.341658}, {MaxwellDistribution[2.92588*10^-6], GammaDistribution[7.1867, 1.37091*10^-6]}], ExtremeValueDistribution[4.80446*10^-6, 2.73831*10^-6], GammaDistribution[2.94963, 2.17794*10^-6], BetaDistribution[2.94962, 459143.], MixtureDistribution[{0.687855, 0.312145}, {MaxwellDistribution[2.82857*10^-6], LogNormalDistribution[-11.5031, 0.28619]}]}*) This happens when we plot all of them: Plot[Evaluate[PDF[#, x] & /@ dis], {x, 0, 0.00003}, PlotRange -> All, LabelStyle -> Directive[Bold, 16]] If you compare that with the histogram: Show[Histogram[areas, 200, "PDF"], Plot[Evaluate[PDF[#, x] & /@ dis], {x, 0, 0.00003}, PlotRange -> All,LabelStyle -> Directive[Bold, 16]]] I would say that the second (ExtremeValueDistribution) fits best:but it does not get the small areas quite right.Regarding the boundary effects, one might also (similar to what Sander suggests) use the torus as the geometry: disfun[x_, y_] := Sqrt[Min[Abs[x[[1]] - y[[1]]], 1 - Abs[x[[1]] - y[[1]]]]^2 + Min[Abs[x[[2]] - y[[2]]], 1 - Abs[x[[2]] - y[[2]]]]^2] M = 50; nf = Nearest[Rule @@@ Transpose[{RandomReal[1, {M, 2}], Range[M]}], DistanceFunction -> disfun] DensityPlot[First[nf[{x, y}]], {x, 0, 1}, {y, 0, 1}, PlotPoints -> 100, ColorFunction -> "TemperatureMap"] Cheers,Marco
Posted 11 months ago
@Marco Thiel thank you for the references and insight! I believe FindDistribution, while finding correct distributions, needs some tune up to get the distribution parameters right. To test the waters, I tried using EstimatedDistribution and got a bit better results. Let's run this for 500,000 points: pts=RandomPoint[Disk[],500000]; mesh=VoronoiMesh[pts,PerformanceGoal->"Speed"]; vor=MeshPrimitives[mesh,2] vorInner=Select[vor,RegionWithin[Disk[],#]&]; areas=Area/@vorInner; hist=Histogram[areas,200,"PDF",PlotTheme->"Detailed"]; and then find the parameters assuming distributions are known: disGAM = EstimatedDistribution[areas, GammaDistribution[a, b]] GammaDistribution[3.526166220777205, 1.7809806247800413*^-6] disEXT = EstimatedDistribution[areas, ExtremeValueDistribution[a, b]] ExtremeValueDistribution[4.775094275466527*^-6, 2.5746412541824736*^-6] We see gamma (red) behaves a bit better now, both for zero and for the tail values: Show[hist,Plot[{PDF[disEXT,x],PDF[disGAM,x]},{x,0,.00002},PlotStyle->{Blue,Red}]] But while p-value of gamma is greater than extreme, it is still pretty low, not sure if there is a catch here somewhere: DistributionFitTest[areas,disGAM] 0.000051399258187534436 DistributionFitTest[areas,disEXT] 0.
Posted 11 months ago
Thanks for sharing everybody! Maybe this can add some more information:In the article "Statistical Distributions of Poisson Voronoi Cells in Two and Three Dimensions" by Masaharu Tanemura, mentioned by Marco Thiel, the distribution of the perimeter of the cells (and the number of edges etc.) is also analyzed. Here I adapted the code of Vitaly to perimeters and we see (as in the article) that now we have a normal distribution: perimeters = Perimeter /@ vorInner; hist = Histogram[perimeters, Automatic, "PDF", PlotTheme -> "Detailed"]; dist = FindDistribution[perimeters] Show[hist, Plot[PDF[dist, x], {x, 0, .1}]] And for the number of edges, we find a BinomialDistribution with FindDistribution although the result look almost "normal" as we use EstimatedDitribution numberOfEdges[poly_] := Length[First@poly] edges = numberOfEdges /@ vorInner; histE = Histogram[edges, Automatic, "PDF", PlotTheme -> "Detailed"]; distE = EstimatedDistribution[edges, NormalDistribution[\[Mu], \[Sigma]]] Show[histE, Plot[PDF[distE, x], {x, 0, 12}]] distEB = FindDistribution[edges] Plot[CDF[distEB, x], {x, 0, 12}, Filling -> Bottom]
Posted 11 months ago
Great, @Erik, thanks for exploring further and sharing!
You may be interested in this paper, which takes a semi-empirical approach to come up with a closed-form approximate formula: Jarai, Neda: On the size-distribution of Poisson Voronoi cells https://arxiv.org/abs/cond-mat/0406116 At the time when it was written, Mathematica did not have VoronoiMesh, but one could "trick" it to quickly compute a Voronoi tessellation by using ListDensityPlot with InterpolationOrder -> 0 and extracting the polygons. | 2020-01-17T22:52:47 | {
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https://math.stackexchange.com/questions/2957549/find-the-coefficient-of-x10 | # Find the coefficient of $x^{10}$
We have been given the following function.
$$f(x)$$= $$x$$ +$$x^2$$ + $$x^4$$ + $$x^8$$ + $$x^{16}$$ + $$x^{32}$$ + ...upto infinite terms
The question is as follows:
What is the coefficient of $$x^{10}$$ in $$f(f(x))$$?
I tried solving it myself and I found the answer too but the method of solving was too much time-consuming.
I had solved it manually by only considering the first four terms of $$f(f(x))$$. This method took me about 10 minutes.
But the problem is that this question was asked in a competitive exam called JEE which requires solving the question in max. 3-4 minutes.
So, I wanted to know if there was a faster method to solve this problem.
Note that we may at any point in our calculation ignore any terms of degree $$11$$ or higher. Therefore we may also ignore any terms whose inclusion will only lead to degree $$11$$ or higher terms. We have: $$f(f(x)) = (x + x^2 + x^4 + x^8 + \cdots) + (x + x^2 + x^4 + x^8 +\cdots)^2 \\+ (x + x^2 + x^4 + \cdots)^4 + (x + x^2 + \cdots)^8 + \cdots$$ From here we may look at each of the brackets and simply extract the ones which lead to degree $$10$$, using the multinomial theorem (basically the binomial theorem) for what it's worth:
• $$x + x^2 + x^4 + x^8 + \cdots$$: no terms
• $$(x + x^2 + x^4 + x^8 + \cdots)^2$$: we get $$2x^2\cdot x^8$$
• $$(x + x^2 + x^4 + \cdots)^4$$: we get $$6 (x)^2\cdot (x^4)^2 + 4(x^2)^3\cdot x^4$$
• $$(x + x^2 + \cdots)^8$$: we get $$28(x)^6\cdot(x^2)^2$$
where I've used brackets to clarify which terms I've picked in each case.
• From writing up my expression of $f(f(x))$ and listing the degree-10 terms from each bracket, this shouldn't take that much time. It really depends on how much you have to write in your answer. If you have to write it out like here (or even more detailed), then 10 minutes to solve this problem isn't unreasonable. If you only have to write the final answer, then much of what I've written here may be skipped in your own notes, and you will get at the answer that much quicker. – Arthur Oct 16 '18 at 7:21
• Thanks a lot for such a fast answer. I did "almost" exactly what you did. When I said ten minutes, it included the time taken to come up with a method as well. – Vaibhav Oct 16 '18 at 17:21
• Also, I am new to Stack Exchange. I didn't expect my question to be a answered within an hour of posting the question. It would have taken weeks if I had asked this on Quora. – Vaibhav Oct 16 '18 at 17:23
$$f(x)$$ contains no tenth power.
$$f^2(x)$$ has $$x^2\cdot x^8$$, with a coefficient $$2$$.
$$f^4(x)$$ has $$x\cdot x\cdot x^4\cdot x^4$$ with a coefficient $$\dfrac{4!}{2!2!}=6$$ and $$x^2\cdot x^2\cdot x^2\cdot x^4$$ with a coefficient $$\dfrac{4!}{3!1!}=4$$.
$$f^8(x)$$ has $$x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x^2\cdot x^2$$ appearing $$\dfrac{8!}{6!2!}=28$$ times.
Total $$40.$$
Using a CAS,
$$\cdots+40x^{10}+22x^9+16x^8+8x^7+8x^6+6x^5+3x^4+2x^3+2x^2+x$$
• Crossed with Arthur. – Yves Daoust Oct 16 '18 at 7:22
Since you are looking for $$x^{10}$$ term, you should ask the question: when would I get $$x^{10}$$ from $$f(x)^n$$? When the power $$n$$ of $$x^n$$ is too large, it would not work. So considering the first 4 terms is reasonable. But you don't need to expand everything, for example, if you would consider $$f(x)+f(x)^2+f(x)^4+f(x)^8$$, note that $$f(x)=x+x^2+...$$ does not contain $$x^{10}$$. Next we look at $$f(x)^2$$, which has the form $$(x+x^2+...)(x+x^2+...)$$, expanding this would give $$x^{10}$$ if 2 powers add up to 10, this happens for $$x^2\cdot x^8=x^{10}$$, and we also have the other term from $$x^8\cdot x^2$$. As for $$f(x)^4$$, you are always multiplying 4 terms together, so you just need to see if you can get 10 from adding 4 powers, this is possible when $$2+2+2+4=10$$, but the 4 can be chosen from any other "bracket", so there are 4 such terms. If you understand what I meant, try to figure out whether $$f(x)^8$$ contains any term of $$x^{10}$$. | 2019-11-16T22:29:38 | {
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https://math.stackexchange.com/questions/1622928/what-is-the-probability-that-an-integer-between-0-and-9-999-has-exactly-one-8-an | # What is the probability that an integer between 0 and 9,999 has exactly one 8 and one 9?
I Googled this question and found some answers but they were all different from each other, so I don't know which one is correct.
Question: What is the probability that an integer between 0 and 9,999 has exactly one 8 and one 9?
I tried by splitting it into 3 cases.
Case #1 (2 digits): 98 or 89, so only 2!=2 possible ways of arranging 2 digits.
Case #2 (3 digits): One possible outcome is 98y where y can be integers from 0 to 7 (so there's 8 possible values for y). 3!=6 ways of arranging it. So there's 6*8-2 = 46 possible 3 digit numbers. I subtracted 2 from it because 098 and 089 are not 3 digit numbers.
Case #3 (4 digits): A possible outcome is 98xy where x and y can both take on 8 possible values ranging from 0 to 7. 4!/2! = 12 ways of arranging it because if x=y, 98xy is the same as 98yx. So we have: 12*8*8-48 = 720 possible 4 digit numbers. 48 is the result of 46 + 2 from case #2 because we have to subtract outcomes where the first two digits are 0 or the first digit is a 0 eg: 0098.
Therefore, the total is 2 + 46 + 720 = 768 numbers with exactly 8 and 9 in it. The probability is 768/10000.
• Note that if you allow the four digit numbers to have leading zeroes (i.e. allow numbers like $0098$), then that takes care of two- and three-digit numbers automatically, without having to treat those as a special case, and without having to treat $0$ as a special digit. – Arthur Jan 22 '16 at 21:20
• Your result is correct. It is a much harder approach than the answers, but you were careful to get them all and once each. – Ross Millikan Jan 22 '16 at 21:29
• leading zeros are allowed because associating strictly 4 digit numbers with leading zeros allowed with the corresponding 1, 2, 3 or 4 digit numbers without leading zeros is a 1-1 association. e.g. 0071 ~ 71 and nothing else. ... or another way of putting it 0071 and 71 are simply two different ways of writing the same thing. And as the two answers show, using leading zeros makes the calculations much easier. – fleablood Jan 22 '16 at 21:45
Let us ask the related question of how many 4-digit strings have exactly one eight and one nine. (Strings are allowed to have leading zeroes, whereas numbers are not) Note that the 4-digit strings are in direct bijection with the integers from $0$ to $9999$.
• Pick the location of the nine ($4$ choices)
• Pick the location of the eight (It can't be where the nine is, so $3$ choices)
• For each remaining position, pick a digit other than eight or nine ($8$ choices each time for a total of $8\cdot 8$ choices)
Thus, the number of four-digit strings with exactly one eight and one nine is $4\cdot 3\cdot 8\cdot 8$
Since there are $10^4$ different four-digit strings regardless, the probability is then $\dfrac{4\cdot 3\cdot 8\cdot 8}{10^4} = \dfrac{768}{10000} = .0768$
If you make all your numbers four digits by allowing leading zeros it is easier. You have four places to put the $9$, then three left to put the $8$, then $8$ choices for each of the other two places. $4 \cdot 3 \cdot 8^2=768$ good numbers out of $10000$, so the probability is $0.0768$ | 2019-11-17T00:07:13 | {
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Power Series Ordinary Differential Equations Esteban Arcaute1 We then need partial fractions to calculate Z dy N(y) = Z Q(y) P(y) dy. For example, diff(y,x) == y represents the equation dy/dx=y. A Differential Equation is a n equation with a function and one or more of its derivatives:. SOLUTION We assume there is a solution of the form. If the function is of only one variable, we call the equation an ordinary differential equation (ODE). So, why are we worried about the convergence of power series? Well in order for a series solution to a differential equation to exist at a particular x it will need to be convergent at that x. This might introduce extra solutions. 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 50 49 48 47 46 45 44 43 42 41. In following section, 2. The number ρ is at least 0, as taking x = x0 gives P 0 which is clearly converging to 0; On the other hand, when the power series is convergent for all x, we say its radius of convergence is infinity,. Sturm-Liouville problems. The differential equation in Example 2 cannot be solved by any of the methods discussed in previous sections. Use the power series method to find 2 linearly independent series solutions to differential equation (x^2 - 2x + 2)y" + 2(x-1)y' - 2y = 0 about the ordinary point xo=1. Sage Quickstart for Differential Equations¶. Pourhabib Yekta1, A. Browse other questions tagged sequences-and-series ordinary-differential-equations power-series or ask your own question. The differential equation can be writ-ten in. After a promising start to his mathematical career, Nash began to suffer from schizophrenia around his 30th year, an illness from which he has. An older book that has a lot of nice material on power series and other numerical methods for ODE's is Einar Hille's Lectures On Ordinary Differential Equations. Let’s consider the equation: 2 ′′ + + ′ x y x x y y − = 2 7 ( 1) 3 0 (1). With the exception of special types, such as the Cauchy equations, these will generally require the use of the power series techniques for a solution. The Bessel differential equation has the form x 2 y+xy'+(x 2-n 2)y=0. Power Series Method for Nonlinear Partial Differential Equations Power series is an old technique for solving linear ordinary differential equations [7,20]. Approximate solutions of first-order differential equations using Euler and/or Runge-Kutta methods. The general Airy differential equation is given by :$D^2y \pm m^2 x y = 0$or equivalently$y’’ \pm m^2 x y = 0$The differential equation in the question is a form of the Airy differential equation with the minus sign preceding $m^2$. Chapter 7 Power series methods 7. 1 in [BD] Many functions can be written in terms of a power series X1 k=0 a k(x x 0)k: If we assume that a solution of a di erential equation is written as a power series, then perhaps we can use a method reminiscent of undetermined coe cients. In this section we learn how to extend series solutions to a class of differential equations that appear at first glance to diverge in our region of interest. Solution of linear equations by power series Def. Use the power series method to find 2 linearly independent series solutions to differential equation (x^2 - 2x + 2)y" + 2(x-1)y' - 2y = 0 about the ordinary point xo=1. Given three points, A, , , B, , , and C, , : a Specify the vector A extending from the origin to the point A. An infinite series of this type is called a power series. An excellent article in the American Journal of Physics, by Fairen, Lopez, and Conde develops power series approximations for various systems of nonlinear differential equations. With the exception of special types, such as the Cauchy equations, these will generally require the use of the power series techniques for a solution. 8 Nonlinear systems. Linear and separable first order differential equations. math 230 psu reddit, Review of calculus, properties of real numbers, infinite series, uniform convergence, power series. Finding coefficients in a power series expansion of a rational function. shirin setayesh 55,974 views. 79: conditions calculate Cauchy-Euler equation portrait population pounds power series predictions recurrence. 3 - Recursively Defined Sequences. This method aims to find power series for the solution functions to a differential equation. Equations relating the partial derivatives (See: Vector calculus ) of a function of several variables are called partial differential equations (PDEs). Example $$\PageIndex{2}$$ Find the the first three nonzero terms of two linearly independent solutions to $$xy'' + 2y = 0$$. Equate coefficients of like powers of $$x$$ to determine values for the coefficients $$a_n$$ in the power series. If initial conditions are given, determine the particular solution. Home PDF Paperback Index PrevUp Next. Power series solution is a method to solve the differential equations. • Use power series to find solutions to higher order linear differential equation with nonconstant coefficients at any regular singular point; and • Use Laplace transforms to solve initial value problems. Comprehensive & Detailed COMPLETE note package for the course Math 128 (Calculus II). This RPS method gives approximate solutions in convergent series formula with surely computable components. The differential equation can be writ-ten in. Shifting the Index for Power Series - Duration: 14:49. Differential Equations for Engineers. ’s need to be. Research Article Power Series Extender Method for the Solution of Nonlinear Differential Equations HectorVazquez-Leal 1 andArturoSarmiento-Reyes 2 Electronic Instrumentation and Atmospheric Sciences School, Universidad Veracruzana, Cto. An ordinary differential equation (ODE) is an equation containing an unknown function of one real or complex variable x, its derivatives, and some given functions of x. If the function is of only one variable, we call the equation an ordinary differential equation (ODE). An important application of power series in the field of engineering is spectrum analysis. Power series neural network solution for ordinary differential equations with initial conditions Abstract: Differential equations are very common in most academic fields. {image} {image} {image} {image} 3. Now We have two components R and L connected in Series and a voltage source to those components as shown below. For the differential equation show below:y(1) = 1y'(1) = 0a) Write down a general expansion for a power series solution y(x) about the point Xo = 1b)Find a recurrence relationshi between the coefficients of your power series expansion. Notice that 0 is a singular point of this differential equation. Get this from a library! Formal power series and linear systems of meromorphic ordinary differential equations. 3 Power Series Solutions to Linear Differential Equations 442 8. We will only be able to do this if the point x = x0. Since cos x = Σ(n=0 to ∞) (-1)^n x^(2n)/(2n)!, so u can get by potential of heart a million qn and remedy each and every) Non-linear partial differential equation, Homogenous and non-homogeneous. In this video from PatrickJMT we show how to use power series to find a solution of a differential equation. In the equation, represent differentiation by using diff. 3 - Taylor Series. Fourier series with Ti84: Calculate for all different periodic signals the spectrum. y'' − y' = 0. ; Dig deeper into specific steps Our solver does what a calculator won't: breaking down key steps. 2) By insertion of y = n=0 anx n,y = n=1 nanx n 1,y = n=2 n(n 1)anx n 2, Solution of differential equations by the power series method. Illustrative numerical example is included to demonstrate efficiency. equation is given in closed form, has a detailed description. c) Write down the first few terms of your power series (up to the fourth. We have step-by-step solutions for your textbooks written by Bartleby experts!. 1126 CHAPTER 15 Differential Equations In Example 1, the differential equation could be solved easily without using a series. Full curriculum of exercises and videos. A power series solution to a differential equation is a function with infinitely many terms, each term containing a different power of the dependent variable. In following section, 2. Simple Ordinary Differential Equations may have solutions in terms of power series whose coefficients grow at such a rate that the series has a radius of convergence equal to zero. AMS30, 151-156 (1971). Underdamped Overdamped Critically Damped. Use MathJax to format equations. Conic Sections Trigonometry. An older book that has a lot of nice material on power series and other numerical methods for ODE's is Einar Hille's Lectures On Ordinary Differential Equations. Since many physical laws and relations appear mathematically in the form of differential equations, such equations are of fundamental importance in engineering mathematics. Derivatives Derivative Applications Limits Integrals Integral Applications Series ODE Laplace Transform Taylor/Maclaurin Series Fourier Series. Learn AP®︎ Calculus BC for free—everything from AP®︎ Calculus AB plus a few extra goodies, such as Taylor series, to prepare you for the AP®︎ test. We’ll assume mechanical boresight (θ = 0º), N = 8, and d = λ/2. Now We have two components R and L connected in Series and a voltage source to those components as shown below. The validity of term‐by‐term differentiation of a power series within its interval of convergence implies that first‐order differential equations may be solved by assuming a solution of the form. The Differential Equations diagnostic test results highlight how you performed on each area of the test. Related Calculators. I am now at this point where I have got: \displaystyle \sum_{n = 0}^&#. It often happens that a differential equation cannot be solved in terms of elementary functions (that is, in closed form in terms of polynomials, rational functions, e x, sin x, cos x, In x, etc. Covers material on integration methods (trig, partial fractions, etc. Video tutorial on Power Series Solutions of Differential Equations - In this video, I show how to use power series to find a solution of a differential equation. This might introduce extra solutions. Differential equation or system of equations, specified as a symbolic equation or a vector of symbolic equations. In dissertation we discuss power series characteristics that we use for solving the equations in question. Generalized series expansions involving integer powers and fractional powers in the independent variable have recently been shown to provide solutions to certain linear fractional order differential equations. ; Dig deeper into specific steps Our solver does what a calculator won't: breaking down key steps. Re-index sums as necessary to combine terms and simplify the expression. In our world things change, and describing how they change often ends up as a Differential Equation: an equation with a function and one or more of its derivatives: Introduction to Differential Equations; Differential Equations Solution Guide; Separation of Variables. View and Download PowerPoint Presentations on Solution Differential Equation By Power Series PPT. Even if you don't know how to find a solution to a differential equation, you can always check whether a proposed solution works. Mathematics > Calculus and Analysis > Differential Equations Keywords Calculus, series expansion, Taylor Series, Ordinary Differential Equation, ODE, , Power Series. Order Differential Equations with non matching independent variables (Ex: y'(0)=0, y(1)=0 ) Step by Step - Inverse LaPlace for Partial Fractions and linear numerators. Determining the value of a definite integral on the graphing calculator. Find the singular points (if any) for the following equations. How calculators calculate is by power series. Xavier Sigaud, 150 CEP 22290-180, Rio de Janeiro, RJ, Brazil. Exercises 8. Given a linear differential equation with polynomial coefficients a point x = x 0 is called an ordinary point if b 0 (x 0) 0. Definition 5. Taylor's Series method. 5 The Power Series Method. To leave a comment or report an error, please use the auxiliary blog. Each Differential Equations problem is tagged down to the core, underlying concept that is being tested. = 1 for y at x = 1 with step length 0. This Sage quickstart tutorial was developed for the MAA PREP Workshop "Sage: Using Open-Source Mathematics Software with Undergraduates" (funding provided by NSF DUE 0817071). Special Functions The power series method gives solutions of linear ODEs (1) y” + p(x)y’ + q(x)y = 0 with variable coefficients p and q in the form of a power series (with any center x0, e. lol this is a question you'll look back on after doing more math/physics and laugh. I Check the endpoints, jx x 0j= ˆ, separately to nd the INTERVAL of CONVERGENCE. {image} {image} {image} {image} 3. Solutions by separation of variables and expansion in Fourier Series or other appropriate orthogonal sets. 1 - Sequences; Lesson 21. In every upper division physics class you will use a power series. The unknown function is generally represented by a variable (often denoted y), which, therefore, depends on x. We propose a power series extender method to obtain approximate solutions of nonlinear differential equations. Hermite’s differential equation shows up during the solution of the Schrödinger equation for the harmonic oscillator. the menu option in Differential Equations Made Easy from www. AN EXAMPLE. A differential equation, shortly DE, is a relationship between a finite set of functions and its derivatives. Calculus: Difference Equations to Differential Equations ADD. 3 - Recursively Defined Sequences. 2 Power Series, Analytic Functions, and the Taylor Series Method 431 8. In our world things change, and describing how they change often ends up as a Differential Equation: an equation with a function and one or more of its derivatives: Introduction to Differential Equations; Differential Equations Solution Guide; Separation of Variables. 1 Power series Note: 1 or 1. Solution of linear equations by power series Def. Get this from a library! Formal power series and linear systems of meromorphic ordinary differential equations. In dissertation we discuss power series characteristics that we use for solving the equations in question. The general Airy differential equation is given by :$D^2y \pm m^2 x y = 0$or equivalently$y’’ \pm m^2 x y = 0$The differential equation in the question is a form of the Airy differential equation with the minus sign preceding $m^2$. Q3: Find the series solution for the following ordinary differential equation using the Frobenius method: 𝑥 𝑦 + 𝑥 𝑥 + 1 2 𝑦 − 𝑥 + 1 2 𝑦 = 0. 1 of 3 Go to page. In general, such a solution assumes a power series with unknown coefficients, then substitutes that solution into the differential equation to find a recurrence relation for the coefficients. order Power series solution of differential-algebraic equations in (1. In this lecture, we will study the solution of the second-order linear differential equations in terms of Power Series. Nonlinear Differential Equations Fourier series For a periodicfunction one may write The Fourier series is a "best fit" in the least square sense of data fitting y(t +T) =y(t) ()cos( ) sin( ), a plot of versus is called the power spectrum ∫ +∞ −∞. We also discuss more about initial conditions and how they determine the first two coefficients in the power series solution. Next enter the c value and view the Laplace transform below the entry box. The general idea is as follows: Assume that the solution function has a power series that converges to it. This method aims to find power series for the solution functions to a differential equation. Video # Video Tutorial Title: Remarks: 1: Classification of Differential Equations. We will only be able to do this if the point x = x0. Give the first four non-zero terms of each of the 2 independent solutions, y1. In this thesis, the reader will be made aware of methods for finding power series solutions to ordinary differential equations. S = dsolve(eqn) solves the differential equation eqn, where eqn is a symbolic equation. There is a Review Sheet (with Solutions). Module 25 - Parametric. But first: why?. 1 - Power Series; Lesson 22. It's more plug-and-chug and you should do well if you can match up the differential equation to the approach used to solve it. Topic: Differential Equations, Sequences and Series. Module 21 - Sequences and Series; Lesson 21. 2 Linear Ordinary Differential Equations with Constant Coefficients A174 A. In some cases, these power series representations can be used to find solutions to differential equations. The method works analogously for higher order equations as well as for systems. In the equation, represent differentiation by using diff. Convergent Power Series of sech ( x ) and Solutions to Nonlinear Differential Equations Article (PDF Available) in International Journal of Differential Equations 2018(1-2):1-10 · February. However, you can specify its marking a variable, if write, for example, y(t) in the equation, the calculator will automatically recognize that y is a function of the variable t. Taylor Series Calculator - an Introduction. The dsolve function finds a value of C1 that satisfies the condition. Let's nd a solution as a= 0. Hi and welcome back to the differential equations lectures here on www. Below is one of them. Power Series Solution to Non-Linear Partial Differential equations of Mathematical Physics. delay differential equations using the residual power series method (RPSM), which obtains. In the first part of this course, the student learns to solve the most common types of differential equations. Derivative Calculator Integral Calculator Limit Calculator. Fourier series with Ti84: Calculate for all different periodic signals the spectrum. , x0 = 0) (2) y( x) am ( x x0 )m a0 a1 ( x x0 ) a2 ( x x0 )2. What is more, we present the post-treatment of the power series. Access course-tailored video lessons, exam-like quizzes, mock exams & more for MATH 118 at Waterloo. The dsolve function finds a value of C1 that satisfies the condition. Example: t y″ + 4 y′ = t 2 The standard form is y t t. Differential Equation Calculator. Solved Examples of Differential Equations Sunday, July 9, 2017 Find the first 6 non-zero terms of the power series expansion about x = 0 for a general solution to the given differential equation y'' - x^2y' - xy = 0. Hi and welcome back to the differential equations lectures here on www. Simple Ordinary Differential Equations may have solutions in terms of power series whose coefficients grow at such a rate that the series has a radius of convergence equal to zero. The validity of term‐by‐term differentiation of a power series within its interval of convergence implies that first‐order differential equations may be solved by assuming a solution of the form. Let's study about the order and degree of differential equation. KEYWORDS: Course Materials, Lecture notes, Laboratories, HW Problems SOURCE: Joseph M. The unknown function is generally represented by a variable (often denoted y), which, therefore, depends on x. 2 using Taylor series method of order four. 100, 61111 Ljubljana, Slovenia and Racah Institute of Physics,. 2 Power Series, Analytic Functions, and the Taylor Series Method 431 8. Answer all the questions. This method aims to find power series for the solution functions to a differential equation. Differential equation or system of equations, specified as a symbolic equation or a vector of symbolic equations. Illustrative numerical example is included to demonstrate efficiency. ), probability functions, Taylor polynomials/series approximations, power series’, differential equations (linear and separable), partial derivatives, multivariable functions (and their real-world applications), and double integrals in. {image} {image} {image} {image} 3. Modeling with Differential Equations Solve a problem in the physical sciences (such as a growth or decay problem, a mixture problem, or a Newton’s Law of Cooling problem) whose solution utilizes a first-order linear differential equation. 3 - Taylor Series. The coefficient functions in these series are obtained by recursively iterating a simple integration process, begining with a solution system for $\lambda=0$. This paper presents a Modified Power Series Method (MPSM) for the solution of delay differential equations. By solving such equations, we mean computing a vector F of power series such that (1) holds modulo xN. In the equation, represent differentiation by using diff. 2) for all x implies, by the nth term test for diver-gence, that lim n→∞ n! = 0 (5. 5 Cauchy-Euler (Equidimensional) Equations Revisited 459. Here we have discussed an Ordinary and singular point for linear Second Order Differential Equations, classification of Singular Point and method to solve Differential Equations about an Ordinary Point. Print Book & E-Book. Solving Separable First Order Differential Equations – Ex 1 Homogeneous Second Order Linear Differential Equations Power Series Solutions of Differential Equations. try to explain any differences between the two forms of the solution. 1 - Sequences; Lesson 23. Stefan, Jamova 39, P. EXCHANGE RATE MISALIGNMENT AND CAPITAL INFLOWS: AN ENDOGENOUS THRESHOLD ANALYSIS FOR MALAYSIAABSTRACTThis study presents an attempt to investigate the impact of exchange rate misalignment on capital inflows in Malaysia. Find more Mathematics widgets in Wolfram|Alpha. $y^\prime = 2|x|$ is kind of an artificial example. HERMITE DIFFERENTIAL EQUATION - GENERATING FUNCTIONS Link to: physicspages home page. Why most of Hille's texts-which are all wonderful-are out of print mystifies me. Review of Series and Power Series. This Sage quickstart tutorial was developed for the MAA PREP Workshop "Sage: Using Open-Source Mathematics Software with Undergraduates" (funding provided by NSF DUE 0817071). However, you can specify its marking a variable, if write, for example, y(t) in the equation, the calculator will automatically recognize that y is a function of the variable t. Initialization. In a few upper div math classes, like differential equations, real and complex analysis you'll see them. In Introduction to Power Series , we studied how functions can be represented as power series, We also saw that we can find series representations of the derivatives of such functions by differentiating the power series term by term. It is interesting to know whether sympy supports such equations along with usual ones. Why most of Hille's texts-which are all wonderful-are out of print mystifies me. Initialization. Sympy: how to solve algebraic equation in formal. 5 Cauchy-Euler (Equidimensional) Equations Revisited 459. Equations Inequalities System of Equations System of Inequalities Polynomials Rationales Coordinate Geometry Complex Numbers Polar/Cartesian Functions Arithmetic & Comp. Differential equation. It is shown how to obtain such an initializing system working upwards from equations of lower order. Sage Quickstart for Differential Equations¶. He shared the 1994 Nobel Prize for economics with two other game theorists, Reinhard Selten and John Harsanyi. Order Differential Equations with non matching independent variables (Ex: y'(0)=0, y(1)=0 ) Step by Step - Inverse LaPlace for Partial Fractions and linear numerators. My class, and many other's, continue onto power series solutions of differential equations. Convergent Power Series of sech ( x ) and Solutions to Nonlinear Differential Equations Article (PDF Available) in International Journal of Differential Equations 2018(1-2):1-10 · February. Power Series Method for Linear Partial Differential Equations of Fractional Order 73 Lemma 2. Practice your math skills and learn step by step with our math solver. Solve y0 +(2x 1)y = 0 with initial conditions y(0) = 2. SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS— SOME WORKED EXAMPLES First example Let’s start with a simple differential equation: ′′− ′+y y y =2 0 (1) We recognize this instantly as a second order homogeneous constant coefficient equation. Finding coefficients in a power series expansion of a rational function. Answers to Solving Ordinary Differential Equations with Power Series Here are the answers to the practice questions I provide throughout this chapter. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. In fact, every linear meromorphic system has a formal solution of a certain form, which can be relatively easily computed, but which generally involves such power. Power series representations of functions can sometimes be used to find solutions to differential equations. Differential Equations. Mathematical concepts and various techniques are presented in a clear, logical, and concise manner. Learn AP®︎ Calculus BC for free—everything from AP®︎ Calculus AB plus a few extra goodies, such as Taylor series, to prepare you for the AP®︎ test. Initial conditions are also supported. This page covers two areas related to Fourier Series. We will only be able to do this if the point x = x0. If m 1 < 0 The Caputo fractional derivative is considered here because it allows traditional initial and boundary conditions to be included in the formulation of. Solving Separable First Order Differential Equations – Ex 1 Homogeneous Second Order Linear Differential Equations Power Series Solutions of Differential Equations. One-variable linear equations Calculator; One-variable linear inequalities Calculator; Operations with infinity Calculator; Perfect square trinomial Calculator; Polynomial factorization Calculator; Polynomial long division Calculator; Polynomials Calculator; Power of a product Calculator; Power rule Calculator; Power series Calculator; Powers. Homogeneous Differential Equations Calculator. equation, method of Laplace transforms for solving ordinary differential equations, series solutions (power series, Frobenius method); Legendre and Bessel functions and their orthogonal properties; Systems of linear first order ordinary differential equations. This RPS method gives approximate solutions in convergent series formula with surely computable components. 5 lecture , §3. 1 First order equations. The Organic. Module 23 - Parametric. In general, such a solution assumes a power series with unknown coefficients, then substitutes that solution into the differential equation to find a recurrence relation for the coefficients. In this work, we studied that Power Series Method is the standard basic method for solving linear differential equations with variable coefficients. MA 401 Applied Differential Equations II (Wave, heat and Laplace equations. I Check the endpoints, jx x 0j= ˆ, separately to nd the INTERVAL of CONVERGENCE. {image} {image} {image} {image} 3. You can view the Laplace Table that you will be given on the Exam. Practice 2610. Specifically, a precise threshold value is estimated to examine when exchange rate misalignment suppresses capital inflows. You can view the Laplace Table that you will be given on the Exam. This is simply a matter of plugging the proposed value of the dependent variable into both sides of the equation to see whether equality is maintained. The solution diffusion. But first: why?. John Forbes Nash Jr Essay John Forbes Nash Jr. lol this is a question you'll look back on after doing more math/physics and laugh. Example The differential equation ay00 +by0 +cy = 0 can be solved by seeking exponential solutions with an unknown exponential factor. Power Series; Method of series solutions; 6 The Laplace. After finding the constants a 2 ,a 3 ,a 4 , etc I replaced them in y(x) and factored out the undetermined coefficients a 0 and a 1. 7 Power series methods. using traditional way with pencil and paper. Department of Mathematics, Rasht Branch, Islamic Azad University, Rasht, Iran2 Abstract: In this article power series method, as well-known method for. The extension of fractional power series solutions for linear fractional differential equations with variable coefficients is considered. Since cos x = Σ(n=0 to ∞) (-1)^n x^(2n)/(2n)!, so u can get by potential of heart a million qn and remedy each and every) Non-linear partial differential equation, Homogenous and non-homogeneous. SOLVING DIFFERENTIAL EQUATIONS USING POWER SERIES 4 (2) Plug the expression (1) for y(x) into the di erential equation; (3) Manipulate the resulting equation to obtain an equation in which single power series expression (rather that a sum of several power series) is set equal to zero. In this work, we studied that Power Series Method is the standard basic method for solving linear differential equations with variable coefficients. We propose a power series extender method to obtain approximate solutions of nonlinear differential equations. 2, the power series method is used to derive the wave function and the eigenenergies for the quantum harmonic oscillator. THE METHOD OF FROBENIUS We have studied how to solve many differential equations via series solutions. The coe cient functions here are constants, so the power series solution can be computed at any point aand the radius of convergence will be R=1. Houston Math Prep 245,356 views. Conic Sections Trigonometry. After finding the constants a 2 ,a 3 ,a 4 , etc I replaced them in y(x) and factored out the undetermined coefficients a 0 and a 1. Stefan, Jamova 39, P. using traditional way with pencil and paper. Together we will learn how to express a combination of power series as a single power series. Applications of Fourier Series to Differential Equations Fourier theory was initially invented to solve certain differential equations. I used the power series method to solve the differential equation y''+y=0 with. So, the convergence of power series is fairly important. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. To pursue these objectives, this study relies on the. James Stewart's Calculus, Metric series is the top-seller in the world because of its problem-solving focus, mathematical precision and accuracy, and outstan. This particular number ρ is called the r adiu s of c onv er ge nc e. Equations Inequalities System of Equations System of Inequalities Polynomials Rationales Coordinate Geometry Complex Numbers Polar/Cartesian Functions Arithmetic & Comp. We will only be able to do this if the point x = x0. Question: Find two power series solutions of the given differential equations about the ordinary point {eq}x = 0 {/eq}. Solve differential equation using "Power Series" Thread starter EmilyL; Start date Jul 23, 2010; Tags differential differential equations equation power series solve; Home. ; Dig deeper into specific steps Our solver does what a calculator won't: breaking down key steps. HAFTEL Code 6651, Naval Research Laboratory, Washington, DC 20735-5345 R. Browse other questions tagged ordinary-differential-equations power-series or ask your own question. Motivation: Following this discussion about using asymptotic expansions (i. ODEs Summer08 Esteban Arcaute Introduction First Order ODEs Separation of Power Series Exact Equation End Thus, if the equation is exact, we have f(x,y) = c. Module 25 - Parametric. 307) than what Sal got by raising both sides to the power of e: 2. , in the form. KEYWORDS: Course Materials Calculus for Biology I ADD. Use power series to solve the differential equation. The chapter discusses a method by which the coefficients in the power series expansions of the solutions can be calculated. Those I have learnt in lecture and online are mostly with only one part of summation or two parts with two distinctive roots and two constants. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Linear and separable first order differential equations. Xavier Sigaud, 150 CEP 22290-180, Rio de Janeiro, RJ, Brazil. AN EXAMPLE. Up to 25 % of the generating costs relate to mainte- nance. Mathematics > Calculus and Analysis > Differential Equations Keywords Calculus, series expansion, Taylor Series, Ordinary Differential Equation, ODE, , Power Series. 4 Fourier series and PDEs. Substituting y = ert into the equation gives a solution if the quadratic equation ar2 +br+c = 0 holds. compare the series solutions with the solutions of the differential equation obtained using the method of section 4. Welcome! This is one of over 2,200 courses on OCW. Depending upon the domain of the functions involved we have ordinary differ-ential equations, or shortly ODE, when only one variable appears (as in equations (1. Let's consider the power series solution of the Hermite differential equation: ${\displaystyle u''-2xu'=-2 n u}$ ${\displaystyle u''-2xu'+2 n u =0 \qquad (1)}$ The solutions to the Hermite differential equation ca be expresse. EXAMPLE2 Power Series Solution Use a power series to solve the differential equation Solution Assume that is a. 5 lecture , §3. Solving the Systems of Differential Equations by a Power Series Method A. Conic Sections Trigonometry. An infinite series of this type is called a power series. form a fundamental system of solutions for Airy's Differential Equation. Shifting the Index for Power Series - Duration: 14:49. Textbook solution for Calculus: Early Transcendentals 8th Edition James Stewart Chapter 17. Such an expression is nevertheless an entirely valid solution, and in fact, many specific power series that arise from solving particular differential equations have been extensively studied and hold prominent places in mathematics and physics. ; Dig deeper into specific steps Our solver does what a calculator won't: breaking down key steps. 3 - Second-Order Differential Equations. We will solve this using power series technique. Recently, the first author applied the power series method to studying the Hyers-Ulam stability of several types of linear differential equations of second order (see [26-34]). Worked example: exponential solution to differential equation. Solving Differential Equations with Power Series - Duration: 18:29. Do the differential equation solvers - Support ordinary differential equations; systems of differential equations, and boundary value problems both at the command line and in solve blocks that use natural notation to specify the DiffEQs and constraints. 4 The Power Series Method, Part I A187 A. Finding coefficients in a power series expansion of a rational function. For example, here's a differential equation […]. Let's consider the power series solution of the Hermite differential equation: ${\displaystyle u''-2xu'=-2 n u}$ ${\displaystyle u''-2xu'+2 n u =0 \qquad (1)}$ The solutions to the Hermite differential equation ca be expresse. In the cases where series cannot be reduced to a closed form expression an approximate answer could be obtained using definite integral calculator. lol this is a question you'll look back on after doing more math/physics and laugh. For example, diff(y,x) == y represents the equation dy/dx=y. Find the Taylor series expansion of any function around a point using this online calculator. We will only be able to do this if the point x = x0. 3 - Recursively Defined Sequences. And find the power series solutions of a linear first-order differential equations whose solutions can not be written in terms of familiar functions such as polynomials, exponential or trigonometric functions, as SOS Math so nicely states. The Bessel differential equation has the form x 2 y+xy'+(x 2-n 2)y=0. Around the Point a = (default a = 0) Maximum Power of the Expansion: How to Input. 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 50 49 48 47 46 45 44 43 42 41. Definition 5. ’s need to be. Introduction to Differential Equations. Write y' as dy/dx and the answer follows relatively - dy/dx + 3(a million+ x^2)y = 0 => dy/dx = -3(a million+x^2)y => dy/y = -3(a million+x^2)dx Now integrating the two components supplies ln(y) = -3x - x^3 + C (C is unknown integration consistent) => y = ok exp(-3x - x^3) the place ok = exp(C), and can be solved utilising extra suitable education approximately y (as an occasion the fee of y. In advanced treatments of calculus, these power series representations are often used to define the exponential. I used the power series method to solve the differential equation y''+y=0 with y[0]=0 and y'[0]=1 using the following code. Print Book & E-Book. 5 lecture , §3. Linear methods applied to the solution of differential equations. The term "ordinary" is used in contrast with the term. Module 25 - Parametric. The Overflow Blog This week, #StackOverflowKnows molecule rings, infected laptops, and HMAC limits. Differential equation or system of equations, specified as a symbolic equation or a vector of symbolic equations. and then try to determine what the an. 4 Equations with Analytic Coefficients 453 *8. Back to top; 6: Power Series Solutions of Differential Equations; 6. Differential equation. In the first part of this course, the student learns to solve the most common types of differential equations. It is interesting to know whether sympy supports such equations along with usual ones. Even if this is the case, for simplicity we will see how the method works with a problem whose solution is a known function. Pourhabib Yekta1, A. Return to Differential Equations. 0012 Power Series Solution of Coupled Differential Equations in One Variable M. 1 in [EP], §5. MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. equation, method of Laplace transforms for solving ordinary differential equations, series solutions (power series, Frobenius method); Legendre and Bessel functions and their orthogonal properties; Systems of linear first order ordinary differential equations. In the previous solution, the constant C1 appears because no condition was specified. Solve y0 = x2y with initial conditions y(0) = 1. The degree of an equation is the power to which the highest order term is raised. This paper presents a Modified Power Series Method (MPSM) for the solution of delay differential equations. which makes calculations very simple and interesting. 3 - Recursively Defined Sequences. We will then move to a problem whose solution can be expressed as a series only. Derivatives Derivative Applications Limits Integrals Integral Applications Series ODE Laplace Transform Taylor/Maclaurin Series Fourier Series. In the first part of this course, the student learns to solve the most common types of differential equations. S = dsolve(eqn) solves the differential equation eqn, where eqn is a symbolic equation. Various visual features are used to highlight focus areas. 1) The equation is linear of second order with polynomial coecients. EXAMPLE 2 Power Series Solution Use a power series to solve the differential equation Solution Assume that is. Notice that 0 is a singular point of this differential equation. A power series represents a function f on an interval of convergence, and you can successively differentiate the power series to obtain a series for and so. The general Airy differential equation is given by :$D^2y \pm m^2 x y = 0$or equivalently$y’’ \pm m^2 x y = 0$The differential equation in the question is a form of the Airy differential equation with the minus sign preceding $m^2$. Power Series Method for Nonlinear Partial Differential Equations Power series is an old technique for solving linear ordinary differential equations [7,20]. You can view the Laplace Table that you will be given on the Exam. 0012 Power Series Solution of Coupled Differential Equations in One Variable M. Power Series Solutions Differential Equations Power Series Solutions Differential Equations Yeah, reviewing a ebook Power Series Solutions Differential Equations could accumulate your near associates listings. Review of Series and Power Series. In this video we solve another differential equation by finding a power series solution. First order numerical / graphical differential equation solver: Transient analysis of RC or RL circuits. The solution diffusion. Each of the following waveform plots can be clicked on to open up the full size graph in a separate window. Solve y00 = xy0 +y with initial conditions y(0) = 1 and y0(0) = 0. AMS30, 151-156 (1971). Covers material on integration methods (trig, partial fractions, etc. The solution of the general differential equation dy/dx=ky (for some k) is C⋅eᵏˣ (for some C). This equation with concrete values of the parameter appeared in the articles by F. DIFFERENTIAL EQUATIONS FOR ENGINEERS This book presents a systematic and comprehensive introduction to ordinary differential equations for engineering students and practitioners. We have step-by-step solutions for your textbooks written by Bartleby experts!. Laplace’s transformation maps a differential equation onto an algebraic equation, which can be solved relatively easily. Home » Supplemental Resources » Calculus Revisited: Complex Variables, Differential Equations, and Linear Algebra » Part II: Differential Equations » Lecture 6: Power Series Solutions Lecture 6: Power Series Solutions. SERIES SOLUTIONS OF DIFFERENTIAL EQUATIONS— SOME WORKED EXAMPLES First example Let’s start with a simple differential equation: ′′− ′+y y y =2 0 (1) We recognize this instantly as a second order homogeneous constant coefficient equation. Find materials for this course in the pages linked along the left. Series of Solutions Review of Power Series The Recurrence Power Solutions about an Ordinary Point Euler Equations Series Solutions Near a Regular Singular Point Equations of Hypergeometric Type Bessel’s Equations Legendre’s Equation Orthogonal Polynomials Review Questions for Chapter 6 Applications of Higher Order Differential Equations. Delay Differential Equations, Power Series, Taylor Series, Newton's Method 1. 7MB) To complete the reading assignments, see the Supplementary Notes in the Study Materials section. First‐order equations. In the next section, Section 8. Type: Artigo de periódico: Title: Group Classification Of A Generalized Black-scholes-merton Equation: Author: Bozhkov Y. This is a simple example and the final solution is very nice compared to what would normally happen with a more complicated differential equation, so please be aware of that!. order Power series solution of differential-algebraic equations in (1. My longest video yet, power series solution to differential equations, solve y''-2xy'+y=0, www. which makes calculations very simple and interesting. Power Series Method for Linear Partial Differential Equations of Fractional Order 73 Lemma 2. polynomial differential equations has a solution of this form. Differential equation. The ideas that you guess a power series solution to a differential equation and then you plug it in and in order to plug it in, you got to calculate its derivatives. First order numerical / graphical differential equation solver: Transient analysis of RC or RL circuits. The governing equation is also based on Kirchoff's law as described below. Back to top; 6: Power Series Solutions of Differential Equations; 6. Applications of first and second order equations; Sequences and infinite series. Write y' as dy/dx and the answer follows relatively - dy/dx + 3(a million+ x^2)y = 0 => dy/dx = -3(a million+x^2)y => dy/y = -3(a million+x^2)dx Now integrating the two components supplies ln(y) = -3x - x^3 + C (C is unknown integration consistent) => y = ok exp(-3x - x^3) the place ok = exp(C), and can be solved utilising extra suitable education approximately y (as an occasion the fee of y. {image} {image} {image} {image} 3. Making statements based on opinion; back them up with references or personal experience. Research Article Power Series Extender Method for the Solution of Nonlinear Differential Equations HectorVazquez-Leal 1 andArturoSarmiento-Reyes 2 Electronic Instrumentation and Atmospheric Sciences School, Universidad Veracruzana, Cto. Yes, y(x) is the general solution of the differential equation represented as a power series. Example: an equation with the function y and its derivative dy dx. 3, Issue 4, April 2014 Solving the Systems of Differential Equations by a Power Series Method A. Differential Equations. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. Finding coefficients in a power series expansion of a rational function. 2 using Taylor series method of order four. Even if this is the case, for simplicity we will see how the method works with a problem whose solution is a known function. The TI-89 cannot solve second order linear differential equations with variable coefficients. This is the currently selected item. Solve the equation with the initial condition y(0) == 2. A differential equation (DE) is an equation relating a function to its derivatives. 1) The equation is linear of second order with polynomial coecients. Tìm kiếm power series solution of differential equations calculator , power series solution of differential equations calculator tại 123doc - Thư viện trực tuyến hàng đầu Việt Nam. Use MathJax to format equations. This note explains the following topics: First-Order Differential Equations, Second-Order Differential Equations, Higher-Order Differential Equations, Some Applications of Differential Equations, Laplace Transformations, Series Solutions to Differential Equations, Systems of First-Order Linear Differential Equations and Numerical Methods. This gives a recurrence formula for the coefficients. In the equation, represent differentiation by using diff. EXCHANGE RATE MISALIGNMENT AND CAPITAL INFLOWS: AN ENDOGENOUS THRESHOLD ANALYSIS FOR MALAYSIAABSTRACTThis study presents an attempt to investigate the impact of exchange rate misalignment on capital inflows in Malaysia. Power series solution (PSS) method is an old method that has been limited to solve linear differential equations, both ordinary differential equations (ODE) [1, 2] and partial differential equations (PDE) [3, 4]. m 0 Such a solution is obtained by substituting (2) and its derivatives into (1). = 1 for y at x = 1 with step length 0. For Example (i): $$\frac{d^3 x}{dx^3} + 3x\frac{dy}{dx} = e^y$$ In this equation the order of the highest derivative is 3 hence this is a third order. Next enter the c value and view the Laplace transform below the entry box. General Differential Equation Solver. Course summary; Differential equations Verifying solutions for differential equations: Series Power series intro: Series Function as a geometric series: Series Maclaurin series of eˣ, sin(x), and cos(x): Series Representing functions as power series: Series Telescoping series: Series Proof videos: Series. 2 - Series and Sequences of Partial Sums; Lesson 23. Initial conditions are also supported. Thanks for contributing an answer to Mathematics Stack Exchange! Please be sure to answer the question. 2 Power Series Section 5. compare the series solutions with the solutions of the differential equation obtained using the method of section 4. Calculator: A calculator such as TI83/84 plus will be allowed to use on tests and final exam. The first differential equation, , is rather easy to solve, we simply integrate both sides. Convergence of Sequences; Convergence of series using geometric series, the comparison tests, the alternating series test, the root test, and the ratio test. y'' − y' = 0. MA 401 Applied Differential Equations II (Wave, heat and Laplace equations. First, we present an introduction to Fourier Series, then we discuss how to solve differential equations using Fourier Series. In this work we present a power series method for solving ordinary and partial differential equations. Every project on GitHub comes with a version-controlled wiki to give your documentation the high level of care it deserves. It is licensed under the Creative Commons Attribution-ShareAlike 3. Exact Differential Equation Non-Exact Differential Equation M(x,y)dx+N(x,y)dy=0 N(x,y)y'+M(x,y)=0 Linear in x Differential Equation Linear in y Differential Equation RL Circuits Logistic Differential Equation Bernoulli Equation Euler Method Runge Kutta4 Midpoint method (order2) Runge Kutta23 2. 03/26/18 - We propose a computational method to determine when a solution modulo a certain power of the independent variable of a given algeb. The general Airy differential equation is given by :$D^2y \pm m^2 x y = 0$or equivalently$y’’ \pm m^2 x y = 0$The differential equation in the question is a form of the Airy differential equation with the minus sign preceding $m^2$.
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Can I use math markup in code blocks ? Apparently, many others ( 1, 2, 3 ), have asked the same question. Using HTML is also helpful when you need to change the attributes of an element, like specifying the color of text or changing the width of an image. Markdown. plain text: ‰ (does render properly in PDF, does in Word) HTML: ‰ (does renders properly in PDF, does in Word) ex) $\sigma = \frac{1/\lambda}{\sqrt{n}}$ r math markdown. Microsoft Word does support large sized brackets, so I am unsure if this is expected behavior or a bug. It can be used on some websites like Stack Overflow or to write documentations (essentially on GitHub). share | improve this answer | follow | answered Mar 31 '19 at 18:18. itsmysterybox itsmysterybox. Turn your analyses into high quality documents, reports, presentations and dashboards with R Markdown. I have a bookdown book with a lot of greek symbols inline. Use multiple languages including R, Python, and SQL. 775 1 1 gold badge 8 8 silver badges 19 19 bronze badges. Yes. R Markdown adds a few features which include R code and results in the formatted document. This allow you write documents which integrate results from your analysis. Use the Markdown: Clip Markdown+Math to Html command or the key binding Ctrl+K.. If I render to gitbook or other HTML it looks as expected. 2.1 Overview; 2.2 Questions; 2.3 Objectives; 2.4 Your Turn; 2.5 Reproducibility is a problem; 2.6 Literate programming is a partial solution; 2.7 Markdown as a new player to legibility. The first official book authored by the core R Markdown developers that provides a comprehensive and accurate reference to the R Markdown ecosystem. Can I use math markup in code blocks ? I need a tool that convert equation that convert into r markdown rules. Many Markdown applications allow you to use HTML tags in Markdown-formatted text. See all the Latex Math Symbols and difference between dot characters. Scroll down to #List of mathematical symbols for a complete list of Greek symbols. Provide details and share your research! List of Greek letters and math symbols. Ideally, what we want a free, cross-platform Markdown editor that comes packaged with the ability to input mathematical symbols, has support for tables, and allows for exporting as PDF. Da es praktisch unmöglich ist, alle jemals in der Mathematik verwendeten Symbole aufzuführen, werden in dieser Liste nur diejenigen Symbole angegeben, die häufig im Mathematikunterricht oder im Mathematikstudium auftreten. After reading this book, you will understand how R Markdown documents are transformed from plain text and how you may customize nearly every step of this processing. This book showcases short, practical examples of lesser-known tips and tricks to helps users get the most out of these tools. Use TinyTex. 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https://math.stackexchange.com/questions/1529339/domain-of-the-n-composed-logarithms-on-x | # Domain of the n composed logarithms on x.
As we know, the real logarithm has the domain
$$D_1 = \{x : x \in \mathbb{R}, x > 0\}$$
What is the logarithmic domain of "higher order" logarithms, at index n? For example, it seems that
$$D_2 = \{x : x \in \mathbb{R}, x > 1\}$$
which would be the domain of
$$f(x) = \log(\log(x))$$
$D_3$ is a little hard to imagine. I think my real question deals with formalization, and the extended case of $D_n$ however.
There may be connections to the iterated logarithm, which says how many iterations are necessary before a value breaks in a certain base. See the wikipedia article.
• Yes, there is a connection. Which base did you have in mind? – hardmath Nov 14 '15 at 23:08
• base e? I was kind of asking abstractly, if you have an answer in an arbitrary base that would be impressive and helpful. – theREALyumdub Nov 14 '15 at 23:08
• Hint: If $\ln^{(n)}(x)$ denotes the iterated logarithm and ${^n}e=e^{e^{\ldots{e}}}$ ($n$-times) denotes tetration, then the following identity holds for all $n\in\mathbb{N}$: $\ln^{(n)}({^n}e)=1$. – Yiannis Galidakis Nov 14 '15 at 23:34
• @YiannisGalidakis, do you mean the super logarithm (inverse of tetration)? – theREALyumdub Nov 14 '15 at 23:37
• I think so, yes, although I am not very familiar with the actual definition Andrew Robbins uses. It's bullet 3 on that page :As the number of times a logarithm must be iterated to get to 1 (the Iterated logarithm) @theREALyumdub – Yiannis Galidakis Nov 14 '15 at 23:42
If you denote by $\ln^{(n)}(x)$ the iterated logarithm and by ${^n}e=e^{e^{\ldots^{e}}}$ (height $n$) iterated exponentiation of the base $e$ (as per the comment), we have by definition:
$$\ln^{(n)}({^n}e)=1\Rightarrow$$ $$\ln^{(n+1)}({^n}e)=\ln(1)=0$$
Apply for $n=1$ and we get:
$$\ln(\ln(e))=0$$
So $D_2$ should be $D_2=\{x\in\mathbb{R}\colon x\ge e\}$. This however breaks the pattern, because the range of $\ln(\ln(x))$ can be negative for this case, if we extend the domain of $x$ to be $x\gt 1$.
You can't do this for higher iterates however, because negative ranges are not allowed in the domain of $\ln$. Therefore:
$$D_1=\{x\in\mathbb{R}\colon x\gt 0\}$$ $$D_2=\{x\in\mathbb{R}\colon x\gt 1\}$$ $$D_3=\{x\in\mathbb{R}\colon x\ge {^2}e\}$$ $$D_4=\{x\in\mathbb{R}\colon x\ge {^3}e\}$$
and in general ($n\ge 3$):
$$D_n=\{x\in\mathbb{R}\colon x\ge {^{n-1}}e\}$$
• Much more formal than my answer; I was so busy trying to edit mine to be correct that I missed this. – theREALyumdub Nov 15 '15 at 1:35
Actually, now that I've looked into it, the table on the page seems to suggest that the iterated logarithm values are the way you compute the bounds on the $D_n$ domains, and it switches at the tetration integer values.
## The tetration operation on the base defines the bounds of the domains $D_n$.
Let b be the base of the logarithm in question, so that $f_1(x) = \log_b(x)$
We can say
$$D_1 = \{x : x \in \mathbb{R}, x > 0\}$$ $$D_2 = \{x : x \in \mathbb{R}, x > b\}$$
And for n > 2, we apply the iterated logarithm to say that
$$D_3 = \{x : x \in \mathbb{R}, x > b^b\}$$ $$D_4 = \{x : x \in \mathbb{R}, x > (b^b)^b\}$$ $$...$$ $$D_n = \{x : x \in \mathbb{R}, x > tetra(b,n - 1)\}$$
• Yes, that looks ok. – Yiannis Galidakis Nov 15 '15 at 0:03
• The exponent should probably be $n-1$. Your $D_2$ is wrong, since $\ln(\ln(1))$ has problems. It should be $x>e$ for base $e$. Then $D_n=\{x:x\in\mathbb{R},x>tetra(e,n-1)\}$ for base $e$. – Yiannis Galidakis Nov 15 '15 at 0:14
• Does that look better? – theREALyumdub Nov 15 '15 at 1:32
• Your $D_2$ needs correction. It is $x>1$ (it is the only one that breaks the pattern, see my answer. My comment above has a typo) and when you notate tetration it's prettier to stick to one of the two forms you have: $tetra(b,3)=b^{b^{b}}$, but parentheses go from top to bottom. So it'd be $b^{(b^b)}$, $b^{(b^{(b^b)})}=tetra(b,4)$, etc. – Yiannis Galidakis Nov 15 '15 at 1:55 | 2019-04-23T18:31:59 | {
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https://math.stackexchange.com/questions/3594010/a-five-digit-number-minus-a-four-digit-number-equals-33333-what-are-the-two-n | # A five digit number minus a four digit number equals $33333$. What are the two numbers, if you are only allowed to use the numbers $1-9$
Question: A five digit number minus a four digit number equals $$33333$$. What are the two numbers, if you are only allowed to use the numbers $$1-9$$ once? More precisely, \begin{align} & & A_1 \, A_2 \, A_3 \, A_4 \, A_5 \\ & - & A_6 \, A_7 \, A_8 \, A_9 \\ & & \hline 3\,\,\,\,\,3 \,\,\,\,\,3 \,\,\,\,\,\,3 \,\,\,\,\,\,3\\ & & \hline \end{align} where $$A_1,A_2,A_3,A_4,A_5, A_6, A_7, A_8, A_9 \in \{1,2,3,4,5,6,7,8,9\}$$ and they form a pairwise distinct set.
For me, I would guess $$A_1=3$$ or $$A_1 = 4.$$ But that is all I got. I am interested to know its thought process.
• Simplify the problem: A two digit number minus a one digit number equals $33$. What do you notice? Then move on to three digit minus two digit and so on. What are your thoughts on this? Mar 25 '20 at 4:51
• A quick Google search returns this reddit post from 6 years ago; it contains two answers but no solution. Mar 25 '20 at 4:53
• @AndrewChin For simpler problem like A two digit number minus a one digit number equals 33, what digits can I use? Mar 25 '20 at 5:16
• Note that these are digits, not numbers. Also, an individual $A_i$ cannot be pairwise distinct; it is the $A_i$ in the plural that are pairwise distinct. Mar 25 '20 at 6:50
• @joriki I edited my question, Is it okay now? Mar 25 '20 at 6:52
The sum of the digits $$1$$ through $$9$$ is odd. They contribute to the parity of the digit sum of the result no matter which row they’re in. The digit sum of the result is odd. Thus there must be an even number of borrowings.
A column that causes borrowing must have a $$7$$, $$8$$ or $$9$$ in the bottom row, so we cannot have four borrowings.
On the other hand, if there were no borrowing at all, the possible pairs in a column would be $$9-6-3$$, $$8-5-2$$ and $$7-4-1$$, but we can use at most one from each of these three groups.
It follows that there are exactly two borrowings. Thus the difference between the digit sums of the rows must be $$5\cdot3-2\cdot9=-3$$, and since the sum of all digits is $$\frac{9(9+1)}2=45$$, the top row must sum to $$21$$ and the bottom row to $$24$$.
We need to have exactly two of $$7$$, $$8$$ and $$9$$ in the bottom row to cause the two borrowings.
It can’t be $$7$$ and $$8$$ because then $$7$$ would have to be subtracted from $$1$$ and $$8$$ from $$2$$, so the two borrowing columns would have to be the two lending columns.
If it were $$8$$ and $$9$$, that would leave a sum of $$7$$ for the bottom row, so that could be $$3,4$$ or $$2,5$$ or $$1,6$$. It can’t be $$3,4$$ because one of those needs to be $$A_1$$; it can’t be $$2,5$$ because $$5$$ would need to be subtracted from $$8$$ or $$9$$; and it can’t be $$1,6$$ because $$6$$ would need to be subtracted from $$9$$.
Thus $$7$$ and $$9$$ are in the bottom row. That leaves a sum of $$8$$ for the bottom row, which could be $$3,5$$ or $$2,6$$. But it can’t be $$2,6$$, again because $$6$$ would need to be subtracted from $$9$$.
Thus we have $$3,5,7,9$$ in the bottom row and $$1,2,4,6,8$$ in the top row. So $$4$$ must be $$A_1$$, $$7$$ must be subtracted from $$1$$, $$9$$ from $$2$$, $$3$$ from $$6$$ and $$5$$ from $$8$$. Thus the lenders must be $$4$$ and $$1$$, so the top row must start $$412$$. That leaves two possibilities for the order of the last two columns, so there are two solutions:
41286 41268
-7953 and -7935
----- -----
33333 33333
The solutions are confirmed by this Java code. (Full disclosure: I initially made a mistake in the proof and wrote the code to find it, so I knew the solution before I completed the proof.)
• I do not fully understand the first paragraph. Why would the sum of digits from $1$ to $9$ contribute to the parity of the digit sum of the result? Mar 25 '20 at 12:15
• If you ignore borrowing for the time being, the digit sum of the result is a sum of the digits from $1$ to $9$, each with either a $+$ or a $-$. The sign makes no difference; in either case an odd digit contributes $1$ to the parity and an even digit contributes $0$. Mar 25 '20 at 12:19
• Can you eplain 'Thus the difference between the digit sums of the rows must be $5\cdot3-2\cdot9=-3$'. Where do we get $5\cdot3$ and $2\cdot9$? Mar 27 '20 at 7:02
• @Idonknow: There are five $3$s in the result, so the digit sum of the result is $5\cdot3$. If there were no borrowing, the digit sum of the result would be the difference of the digit sums of the rows. Each borrowing subtracts $1$ from the higher digit and adds $10$ to the lower digit, so the net effect is to add $9$ per borrowing. We found that there are exactly two borrowings. Thus the digit sum of the result is $2\cdot9$ more than the difference between the digit sums of the rows. Mar 27 '20 at 7:06 | 2022-01-28T17:54:04 | {
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http://greenbirdit.com/kmqv9o/tth1uxx.php?id=exponential-distribution-standard-deviation-643a98 | $\endgroup$ – André Nicolas Apr 30 '11 at 18:58 $\begingroup$ @shino: Or else if you are doing everything correctly, and exponential is a poor fit, look for a better fit from one of the Weibull distributions. The probability density function is $$f(x) = me^{-mx}$$. distribution is a discrete distribution closely related to the binomial distribution and so will be considered later. First consider λ = 1 λ = 1. Here e is the mathematical constant e that is approximately 2.718281828. Distributions with CV < 1 (such as an Erlang distribution ) are considered low-variance, while those with CV > 1 (such as a hyper-exponential distribution ) are considered high-variance [ … The ‘moment generating function’ of an exponential random variable X for any time interval t<λ, is defined by; M X (t) = λ/λ-t 0.5. c. 0.25. d. 2.0. e. the means of the two distributions can never be equal. Scientific calculators have the key “e … Therefore, X ~ Exp(0.25). Sample means from an exponential distribution do not have exponential distribution. The exponential distribution is one of the widely used continuous distributions. To describe the time between successive occurrences when all occurrences follow an exponential. μ = σ. Study notes and guides for Six Sigma certification tests. In fact, the mean and standard deviation are both equal to A. f ( x) = e-x/A /A, where x is nonnegative. Exponential Distribution Modelling Of Wet-day Rainfall Totals Assume An Exponential Distribution Can Be Used To Model Precipitation Totals On Wet Days. (Taken from ASQ sample Black Belt exam.). The distribution notation is X ~ Exp(m). Required fields are marked *. Exponential distribution is the time between events in a Poisson process. Exponential distribution is the time between events in a Poisson process. Construct a histogram of the dat The mean or expected value of an exponentially distributed random variable X with rate parameter λ is given by Simply, it is an inverse of Poisson. Log in or Sign up in seconds with the buttons below! Exponential Distribution Variance. http://www.public.iastate.edu/~riczw/stat330s11/lecture/lec13.pdf, Your email address will not be published. The amount of time (may be in months) a car battery lasts. Learn how your comment data is processed. μ = σ. 100% of candidates who complete my study guide report passing their exam! Therefore, $$X \sim Exp(0.25)$$. This section requires you to be logged in. That is, for an Exponential distribution, the mean and the standard deviation are equal, and equal to the reciprocal of the rate parameter. Questions, comments, issues, concerns? E ( X k) = ∫ 0 ∞ x k e − x d x = k! The mean of exponential distribution is 1/lambda and the standard deviation is also 1/lambda. This section requires you to be a Pass Your Six Sigma Exam member. (. it describes the inter-arrival times in a Poisson process.It is the continuous counterpart to the geometric distribution, and it too is memoryless.. This statistics video tutorial explains how to solve continuous probability exponential distribution problems. It can be shown for the exponential distribution that the mean is equal to the standard deviation; i.e., μ = σ = 1/λ Moreover, the exponential distribution is the only continuous distribution that is I have seen this question on one of the websites (I guess ASQ, not sure). 16. Therefore, X ~ Exp(0.25). IASSC Lean Six Sigma Green Belt Study Guide, Villanova Six Sigma Green Belt Study Guide, IASSC Lean Six Sigma Black Belt Study Guide, Villanova Six Sigma Black Belt Study Guide, Where e is base natural logarithm = 2.71828. I’ll investigate the … If it is a negative value, the function is zero only. $$\mu = \sigma$$ The distribution notation is $$X \sim Exp(m)$$. Your instructor will record the amounts in dollars and cents. If the number of occurrences follows a Poisson distribution, the lapse of time between these events is distributed exponentially. Login to your account OR Enroll in Pass Your Six Sigma Exam. It is a number that is used often in mathematics. The exponential distribution can be used to determine the probability that it will take a given number of trials to arrive at the first success in a Poisson distribution; i.e. This site uses Akismet to reduce spam.
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System Engineering Requirements Template, Psalm 111 Kjv Audio, Bugs On Monstera, Spinach Is A Of A Spinach Plant, Raw Egg Ice Cream Recipe, Ffxiv Important Side Quests, | 2021-04-16T16:31:31 | {
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https://math.stackexchange.com/questions/1075054/how-can-i-write-this-power-series-as-a-power-series-representation | # How can I write this power series as a power series representation?
How can I write this power series ($1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8....$) as a power series representation (like $\dfrac{1}{1-x}$ or something neat like that)?
• This is S(1+x) where $S=1+2x^2+3x^4+...$ for a start. – Paul Dec 19 '14 at 23:18
• @paul: What about x? – Mathy Person Dec 19 '14 at 23:21
• @Paul Saying $(1+x)S(x)$ might be more clear. – Tim Raczkowski Dec 19 '14 at 23:25
Hint: using $y=x^2$ and derivative in $y$: $$(1+x)(1+2x^2+3x^4+\ldots)$$ $$=(1+x)(1+2y+3y^2+4y^3 +\ldots)$$ $$= (1+x)(y+y^2+y^3+y^4+\ldots)'$$ $$= (1+x)\left( \frac{y}{1-y}\right)'$$
Edit: $$= (1+x) \frac{1}{(1-y)^2}$$ $$= \frac{1+x}{(1-x^2)^2}$$ $$= \frac{1}{(1-x)(1+x^2)}.$$
• I believe $(y+y^2+y^3+y^4+....)$ = $\frac{1}{1-y}$, if I am not mistaken. Then is it: $\frac{1+x}{1-y}$? – Mathy Person Dec 19 '14 at 23:28
• Oh wait, I see that you had $\frac{y}{1-y}$ instead. Is it: $\frac{(1+x)(y)}{1-y}$? – Mathy Person Dec 19 '14 at 23:30
• And $\frac{(1+x)(y)}{1-y}$, and $y=x^2$, then $\frac{(1+x)(x^2)}{1-x^2}$? – Mathy Person Dec 19 '14 at 23:30
• Simplifying would result in: $\frac{x^2}{1-x}$? – Mathy Person Dec 19 '14 at 23:31
• Just take derivative in y of $y/(1-y)$, then replace $y$ with $x^2$. – ir7 Dec 19 '14 at 23:33
I would go about this by first splitting the series up:
$$1+x+2x^2+2x^3+3x^4+3x^5+...=(1+x)(1+2x^2+3x^4+...)$$
Letting $s=1+2x^2+3x^4$ we can do a few tricks:
$$s-x^2s=\begin{array}{c} 1&+2x^2&+3x^4+... \\ &-x^2&-2x^4-...\end{array}$$ $$=1+x^2+x^4+...$$
Which converges to $\frac{1}{1-x^2}$ for $-1 < x < 1$ (proving this is not hard, and can be done by a technique like the above). This gives
$$s -x^2s=\frac{1}{1-x^2}\Leftrightarrow s=\frac{1}{(1-x^2)^2}=$$
Thus the original series converges to:
$$(1+x)s=\frac{(1+x)}{(1-x^2)^2}$$
For $-1 < x - 1$.
• Hmm..I got $\frac{x^2}{1-x}$ (see what I posted in reply to ir7's comment). Did I make a mistake somewhere? – Mathy Person Dec 19 '14 at 23:33
• Whoops my bad, I forgot a step – SBareS Dec 19 '14 at 23:36
• Apparently I am to tired for maths right now... – SBareS Dec 20 '14 at 0:05
Unless I'm mistaken, it is $$\sum_{n=1}^\infty nx^{2n-2} + \sum_{n=1}^\infty nx^{2n-1}$$ If you can compute one of the two terms, e.g. $\sum_{n=1}^\infty nx^{2n-2} = \sum_{n=0}^\infty (n+1)x^{2n} = \sum_{n=0}^\infty (n+1)(x^{2})^n$ (see e.g. this, then you'll also get the other term (by multiplying it by $x$), and thus the sum.
One way is to look at $$1+2x^2+3x^4+4x^6+5x^8+\dots$$ as $$1+2t+3t^2+4t^3+5t^4+\dots$$ where $t=x^2$. The last series is the derivative of $$1+t+t^2+t^3+t^4+t^5+\dots=\frac1{1-t}$$ Therefore, $$1+2t+3t^2+4t^3+5t^4+\dots=\frac1{(1-t)^2}$$ and $$1+2x^2+3x^4+4x^6+5x^8+\dots=\frac1{(1-x^2)^2}$$ Now, just multiply by $1+x$: \begin{align} 1+x+2x^2+2x^3+3x^4+3x^5+4x^6+4x^7+5x^8+5x^9+\dots &=\frac{1+x}{(1-x^2)^2}\\ &=\frac1{(1-x)(1-x^2)} \end{align} | 2019-12-12T19:23:08 | {
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http://math.stackexchange.com/questions/284513/how-to-expand-binomials?answertab=active | # How to expand binomials?
I'm working on a few proofs and am missing how this algebra works....
So, how does one expand $(k+1)^3\,$? Can I use FOIL? What does it expand to?
And how to expand $(k+1)^5\,$?
Thanks!
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look up Pascal's Triangle, and Binomial Theorem. Also no, you can't use FOIL, at very least you must use distributive property. – Joseph Skelton Jan 22 '13 at 20:59
Check out the entry Binomial Theorem in Wikipedia.
Putting $y = 1$, that will give you the tools you need to expand $(k+1)^3,\; (k+1)^5, \;$ and $\,(k + 1)^n\,$ for any non-negative integer $n$.
FOIL works fine for $(k + 1)^2 = k^2 + 2k + 1$
One can go a step further by distributing $(k+1)$ over $(k^2 + 2k + 1)$ to get $$(k^3 + 3k^2 + 3k + 1) = (k + 1)^3.$$
But for large exponents, it's handy to know the pattern of coefficients that correspond to different powers of $k$ in the expansion of $(k+1)^n$: Pascal's triangle shows this handy relationship.
I'll include an animation and image of "Pascal's Triangle" which displays the coefficients of expansions of a binomial $(k + 1)$ (these coefficients are referred to as: binomial coefficients): up to and including fourth and fifth degree binomials, respectively:
$\quad\quad\quad\quad$ $\quad\quad\quad\quad$
$$\text{Each number in the triangle is the sum of the two directly above it.}$$
To see how this "plays out" in the expansion of $(x + 1)^n,\;0 \le n \le 6$:
$$(x + 1)^0 = \color{blue}{\bf{1}}$$ $$(x + 1)^1 = \color{blue}{\bf{1}}\cdot x +\color{blue}{\bf{1}}$$ $$(x + 1)^2 = \color{blue}{\bf{1}}\cdot x^2 + \color{blue}{\bf{2}}x + \color{blue}{\bf{1}}$$ $$(x+1)^3 = \color{blue}{\bf{1}}\cdot x^3 + \color{blue}{\bf{3}}x^2 + \color{blue}{\bf{3}}x + \color{blue}{\bf{1}}$$ $$(x+1)^4 = \color{blue}{\bf{1}}\cdot x^4 + \color{blue}{\bf{4}} x^3+ \color{blue}{\bf{6}}x^2 + \color{blue}{\bf{4}}x +\color{blue}{\bf{1}}$$ $$(x+1)^5 = \color{blue}{\bf{1}}\cdot x^5 + \color{blue}{\bf{5}}x^4 + \color{blue}{\bf{10}} x^3 + \color{blue}{\bf{10}} x^2 + \color{blue}{\bf{5}}x + \color{blue}{\bf{1}}$$ $$(x + 1)^6 = \color{blue}{\bf{1}}\cdot x^6 + \color{blue}{\bf{6}}x^5 +\color{blue}{\bf{15}}x^4 + \color{blue}{\bf{20}}x^3 +\color{blue}{\bf{15}}x^2 + \color{blue}{\bf{6}}x + \color{blue}{\bf{1}}$$ $${\bf{\vdots}}$$
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Image source: Wikipedia, found at link to "Pascal's Triangle". – amWhy Jan 22 '13 at 22:17
Nice animated GIF! The answer is good, too :-) (+1) – robjohn Jan 22 '13 at 22:53
Thanks, @robjohn! It says more concisely, in animation, what would take a bit of explaining to write! – amWhy Jan 22 '13 at 22:56
@amWhy: It is unbelievable!! Great!! I cannot say anything. Nothing such this can describe the problem soooo well. +10! – Babak S. Jan 23 '13 at 15:50
This is incredible! Perfect explanation. Thank you! – user56763 Jan 23 '13 at 16:44
Before you jump to the binomial theorem (still the best way to go, in general, for expressions of the form $(a+b)^n$), let's start at the beginning. You undoubtedly know that $$(k+1)^3=(k+1)(k+1)(k+1)$$ We'll start by expanding $(k+1)(k+1)$. You can use FOIL here so we have $$(k+1)(k+1)=k^2+2k+1$$ We're two-thirds of the way to the answer. Now we have $$(k+1)^3=(k+1)(k^2+2k+1)$$ and by the distributive property, namely that $(a+b)c=ac+bc$, we have \begin{align} (k+1)^3=(k+1)(k^2+2k+1)&=(k)(k^2+2k+1)+(1)(k^2+2k+1)\\ &=(k^3+2k^2+k)+(k^2+2k+1)\\ &=k^3+3k^2+3k+1 \end{align} This will work for any positive integer exponent but, as Michael notes, you wouldn't want to do this for $(a+b)^n$.
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Thanks, makes sense. – user56763 Jan 23 '13 at 16:45
@ user56763: From what Rick Decker posted you can see that $(k+1)^3 = (k+1)(k+1)(k+1)$ can be expanded in the following way. Write down all possible 3-fold products of the form $ABC,$ where $A$ is a choice of one of the terms in the first $(k+1)$ factor, $B$ is a choice of one of the terms in the second $(k+1)$ factor, and $C$ is a choice of one of the terms in the third $(k+1)$ factor. Then add the 8 products. Thus, among the things you'll add will be $(k)(k)(1)$ and $(1)(k)(k)$ and $(k)(1)(k).$ Note that there are 3 of these with exactly two $k$'s and exactly one $1,$ so you get $3k^2$ ... – Dave L. Renfro Jan 28 '13 at 19:30
You can use "FOIL" twice. You should get $$k^3+3k^2+3k+1.$$ More generally, $$(a+b)^3 = a^3 + 3a^2b+3ab^2+ b^3.$$
Please don't vacilate between lower-case $k$ and capital $K$ in mathematical notation. Pick one and stick to it. Mathematical notation is case sensitive. Sometimes one uses lower-case $k$ and capital $K$ for two different things in the same problem, and you need to be clear about which is which.
But it would take a while to use FOIL to get $$(a+b)^9 = a^9+9a^8b+36a^7b^2+84a^6b^3+126a^5b^4+126a^4b^5+84a^3b^6+36a^2b^7+9ab^8+b^9.$$ That's one reason to be aware of the binomial theorem, which explains the pattern.
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What you're looking for is the binomial theorem, where y = 1.
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This is very scant. Something more would be good: some explanation or at least a link to an article. Of course, an answer should be more than simply a link. Look at the other answers for examples. – robjohn Jan 22 '13 at 22:56
Yeah, I was going to, but was having trouble typing out the math for a more exhaustive answer. – MITjanitor Jan 23 '13 at 2:10 | 2015-12-01T05:57:36 | {
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https://www.physicsforums.com/threads/this-is-not-a-necessary-part-of-the-course-im-taking-but-an-interesting-problem.633410/ | # This is not a necessary part of the course i'm taking, but an interesting problem.
## Homework Statement
Let a; b; c $\in$ (1,∞) and m; n $\in$ (0,∞). Prove that
\log_{b^mc^n} a + \log_{c^ma^n} b +\log_{a^mb^n} c \ge \frac 3 {m + n}
## The Attempt at a Solution
I do not even know where to start. A coherent explanation and possible solutions would greatly farther my knowledge of mathematics. Thanks for any and all help.
Last edited:
Do you mean $$\log_{b^mc^n} a + \log_{c^ma^n} b +\log_{a^mb^n} c \ge \frac 3 {m + n}$$
Or something else?
And what does "a; b; c (1;1) and m; n (0;1)" mean?
To your first post yes and thank you. Also, Let a; b; c $\in$ (1,∞) and m; n $\in$ (0,∞).
Last edited:
SammyS
Staff Emeritus
Homework Helper
Gold Member
## Homework Statement
Let a; b; c $\in$ (1,∞) and m; n $\in$ (0,∞). Prove that
$\log_{b^mc^n} a + \log_{c^ma^n} b +\log_{a^mb^n} c \ge \frac 3 {m + n}$
## The Attempt at a Solution
I do not even know where to start. A coherent explanation and possible solutions would greatly farther my knowledge of mathematics. Thanks for any and all help.
See how you might use the "Change of base" formula:
$\displaystyle \log_T(P)=\frac{\log_R(P)}{\log_R(T)}=\frac{\log_{10}(P)}{\log_{10}(T)}=\frac{\ln(P)}{\ln(T)}$
I think you could use this property $$\log_b x = \frac {\log_k x} {\log_k b}$$ Using that, for example, $$\log_{b^mc^n} a = \frac {\log_a a}{\log_a b^mc^n} = \frac {1}{\log_a b^mc^n}$$ You could deal with the other logs similarly, so you would get some rational expression involving sums and products of $m\log_a b$ with permutations of a, b, c, m and n. Another consequence of the formula above is that $$\log_b a = \frac {\log_a a} {\log_a b} = \frac 1 {\log_a b}$$ so you should be able to express everything in terms of $A = \log_a b$, $B = \log_a c$ and $C = \log_b c$ and m and n. After some algebra (probably, quite some algebra) you should be able to simplify that into something manageable and prove the inequality.
Well, that certainly clears a lot up for me, but I would very much appreciate some examples of where to go from there. (As you can tell, I'm clearly not versed well enough in mathematics to attempt such a problem). Thank you for any more input and helping me get a better grasp on mathematics.
SammyS
Staff Emeritus
Homework Helper
Gold Member
M4th,
Where did you get this problem from? What is the level of the mathematics that might be used to solve this problem?
BTW: Notice what you get if a = b = c .
I am in a pre-calculus course but my teacher chose me to be part of a separate "group" of students who solves problems outside of the classroom. I would just like to be able to simply understand what is going on in some of these problems.
Last edited:
Using the properties of the logarithm as we already discussed, you can show that $$\\ \log_{b^mc^n} a + \log_{c^ma^n} b + \log_{a^mb^n} c = \frac {1} {mA + nB} + \frac {1} {mB/A + n/B} + \frac {1} {m/A + nA/B}$$ where $$1 \le A \le B$$ Then you will need to prove $$\frac {1} {mA + nB} + \frac {1} {mB/A + n/B} + \frac {1} {m/A + nA/B} \ge \frac 3 {m + n}$$
This is a great forum, I already have learned a bit more about logarithms just watching how you've broken this problem up into its subsequent pieces. I thank you for such rapid responses and hospitality. Im looking forward to learning more about mathematics from you all and would greatly appreciate some more information on this and many more problems to come in my quest to broaden my grasp on this subject. I would also greatly appreciate a worded summary as to how you get to these results. I understand alot of it, but am still missing some peices and any more details would be way more than I expected to get but would also be extremely helpful.
Thank you again
This forum is great because it helps you understand the problem and learn how to solve it. Learn, but not just copy a complete solution. That would be against the rules. The rules require that you make an attempt at solving the problem. Here that would be trying to get the expression to the form given above. You have been given enough information to do so. Now you should try and tell us whether that worked and when not, what exactly went wrong.
Yes my point exactly. My goal is to be able to arrive at, and transform equations with ease into more workable ones as you have above and really have a good grasp at how to do so with any other similar problems.
Now one problem in my understanding is I grasp how logba=logaa/logab and thus I see where
\\log_{b^mc^n} a = \frac {\log_a a}{\log_a b^mc^n} = \frac {1}{\log_a b^mc^n}
but how is that transformed into the last mathematical statement you made
explain how you can transform logb^mc^na = 1/logabmcn into the statement below
Using the properties of the logarithm as we already discussed, you can show that $$\\ \log_{b^mc^n} a + \log_{c^ma^n} b + \log_{a^mb^n} c = \frac {1} {mA + nB} + \frac {1} {mB/A + n/B} + \frac {1} {m/A + nA/B}$$ where $$1 \le A \le B$$ Then you will need to prove $$\frac {1} {mA + nB} + \frac {1} {mB/A + n/B} + \frac {1} {m/A + nA/B} \ge \frac 3 {m + n}$$
explain how you can transform logb^mc^na = 1/logabmcn into the statement below
Basic logarithm properties:
$$\log xy = \log x + \log y \\ \log x^y = y \log x$$
Apply them to $\log_a b^mc^n$ and see where that gets you.
Of course, now it's all starting to come together. Thanks again for your help. | 2022-05-18T06:23:45 | {
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http://e-pres.di.uoa.gr/fmfjp/594efb-set-notation-symbols | Cardinality and ordinality Purplemath. We attempt to follow the standards set out in Norman Weinberg’s Guide to Standardized Drumset Notation. If you want to get in on their secrets, you'll want to become familiar with these Venn diagram symbols. Note that it's unnecessary to load amsmath if you load mathtools. The following table gives a summary of the symbols use in sets. While crow's foot notation is often recognized as the most intuitive style, some use OMT, IDEF, Bachman, or UML notation, according to their preferences. Set Theory • A mathematical model that we will use often is that of . Compact set notation is a useful tool to describe the properties of each element of a set, rather than writing out all elements of a set. Sometimes the set is written with a bar instead of a colon: {x¦ x > 5}. Basic set notation. State whether each … Because null set is not equal to A. Basic set operations. Thankfully, there is a faster way. Inequalities can be shown using set notation: {x: inequality}where x: indicates the variable being described and inequality is written as an inequality, normally in its simplest form. Typing math symbols into Word can be tedious. Basic Set Theory . The Universal Set … Set Notation. Which is why the bulk of this follow-up piece covers the very basics of set theory notation, operations & visual representations extensively. Also, check the set symbols here.. ... Set Language And Notation. This section is to introduce the notation to the reader and explain its usage. Illustration: In this section, we will introduce the standard notation used to define sets, and give you a chance to practice writing sets in three ways, inequality notation, set-builder notation, and interval notation. Use set notation to describe: (a) the area shaded in green (b) the area shaded in red : Look at the venn diagrams on the left. Symbol Symbol Name Meaning / definition Example { } set: a collection of elements: A = {3,7,9,14}, Set notation and Venn diagrams questions. On the Insert tab, in the Symbols group, click the arrow under Equation, and then click Insert New Equation. Lots symbols look similar but mean different things. The guide you are now reading is a “legend” to how we notate drum and percussion parts when we engrave music at Audio Graffiti. Symbolab: equation search and math solver - solves algebra, trigonometry and calculus problems step by step Example: Set-Builder Notation: Read as: Meaning: 1 {x : x > 0}the set of all x such that x is greater than 0. any value greater than 0: 2 {x : x ≠ 11}the set of all x such that x is any number except 11. any value except 11: 3 {x : x < 5}the set of all x such that x is any number less than 5. any value less than 5 Use set notation to describe: (a) the area shaded in blue (b) the area shaded in purple. Email. The domains and ranges used in the discrete function examples were simplified versions of set notation. Sets, in mathematics, are an organized collection of objects and can be represented in set-builder form or roster form.Usually, sets are represented in curly braces {}, for example, A = {1,2,3,4} is a set. Universal set and absolute complement. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions (area hyperbolic functions), even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. Author: Created by Maths4Everyone. Set notation practice. The table below lists all of the necessary symbols for compact set notation. The default way of doing it is to use the Insert > Symbols > More Symbols dialog, where you can hunt for the symbol you want. Mathematical Set Notation. Shading task. elements . When using set notation, we use inequality symbols to describe the domain and range as a set of values. This quiz and attached worksheet will help gauge your understanding of set notation. Below is the complete list of Windows ALT codes for Math Symbols: Set Membership & Empty Sets, their corresponding HTML entity numeric character references, and when available, their corresponding HTML entity named character references, and Unicode code points. There are many different symbols used in set notation, but only the most basic of structures will be provided here. They are { } and { 1 }. Crow's foot notation, however, has an intuitive graphic format, making it the preferred ERD notation for Lucidchart. Look at the venn diagram on the left. Click the arrow next to the name of the symbol set, and then select the symbol set that you want to display. Let us discuss the next stuff on "Symbols used in set theory" If null set is a super set Topics you will need to know in order to pass the quiz include sets, subsets, and elements. Let’s kick off by introducing the two most basic symbols for notating a set & it’s corresponding elements. Null set is a proper subset for any set which contains at least one element. Set notation is an important convention in computer science. Here null set is proper subset of A. For example, a set F can be specified as follows: = {∣ ≤ ≤}. take the previous set S ∩ V ; then subtract T: This is the Intersection of Sets S and V minus Set T (S ∩ V) − T = {} Hey, there is nothing there! In set-builder notation, the set is specified as a selection from a larger set, determined by a condition involving the elements. Subset, strict subset, and superset. Set notation. MS Word Tricks: Typing Math Symbols 2015-05-14 Category: MS Office. mathematical sets • A (finite) set can be thought of as a collection of zero or more . of . The Wolfram Language has the world's largest collection of consistent multifont mathematical notation characters\[LongDash]all fully integrated into both typesetting and symbolic expression construction . A set is a well-defined collection of distinct objects. Consider the set $\left\{x|10\le x<30\right\}$, which describes the behavior of $x$ in set-builder notation. Preview. Set notation is used to help define the elements of a set. Set theory is one of the foundational systems for mathematics, and it helped to develop our modern understanding of infinity and real numbers. Under Equation Tools , on the Design tab, in the Symbols group, click the More arrow. It is still a set, so we use the curly brackets with nothing inside: {} The Empty Set has no elements: {} Universal Set. Intersection and union of sets. You never know when set notation is going to pop up. Sets. For example, let us consider the set A = { 1 } It has two subsets. Bringing the set operations together. Usually, you'll see it when you learn about solving inequalities, because for some reason saying "x < 3" isn't good enough, so instead they'll want you to phrase the answer as "the solution set is { x | x is a real number and x < 3 }".How this adds anything to the student's understanding, I don't know. Set theory starter. Set Theory Symbols Posted in engineering by Christopher R. Wirz on Wed Feb 08 2017. This carefully selected compilation of exam questions has fully-worked solutions designed for students to go through at home, saving valuable time in class. The following list documents some of the most notable symbols in set theory, along each symbol’s usage and meaning. The individual objects in a set are called the members or elements of the set. They wrote about it on the chalkboard using set notation: P = {Kyesha, Angie and Eduardo} When Angie's mother came to pick her up, she looked at the chalkboard and asked: What does that mean? If you … ALT Codes for Math Symbols: Set Membership & Empty Sets Read More » Admin Igcse Mathematics Revision Notes, O Level Mathematics Revision Notes 2 Comments 12,074 Views. In, sets theory, you will learn about sets and it’s properties. When picking a symbol, best to trust the symbol's unicode name for its meaning, not appearance. Occasionally we will introduce a new symbol to cater for an unusual requirement of a client. Created: Jan 19, 2018 | Updated: Feb 6, 2020. Symbols Used in this Book; Glossary; While a comprehensive list of notation is included in the appendix, that is meant mostly as a reference tool to refresh the reader of what notation means. Researchers and mathematicians have developed a language and system of notation around set theory. To fully embrace the world of professional Venn diagrams, you should have a basic understanding of the branch of mathematical logic called ‘set theory’ and its associated symbols and notation. The table below contains one example set… 8 February 2019 OSU CSE 1. Solution: Let P be the set of all members in the math Problem 1: Mrs. Glosser asked Kyesha, Angie and Eduardo to join the new math club. A variant solution, also based on mathtools, with the cooperation of xparse allows for a syntax that's closer to mathematical writing: you just have to type something like\set{x\in E;P(x)} for the set-builder notation, or \set{x_i} for sets defined as lists. Google Classroom Facebook Twitter. The colon means such that.. For example: {x: x > 5}.This is read as x such that x is greater than > 5.. and symbols. Some notations for sets are: {1, 2, 3} = set of integers greater than 0 … Because rarely used symbol may look very different on another computer. Probability and statistics symbols table and definitions - expectation, variance, standard deviation, distribution, probability function, conditional probability, covariance, correlation After school they signed up and became members. CCSS.Math: HSS.CP.A.1. Quiz & Worksheet Goals A set is a collection of objects, things or symbols which are clearly identified.The individual objects in the set are called the elements or members of the set. Demo. In this notation, the vertical bar ("|") means "such that", and the description can be interpreted as "F is the set of all numbers n, such that n is an integer in the range from 0 to 19 inclusive". IGCSE 9-1 Exam Question Practice (Sets + Set Notation) 4.9 34 customer reviews. Relative complement or difference between sets. The symbols shown in this lesson are very appropriate in the realm of mathematics and in mathematical logic. That is OK, it is just the "Empty Set". 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Using set notation symbols notation, operations & visual representations extensively sets and it helped to develop modern... To the reader and explain its usage { ∣ ≤ ≤ } selected compilation exam! Set of values Equation Tools, on the Design tab, in the shown. The area shaded in purple get in on their secrets, you will about! Inequality symbols to describe: ( a ) the area shaded in blue ( b the... To develop our modern understanding of infinity and real numbers in on secrets... Off by introducing the two most basic symbols for notating a set written! Ms Word Tricks: Typing Math symbols 2015-05-14 Category: ms Office in blue ( b ) the shaded! Preferred ERD notation for Lucidchart in on their secrets, you will need know! To become familiar with these Venn diagram symbols developed a language and system of notation set! Will be provided here the foundational systems for Mathematics, and then select the symbol set that you to... ≤ ≤ } appropriate in the discrete function examples were simplified versions of set theory its meaning not! Using set notation is an important convention in computer science Igcse Mathematics Revision Notes, O Mathematics... Load mathtools ms Word Tricks: Typing Math symbols 2015-05-14 Category: Office., operations & visual representations extensively Empty set '' graphic format, making it the preferred ERD for. 19, 2018 | Updated: Feb 6, 2020 be thought of as a selection from a set... S kick off by introducing the two most basic of structures will be provided here Weinberg ’ s.... Join the new Math club a larger set, and elements symbols use sets. | 2021-03-05T20:47:51 | {
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http://math.stackexchange.com/questions/172543/why-do-these-two-methods-of-calculating-the-probability-of-winning-a-best-of-7-s/172550 | # Why do these two methods of calculating the probability of winning a best-of-7 series give the same answer?
I was having a discussion with a friend about the probability, and we came up with very different methods to solve it that lead to the same answer. The problem is pretty simple: you have two teams A and B playing a best of 7 series where wins are independent and the probability that team B wins is .35. What's the probability of team B winning th series? Friend says, "this is just a binomal random variable $X$ with $n=7$ and $p=.35$ and we are looking for $P(X\geq 4)$, which according to my TI-83 here is .1998".
I was rather convinced that this must be wrong. If we think of the series as a sequence of A's and B's, then we are looking for the probability of obtaining a sequence of length $n=4,5,6,7$ with 4 B's and the last element is a B. I was sure that his method will include the probability of obtaining, say, {B,A,B,B,B,A,A}, which we are not interested in and in fact couldn't even ever occur. So I figure that what we really want for a sequence of length $n$ is the probability of obtaining a sequence of length $n-1$ with exactly 3 B's, and then tacking a B onto the end. So for a sequence of length $n$, the probability should be $\binom{n-1}{3}.35^4 .65 ^{n-4}$, and then the answer should be $\sum_{n=4}^{7}\binom{n-1}{3}.35^4 .65 ^{n-4}$. I was confident that I was right and he was wrong, but then I plugged that into Wolfram Alpha and got... .1998.
What's going on here? Is it a coincidence?
-
Why do you say that {B,A,B,B,B,A,A} could not ever occur? This looks like a valid sequence of wins to me... – Code-Guru Jul 18 '12 at 19:22
Once team B wins 4 games the series is over. – crf Jul 18 '12 at 19:24
See David Spencer's answer below. – Code-Guru Jul 18 '12 at 19:32
To make more money, the league has decided that the series will go $7$ games, but the usual rules for determining the winner (first to win $4$) apply. Then Team B wins the modified series iff it wins the real series. – André Nicolas Jul 18 '12 at 20:54
This is not a coincidence, you are looking at the same problem in two different ways. The situation {B,A,B,B,B,A,A} is irrelevant in your friend's argument, because you are not counting the number of different possible sequences of events.
Rather, you are attempting to estimate the probability of at least $n$ successes in $k$ trials, which is exactly what the binomial distribution does.
In your second approach, you are essentially estimating the number of possible legal sequences terminating in a victory condition, and then computing the total fraction out of all possible sequences.
Both approaches are equivalent. One is a little easier to implement ;)
-
He is correct. Although this does include series that would be over after fewer than $7$ games "in real life", they would still all result in team B winning, no matter what happens after team B gets $4$ wins.
In your example, the probability that one of
{B,A,B,B,B,A,A}
{B,A,B,B,B,A,B}
{B,A,B,B,B,B,A}
{B,A,B,B,B,B,B}
happening without terminating series after four wins is equivalent to the probability of {B,A,B,B,B} happening across the first five games because later events do not affect anything.
- | 2015-07-30T00:14:22 | {
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https://math.stackexchange.com/questions/2422782/how-to-find-the-argument-of-cos-2-i-sin-2 | How to find the argument of $\cos 2 - i\sin 2$
How would you find the argument of the following number. $$\cos{2}-i\sin{2}$$
I'm aware that complex numbers in the form $r(\cos{\theta}+i\sin{\theta})$ have an argument of $\theta$, but what do you do with the $-$ sign?
Regardless of method, you will demonstrate some knowledge of Euler's identity and/or trigonometric identities.
Method I: Arctangent of slope
The argument is the arctangent of the ratio of the imaginary component to the real component, accounting for quadrant. If you forget to account for quadrant, this gives arguments in the interval $(-\pi/2, \pi/2)$, "half" of which are wrong. (Recall that arctangent is not a single-valued inverse. This is obvious when one remembers that tangent is periodic.)
Since $\cos 2 < 0$ and $-\sin 2 < 0$, this is in quadrant III.
\begin{align*} \arctan \frac{-\sin 2}{\cos 2} &= \arctan(-\tan 2)) \\ &= - \arctan \tan 2 &&\text{$\arctan$ is an odd function}\\ &= - \arctan \tan (2 - \pi) &&\text{$\tan$ has period $\pi$}\\ &= -(2 - \pi +\pi k), k \in \mathbb{Z} &&-\pi/2 < 2 - \pi < \pi/2 \text{.} \end{align*} Choosing $k=1$, this lands in quadrant III, giving the argument $-2$. If we had just mechanically evaluated, say with a calculator, \begin{align*} \arctan \frac{-\sin 2}{\cos 2} &= \arctan(2.18504\dots) \\ &= 1.14159\dots{} + \pi k, k \in \mathbb{Z} \text{.} \end{align*} The subset of these in quadrant III have $k$ odd. Picking $k = -1$, we get $-2$.
Of all the $k$ that give a result in the correct quadrant, which do you pick? You pick the one that conforms to your conventions for the range of value of an argument, if you have such a convention. If you do not, pick the one that makes your subsequent steps easier (or don't pick and leave the result as an equivalence class $\mod 2 \pi$).
Method II: Conjugation
If your only problem is the wrong sign of imaginary component, use conjugation. \begin{align*} \arg(\cos 2 - \mathrm{i} \sin 2) &= \arg(\overline{\cos 2 + \mathrm{i} \sin 2}) \\ &= \arg(\overline{\mathrm{e}^{\mathrm{i} 2}}) \\ &= -\arg(\mathrm{e}^{\mathrm{i} 2}) \\ &= -2 \text{.} \end{align*}
Method III: Even-odd identities
Sine is odd. Cosine is even. This is expressed in the even-odd identities. So $\cos(-2) = \cos(2)$ and $\sin(-2) = -\sin(2)$. Consequently, \begin{align*} \cos 2 - \mathrm{i} \sin 2 &= \cos -2 + \mathrm{i} \sin -2 \\ &= \mathrm{e}^{\mathrm{i}(-2)} \text{,} \end{align*} having argument $-2$.
Brahadeesh's and MrYouMath's answers use this method, without identifying what was done. Michael Rozenberg's answer combines this with the periodicity (by $2\pi$) identity, again without identifying what was done.
• Thank you very much. This has really cleared things up for me! – Ewan Miller Sep 9 '17 at 20:53
You can write this complex number as $\cos(-2) + i \sin(-2)$. Now you can see that the argument is $-2$.
$z=r(\sin\theta+i\sin\theta)$, where $r\geq0$ and $\theta=\arg{z}\in[0,2\pi)$.
$$\cos2-i\sin2=\cos(2\pi-2)+i\sin(2\pi-2).$$
Thus, $\arg(\cos2-i\sin2)=2\pi-2$.
• Thank you. Am I right in thinking then that there is a general rule that r(cosA-isinA)=r(cos(2π-A)+isin(2π-A))? – Ewan Miller Sep 9 '17 at 16:05
• @Ewan Miller It depends on definition. Sometimes $\arg{z}\in[0,2\pi)$ and sometimes $\arg{z}\in[-\pi,\pi)$. You need to ask your teacher. I like the first versa. – Michael Rozenberg Sep 9 '17 at 16:13
• In the case where $\arg{z}\in[-\pi,\pi)$, would it be r(cosA-isinA)=r(cos(-A)+isini(-A))? – Ewan Miller Sep 9 '17 at 16:17
• Also, I haven't got a teacher as I have just left school and am teaching this to myself in my gap year – Ewan Miller Sep 9 '17 at 16:19
• @Ewan Miller It's not so easy. If $A\in(-\pi,\pi)$ then you are right:$r(\cos(-A)+i\sin(-A))$ , but for $A=-\pi$ we get $r(\cos{A}+i\sin{A})$. All this depend on the level of your teacher. Maybe he don't see here problems. – Michael Rozenberg Sep 9 '17 at 16:23 | 2019-10-16T04:19:55 | {
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http://mathhelpforum.com/statistics/17095-counting-permutation-combinations.html | # Thread: Counting, Permutation, and Combinations
1. ## Counting, Permutation, and Combinations
Guys I have several questions that I would like to ask about:
1. In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, if the bride and the groom are among these 10 people, if:
a. the bride must be in the picture
b. exactly one of the bride and the groom is in the picture
2. How many ways are there for 10 women and six men to stand in a line so that no two men stand next to each other?
3. A professor writes 40 discrete mathematics true/false questions. Of these statements in these questions, 17 are true. If the questions can be positioned in any order, how many different answer keys are possible.
1a. because the bride is already included among 6 people, therefore there are 5 more people that we can choose and arrange from, so it's 5!. I am not sure if this is right
1b. I don't quite understand the question
2. P(10,10) x P(11,6)
3. C(40,17)
I am not quite sure of all these answers, but suggestions/help to my questions are highly appreciated
2. a. the bride must be in the picture
Let us divide this problem into cases.
Code:
BXXXXX
XBXXXX
XXBXXX
XXXBXX
XXXXBX
XXXXXB
Where "X" stands for any other person. And "B" stands for bride. We will find the number of such cases and then add all of them up together.
In the first case there are $\displaystyle (1)(9)(8)(7)(6)(5) = 15120$. Similarly in each case we have the same number of arrangements. So in total we have $\displaystyle 6\cdot 15120 = 90720$.
b. exactly one of the bride and the groom is in the picture
Again divide this into cases. Case 1 is that bride is in picture and groom is not. Case 2 is that bride is not in picture and groom is. Then add those results together. Each subcase is similar to the problem just did above.
3. Originally Posted by ThePerfectHacker
Let us divide this problem into cases.
Code:
BXXXXX
XBXXXX
XXBXXX
XXXBXX
XXXXBX
XXXXXB
Where "X" stands for any other person. And "B" stands for bride. We will find the number of such cases and then add all of them up together.
In the first case there are $\displaystyle (1)(9)(8)(7)(6)(5) = 15120$. Similarly in each case we have the same number of arrangements. So in total we have $\displaystyle 6\cdot 15120 = 90720$.
Again divide this into cases. Case 1 is that bride is in picture and groom is not. Case 2 is that bride is not in picture and groom is. Then add those results together. Each subcase is similar to the problem just did above.
as for case 1:
the bride is in the picture but the groom is not, 1 x 8 x 7 x 6 x 5 x 4 = 6,720. 6720 x 6 = 13,440. I choose 8 because therefore the number of possibilities of people chosen is now 9 instead of 10 because the groom is excluded here.
and for case 2:
the bride is not in picture but the groom is has the same possibilities as case 1 right?
then we just add case 1 and case 2?
4. Originally Posted by TheRekz
as for case 1:
the bride is in the picture but the groom is not, 1 x 8 x 7 x 6 x 5 x 4 = 6,720. 6720 x 6 = 13,440. I choose 8 because therefore the number of possibilities of people chosen is now 9 instead of 10 because the groom is excluded here.
and for case 2:
the bride is not in picture but the groom is has the same possibilities as case 1 right?
then we just add case 1 and case 2?
I agree with you.
5. what if now the question is both the bride and the groom must be in the picture:
(1 x 1 x 8 x 7 x 6 x 5) = 1,680 x 6 = 10,080
is this right?
6. Hello, TheRekz!
1. In how many ways can a photographer at a wedding
arrange 6 people in a row from a group of 10 people,
if the bride and the groom are among these 10 people, and:
a. the bride must be in the picture
b. exactly one of the bride and the groom is in the picture
a) The bride is in the picture.
Select five more from the other nine people: .$\displaystyle C(9,5)$ ways.
These six people can be arranged in $\displaystyle 6!$ ways.
Therefore, there are: .$\displaystyle C(9,5) \times 6!$ ways.
b) Either bride or the groom is in the picture (but not both).
There are 2 choices for the newlywed to be in the picture.
The other 5 people are chosen from the remaining 8 people (not the other newlywed).
. . There are: .$\displaystyle C(8,5)$ ways.
The six people can be arranged in $\displaystyle 6!$ ways.
Therefore, there are: .$\displaystyle 2 \times C(8,5) \times 6!$ ways.
2. How many ways are there for 10 women and six men to stand in a line
so that no two men stand next to each other?
Arrange the tenwomen in a row. .There are $\displaystyle P(10,10) = 10!$ possible arrangements.
Leave spaces between the women: .$\displaystyle \_\;W\;\_\;W\;\_\;W\;\_\;W\;\_\;W\;\_\;W\;\_\;W\;\ _\;W\;\_\;W\;\_\;W\;\_$
Arrange the six men in any six of the eveleven spaces.
. . There are: .$\displaystyle P(11,6)$ ways.
Therefore, there are: .$\displaystyle P(10,10) \times P(11,6)$ ways.
3. A professor writes 40 discrete mathematics true/false questions.
Of these statements in these questions, 17 are true.
If the questions can be positioned in any order,
how many different answer keys are possible?
Your answer is correct . . . $\displaystyle C(40,17)$
7. Originally Posted by TheRekz
what if now the question is both the bride and the groom must be in the picture:
(1 x 1 x 8 x 7 x 6 x 5) = 1,680 x 6 = 10,080
is this right?
If both the bride and groom are in the picture then:
fix the bride in the first position:
BGXXXX = 1*1*8*7*6*5
BXGXXX = 1*8*1*7*6*5
.
.
.
BXXXXG = 1*8*7*6*5*1
Then there are 5*(1*1*8*7*6*5) ways to arrange them with the bride in the first position.
Now repeat with the bride in the second, third,....,sixth position and you get the same result each time. Therefore, if you add these all up you get (5*(1*1*8*7*6*5))*6 = 50,400
,
,
,
# A professor writes 40 discrete mathematics true/false
Click on a term to search for related topics. | 2018-06-21T20:34:03 | {
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https://math.stackexchange.com/questions/4016228/integration-by-trigonometric-substitution-vs-table-of-integral-solution | # Integration by Trigonometric Substitution vs Table of Integral Solution
I'm not sure how to phrase this question, but I find myself confused over the correctness of a particular solution in the table of integral, specifically: $$\int\frac{dx}{\sqrt{x^2-a^2}}=\ln|\sqrt{x^2-a^2}+x|+C$$
I find that if you try to solve it with trigonometric substitution, you get a different answer:
So, for the example, since the form $$\sqrt{x^2-a^2}$$ is present, we use the substitution $$x=asec \theta$$, and $$dx=asec\theta tan \theta d\theta$$. Doing so:
$$\int\frac{asec\theta tan \theta d\theta}{\sqrt{(asec \theta)^2-a^2}}$$ $$\int\frac{asec\theta tan \theta d\theta}{\sqrt{a^2sec^2 \theta-a^2}}$$ $$\int\frac{asec\theta tan \theta d\theta}{a\sqrt{sec^2 \theta-1}}$$ Using the trigonometric identity $$tan^2 \theta=sec^2 \theta -1$$ $$\int\frac{asec\theta tan \theta d\theta}{atan \theta}$$
Cancelling like terms, we get $$\int sec \theta$$, which finally integrates to $$ln|sec \theta + tan \theta|+C$$
Undoing the substitution, using this triangle:
https://i.stack.imgur.com/6tkR7.png
With $$sec \theta = \frac{x}{a}$$ and $$tan \theta = \frac{\sqrt{x^2-a^2}}{a}$$, the final answer then is:
$$ln\biggl|\frac{x}{a} + \frac{\sqrt{x^2-a^2}}{a}\biggl|+C$$
Which is evidently different from the one from the table of integrals. Is my solution wrong or something?
• looks the same to me, since $-\ln |a|$ can be absorbed into $C$? – Calvin Khor Feb 7 at 13:48
• Oh right, that can be done. Thanks for pointing that out – Xyzar Feb 7 at 13:52
$$\ln \left|\dfrac{x+\sqrt{x^2-a^2}}{a}\right| +c= \ln |x+\sqrt{x^2-a^2}| - \ln|a| +c = \ln(x+\sqrt{x^2-a^2}) +k$$
$$k = c-\ln |a| =$$ constant | 2021-04-14T16:21:26 | {
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http://math.stackexchange.com/questions/495159/different-ways-of-integrating-3-sin-x-cos-x | # Different Ways of Integrating $3\sin x\cos x$
I am asking this question for my son who is in (equivalent) twelfth grade and I failed to answer his query.
When he tries to integrate $3\sin x\cos x$, he finds that this can be done in at least following three ways. And these three ways do not produce equivalent results.
ONE
Let us assume, $\sin x = z$.
This gives, \begin{align*} \cos x &= \frac{dz}{dx}\\ \cos x dx &= dz \end{align*}
So, we can write,
\begin{align*} \int 3\sin x\cos x dx &=3 \int zdz\\ &=3 \frac{z^2}{2}\\ &=\frac{3}{2} \sin^2 x\\ &=\frac{3}{4}\times 2\sin^2 x\\ &=\frac{3}{4} (1 -\cos 2x)\\ \end{align*}
TWO
Let us assume, $\cos x = z$.
This gives, \begin{align*} -\sin x &= \frac{dz}{dx}\\ \sin x dx &= -dz \end{align*}
So, we can write,
\begin{align*} \int 3\sin x\cos x dx &=-3 \int zdz\\ &=-3 \frac{z^2}{2}\\ &=-\frac{3}{2} \cos^2 x\\ &=-\frac{3}{4}\times 2\cos^2 x\\ &=-\frac{3}{4} (1 +\cos 2x)\\ \end{align*}
THREE
\begin{align*} \int 3\sin x\cos x dx &=\frac{3}{2}\int 2\sin x\cos x dx\\ &=\frac{3}{2}\int \sin 2x dx\\ &=-\frac{3}{2}\times\frac{1}{2} \cos 2x\\ &=-\frac{3}{4} \cos 2x\\ \end{align*}
The results found in above three methods are not the same.
If we try a simple approach of evaluating the integration results at, $x = \frac{\pi}{6}$, we get as follows.
From the first one,
$\frac{3}{4} (1 -\cos 2x) = \frac{3}{4} (1 -\cos \frac{2\pi}{6}) = \frac{3}{4} (1 -\cos \frac{\pi}{3}) = \frac{3}{4} (1 - \frac{1}{2}) = \frac{3}{4}\times\frac{1}{2} = \frac{3}{8}$
From the second one,
$-\frac{3}{4} (1 +\cos 2x) = -\frac{3}{4} (1 +\cos \frac{2\pi}{6}) = -\frac{3}{4} (1 +\cos \frac{\pi}{3}) = -\frac{3}{4} (1 + \frac{1}{2}) = -\frac{3}{4}\times\frac{3}{2} = -\frac{9}{8}$
From the third one,
$-\frac{3}{4} \cos 2x=-\frac{3}{4} \cos \frac{2\pi}{6} = -\frac{3}{4} \cos \frac{\pi}{3} = -\frac{3}{4} \times \frac{1}{2} = -\frac{3}{8}$
Clearly, we are getting some nonequivalent results. We have failed to find the mistakes or explanations behind this. Your help will be appreciated.
-
You can locate an several primitives of a function, they differ in a constant, but that does not mean that the results are wrong. – Hiperion Sep 16 '13 at 5:50
Indefinite integration can be 'misleading' in this sense. – copper.hat Sep 16 '13 at 5:52
A great question to ask students to see if they can come up with Zev's answer (or equivalent). – nbubis Sep 16 '13 at 6:01
– Jonas Meyer Sep 16 '13 at 6:07
When I was doing A-levels, I came across a similar example, but the two "different" answers to the integral involved respectively $\tan^2(x)$ and $\sec^2(x)$. Of course, these two expressions differ by a constant, but the explanation was non-obvious to me (and my dad) then, and I had to get my teacher to explain it. – Hammerite Sep 16 '13 at 9:29
You're forgetting that an indefinite integral must include a constant of integration; for any chosen constant $C$, we have that $$\frac{d}{dx}\left(-\frac{3}{4}\cos(2x)+C\right)=3\sin(x)\cos(x),$$ and that is precisely the relationship captured by the statement that
$$\int 3\sin(x)\cos(x)\,dx=-\frac{3}{4}\cos(2x)+C.$$
-
Indeed this is a great demonstration of that's so important. – David H Sep 16 '13 at 5:50
So, if is asked to solve the problem in the examination, which approach he should follow? His textbook contains only one of the answers. And as you can see, some graders may like only the conventional answers. No disrespect intended. – Masroor Sep 16 '13 at 6:01
@MMA - All the answers are equally "wrong" since non of them contain a constant. When you let the constant C "swallow" the other constant term, all three give the same result. – nbubis Sep 16 '13 at 6:02
@nbubis But his text book contains only one of the answers, (I can not get the book now), and given the prevailing circumstances, it will be difficult to modify the system. I can put down the answer which is the book in six/seven hours when the book comes back from school. – Masroor Sep 16 '13 at 6:07
@MMA I sympathize with your situation, especially if your particular school has you feeling that cynical towards the system. But I have no way of predicting what which technique a grader would consider most appropriate, because there isn't one (and if the grader doesn't know it then your son knows more calculus than his teacher). He should learn all three methods. Solving integrals is like choosing a golf-club for a given shot. The best one is the one that gets the job done. – David H Sep 16 '13 at 6:23
All three answers are correct provided you add a constant to each one of those.
Because from the very definition of integration, it is the area under the curve, so it requires bounds to give a unique value. You can't evaluate the value of an indefinite integral without including constant.
And I am sure that in the examination, your son won't be asked to evaluate the value of an integral without providing limits of integration or providing its value at some other point.
For instance, in question it may be mentioned that evaluate the value of expression at x=π/6 , given its value at x=0 is 1. So in this case, all three answers will give the correct value i.e. 11/8
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https://mathhelpboards.com/threads/chain-rule-and-e-u.1637/ | # [SOLVED]chain rule and e^(u)
#### DeusAbscondus
##### Active member
Hi folks,
I don't know if my experience is at all common (and I would like some feedback on this if possible), but I can't seem to nail down the properties of euler's number in the context of chain rule problems.
Here is the nub of my difficulty:
1. $\text{If }f(x)=e^x \text{then }f'(x)=e^x$
This I accept, though, not having seen a formal proof of it, and since it is counter-intuitive, I must take it on faith.
But the following I do not understand; could someone help me towards understanding?
2. $\text{If }g(x)=e^{x^{4}} \text{then }g'(x)=4e^{x^4}x^3$
Am I on the right track to observe that, if 2. is correct, then g(x) is a composite function, hence subject to the chain rule?
And if so, is the following a generally valid way to work through this and all such problems:
$\text{If }g(x)=e^{x^{4}} \text{find }g'(x)$
$\text{Now}g'(x)=u'v' \text{by Chain Rule}$
$\text{So, let }u=x^4 \text{and }v=e^u$
$\text{Then }u'=4x^3 \text{and }v'=e^u \text{ (by some rule which currently exceeds my understanding)}$
$\text{Therefore }g'(x)=u'v'=4x^3*e^u=\text{(via substitution) }4x^3*e^{x^{4}}$
$\text{Which, simplified }=4e^{x^{4}}x^3$
Finally, I have a similar hesitation/scruple/sense of vertigo when it comes to dealing with another unusual derivative, that of:
$ln(x)$
If anyone can see why, having the read the foregoing, I would feel unsure of myself around this animal, could they possibly add some notes to help me tame it?
Regs,
DeusAbs
Last edited:
#### Jameson
Staff member
Hi DeusAbscondus,
1) Let's try to derive the derivative of $e^x$. Recall that for $f(x)$ it follows that $$\displaystyle f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$.
In this case that becomes $$\displaystyle \lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h}=\frac{e^x \cdot e^h-e^x}{h}=\frac{e^x(e^h-1)}{h}$$.
Ok, from here we should notice that $e^x$ isn't dependent on h in this limit thus can be brought outside of the computation as long as we multiply through at the end.
So the last bit becomes $$\displaystyle \left( e^x \cdot \lim_{h \rightarrow 0} \frac{e^h-1}{h} \right)$$.
From here you need to compute that limit and you can do it a few ways. For a less rigorous approach that is more intuitive as well, I suggest graphing $$\displaystyle \frac{e^h-1}{h}$$ on a graphing calculator and you'll see that as h approaches zero, then the y value approaches 1.
So at the end we get that $$\displaystyle \left( e^x \cdot \lim_{h \rightarrow 0} \frac{e^h-1}{h} \right) = e^x \cdot (1) = e^x$$, therefore $$\displaystyle \frac{d}{dx} e^x = e^x$$
Please let me know if any of that is unclear. Again, it's not a rigorous proof by any means, more of a "walk through".
2) You are completely correct that $$\displaystyle \frac{d}{dx} e^{f(x)} = e^{f(x)} \cdot f'(x)$$. As you said this is an application of the chain rule.
#### Sudharaka
##### Well-known member
MHB Math Helper
$\text{Then }u'=4x^3 \text{and }v'=e^u \text{ (by some rule which currently exceeds my understanding)}$
Hi DeusAbscondus,
The only rule that is used here is the chain rule.
\begin{eqnarray}
g(x)&=&e^{x^{4}}\\
g'(x)&=&\frac{d}{d(x^4)}e^{x^{4}}\frac{d}{dx}x^4\\
&=&e^{x^{4}}(4x^3)\\
\therefore g'(x)&=&4x^3e^{x^{4}}
\end{eqnarray}
Kind Regards,
Sudharaka.
#### Fantini
##### "Read Euler, read Euler." - Laplace
MHB Math Helper
Perhaps to avoid particularizing and help you think of the exponential as any other function, think of $e^{x^4}$ as $h(x) = e^x$ and $g(x) = x^4$, then $(h \circ g)(x) = h(g(x)) = e^{x^4}$ and $(h \circ g)' (x) = h'(g(x)) \cdot g'(x) = e^{x^4} \cdot (4x^3) = 4 e^{x^4} x^3.$
#### chisigma
##### Well-known member
Hi DeusAbscondus,
1) Let's try to derive the derivative of $e^x$. Recall that for $f(x)$ it follows that $$\displaystyle f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$$.
In this case that becomes $$\displaystyle \lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h}=\frac{e^x \cdot e^h-e^x}{h}=\frac{e^x(e^h-1)}{h}$$.
Ok, from here we should notice that $e^x$ isn't dependent on h in this limit thus can be brought outside of the computation as long as we multiply through at the end.
So the last bit becomes $$\displaystyle \left( e^x \cdot \lim_{h \rightarrow 0} \frac{e^h-1}{h} \right)$$.
From here you need to compute that limit and you can do it a few ways...
One possible 'rigorous' way to demonstrate that...
$\displaystyle \lim_{h \rightarrow 0} \frac{e^{h}-1}{h}=1$ (1)
... is the following...
a) You start from the definition of exponential...
$\displaystyle e^{h}= \lim_{n \rightarrow \infty} (1+\frac{h}{n})^{n}$ (2)
b) You demonstrate the identity...
$\displaystyle \lim_{n \rightarrow \infty} (1+\frac{h}{n})^{n}= \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac {h^{k}}{k!} = 1 + h + \frac {h^{2}}{2}+... \frac{h^{n}}{n!}+...$ (3)
c) from (3) You derive...
$\displaystyle \frac{e^{h}-1}{h}= 1 + \frac{h}{2} +... + \frac{h^{n-1}}{n!}+...$ (4)
... and the (1) follows immediately...
Kind regards
$\chi$ $\sigma$
#### DeusAbscondus
##### Active member
thanks: Jameson, Sudharaka and Fantini: Jameson for the informal proof of $e^x$, Sudharaka for reminding me that there is no mystery here, just the chain rule and Fantini for the code behind $(h \circ g)$ and some other stuff besides.
Just a quick follow up:
can one of you gents refer me to a gentle approach to understanding a delta/epsilon proof, via video preferably, so as I can see it being worked out, pause it and study it?
and, Jameson, is this proof: the delta/epsilon proof, the more formal proof of the first derivative to which your notes allude?
thx again gentelmen,
DeusAbs
#### Jameson
Staff member
It's not actually completely informal the way I did it, just not rigorous. By applying chisigma's technique as well as many other possibilities, such as L'Hopital's Rule you can soundly demonstrate that this limit is 1. A delta-epsilon proof of this limit wouldn't be asked in basic calculus course as far as I know.
Anyway, as to your question. I'm not sure about videos but here's a link that's part of a great Calculus help site I would recommend for anyone, Paul's Online Notes. The link I gave you specifically deals with delta-epsilon proofs and has nice diagrams plus examples that cover both the theory and show you how it's done in practice.
Let me know what you think!
#### Sudharaka
##### Well-known member
MHB Math Helper
thanks: Jameson, Sudharaka and Fantini: Jameson for the informal proof of $e^x$, Sudharaka for reminding me that there is no mystery here, just the chain rule and Fantini for the code behind $(h \circ g)$ and some other stuff besides.
Just a quick follow up:
can one of you gents refer me to a gentle approach to understanding a delta/epsilon proof, via video preferably, so as I can see it being worked out, pause it and study it?
and, Jameson, is this proof: the delta/epsilon proof, the more formal proof of the first derivative to which your notes allude?
thx again gentelmen,
DeusAbs
Hi DeusAbscondus,
A good place filled with video lessons is Khan Academy. Here is a link to the Epsilon Delta limit definition introduction video.
Kind Regards,
Sudharaka.
#### CaptainBlack
##### Well-known member
Hi folks,
I don't know if my experience is at all common (and I would like some feedback on this if possible), but I can't seem to nail down the properties of euler's number in the context of chain rule problems.
Here is the nub of my difficulty:
1. $\text{If }f(x)=e^x \text{then }f'(x)=e^x$
This I accept, though, not having seen a formal proof of it, and since it is counter-intuitive, I must take it on faith.
But the following I do not understand; could someone help me towards understanding?
From the way this is worded it seems that you think that there is some universally agreed definition of the exponential function. The truth is that there is not, you may have been told it is defined as:
$\exp(x)=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n$
but it is often defined as the solution to the differential equation initial value problem: $$f'(x)=f(x), f(0)=1$$, or as the inverse of the natural logarithm, ..
So to prove that the derivative of the exponential function is itself we really need to know what definition you are starting from (and also what you already know).
CB
#### DeusAbscondus
##### Active member
From the way this is worded it seems that you think that there is some universally agreed definition of the exponential function. The truth is that there is not, you may have been told it is defined as:
$\exp(x)=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n$
but it is often defined as the solution to the differential equation initial value problem: $$f'(x)=f(x), f(0)=1$$, or as the inverse of the natural logarithm, ..
So to prove that the derivative of the exponential function is itself we really need to know what definition you are starting from (and also what you already know).
CB
I just had a look at graphs of these functions: given that they are mirror images of each other graphically, I would have thought their inter-definability (if i can put it thus) bordered the trivial.
I'd also have thought my post here (indeed, the combined oeuvre of my posts to date ) makes perfectly clear my starting position in the knowledge of mathematics stakes: zero + epsilon.
Prescinding from that, any light you can throw on it would seem to me to be, well - in a dark place - illuminating, n'est-ce pas?
I always enjoy your sparse, ascetical take on things, Cap'n.
So, thanks for the input,
Deus Abs
#### DeusAbscondus
##### Active member
It's not actually completely informal the way I did it, just not rigorous. By applying chisigma's technique as well as many other possibilities, such as L'Hopital's Rule you can soundly demonstrate that this limit is 1. A delta-epsilon proof of this limit wouldn't be asked in basic calculus course as far as I know.
Anyway, as to your question. I'm not sure about videos but here's a link that's part of a great Calculus help site I would recommend for anyone, Paul's Online Notes. The link I gave you specifically deals with delta-epsilon proofs and has nice diagrams plus examples that cover both the theory and show you how it's done in practice.
Let me know what you think!
Thanks Jameson; just spent some hours there! I suffer from a dearth of really first-rate material and my course notes are a nest of errors and poorly edited material, so I appreciate such a rich fund of examples, proofs, worked examples and tips.
I'll be sure to make a fuller response once I've had more time to check it.
DeusAbs
#### chisigma
##### Well-known member
From the way this is worded it seems that you think that there is some universally agreed definition of the exponential function. The truth is that there is not, you may have been told it is defined as:
$\exp(x)=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n$
but it is often defined as the solution to the differential equation initial value problem: $$f'(x)=f(x), f(0)=1$$, or as the inverse of the natural logarithm, ..
So to prove that the derivative of the exponential function is itself we really need to know what definition you are starting from (and also what you already know).
CB
More that 280 years ago [!] Leonhard Euler defined in such way the exponential and natural log functions...
$\displaystyle e^{z}= \lim_{n \rightarrow \infty} (1+\frac{z}{n})^{n}$
$\displaystyle \ln z= \lim_{n \rightarrow \infty} n\ (z^{\frac{1}{n}}-1)$ (1)
... and also demonstrated that the two function are inverse to one other. In my [humble ...] opinion the definitions (1) are also today 'the best' mainly because they are valid 'without additions' for any real or complex value of z...
Kind regards
$\chi$ $\sigma$
#### CaptainBlack
##### Well-known member
I'd also have thought my post here (indeed, the combined oeuvre of my posts to date ) makes perfectly clear my starting position in the knowledge of mathematics stakes: zero + epsilon.
Prescinding from that, any light you can throw on it would seem to me to be, well - in a dark place - illuminating, n'est-ce pas?
You must have at least some idea of how the exponential function is defined, or at the very least how "e" is defined, otherwise you are starting in the middle.
(I would recommend you get a copy of Morris Kline's book "Calculus; An intuitive and physical approach", the only flaw of which is its use of US Customary Units)
CB
#### Opalg
##### MHB Oldtimer
Staff member
More than 280 years ago [!] Leonhard Euler defined in such way the exponential and natural log functions...
$\displaystyle e^{z}= \lim_{n \rightarrow \infty} \Bigl(1+\frac{z}{n}\Bigr)^{n}$
$\displaystyle \ln z= \lim_{n \rightarrow \infty} n\ (z^{\frac{1}{n}}-1)$ (1)
... and also demonstrated that the two function are inverse to one other. In my [humble ...] opinion the definitions (1) are also today 'the best' mainly because they are valid 'without additions' for any real or complex value of z...
In my [equally humble ...] opinion, the definitions (1) are not the best, because they are hard to work with. For example, starting from (1) how would you prove the index law $e^{x+y} = e^xe^y$, or the derivative rule $\frac d{dx}e^x = e^x$? Those proofs can surely be given, but the limit definition does not strike me as a natural or convenient one to start from. I much prefer the power series definition $e^x = \sum_{n=0}^\infty\frac{x^n}{n!}$ for the (complex) exponential. The basic properties follow quite easily from that, as does the fact that the real exponential $e^x:\mathbb{R}\to(0,\infty)$ increases monotonically and hence has an inverse, which defines the real logarithm function. (The complex logarithm is not a single-valued function, and must therefore be defined more indirectly.)
#### chisigma
##### Well-known member
In my [equally humble ...] opinion, the definitions (1) are not the best, because they are hard to work with. For example, starting from (1) how would you prove the index law $e^{x+y} = e^xe^y$, or the derivative rule $\frac d{dx}e^x = e^x$? Those proofs can surely be given, but the limit definition does not strike me as a natural or convenient one to start from. I much prefer the power series definition $e^x = \sum_{n=0}^\infty\frac{x^n}{n!}$ for the (complex) exponential. The basic properties follow quite easily from that, as does the fact that the real exponential $e^x:\mathbb{R}\to(0,\infty)$ increases monotonically and hence has an inverse, which defines the real logarithm function. (The complex logarithm is not a single-valued function, and must therefore be defined more indirectly.)
On the basis of the [easily enough demonstrable...] identity...
$\displaystyle \lim_{n \rightarrow \infty} (1+\frac{z}{n})^{n} = \lim_{n \rightarrow \infty} \sum_{k=0}^{n} \frac{z^{k}}{k!}$ (1)
... the two definitions must be considered fully equivalent. On the purely conceptual basis however the original Euler's definition requires only the concept of limit and is 'more elementary' than any other definition and that is an excellent reason for defining it as 'the best'... at least for me..
Kind regards
$\chi$ $\sigma$
#### QuestForInsight
##### Member
1. $\text{If }f(x)=e^x \text{then }f'(x)=e^x$
This is very easy to get using the power series definition from Opalg's post.
\begin{aligned} & (e^x)' = \bigg(\sum_{k \ge 0}\frac{x^k}{k!}\bigg)' = \sum_{k \ge 0}\frac{kx^{k-1}}{k!} = \sum_{k \ge 1}\frac{kx^{k-1}}{k!} = \sum_{k+1 \ge 1}\frac{(k+1)x^{k}}{(k+1)!} = \sum_{k \ge 0}\frac{x^{k}}{k!}.\end{aligned}
#### chisigma
##### Well-known member
This is very easy to get using the power series definition from Opalg's post.
\begin{aligned} & (e^x)' = \bigg(\sum_{k \ge 0}\frac{x^k}{k!}\bigg)' = \sum_{k \ge 0}\frac{kx^{k-1}}{k!} = \sum_{k \ge 1}\frac{kx^{k-1}}{k!} = \sum_{k+1 \ge 1}\frac{(k+1)x^{k}}{(k+1)!} = \sum_{k \ge 0}\frac{x^{k}}{k!}.\end{aligned}
That is true of course but [conceptually...] the preliminary demonstration of the convergence theorem of the series of derivatives is required. The computation of derivative of $e^{x}$ using the standard derivative definition require only to demonstrate that...
$\displaystyle \lim_{h \rightarrow 0} \frac {e^{h}-1}{h}=1$ (1)
... and that requires elementary instruments...
Kind regards
$\chi$ $\sigma$ | 2020-09-30T14:27:28 | {
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https://www.coursehero.com/file/p5ubjj9/77-To-use-the-alternating-series-test-consider-a-n-f-n-where-f-x-arctan-ax-We/ | # 77 to use the alternating series test consider a n f
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Chapter 11 / Exercise 77
Single Variable Calculus
Stewart
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77. To use the alternating series test, consider a n = f ( n ) , where f ( x ) = arctan( a/x ) . We need to show that f ( x ) is decreasing. Since f 0 ( x ) = 1 1 + ( a/x ) 2 - a x 2 , we have f 0 ( x ) < 0 for a > 0 , so f ( x ) is decreasing for all x . Thus a n +1 < a n for all n , and as lim n →∞ arctan( a/n ) = 0 for all a , by the alternating series test, X n =1 ( - 1) n arctan( a/n ) converges. 78. The n th partial sum of the series is given by S n = 1 - 1 2 + 1 3 - · · · + ( - 1) n - 1 n , so the absolute value of the first term omitted is 1 / ( n + 1) . By Theorem 9.9, we know that the value, S , of the sum differs from S n by less than 1 / ( n +1) . Thus, we want to choose n large enough so that 1 / ( n +1) 0 . 01 . Solving this inequality for n yields n 99 , so we take 99 or more terms in our partial sum. 79. The n th partial sum of the series is given by S n = 1 - 2 3 + 4 9 - · · · + ( - 1) ( n - 1) 2 3 ( n - 1) ,
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Chapter 11 / Exercise 77
Single Variable Calculus
Stewart
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750 Chapter Nine /SOLUTIONS so the absolute value of the first term omitted is (2 / 3) n . By Theorem 9.9, we know that the value, S , of the sum differs from S n by less than (2 / 3) n . Thus, we want to choose n large enough so that (2 / 3) n 0 . 01 . Solving this inequality for n yields n 11 . 358 , so taking 12 or more terms in our partial sum is guaranteed to be within 0 . 01 of the sum of the series. Note: Since this is a geometric series, we know the exact sum to be 1 / (1+2 / 3) = 0 . 6 . The partial sum S 12 is 0 . 595 , which is indeed within 0 . 01 of the sum of the series. Note, however, that S 11 = 0 . 6069 , which is also within 0 . 01 of the exact sum of the series. Theorem 9.9 gives us a value of n for which S n is guaranteed to be within a small tolerance of the sum of an alternating series, but not necessarily the smallest such value. 80. The n th partial sum of the series is given by S n = 1 2 - 1 24 + 1 720 - · · · + ( - 1) n - 1 (2 n )! , so the absolute value of the first term omitted is 1 / (2 n + 2)! . By Theorem 9.9, we know that the value, S , of the sum differs from S n by less than 1 / (2 n +2)! . Thus, we want to choose n large enough so that 1 / (2 n +2)! 0 . 01 . Substituting n = 2 into the expression 1 / (2 n + 2)! yields 1 / 720 which is less than 0.01. We therefore take 2 or more terms in our partial sum. 81. Since 0 c n 2 - n for all n , and since 2 - n is a convergent geometric series, c n converges by the Comparison Test. Similarly, since 2 n a n , and since 2 n is a divergent geometric series, a n diverges by the Comparison Test. We do not have enough information to determine whether or not b n and d n converge. 82. (a) The sum a n · b n = 1 /n 5 , which converges, as a p -series with p = 5 , or by the integral test: Z 1 1 x 5 dx = lim b →∞ x - 4 ( - 4) b 1 = lim b →∞ b - 4 ( - 4) + 1 4 = 1 4 . Since this improper integral converges, a n · b n also converges. (b) This is an alternating series that satisfies the conditions of the alternating series test: the terms are decreasing and have limit 0 , so ( - 1) n / n converges. (c) We have a n b n = 1 /n , so a n b n is the harmonic series, which diverges. | 2021-12-01T16:38:49 | {
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https://math.stackexchange.com/questions/1407002/are-these-two-optimization-problems-equivalent-to-each-other | # Are these two optimization problems equivalent to each other?
Let $\mathbf{x}=[x_1,\ldots,x_K]^T$. For a fixed vector $\mathbf{a}$, I have the following optimization problem : \begin{array}{rl} \min \limits_{\mathbf{x}} & | \mathbf{a}^T \mathbf{x} | \\ \mbox{s.t.} & x_k\ge 0, \forall k \\ & \sum_k x_k=1 \end{array} The second optimization problem is: \begin{array}{rl} \min \limits_{\mathbf{x}} & ( \mathbf{a}^T \mathbf{x})^2\\ \mbox{s.t.} & x_k\ge 0, \forall k \\ & \sum_k x_k=1 \end{array}
My question: are these two problems equivalent to each other? if yes, how to solve them ? and which one is easier to solve ?
• yes they are equivalent. have you tried the Lagrange multiplier method? – user251257 Aug 23 '15 at 16:07
• The objective function of the problem 1 is $\ell^1$-norm and problem 2 is $\ell^2$-norm, so although equivalent, the objective function of the second problem is continuously differentiable, therefore the second problem is much easier to solve. – guille_NP Aug 23 '15 at 21:22
• @guille_NP: The second problem is indeed much easier to solve, but definitely not for the reason you provide. Your statements about norms are simply wrong (for example, the second objective is not a norm...). Just say $t \mapsto t^2$ is increasing on $\mathbb{R}_+$ (so that the problems are equivalent) and differentiable (so that the second problem is easier). – dohmatob Aug 24 '15 at 2:35
• @mat: For solving the second problem, checkout this thread which treats a slightly more general problem math.stackexchange.com/q/1309671/168758 – dohmatob Aug 24 '15 at 3:07
Which is easier? Neither. Both of these models can be solved analytically, and in the exact same way. First, let's knock out the easy cases:
1. If any $a_i=0$ for some $i$, then the optimal value of either model is clearly $0$, as demonstrated by selecting setting $x$ to be the $i$th unit vector (the vector with $1$ at position $i$ and zeros everywhere else).
2. If $a_i>0$ and $a_j<0$ for some pair $(i,j)$, then again the optimal value is zero. Just set $$x_i=-a_j/(a_i-a_j), \quad x_j = a_i/(a_i-a_j)$$ and the other elements of $x$ to zero.
3. The cases that remain are where $a$ is entirely positive or entirely negative. But if $a$ is negative, then substituting $a$ for $-a$ will not change the optimal value or the values of $x$ that achieve this value, thanks to the presence of the absolute value or square.
4. So we now have one case remaining: when $a$ is a positive vector. But in this case, $a^Tx$ is positive over the simplex, allowing us to drop the absolute value! \begin{array}{ll} \text{minimize} & a^T x \\ \text{subject to} & \mathbf{1}^T x = 1 \\ & x \geq 0 \end{array} We can solve this by inspection: the optimal value is $\min_i a_i$ for the first model, and $\min_i a_i^2$ for the second. If the minimizing $a_i$ is unique, then the unique solution is the $i$th unit vector. But if there are multiple elements of $a$ with the same magnitude, any convex combination of those corresponding unit vectors is a solution.
Putting it all together, the solution is $\min_i |a_i|$ or $\min_i a_i^2$ if $a\succeq 0$ or $a\preceq 0$, and $0$ otherwise. There really was no need for case 1, but it was easy to see.
• I am interested in the case where the elements of $\mathbf{a}$ are $\ge 0$. I suppose that adding a constant $b$ to the objective function, i.e. $\mathbf{a}^T\mathbf{x}+b$, will not affect the solution ($i$th unit vector) but it will modify the optimal value to $\min_i a_i +b$. Am I right? – tam Aug 24 '15 at 9:37
• Yes, that is right. – Michael Grant Aug 24 '15 at 10:56
• Please, I have an additional question (I don't know if I have to post it in a seperate question): I want to maximize $|\mathbf{a}^T\mathbf{x}-b|$, with $\mathbf{1}^T\mathbf{x}=1$, $\mathbf{x} >0$ and $\mathbf{a}^T\mathbf{x} >b$. How to solve this kind of problems. (note that $b>0$) – tam Aug 29 '15 at 10:42
• I think that this problem can be re-written as: maximize $\mathbf{a}^T \mathbf{x}-b$, with $\mathbf{1}^T\mathbf{x}=1$ and $\mathbf{x}>0$. So the solution is similar to that in 4), with the difference of putting $\max_i a_i-b$ instead of $\min_i |a_i-b|$. Is it right ? – tam Aug 29 '15 at 16:03
Let's go for an explicit analytic construction of the set of all the solutions to your problem.
Basic notation: $e_K := \text{ column vector of }K\text{ }1'$s. $\langle x, y \rangle$ denotes the inner product between two vectors $x$ and $y$. For example $\langle e_K, x\rangle$ simply amounts to summing the components of $x$.
Recall the following definitions, to be used without further explanation.
Preimage: Given abstract sets $X$, $Y$, $Z \subseteq Y$., the preimage of $Z$ under $f$ is defined by $f^{-1}Z := \{x \in X | f(x) \in Z\}$. This has nothing to do with function inversion.
Convex hull: Given a subset $C$ of $\mathbb{R}^K$, its convex hull is defined by $\textit{conv }C :=$ smallest convex subset of $\mathbb{R}^K$ containing $C$.
Indicator function: Given a subset $C$ of $\mathbb{R}^K$ its indicator function $i_C:\mathbb{R}^K \rightarrow (-\infty, +\infty]$ is defined by $i_C(x) = 0$ if $x \in C$; otherwise $i_C(x) = +\infty$.
Simplex: The $K$-dimensional simplex is defined by $\Delta_K := \{x \in \mathbb{R}^K|x\ge 0, \langle e_K, x\rangle = 1\}$. Equivalently, $\Delta_K = \{\text{rows of the }K \times K\text{ identity matrix }I_K\}$. Each row of $I_K$ is a vertex of $\Delta_K$.
Faces of a simplex: Given a subset of indices $I \subseteq \{1,2,...,K\}$, the face of $\Delta_K$ spanned by $I$, denoted $F_K(I)$, is defined to be the convex hull of those rows of $I_k$ indexed by $I$. Trivially, $\Delta_K$ is a face of itself. Geometrically, $F_K(I)$ is a $\#I$-dimensional simplex whose vertices are those vertices of $\Delta_K$ which labelled by $I$.
Let $f:\mathbb{R}^K \rightarrow (-\infty,+\infty]$ be an extended real-value function.
Subdifferential: $\partial f(x) := \{g \in \mathbb{R}^K| f(z) \ge f(s) + \langle g, z - x\rangle, \forall z \in \mathbb{R}^K\}$.
Fenchel-Legengre transform: $f^*(x) := \underset{z \in \mathbb{R}^K}{\text{max }}\langle x, z\rangle - f(z)$.
The optimal value of your objective under the given constraints can be conveniently written as \begin{eqnarray} v = \underset{x \in \mathbb{R}^K}{\text{min }}g(x) + f(Dx), \end{eqnarray} where $D := a^T \in \mathbb{R}^{1 \times K}$, $g := i_{\Delta_K}$ and $f:s \mapsto \frac{1}{2}s^2$. One recognizes the above problem as the primal form of the saddle-point problem \begin{eqnarray*} \underset{x \in \mathbb{R}^K}{\text{min }}\underset{s \in \mathbb{R}}{\text{max }}\langle s, Dx\rangle + g(x) - f^*(s) \end{eqnarray*} Given a dual solution $\hat{s} \in \mathbb{R}$, the entire set of primal solutions $\hat{X}$ is given by (one can check that sufficient conditions that warrant the strong Fenchel duality Theorem) \begin{eqnarray*} \hat{X} = \partial g^*(-D^T\hat{s}) \cap D^{-1}\partial f^*(\hat{s}). \end{eqnarray*} For your problem, one easily computes, $g^*(y) = \underset{x \in \Delta_K}{\text{max }}\langle x, y\rangle$, and so by the Bertsekas-Danskin theorem (see Proposition A.22 of Bertsekas' PhD thesis) for subdifferentials, we have \begin{eqnarray*} \begin{split} \partial g^*(-D^T\hat{s}) &= \partial g^*(-\hat{s}a) = ... \text{( some computations masked )} \\ &= F_K\left(\{1 \le i \le K | \hat{s}a_i\text{ is a minimal component of }\hat{s}a\}\right). \end{split} \end{eqnarray*} Also $\partial f^*(\hat{s}) = \{\hat{s}\}$, and so $D^{-1}\partial f^*(\hat{s}) = \{x \in \mathbb{R}^K|\langle a, x\rangle = \hat{s}\}$. Thus the set of all primal solutions is \begin{eqnarray*} \hat{X} = F_K\left(\{1 \le i \le K | \hat{s}a_i\text{ is a minimal component of }\hat{s}a\}\right) \cap \{x \in \mathbb{R}^K|\langle a, x\rangle = \hat{s}\}. \end{eqnarray*} It remains now to find a dual solution. Define $\alpha := \underset{1 \le i \le K}{\text{min }}a_i \le \beta := \underset{1 \le i \le K}{\text{max }}a_i$. One computes \begin{eqnarray*} \begin{split} v &= \underset{x \in \Delta_K}{\text{min }}\underset{s \in \mathbb{R}}{\text{max }}s\langle a, x\rangle -\frac{1}{2}s^2 = \underset{s, \lambda \in \mathbb{R}}{\text{max }}\underset{x \ge 0}{\text{min }}\langle sa + \lambda e_K, x\rangle - \frac{1}{2}s^2 - \lambda\\ &=\underset{s, \lambda \in \mathbb{R}}{\text{max }}-\frac{1}{2}s^2 - \lambda\text{ subject to }sa + \lambda e_K \ge 0\\ &=-\underset{s, \lambda \in \mathbb{R}}{\text{min }}\frac{1}{2}s^2 + \lambda\text{ subject to }\lambda \ge -\underset{1 \le i \le K}{\text{min }}sa_i\\ &= -\underset{s \in \mathbb{R}}{\text{min }}\frac{1}{2}s^2 -\underset{1 \le i \le K}{\text{min }}sa_i = -\frac{1}{2}\underset{s \in \mathbb{R}}{\text{min }}\begin{cases}(s - \alpha)^2 - \alpha^2, &\mbox{ if }s \ge 0,\\(s - \beta)^2 - \beta^2, &\mbox{ otherwise}\end{cases} \end{split} \end{eqnarray*} Thus \begin{eqnarray*} \hat{s} = \begin{cases}\beta, &\mbox{ if }\beta < 0,\\0, &\mbox{ if }\alpha \le 0 \le \beta,\\\alpha, &\mbox{ if }\alpha > 0.\end{cases} \end{eqnarray*} Therefore the set of all solutions to the original / primal problem is \begin{eqnarray*} \hat{X} = \begin{cases}F_K\left(\{1 \le i \le K | a_i = \beta\}\right), &\mbox{ if }\beta < 0,\\ \Delta_K \cap \{x \in \mathbb{R}^K| \langle a, x\rangle = 0\}, &\mbox{ if } \alpha \le 0 \le \beta,\\ F_K\left(\{1 \le i \le K | a_i = \alpha\}\right), &\mbox{ if }\alpha > 0,\end{cases} \end{eqnarray*}
Now that you have the hammer, find the nails...
• Case (1): $\beta < 0$. Choose any $k \in \{1,2,...,K\}$ such that $a_k = \beta$, and take $\hat{x} = k$th row of $I_K$.
• Case (2a): $a_k = 0$ for some $k$. Take $\hat{x} = k$th row of $I_K$.
• Case (2b): $\alpha < 0 < \beta$, and $a_k \ne 0 \forall k$. Choose $k_1, k_2 \in \{1,2,...,k\}$ such that $a_{k_1} < 0 < a_{k_2}$, and take $\hat{x}_k = \frac{1}{{a_{k_2} - a_{k_1}}}\begin{cases}a_{k_2}, &\mbox{ if }k = k_1,\\-a_{k_1},&\mbox{ if }k = k_2,\\0, &\mbox{ otherwise.}\end{cases}$
• Case (3): $\alpha > 0$. Choose any $k \in \{1,2,...,K\}$ such that $a_k = \alpha$, and take $\hat{x} = k$th row of $I_K$.
• Dude. If I were a professor I'd want you in my research group cranking out papers. This is impressive, and you have my vote. But don't you think this is like bringing a fire truck to put out a campfire? – Michael Grant Aug 24 '15 at 14:41
• @MichaelGrant: Your solution is actually more elegant in many respects. Mine is indeed probably an overkill. Mindful of this, I tried to be particularly pedagocial here :) On such toy problems ($D$ is a vector, $g^*$ is a polyhedral function, etc.), it turns out that one can explicitly construct the entire solution set using Fenchel's machinery. Unfortunately, this is not always the case. :) – dohmatob Aug 25 '15 at 6:33
Your problem can be equivalently modeled to LP with an extra variable, say $\epsilon$ and two more simple linear constraints, $| \mathbf{a}^T \mathbf{x} |\leq \epsilon$. Now your problem can be rewritten as $$\begin{array}{rl} \min \limits_{\mathbf{x}} & \epsilon \\ \mbox{s.t.} & \mathbf{a}^T \mathbf{x} \leq \epsilon \\&- \mathbf{a}^T \mathbf{x} \leq \epsilon \\ & x_k\ge 0, \forall k \\ & \sum_k x_k=1 \end{array}$$
Now you can solve this problem using simplex method or any other method you like! | 2019-10-17T17:57:13 | {
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https://mathhelpboards.com/threads/for-which-parameter-values-the-function-is-continuous-and-differentiable.3380/ | # For which parameter values the function is continuous and differentiable
#### Yankel
##### Active member
For which values of a,b and c, the next function is continuous and differentiable at x=2 ?
$$\left\{\begin{matrix} 3x-1 & x\leq 2\\ ax^{2}+bx+c & x>2 \end{matrix}\right.$$
1. b=2-c
2. b=6+2c+2a
3. 7+c-2a
4. b=3-a-(3/4)c
I know that f(2)=5, and so is the limit of f when x goes to 2 from the left side.
I have calculated the limit when f goes to 2 from the right side, and I got:
4a+2b+c
and so for continuously I need:
4a+2b+c = 5
but here I got stuck...
#### Bacterius
##### Well-known member
MHB Math Helper
The only point you need to look at is $x = 2$, as all the others must be continuous and differentiable (remember all polynomials are well-behaved).
Recall that for a function to be continuous at $a$, you need:
$$\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)$$
So in your case, you need to show that the following is true:
$$\lim_{x \to 2^{-}} 3x - 1 = \lim_{x \to 2^{+}} ax^2 + bx + c ~ ~ ~ \Longleftrightarrow ~ ~ ~ 5 = 4a + 2b + c$$
Which is what you got, so this is correct. But, you also need differentiability.
Recall that for a function $f(x)$ to be differentiable at $a$, you need:
$$\lim_{x \to a^{-}} f'(x) = \lim_{x \to a^{+}} f'(x)$$
So, differentiate both pieces of the function, and show that the following is true:
$$\lim_{x \to 2^{-}} 3 = \lim_{x \to 2^{+}} 2ax + b ~ ~ ~ \Longleftrightarrow ~ ~ ~ 3 = 4a + b$$
You are now left with two equations in three unknowns:
$$4a + 2b + c = 5$$
$$4a + b = 3$$
Subtract the second from the first, to obtain:
$$b + c = 2 ~ ~ ~ \Longleftrightarrow b = 2 - c$$
And substitute this back into the second:
$$4a + 2 - c = 3 ~ ~ ~ \Longleftrightarrow ~ ~ ~ a = \frac{1}{4} \left ( c + 1 \right )$$
This gives you $a$ and $b$ in terms of $c$. Choose any $c$, for instance $c = 2$, and you get:
$$a = \frac{3}{4} ~ ~ ~ \text{and} ~ ~ ~ b = 0$$
That is one solution, out of infinitely many. This is also the only class of solutions.
Last edited:
#### chisigma
##### Well-known member
For which values of a,b and c, the next function is continuous and differentiable at x=2 ?
$$\left\{\begin{matrix} 3x-1 & x\leq 2\\ ax^{2}+bx+c & x>2 \end{matrix}\right.$$
1. b=2-c
2. b=6+2c+2a
3. 7+c-2a
4. b=3-a-(3/4)c
I know that f(2)=5, and so is the limit of f when x goes to 2 from the left side.
I have calculated the limit when f goes to 2 from the right side, and I got:
4a+2b+c
and so for continuously I need:
4a+2b+c = 5
but here I got stuck...
You have to impose first that $\displaystyle \lim_{x \rightarrow 2 +} f(x) = \lim_{x \rightarrow 2 -} f(x)$ and after that $\displaystyle \lim_{x \rightarrow 2 +} f ^{\ '} (x) = \lim_{x \rightarrow 2 -} f^{\ '}(x)$...
Kind regards
$\chi$ $\sigma$
#### HallsofIvy
##### Well-known member
MHB Math Helper
The only point you need to look at is $x = 2$, as all the others must be continuous and differentiable (remember all polynomials are well-behaved).
Recall that for a function to be continuous at $a$, you need:
$$\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x)$$
So in your case, you need to show that the following is true:
$$\lim_{x \to 2^{-}} 3x - 1 = \lim_{x \to 2^{+}} ax^2 + bx + c ~ ~ ~ \Longleftrightarrow ~ ~ ~ 5 = 4a + 2b + c$$
Which is what you got, so this is correct. But, you also need differentiability.
Recall that for a function $f(x)$ to be differentiable at $a$, you need:
$$\lim_{x \to a^{-}} f'(x) = \lim_{x \to a^{+}} f'(x)$$
This is true, but not an obvious statement. It looks like the definition of "continuous" but derivatives are NOT necessarily contuinuous. What is true is that any derivative, even if not continuous, satisfies the "intermediate value theorem".
So, differentiate both pieces of the function, and show that the following is true:
$$\lim_{x \to 2^{-}} 3 = \lim_{x \to 2^{+}} 2ax + b ~ ~ ~ \Longleftrightarrow ~ ~ ~ 3 = 4a + b$$
You are now left with two equations in three unknowns:
$$4a + 2b + c = 5$$
$$4a + b = 3$$
Subtract the second from the first, to obtain:
$$b + c = 2 ~ ~ ~ \Longleftrightarrow b = 2 - c$$
And substitute this back into the second:
$$4a + 2 - c = 3 ~ ~ ~ \Longleftrightarrow ~ ~ ~ a = \frac{1}{4} \left ( c + 1 \right )$$
This gives you $a$ and $b$ in terms of $c$. Choose any $c$, for instance $c = 2$, and you get:
$$a = \frac{3}{4} ~ ~ ~ \text{and} ~ ~ ~ b = 0$$
That is one solution, out of infinitely many. This is also the only class of solutions. | 2021-09-25T07:18:38 | {
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https://math.stackexchange.com/questions/943384/how-to-compute-sum-k-1100-1k | How to compute $\sum_{k =1}^{100}(-1)^k$
Today I tried to compute $$\sum_{k =1}^{100}(-1)^k$$
Is there a way to find the result more quickly ?
Below if my attempt to find the result.
Especially without considering the case of odd and even numbers like I did ?
Let's consider the following sum:
$$\sum_{k=1}^{n}(-1)^k = (-1)^1 + (-1)^2 + (-1)^3 + ... +(-1)^n$$
If $n$ is even then $\frac n2$ terms have an even exponent and $\frac n2$ terms have an odd exponent.
$$(-1)^p = -1 \tag{when p is odd}$$ $$(-1)^p = 1 \tag{when p is even}$$
Then when $n$ is even
\begin{align} \sum_{k=1}^{n}(-1)^k &= \frac n2 \times(-1) + \frac n2 \times 1 \\ & = \frac n2 - \frac n2 \\ & = 0 \\ \end{align}
100 is an even number, so
$$\sum_{k =1}^{100}(-1)^k = 0$$
• Call the sum $S$, so that $S=(-1)+(1)+\dotsb+(-1)+(1)$. Notice that $-S=(1)+(-1)+\dotsb+(1)+(-1)$—that is, $-S$ is just $S$ reversed. So, $S=-S$, which means that $S=0$. – Akiva Weinberger Sep 23 '14 at 20:15
• If you say that $n/2$ terms have an even exponent, then you are already assuming that $n$ is even. – Dietrich Burde Sep 23 '14 at 20:15
• The quick way is to cancel $1$s with $-1$s. You have equally many of both, and if you didn't, you'd have just one term left after canceling. But of course it's also a finite geometric series, and there's a standard formula for that. – Michael Hardy Sep 23 '14 at 20:57
yes the series has a direct summation formula: $$S_k=\sum_{i=1}^k (-1)^i={1 \over 2}\Big((-1)^k-1\Big)$$ you can prove this easily using induction. \begin{aligned} \mbox{k=1: }\ \ \ &true\\ \mbox{k }\to\mbox{ k+1: }\ \ \ &S_{k+1}=S_{k}+(-1)^{k+1}&={1 \over 2}\Big((-1)^k-1\Big)+(-1)^{k+1} =\\ & &={1 \over 2}\Big((-1)^k+2(-1)^{k+1}-1\Big) =\\ & &={1 \over 2}\Big((-1)^k(1+2(-1)^{1})-1\Big) =\\ & &={1 \over 2}\Big((-1)^k(1-2)-1\Big) =\\ & &={1 \over 2}\Big((-1)^k(-1)-1\Big) =\\ & &={1 \over 2}\Big((-1)^{k+1}-1\Big) \ \ \ \square \end{aligned}
Note that $(-1)^{2k-1}=-1$ and $(-1)^{2k}=1$ for all $k\geq1$. Hence the sum upto any even power should equal $0$.
The easiest way to find the result is to look at the first few partial sums
$$-1, 0, -1, \ldots$$
and identify the pattern. Everything beyond that is just seeking to give a more rigorous justification of it.
• Sure. I think that's what I am looking for. A rigorous justification but explained quickly. – alexandrekow Sep 23 '14 at 20:37
Since $(-1)^{2n}=1$ and $(-1)^{2n-1}=-1$, we have $$\sum_{k=1}^n (-1)^k = -1+1-1+1-\cdots \pm 1=\left\{\begin{array}{ll}-1 & \text{if}\,n\,\text{is odd}\\ 0 & \text{if}\,n\,\text{is even}\end{array}\right.$$
$$\sum_{k=1}^{n}(-1)^k=\sum_{k=1}^{n}(-1)(-1)^{k-1}=-\sum_{k=1}^{n}(-1)^{k-1}=$$ $$=-\sum_{k=0}^{n-1}(-1)^{k}=-\frac{1-(-1)^{n}}{1-(-1)}=\frac{(-1)^{n}-1}{2}$$ for $n=100$ $$\sum_{k=1}^{100}(-1)^k=\frac{(-1)^{100}-1}{2}=\frac{1-1}{2}=0$$
• How do you do to transform the third part of the first line into the fourth one? – alexandrekow Sep 23 '14 at 20:39
• This is of course the hard way. – Michael Hardy Sep 23 '14 at 20:57
• you can see my edit! – Adi Dani Sep 23 '14 at 20:59 | 2019-08-19T01:35:08 | {
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https://byjus.com/question-answer/assertion-for-n-ge-4-let-displaystyle-a-n-sum-j-1-n-sum-k/ | Question
# Assertion :For $$n\ge 4$$. Let $$\displaystyle{ a }_{ n }=\sum _{ j=1 }^{ n }{ \sum _{ k=j }^{ n }{ \begin{pmatrix} n \\ k \end{pmatrix} } } \begin{pmatrix} k \\ j \end{pmatrix}$$ and $${ b }_{ n }={ a }_{ n }+{ 2 }^{ n+1 }$$;$${ b }_{ n+1 }=5{ b }_{ n }-6{ b }_{ n-1 }$$ for all $$n\ge 2$$ Reason: $${ a }_{ n }=5{ a }_{ n }-6{ a }_{ n }$$ for all $$n\ge 2$$
A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C
Assertion is correct but Reason is incorrect
D
Assertion is incorrect but Reason is correct
Solution
## The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion$$\displaystyle \begin{pmatrix} n \\ k \end{pmatrix}\begin{pmatrix} k \\ j \end{pmatrix}=\frac { n! }{ k!\left( n-k \right) ! } .\frac { k! }{ j!\left( k-j \right) ! }$$$$\displaystyle=\frac { n! }{ j!\left( n-j \right) ! } \frac { \left( n-j \right) ! }{ \left( n-k \right) !\left( k-j \right) ! } =\begin{pmatrix} n \\ k \end{pmatrix}\begin{pmatrix} n\quad - & j \\ n\quad - & k \end{pmatrix}$$Thus $$\displaystyle { a }_{ n }=\sum _{ j=1 }^{ n }{ \begin{pmatrix} h \\ k \end{pmatrix} } \sum _{ j=1 }^{ n }{ \begin{pmatrix} n\quad - & j \\ n\quad - & k \end{pmatrix} }$$$$\displaystyle=\sum _{ j=1 }^{ n }{ \begin{pmatrix} n \\ j \end{pmatrix} } { 2 }^{ n-j }={ \left( 2+1 \right) }^{ n }-{ 2 }^{ n }={ 3 }^{ n }-{ 2 }^{ n }$$$$\Rightarrow { b }_{ 1 }={ 3 }^{ n }{ -2 }^{ n }+2\left( { 2 }^{ n } \right) ={ 3 }^{ n }+{ 2 }^{ n }$$Now $${ a }_{ n+1 }={ 3 }^{ n+1 }-{ 2 }^{ n+1 }=\left( 3+2 \right) \left( { 3 }^{ n }-{ 2 }^{ n } \right) -2\left( { 3 }^{ n } \right) +3\left( { 2 }^{ n } \right)$$$$=5\left( { 3 }^{ n }-{ 2 }^{ n } \right) -6\left( { 3 }^{ n-1 }-{ 2 }^{ n+1 } \right) ={ 5a }_{ n }-{ 6a }_{ n-1 }$$And $${ 5b }_{ n }-{ 6b }_{ n-1 }=5\left( { a }_{ n }+{ 2 }^{ n+1 } \right) -6\left( { a }_{ n-1 }+{ 2 }^{ n } \right)$$$$={ 5a }_{ n }-6{ b }_{ n-1 }+{ 2 }^{ n }\left( 10-6 \right)$$$$\Rightarrow { a }_{ n+1 }+{ 2 }^{ n+2 }={ b }_{ n+1 }$$Mathematics
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https://math.stackexchange.com/questions/3321478/evaluating-lim-k-to-infty-sum-n-1-infty-frac-sin-left-pi-n-k-right | # Evaluating $\lim_{k\to\infty}\sum_{n=1}^{\infty} \frac{\sin\left(\pi n/k\right)}{n}$
Recently, I was asked by a friend to compute the limit of the following series
$$\displaystyle{\lim_{k\to\infty}}\sum_{n=1}^{\infty} \frac{\sin\left(\frac{\pi n}{k}\right)}{n}$$
Having seen a similar problem to this before, Difficult infinite trigonometric series, I used the same complex argument approach as seen in that problem.
Ultimately, for this problem, I obtained $$\frac{\pi}{2}$$ as my answer. However, this limit can also be interpreted as a Riemann Sum, except the answer to the Riemann Sum differs from what I obtained as my answer, and according to Wolfram Alpha, the answer is expressed in terms of $$Si$$, where $$Si$$ is the sine integral.
I'm wondering, does the limit invalidate the argument approach, or is there something else I'm missing, because this limit if I'm not mistaken is a Riemann Sum after all?
• In a Riemann sum we would have both 'variables' tending to infinity simultaneously. In this case we have both $n$ and $k$ tending to infinity seperately and hence the Riemann sum approach should not be possible. This approach seems valid. – Peter Foreman Aug 12 at 21:48
• Nor can you (directly) apply dominated cvg theorem ... – Olivier Aug 12 at 21:51
• You effectively have $\lim_{k\to\infty}\lim_{N\to\infty}f(N,k)$ (where $N$ denotes the maximum summation bound). With a Riemann sum we would have $\lim_{N\to\infty}f(N)$ where $k$ may be used as an index such that the function $f(N)$ is of the form $\frac1N\sum_{k=1}^N g(k/N)\to\int_0^1g(x)\mathrm{d}x$. – Peter Foreman Aug 12 at 21:54
• (Rephrasing of what Peter says) You have $\lim_{k \to \infty} \frac{1}{k} \sum_{n=1}^{k} \frac{\sin(\pi n/k)}{(n/k)} \to \int_{0}^{1} \frac{sin(\pi x)}{x} dx$ by Riemann sum argument; but you have to manage a larger interval here... ($n\ge 1$, not just $n=1...k$) – Olivier Aug 12 at 22:03
• You can do a collection of Riemann sums to deal with n=1...k, then n=k+1...2k, n=2k+1...3k, but you would then need a further argument to put these sums altogether. – Olivier Aug 12 at 22:16
The Riemann Sum would be \begin{align} \lim_{k\to\infty}\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n} &=\lim_{k\to\infty}\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n/k}\frac1k\\ &=\int_0^\infty\frac{\sin(\pi x)}x\,\mathrm{d}x\\ &=\int_0^\infty\frac{\sin(x)}x\,\mathrm{d}x\\[3pt] &=\frac\pi2\tag1 \end{align} However, a cleaner way is to note that \begin{align} \sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n} &=-\mathrm{Im}\!\left(\log\left(1-e^{i\pi/k}\right)\right)\\ &=\frac\pi2-\frac\pi{2k}\tag2 \end{align} and the limit is easy.
Take Care
One must be careful with the convergence of the Riemann Sum. Here is one method to control the remainders.
Because $$|\sin(\pi x)|\le1$$, we have $$\int_m^{m+1}\left|\frac{\sin(\pi x)}x\right|\,\mathrm{d}x \le\frac1m\tag3$$ Furthermore, $$\int_m^{m+2}\sin(\pi x)\,\mathrm{d}x=0$$, thus, \begin{align} \left|\int_m^{m+2}\frac{\sin(\pi x)}x\,\mathrm{d}x\right| &=\left|\int_m^{m+2}\sin(\pi x)\left(\frac1x-\frac1{m+1}\right)\mathrm{d}x\right|\\ &\le\frac1{m(m+1)}+\frac1{(m+1)(m+2)}\\[6pt] &=\frac1m-\frac1{m+2}\tag4 \end{align} Therefore, for any $$N\ge m$$, $$\left|\int_m^N\frac{\sin(\pi x)}x\,\mathrm{d}x\right| \le\frac1m\tag5$$ Because $$|\sin(\pi x)|\le1$$, we have $$\sum_{n=mk}^{(m+1)k}\left|\frac{\sin\left(\frac{\pi n}k\right)}{n}\right| \le\frac1m\tag6$$ Furthermore, $$\sum\limits_{n=mk}^{(m+2)k}\sin\left(\frac{\pi n}k\right)=0$$, thus, \begin{align} \left|\sum_{n=mk}^{(m+2)k}\frac{\sin\left(\frac{\pi n}k\right)}{n}\right| &=\left|\sum_{n=mk}^{(m+2)k}\sin\left(\frac{\pi n}k\right)\left(\frac1n-\frac1{(m+1)k}\right)\right|\\ &\le\frac1{m(m+1)}+\frac1{(m+1)(m+2)}\\[6pt] &=\frac1m-\frac1{m+2}\tag7 \end{align} Therefore, for any $$M\ge mk$$, $$\left|\sum_{n=mk}^M\frac{\sin\left(\frac{\pi n}k\right)}{n}\right|\le\frac1m\tag8$$ For any $$\epsilon\gt0$$, let $$m\ge\frac4\epsilon$$. Then Riemann Sums allow us to choose a $$k$$ large enough so that $$\left|\int_0^m\frac{\sin(\pi x)}x\,\mathrm{d}x-\sum_{n=1}^{mk}\frac{\sin\left(\frac{\pi n}k\right)}{n/k}\frac1k\right|\le\frac\epsilon2\tag9$$ Inequalities $$(5)$$ and $$(8)$$ show that for any $$N\ge m$$ and $$M\ge mk$$, $$\left|\int_m^N\frac{\sin(\pi x)}x\,\mathrm{d}x\right|\le\frac\epsilon4 \quad\text{and}\quad \left|\sum_{n=mk}^M\frac{\sin\left(\frac{\pi n}k\right)}{n/k}\frac1k\right|\le\frac\epsilon4\tag{10}$$ Inequalities $$(9)$$ and $$(10)$$ show that, for the $$k$$ chosen for $$(9)$$, $$\left|\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n}-\frac\pi2\right|\le\epsilon\tag{11}$$ Since $$\epsilon\gt0$$ was arbitrary, $$(11)$$ says that $$\lim_{k\to\infty}\sum_{n=1}^\infty\frac{\sin\left(\frac{\pi n}k\right)}{n}=\frac\pi2\tag{12}$$
• Riemann sums works on a finite intervals; what's the additional argument used here to extend this to (0,\infty) ? – Olivier Aug 12 at 22:11
• I've gotta say, that is one sexy solution and elegant approach! +1 :) – Sanjoy Kundu Aug 12 at 22:16
• @Olivier: In this case, we can take the sum from $1$ to $mk$ to get the integral from $0$ to $m$. Then carefully let $m\to\infty$. – robjohn Aug 12 at 22:27
• @Olivier: I believe one can use Abel's Test to estimate the remainder. – robjohn Aug 12 at 22:42
• @Olivier: I have added a section showing one method to control the remainders. It is a bit long, to include a lot of detail, but quite typical for this kind of Riemann Sum. – robjohn Aug 13 at 8:55 | 2019-11-21T08:03:16 | {
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https://de.mathworks.com/help/matlab/math/solve-system-of-odes-with-multiple-initial-conditions.html | Main Content
# Solve System of ODEs with Multiple Initial Conditions
This example compares two techniques to solve a system of ordinary differential equations with multiple sets of initial conditions. The techniques are:
• Use a `for`-loop to perform several simulations, one for each set of initial conditions. This technique is simple to use but does not offer the best performance for large systems.
• Vectorize the ODE function to solve the system of equations for all sets of initial conditions simultaneously. This technique is the faster method for large systems but requires rewriting the ODE function so that it reshapes the inputs properly.
The equations used to demonstrate these techniques are the well-known Lotka-Volterra equations, which are first-order nonlinear differential equations that describe the populations of predators and prey.
### Problem Description
The Lotka-Volterra equations are a system of two first-order, nonlinear ODEs that describe the populations of predators and prey in a biological system. Over time, the populations of the predators and prey change according to the equations
`$\begin{array}{l}\frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \mathit{x}-\beta \mathrm{xy},\\ \frac{\mathrm{dy}}{\mathrm{dt}}=\delta \mathrm{xy}-\gamma \mathit{y}.\end{array}$`
The variables in these equations are
• $\mathit{x}$ is the population size of the prey
• $\mathit{y}$ is the population size of the predators
• $\mathit{t}$ is time
• $\alpha$, $\beta$, $\delta$, and $\gamma$ are constant parameters that describe the interactions between the two species. This example uses the parameter values $\alpha =\gamma =1$, $\beta =0.01$, and $\delta =0.02$.
For this problem, the initial values for $\mathit{x}$ and $\mathit{y}$ are the initial population sizes. Solving the equations then provides information about how the populations change over time as the species interact.
### Solve Equations with One Initial Condition
To solve the Lotka-Volterra equations in MATLAB, write a function that encodes the equations, specify a time interval for the integration, and specify the initial conditions. Then you can use one of the ODE solvers, such as `ode45`, to simulate the system over time.
A function that encodes the equations is
```function dpdt = lotkaODE(t,p) % LOTKA Lotka-Volterra predator-prey model delta = 0.02; beta = 0.01; dpdt = [p(1) .* (1 - beta*p(2)); p(2) .* (-1 + delta*p(1))]; end ```
(This function is included as a local function at the end of the example.)
Since there are two equations in the system, `dpdt` is a vector with one element for each equation. Also, the solution vector `p` has one element for each solution component: `p(1)` represents $\mathit{x}$ in the original equations, and `p(2)` represents $\mathit{y}$ in the original equations.
Next, specify the time interval for integration as $\left[0,15\right]$ and set the initial population sizes for $\mathit{x}$ and $\mathit{y}$ to 50.
```t0 = 0; tfinal = 15; p0 = [50; 50];```
Solve the system with `ode45` by specifying the ODE function, the time span, and the initial conditions. Plot the resulting populations versus time.
```[t,p] = ode45(@lotkaODE,[t0 tfinal],p0); plot(t,p) title('Predator/Prey Populations Over Time') xlabel('t') ylabel('Population') legend('Prey','Predators')```
Since the solutions exhibit periodicity, plot the solutions against each other in a phase plot.
```plot(p(:,1),p(:,2)) title('Phase Plot of Predator/Prey Populations') xlabel('Prey') ylabel('Predators')```
The resulting plots show the solution for the given initial population sizes. To solve the equations for different initial population sizes, change the values in `p0` and rerun the simulation. However, this method only solves the equations for one initial condition at a time. The next two sections describe techniques to solve for many different initial conditions.
### Method 1: Compute Multiple Initial Conditions with `for-`loop
The simplest way to solve a system of ODEs for multiple initial conditions is with a `for`-loop. This technique uses the same ODE function as the single initial condition technique, but the `for`-loop automates the solution process.
For example, you can hold the initial population size for $\mathit{x}$ constant at 50, and use the `for`-loop to vary the initial population size for $\mathit{y}$ between 10 and 400. Create a vector of population sizes for `y0`, and then loop over the values to solve the equations for each set of initial conditions. Plot a phase plot with the results from all iterations.
```y0 = 10:10:400; for k = 1:length(y0) [t,p] = ode45(@lotkaODE,[t0 tfinal],[50 y0(k)]); plot(p(:,1),p(:,2)) hold on end title('Phase Plot of Predator/Prey Populations') xlabel('Prey') ylabel('Predators') hold off```
The phase plot shows all of the computed solutions for the different sets of initial conditions.
### Method 2: Compute Multiple Initial Conditions with Vectorized ODE Function
Another method to solve a system of ODEs for multiple initial conditions is to rewrite the ODE function so that all of the equations are solved simultaneously. The steps to do this are:
• Provide all of the initial conditions to `ode45` as a matrix. The size of the matrix is `s`-by-`n`, where `s` is the number of solution components and `n` is the number of initial conditions being solved for. Each column in the matrix then represents one complete set of initial conditions for the system.
• The ODE function must accept an extra input parameter for `n`, the number of initial conditions.
• Inside the ODE function, the solver passes the solution components p as a column vector. The ODE function must reshape the vector into a matrix with size `s`-by-n. Each row of the matrix then contains all of the initial conditions for each variable.
• The ODE function must solve the equations in a vectorized format, so that the expression accepts vectors for the solution components. In other words, `f(t,[y1 y2 y3 ...])` must return `[f(t,y1) f(t,y2) f(t,y3) ...]`.
• Finally, the ODE function must reshape its output back into a vector so that the ODE solver receives a vector back from each function call.
If you follow these steps, then the ODE solver can solve the system of equations using a vector for the solution components, while the ODE function reshapes the vector into a matrix and solves each solution component for all of the initial conditions. The result is that you can solve the system for all of the initial conditions in one simulation.
To implement this method for the Lotka-Volterra system, start by finding the number of initial conditions `n`, and then form a matrix of initial conditions.
```n = length(y0); p0_all = [50*ones(n,1) y0(:)]';```
Next, rewrite the ODE function so that it accepts `n` as an input. Use `n` to reshape the solution vector into a matrix, then solve the vectorized system and reshape the output back into a vector. A modified ODE function that performs these tasks is
```function dpdt = lotkasystem(t,p,n) %LOTKA Lotka-Volterra predator-prey model for system of inputs p. delta = 0.02; beta = 0.01; % Change the size of p to be: Number of equations-by-number of initial % conditions. p = reshape(p,[],n); % Write equations in vectorized form. dpdt = [p(1,:) .* (1 - beta*p(2,:)); p(2,:) .* (-1 + delta*p(1,:))]; % Linearize output. dpdt = dpdt(:); end ```
Solve the system of equations for all of the initial conditions using `ode45`. Since `ode45` requires the ODE function to accept two inputs, use an anonymous function to pass in the value of `n` from the workspace to `lotkasystem`.
`[t,p] = ode45(@(t,p) lotkasystem(t,p,n),[t0 tfinal],p0_all);`
Reshape the output vector into a matrix with size `(numTimeSteps*s)`-by-`n`. Each column of the output `p(:,k)` contains the solutions for one set of initial conditions. Plot a phase plot of the solution components.
```p = reshape(p,[],n); nt = length(t); for k = 1:n plot(p(1:nt,k),p(nt+1:end,k)) hold on end title('Predator/Prey Populations Over Time') xlabel('t') ylabel('Population') hold off```
The results are comparable to those obtained by the `for`-loop technique. However, there are some properties of the vectorized solution technique that you should keep in mind:
• The calculated solutions can be slightly different than those computed from a single initial input. The difference arises because the ODE solver applies norm checks to the entire system to calculate the size of the time steps, so the time-stepping behavior of the solution is slightly different. The change in time steps generally does not affect the accuracy of the solution, but rather which times the solution is evaluated at.
• For stiff ODE solvers (`ode15s`, `ode23s`, `ode23t`, `ode23tb`) that automatically evaluate the numerical Jacobian of the system, specifying the block diagonal sparsity pattern of the Jacobian using the `JPattern` option of `odeset` can improve the efficiency of the calculation. The block diagonal form of the Jacobian arises from the input reshaping performed in the rewritten ODE function.
### Compare Timing Results
Time each of the previous methods using `timeit`. The timing for solving the equations with one set of initial conditions is included as a baseline number to see how the methods scale.
```% Time one IC baseline = timeit(@() ode45(@lotkaODE,[t0 tfinal],p0),2); % Time for-loop for k = 1:length(y0) loop_timing(k) = timeit(@() ode45(@lotkaODE,[t0 tfinal],[50 y0(k)]),2); end loop_timing = sum(loop_timing); % Time vectorized fcn vectorized_timing = timeit(@() ode45(@(t,p) lotkasystem(t,p,n),[t0 tfinal],p0_all),2);```
Create a table with the timing results. Multiply all of the results by 1e3 to express the times in milliseconds. Include a column with the time per solution, which divides each time by the number of initial conditions being solved for.
```TimingTable = table(1e3.*[baseline; loop_timing; vectorized_timing], 1e3.*[baseline; loop_timing/n; vectorized_timing/n],... 'VariableNames',{'TotalTime (ms)','TimePerSolution (ms)'},'RowNames',{'One IC','Multi ICs: For-loop', 'Mult ICs: Vectorized'})```
```TimingTable=3×2 table TotalTime (ms) TimePerSolution (ms) ______________ ____________________ One IC 1.6558 1.6558 Multi ICs: For-loop 34.333 0.85832 Mult ICs: Vectorized 5.1357 0.12839 ```
The `TimePerSolution` column shows that the vectorized technique is the fastest of the three methods.
### Local Functions
Listed here are the local functions that `ode45` calls to calculate the solutions.
```function dpdt = lotkaODE(t,p) % LOTKA Lotka-Volterra predator-prey model delta = 0.02; beta = 0.01; dpdt = [p(1) .* (1 - beta*p(2)); p(2) .* (-1 + delta*p(1))]; end %------------------------------------------------------------------ function dpdt = lotkasystem(t,p,n) %LOTKA Lotka-Volterra predator-prey model for system of inputs p. delta = 0.02; beta = 0.01; % Change the size of p to be: Number of equations-by-number of initial % conditions. p = reshape(p,[],n); % Write equations in vectorized form. dpdt = [p(1,:) .* (1 - beta*p(2,:)); p(2,:) .* (-1 + delta*p(1,:))]; % Linearize output. dpdt = dpdt(:); end```
Download ebook | 2021-09-24T11:23:36 | {
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https://mathhelpboards.com/threads/integral-4.8956/ | # Integral #4
#### Random Variable
##### Well-known member
MHB Math Helper
Show that for positive parameters $a$, $b$, and $c$,
$$\int_{0}^{\infty} \left( \frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \frac{e^{-cx}}{x} \ \right) \ dx = b-a + a \ln \left(\frac{a}{c} \right) - b \ln \left(\frac{b}{c} \right)$$
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
$$\displaystyle F = \int_{0}^{\infty} \left( \frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \frac{e^{-cx}}{x} \ \right) \ dx$$
$$\displaystyle F_a =\int_{0}^{\infty} \frac{e^{-cx}-e^{-ax}}{x}\ dx = \ln \left( \frac{a}{c}\right)$$
$$\displaystyle F= a \ln \left( \frac{a}{c}\right) - a + C$$
$$\displaystyle F_b = C_b$$
$$\displaystyle F_b = \int_{0}^{\infty} \frac{e^{-bx}-e^{-cx}}{x} dx = -\ln \left( \frac{b}{c}\right)$$
We have $$\displaystyle C_b = -\ln \left( \frac{b}{c}\right)$$ so
$$\displaystyle C = -b\ln \left( \frac{b}{c}\right)+b$$
$$\displaystyle F = b-a +a \ln \left( \frac{a}{c}\right)-b\ln \left( \frac{b}{c}\right)$$
#### Random Variable
##### Well-known member
MHB Math Helper
@ Zaid
I actually didn't even consider differentiating inside of the integral.
I took the boring approach of finding an antiderivative in terms of the exponential integral and then using the expansion of the exponential integral at $x=0$ to evaluate the antiderivative at the lower limit.
When you integrated to find the constant $C$, how did you know that he constant of integration was zero?
Alternatively what you could do to find $C$ is to let $a=b$ in the original integral.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
When you integrated to find the constant $C$, how did you know that he constant of integration was zero?
Alternatively what you could do to find $C$ is to let $a=b$ in the original integral.
Yeah, I made the evaluations not clear but I actually considered that $C$ as a function of $b$ so I differentiated with respect to $b$. Again putting $a=b$ is by far much clearer.
Can you post your approach ?
#### Random Variable
##### Well-known member
MHB Math Helper
It's easier to work the upper incomplete gamma function.
$$\Gamma(0,a x) = -\text{Ei}(- ax) = \int_{ax}^{\infty} \frac{e^{-t}}{t} \ dt = \int_{x}^{\infty} \frac{e^{-au}}{u} \ du$$
Let's first derive the expansion at $x=0$.
$$\Gamma(0,x) = \int_{x}^{\infty} \frac{e^{-t}}{t} \ dt = \int_{1}^{\infty} \frac{e^{-t}}{t} \ dt - \int_{1}^{x} \frac{e^{-t}}{t} \ dt$$
$$= \int_{1}^{x} \frac{1-e^{-t}}{t} \ dt - \ln x + \int_{1}^{\infty} \frac{e^{-t}}{t} \ dt$$
$$= -\ln x + \Big( -\int_{0}^{1} \frac{1-e^{-t}}{t} \ dt + \int_{1}^{\infty} \frac{e^{-t}}{t} \ dt \Big) + \int_{0}^{x} \frac{1-e^{-t}}{t} \ dt$$
$$= - \ln x - \gamma - \int_{0}^{x} \frac{e^{-t}-1}{t} \ dt$$
$$= - \ln x - \gamma \int_{0}^{x} \Big(-1 + \frac{t}{2} - \frac{t^{2}}{6} + \ldots \Big) \ dt$$
$$= - \ln x - \gamma + x + \mathcal{O}(x^{2})$$
And it's not really needed here, but one can find an asymptotic expansion at $x= \infty$ by integrating by parts.
$$\Gamma(0,x) \sim \frac{e^{-x}}{x} + \mathcal{O}\left( \frac{e^{-x}}{x^{2}} \right) \ \ x \to \infty$$
Then
$$\int \left( \frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \frac{e^{-cx}}{x} \ \right) \ dx = \int \frac{e^{-ax}-e^{-bx}}{x^{2}} \ dx + (a-b) \int \frac{e^{-cx}}{x} \ dx$$
$$= -\frac{e^{-ax} -e^{-bx}}{x} + \int \frac{-ae^{-ax}+be^{-bx}}{x} \ dx -(a-b) \Gamma(0,cx)$$
$$= \frac{e^{-bx} -e^{-ax}}{x} + a \Gamma(0,ax) - b \Gamma(0,bx) - (a-b) \Gamma(0,cx)+C$$
$$\int^{\infty}_{0} \left( \frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \frac{e^{-cx}}{x} \ \right) \ dx$$
$$= 0 - \lim_{x \to 0} \left[ \frac{e^{-bx} -e^{-ax}}{x} + a \left(-\ln ax - \gamma \right) - b \left(-\ln bx - \gamma \right) -(a-b) \left(-\ln cx - \gamma \right) \right]$$
$$= - \lim_{x \to 0} \left[ \frac{e^{-bx} -e^{-ax}}{x} - a \ln \left(\frac{a}{c} \right) + b \left(\frac{b}{c} \right) \right]$$
$$= b-a + a \ln \left(\frac{a}{c} \right) - b \ln \left(\frac{b}{c} \right)$$
Last edited by a moderator: | 2021-07-29T09:46:46 | {
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https://www.themathdoctors.org/why-is-a-linear-equation-called-linear/ | # Why is a Linear Equation Called Linear?
#### (New Question of the Week)
From time to time we get a question that is more about words than about math; usually these are about the meaning or origin of mathematical terms. Fortunately, some of us love words as much as we love math. But the question I want to look at here, which came in last month, is about both the word and the math; in explaining why the word is appropriate, we are learning some things about math itself.
## But it’s not a line …
Here is Christine’s question:
Why is an equation like 2x + 4 = 10 called a linear equation in one variable? Clearly, the solution is a point on one axis, the x-axis, not a line on the two-axis Cartesian coordinate plane? Or are all linear equations in one variable viewed as vertical lines?
Clearly, she is saying, the phrase “linear equation” means “the equation of a line”. And there is a similarity between the linear equation in one variable above, and a linear equation in two variables, such as $$y = 2x + 4$$, which definitely is the equation of a line. But with only one variable, the only way to say that $$2x + 4 = 10$$ is the equation of a line is to plot it on a plane as the vertical line $$x = 3$$. Is that the intent of the term?
No, it goes further than that, because you also have to consider three or more variables. I initially gave just a short answer, to see what response it would trigger before digging in deeper:
The term “linear”, though derived from the idea that a linear equation in two variables represents a line, has been generalized from that to mean that the equation involves a polynomial with degree 1. That is, the variable(s) are only multiplied by constants and added to other terms, with nothing more (squaring, etc.).
So you could say that the term has been taken from one situation that gave it its name, and applied to more general cases with different numbers of variables. A linear equation in three variables, for example, represents a plane.
From my perspective, “linear” means far more than “having a graph that is a line”. My first thought when I see the word (outside of an elementary algebra class) is “first-degree polynomial”. Although one initially connects it to straight lines, when we extend its use (and almost every idea in math is an extension of something simpler), the idea we carry forward is the degree, not the number of dimensions. For example, here is the beginning of the Wikipedia article about linear equations:
A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable (however, different variables may occur in different terms). A simple example of a linear equation with only one variable, x, may be written in the form: ax + b = 0, where a and b are constants and a ≠ 0.
Linear equations can have one or more variables. An example of a linear equation with three variables, x, y, and z, is given by: ax + by + cz + d = 0, where a, b, c, and d are constants and a, b, and c are non-zero.
But why would we call any polynomial equation with degree 1 “linear”, when it is only in two dimensions that it is related to a line? In the one-variable case, as Christine said, the graph is really a point; and in this three-variable case it is a plane:
## But, still – it’s not a line!
Christine wasn’t quite convinced:
Thank you so much. I am very particular with terminology when I teach mathematics. I have to say, I do not like this generalization of the term. I think it is misleading.
Hmmm … is the term “linear” misleading? Not to mathematicians, and I would hope not to students once they get used to it. Yet it’s true that it doesn’t quite mean what it seems to say. And is generalization bad? I think it’s the essence of what math is – and also an integral part of languages,which are always extending the meaning of words to cover new needs.
Here is my response:
Hi, Christine.
Thanks for writing back with additional thoughts. Let’s think a little more deeply about it.
First, this is standard terminology that has been in use for 200 years by many great mathematicians, so we should be very careful about considering it a bad idea. I don’t think you’re likely to convince anyone to change; the word is used not only of a linear equation in itself, but of “systems of linear equations” (in contrast to “non-linear equations”); of the whole major field of “linear algebra”; and for related concepts like “linear combination”, “linear transformation”, and “linear independence”, all of which apply to any number of variables or dimensions. So the term is very well established with a particular but broad meaning, and (at least after the first year) no one is misled by it. We know what it means, and what it means is more than just “line”.
Second, what would you replace it with? If you don’t want to use the word “linear” except in situations that involve actual lines, what word would you use instead to describe the general class of equations involving variables multiplied only by constants, regardless of the number of variables? We need a word for this bigger idea; that word will either be a familiar word whose meaning is stretched to cover a bigger concept, or some made-up word. The tradition in math has always been to take familiar words and give them new meanings (either more specific, like “group” or “combination” or “function”, or more general, like “number” or “multiplication” or “space”). So what we observe here is found throughout mathematics: a word that has grown beyond its humble beginnings. (This is also true of the entire English language! Most words would be “misleading” if you thought too deeply about their origins.)
I could say that, in a sense, all of mathematics is about generalization (or abstraction). I just mentioned “number”; some people do complain about calling anything other than a natural number a “number”, but the logical development from natural numbers, to integers, to rational numbers, to real and complex numbers, involves repeated broadening of the term, which has been extremely useful. We invent new concepts, and give them old names because they are a larger, more powerful version of the old concept.
I mentioned how common it is in English for a word to grow beyond its original meaning. Sitting here at my computer, I look at the mouse – is it misleading to call it that when it doesn’t have legs, and may not even have a tail? And the computer has a screen; at one time, a screen was a flat surface that hid something that wasn’t to be seen (or that kept out bugs); then it was applied to flat surfaces on which pictures were projected; and then to a surface that shows a picture itself. Is that misleading? It would be if you went back a hundred years …
And thinking again about math terms, I’m reminded of this discussion where I pondered what all the different operations that are called “multiplication” have in common.
## So how do we explain it?
But as I wrote this, I realized that I had gone off in a different direction than Christine, and I wanted to relate my answer to her specific context, linear equations in one variable. The real question was pedagogical: How could she explain this to her students, so that (eventually) “linear” would mean to them what they will need it to mean? I continued:
Now, I’ve been mostly thinking of the “enlarging” development of the word “linear”, taking it to more than two dimensions; you’re thinking specifically of the term used with only one variable. So it may help if we focus on that, to fit your particular context. I’d like to explain linear equations in one variable in a way that should make it clear why we use the word, and that it is not a misnomer.
Consider the equation you asked about, 2x+4=10. The left-hand side is an expression; we call it a linear expression, because if you used it in an equation with two variables, y = 2x+4, its graph would be a straight line. So we call 2x+4 a linear expression (or, later, a linear function). A linear equation in one variable is one that says that two linear expressions are equal (or one is equal to a constant).
Or, to look at it another way, one way to solve this equation would be to graph the related equation y = 2x+4, and find where that line intersects the line y = 10. So the linear equation can be thought of very much in terms of a line.
(If this were a student’s first exposure to linear equations in one variable, she likely wouldn’t have seen graphs of lines yet, and wouldn’t be ready for this discussion; either the word linear would be used without explanation yet, or we would hold off on the word until later.)
Does this help?
It’s often hard to be sure what kind of answer will help; but Christine gave the answer I hoped for:
Yes, your response does help very much. In addition, it gives me a lot to think about. It is true, the grade level I am currently teaching does learn to solve linear equations in one variable before they learn about linear equations and functions. Perhaps this is why I question the use of the term. I must remember that I have had experience in mathematics beyond their years, and therefore, am more thoughtful about what terms they are exposed to and more importantly in what sequence the mathematics is presented to them. I thank you again for such an in depth response and will certainly give this discussion much more thought. It is a pleasure speaking with you.
I imagine if I were introducing these equations to students with no exposure to graphs of lines, I might just mention in passing that these are called “linear equations in one variable”, and that we would soon be seeing why the word “linear” is appropriate. For now, what it means is that all we do with the variable is to multiply it by a constant, and to add things. Kids are used to hearing things they’ll understand later …
### 3 thoughts on “Why is a Linear Equation Called Linear?”
1. Very interesting article, Dr.Peterson!
Just wanted to ask, so then why are polynomials of the second degree quadratics? ‘Quad’ reminds me of 4 (quadrilateral is a shape with 4 sides, a quadrant is 1/4 of a circle)
And what’s the difference between an identity and an expression?
1. Dave Peterson
Hi, Sarah.
First, the “quad” issue is another common question; for an answer, start here: Names of Polynomials
Second, an identity is not an expression, but an equation (that is, a statement that two expressions are equal). In particular, it is an equation that is true for any value of the variable(s) it contains. For instance, a+b = b+a is an identity. | 2018-04-26T09:25:37 | {
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https://math.stackexchange.com/questions/528709/how-can-i-prove-by-induction-that-9k-5k-is-divisible-by-4 | # How can I prove by induction that $9^k - 5^k$ is divisible by 4?
Recently had this on a discrete math test, which sadly I think I failed. But the question asked:
Prove that $9^k - 5^k$ is divisible by $4$.
Using the only approach I learned in the class, I substituted $n = k$, and tried to prove for $k+1$ like this:
$$9^{k+1} - 5^{k+1},$$
which just factors to $9 \cdot 9^k - 5 \cdot 5^k$.
But I cannot factor out $9^k - 5^k$, so I'm totally stuck.
• hint: 9 = (5+4) and distribute. – imranfat Oct 16 '13 at 16:56
• Alternatively you can note that $a^n-b^n=(a-b)\sum _{i=0} ^{n-1} a^i b^{n-1-i}$. – pppqqq Oct 16 '13 at 16:58
\begin{align} 9\cdot 9^k - 5\cdot 5^k & = (4 + 5)\cdot 9^k - 5\cdot 5^k \\ \\ & = 4\cdot 9^k + 5 \cdot 9^k - 5\cdot 5^k \\ \\ & = 4\cdot 9^k + 5(9^k - 5^k)\\ \\ & \quad \text{ use inductive hypothesis}\quad\cdots\end{align}
• And don't forget to include verification of the base case in your proof: $n = 0$ or $n = 1$. – Namaste Oct 16 '13 at 17:10
• Okay this seems to make the most sense and is closest to how my prof. teaches it. Dam I really hate theoretical math and proofs! – krb686 Oct 16 '13 at 17:11
• @amWhy: Nice to get a nice answer +1 – Amzoti Oct 18 '13 at 0:57
${\color{White}{\text{Proof without words.}}}$
• Very nice indeed! – Brian M. Scott Oct 17 '13 at 0:02
$9^{k+1}-5^{k+1}=(8+1)9^k-(4+1)5^k=9^k-5^k+4(2\cdot 9^k-5^k)$ The secret to induction proofs is usually to find a way to relate the $k+1$ case to the $k$ case.
Alternately, just note $9\equiv 1 \pmod 4, 5 \equiv 1 \pmod 4$, so $9^k-5^k \equiv 1^k-1^k \pmod 4$
Note that induction is not actually needed: you can instead use the identity
$$x^n-y^n=(x-y)\sum_{k=0}^{n-1}x^ky^{n-1-k}\;.$$
In this case it becomes
$$9^n-5^n=4\sum_{k=0}^{n-1}x^ky^{n-1-k}\;,$$
and since the summation clearly yields an integer, we have $4\mid 9^n-5^n$. The identity is a standard one, but it’s also not hard to prove:
\begin{align*} (x-y)\sum_{k=0}^{n-1}x^ky^{n-1-k}&=x\sum_{k=0}^{n-1}x^ky^{n-1-k}-y\sum_{k=0}^{n-1}x^ky^{n-1-k}\\\\ &=\sum_{k=0}^{n-1}x^{k+1}y^{n-1-k}-\sum_{k=0}^{n-1}x^ky^{n-k}\\\\ &=\sum_{k=1}^nx^ky^{n-k}-\sum_{k=0}^{n-1}x^ky^{n-k}\\\\ &=x^ny^0+\sum_{k=1}^{n-1}x^ky^{n-k}-\sum_{k=1}^{n-1}x^ky^{n-k}-x^0y^n\\\\ &=x^n-y^n\;. \end{align*}
• My test asked to use induction so that's why I narrowed that down, but thank you for all the in depth info! – krb686 Oct 17 '13 at 3:49
• @krb686: You’re welcome! – Brian M. Scott Oct 17 '13 at 3:50
It sounds like you understand the basic idea, but you're having trouble with the "trick" in the inductive step. This is common for proofs by mathematical induction, and you'll get better at it with more experience.
As you said, you're assuming that $9^k-5^k$ is divisible by four, and you want to prove that $9^{k+1}-5^{k+1}$ is divisible by four. You noticed that you can write the last expression as $9\cdot 9^k-5\cdot 5^k$, and you said you're having trouble applying your assumption to it. What happens if you add $9\cdot 5^k-9\cdot 5^k$ to your expression? This is just zero, so you're not changing anything, but you can write $$9^{k+1}-5^{k+1} =9\cdot 9^k-5\cdot 5^k+9\cdot 5^k-9\cdot 5^k =9\cdot (9^k-5^k)+(9-5)\cdot 5^k.$$ By the assumption, the first term, $9\cdot (9^k-5^k)$, is divisible by four, and by observation, the second term, $(9-5)\cdot 5^k$, is divisible by four, so the sum of the two terms and therefore $9^{k+1}-5^{k+1}$ are divisible by four. This is what you wanted to show.
You can always add zero (or multiply by one), and by writing zero (or one) in a particular way, you can sometimes make an argument easier to prove. Also, since your proof is by mathematical induction, you need to show that $9^n-5^n$ is divisible by four for, say, $n=0$ or $n=1$. You may have already done so and not asked about it in your question, but it's worth mentioning. | 2019-11-21T21:25:42 | {
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http://dlmf.nist.gov/1.9 | # §1.9 Calculus of a Complex Variable
## §1.9(i) Complex Numbers
1.9.1 $z=x+iy,$ $x,y\in\mathbb{R}$. ⓘ Symbols: $\in$: element of, $\mathbb{R}$: real line and $z$: variable A&S Ref: 3.7.1 Permalink: http://dlmf.nist.gov/1.9.E1 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9 and 1
### Real and Imaginary Parts
1.9.2 $\displaystyle\Re z$ $\displaystyle=x,$ $\displaystyle\Im z$ $\displaystyle=y.$ ⓘ Defines: $\Im$: imaginary part and $\Re$: real part Symbols: $z$: variable A&S Ref: 3.7.5 3.7.6 Permalink: http://dlmf.nist.gov/1.9.E2 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
### Polar Representation
1.9.3 $\displaystyle x$ $\displaystyle=r\cos\theta,$ $\displaystyle y$ $\displaystyle=r\sin\theta,$ ⓘ Symbols: $\cos\NVar{z}$: cosine function, $\sin\NVar{z}$: sine function, $r$: radius and $\theta$: angle A&S Ref: 3.7.2 Referenced by: §1.9(ii) Permalink: http://dlmf.nist.gov/1.9.E3 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
where
1.9.4 $r=(x^{2}+y^{2})^{1/2},$ ⓘ Symbols: $r$: radius A&S Ref: 3.7.3 Permalink: http://dlmf.nist.gov/1.9.E4 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
and when $z\neq 0$,
1.9.5 $\theta=\omega,\;\;\pi-\omega,\;\;-\pi+\omega,\mbox{ or }-\omega,$ ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter and $\theta$: angle Permalink: http://dlmf.nist.gov/1.9.E5 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
according as $z$ lies in the 1st, 2nd, 3rd, or 4th quadrants. Here
1.9.6 $\omega=\operatorname{arctan}\left(|y/x|\right)\in\left[0,\tfrac{1}{2}\pi\right].$ ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $[\NVar{a},\NVar{b}]$: closed interval, $\in$: element of and $\operatorname{arctan}\NVar{z}$: arctangent function A&S Ref: 3.7.4 Permalink: http://dlmf.nist.gov/1.9.E6 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
### Modulus and Phase
1.9.7 $\displaystyle|z|$ $\displaystyle=r,$ $\displaystyle\operatorname{ph}z$ $\displaystyle=\theta+2n\pi,$ $n\in\mathbb{Z}$. ⓘ Defines: $\operatorname{ph}$: phase Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\in$: element of, $\mathbb{Z}$: set of all integers, $z$: variable, $n$: nonnegative integer, $r$: radius and $\theta$: angle Permalink: http://dlmf.nist.gov/1.9.E7 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
The principal value of $\operatorname{ph}z$ corresponds to $n=0$, that is, $-\pi\leq\operatorname{ph}z\leq\pi$. It is single-valued on $\mathbb{C}\setminus\{0\}$, except on the interval $(-\infty,0)$ where it is discontinuous and two-valued. Unless indicated otherwise, these principal values are assumed throughout the DLMF. (However, if we require a principal value to be single-valued, then we can restrict $-\pi<\operatorname{ph}z\leq\pi$.)
1.9.8 $\displaystyle|\Re z|$ $\displaystyle\leq|z|,$ $\displaystyle|\Im z|$ $\displaystyle\leq|z|,$ ⓘ Symbols: $\Im$: imaginary part, $\Re$: real part and $z$: variable Permalink: http://dlmf.nist.gov/1.9.E8 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
1.9.9 $z=re^{i\theta},$ ⓘ Symbols: $\mathrm{e}$: base of exponential function, $z$: variable, $r$: radius and $\theta$: angle Permalink: http://dlmf.nist.gov/1.9.E9 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
where
1.9.10 $e^{i\theta}=\cos\theta+i\sin\theta;$ ⓘ Symbols: $\cos\NVar{z}$: cosine function, $\mathrm{e}$: base of exponential function, $\sin\NVar{z}$: sine function and $\theta$: angle Permalink: http://dlmf.nist.gov/1.9.E10 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
see §4.14.
### Complex Conjugate
1.9.11 $\displaystyle\overline{z}$ $\displaystyle=x-iy,$ ⓘ Defines: $\overline{\NVar{z}}$: complex conjugate Symbols: $z$: variable A&S Ref: 3.7.7 Permalink: http://dlmf.nist.gov/1.9.E11 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1 1.9.12 $\displaystyle|\overline{z}|$ $\displaystyle=|z|,$ ⓘ Symbols: $\overline{\NVar{z}}$: complex conjugate and $z$: variable A&S Ref: 3.7.8 Permalink: http://dlmf.nist.gov/1.9.E12 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1 1.9.13 $\displaystyle\operatorname{ph}\overline{z}$ $\displaystyle=-\operatorname{ph}z.$ ⓘ Symbols: $\overline{\NVar{z}}$: complex conjugate, $\operatorname{ph}$: phase and $z$: variable A&S Ref: 3.7.9 Permalink: http://dlmf.nist.gov/1.9.E13 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
### Arithmetic Operations
If $z_{1}=x_{1}+iy_{1}$, $z_{2}=x_{2}+iy_{2}$, then
1.9.14 $z_{1}\pm z_{2}=x_{1}\pm x_{2}+\mathrm{i}(y_{1}\pm y_{2}),$ ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E14 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
1.9.15 $z_{1}z_{2}=x_{1}x_{2}-y_{1}y_{2}+i(x_{1}y_{2}+x_{2}y_{1}),$ ⓘ Symbols: $z$: variable A&S Ref: 3.7.10 Permalink: http://dlmf.nist.gov/1.9.E15 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
1.9.16 $\frac{z_{1}}{z_{2}}=\frac{z_{1}\overline{z}_{2}}{|z_{2}|^{2}}=\frac{x_{1}x_{2}% +y_{1}y_{2}+i(x_{2}y_{1}-x_{1}y_{2})}{x_{2}^{2}+y_{2}^{2}},$ ⓘ Symbols: $\overline{\NVar{z}}$: complex conjugate and $z$: variable A&S Ref: 3.7.13 Permalink: http://dlmf.nist.gov/1.9.E16 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
provided that $z_{2}\neq 0$. Also,
1.9.17 $|z_{1}z_{2}|=|z_{1}|\;|z_{2}|,$ ⓘ Symbols: $z$: variable A&S Ref: 3.7.11 Permalink: http://dlmf.nist.gov/1.9.E17 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
1.9.18 $\operatorname{ph}\left(z_{1}z_{2}\right)=\operatorname{ph}z_{1}+\operatorname{% ph}z_{2},$ ⓘ Symbols: $\operatorname{ph}$: phase and $z$: variable A&S Ref: 3.7.12 Referenced by: §1.9(i) Permalink: http://dlmf.nist.gov/1.9.E18 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
1.9.19 $\left|\frac{z_{1}}{z_{2}}\right|=\frac{|z_{1}|}{|z_{2}|},$ ⓘ Symbols: $z$: variable A&S Ref: 3.7.14 Permalink: http://dlmf.nist.gov/1.9.E19 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
1.9.20 $\operatorname{ph}\frac{z_{1}}{z_{2}}=\operatorname{ph}z_{1}-\operatorname{ph}z% _{2}.$ ⓘ Symbols: $\operatorname{ph}$: phase and $z$: variable A&S Ref: 3.7.15 Referenced by: §1.9(i) Permalink: http://dlmf.nist.gov/1.9.E20 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
Equations (1.9.18) and (1.9.20) hold for general values of the phases, but not necessarily for the principal values.
### Powers
1.9.21 $z^{n}=\left(x^{n}-\genfrac{(}{)}{0.0pt}{}{n}{2}x^{n-2}y^{2}+\genfrac{(}{)}{0.0% pt}{}{n}{4}x^{n-4}y^{4}-\cdots\right)+i\left(\genfrac{(}{)}{0.0pt}{}{n}{1}x^{n% -1}y-\genfrac{(}{)}{0.0pt}{}{n}{3}x^{n-3}y^{3}+\cdots\right),$ $n=1,2,\dots$. ⓘ Symbols: $\genfrac{(}{)}{0.0pt}{}{\NVar{m}}{\NVar{n}}$: binomial coefficient, $z$: variable and $n$: nonnegative integer A&S Ref: 3.7.22 Permalink: http://dlmf.nist.gov/1.9.E21 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
### DeMoivre’s Theorem
1.9.22 $\cos n\theta+i\sin n\theta=(\cos\theta+i\sin\theta)^{n},$ $n\in\mathbb{Z}$.
### Triangle Inequality
1.9.23 $\left|\left|z_{1}\right|-\left|z_{2}\right|\right|\leq\left|z_{1}+z_{2}\right|% \leq\left|z_{1}\right|+\left|z_{2}\right|.$ ⓘ Symbols: $z$: variable A&S Ref: 3.7.29 Permalink: http://dlmf.nist.gov/1.9.E23 Encodings: TeX, pMML, png See also: Annotations for 1.9(i), 1.9(i), 1.9 and 1
## §1.9(ii) Continuity, Point Sets, and Differentiation
### Continuity
A function $f(z)$ is continuous at a point $z_{0}$ if $\lim\limits_{z\to z_{0}}f(z)=f(z_{0})$. That is, given any positive number $\epsilon$, however small, we can find a positive number $\delta$ such that $|f(z)-f(z_{0})|<\epsilon$ for all $z$ in the open disk $|z-z_{0}|<\delta$.
A function of two complex variables $f(z,w)$ is continuous at $(z_{0},w_{0})$ if $\lim\limits_{(z,w)\to(z_{0},w_{0})}f(z,w)=f(z_{0},w_{0})$; compare (1.5.1) and (1.5.2).
### Point Sets in $\mathbb{C}$
A neighborhood of a point $z_{0}$ is a disk $\left|z-z_{0}\right|<\delta$. An open set in $\mathbb{C}$ is one in which each point has a neighborhood that is contained in the set.
A point $z_{0}$ is a limit point (limiting point or accumulation point) of a set of points $S$ in $\mathbb{C}$ (or $\mathbb{C}\cup\infty$) if every neighborhood of $z_{0}$ contains a point of $S$ distinct from $z_{0}$. ($z_{0}$ may or may not belong to $S$.) As a consequence, every neighborhood of a limit point of $S$ contains an infinite number of points of $S$. Also, the union of $S$ and its limit points is the closure of $S$.
A domain $D$, say, is an open set in $\mathbb{C}$ that is connected, that is, any two points can be joined by a polygonal arc (a finite chain of straight-line segments) lying in the set. Any point whose neighborhoods always contain members and nonmembers of $D$ is a boundary point of $D$. When its boundary points are added the domain is said to be closed, but unless specified otherwise a domain is assumed to be open.
A region is an open domain together with none, some, or all of its boundary points. Points of a region that are not boundary points are called interior points.
A function $f(z)$ is continuous on a region $R$ if for each point $z_{0}$ in $R$ and any given number $\epsilon$ ($>0$) we can find a neighborhood of $z_{0}$ such that $\left|f(z)-f(z_{0})\right|<\epsilon$ for all points $z$ in the intersection of the neighborhood with $R$.
### Differentiation
A function $f(z)$ is differentiable at a point $z$ if the following limit exists:
1.9.24 $f^{\prime}(z)=\frac{\mathrm{d}f}{\mathrm{d}z}=\lim_{h\to 0}\frac{f(z+h)-f(z)}{% h}.$ ⓘ Symbols: $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$ and $z$: variable Permalink: http://dlmf.nist.gov/1.9.E24 Encodings: TeX, pMML, png See also: Annotations for 1.9(ii), 1.9(ii), 1.9 and 1
Differentiability automatically implies continuity.
### Cauchy–Riemann Equations
If $f^{\prime}(z)$ exists at $z=x+iy$ and $f(z)=u(x,y)+iv(x,y)$, then
1.9.25 $\displaystyle\frac{\partial u}{\partial x}$ $\displaystyle=\frac{\partial v}{\partial y},$ $\displaystyle\frac{\partial u}{\partial y}$ $\displaystyle=-\frac{\partial v}{\partial x}$ ⓘ Symbols: $\frac{\partial\NVar{f}}{\partial\NVar{x}}$: partial derivative of $f$ with respect to $x$, $\partial\NVar{x}$: partial differential of $x$, $u(x,y)$: function and $v(x,y)$: function A&S Ref: 3.7.30 Referenced by: §1.9(ii) Permalink: http://dlmf.nist.gov/1.9.E25 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(ii), 1.9(ii), 1.9 and 1
at $(x,y)$.
Conversely, if at a given point $(x,y)$ the partial derivatives $\ifrac{\partial u}{\partial x}$, $\ifrac{\partial u}{\partial y}$, $\ifrac{\partial v}{\partial x}$, and $\ifrac{\partial v}{\partial y}$ exist, are continuous, and satisfy (1.9.25), then $f(z)$ is differentiable at $z=x+iy$.
### Analyticity
A function $f(z)$ is said to be analytic (holomorphic) at $z=z_{0}$ if it is differentiable in a neighborhood of $z_{0}$.
A function $f(z)$ is analytic in a domain $D$ if it is analytic at each point of $D$. A function analytic at every point of $\mathbb{C}$ is said to be entire.
If $f(z)$ is analytic in an open domain $D$, then each of its derivatives $f^{\prime}(z)$, $f^{\prime\prime}(z)$, $\dots$ exists and is analytic in $D$.
### Harmonic Functions
If $f(z)=u(x,y)+iv(x,y)$ is analytic in an open domain $D$, then $u$ and $v$ are harmonic in $D$, that is,
1.9.26 $\frac{{\partial}^{2}u}{{\partial x}^{2}}+\frac{{\partial}^{2}u}{{\partial y}^{% 2}}=\frac{{\partial}^{2}v}{{\partial x}^{2}}+\frac{{\partial}^{2}v}{{\partial y% }^{2}}=0,$
or in polar form ((1.9.3)) $u$ and $v$ satisfy
1.9.27 $\frac{{\partial}^{2}u}{{\partial r}^{2}}+\frac{1}{r}\frac{\partial u}{\partial r% }+\frac{1}{r^{2}}\frac{{\partial}^{2}u}{{\partial\theta}^{2}}=0$
at all points of $D$.
## §1.9(iii) Integration
An arc $C$ is given by $z(t)=x(t)+iy(t)$, $a\leq t\leq b$, where $x$ and $y$ are continuously differentiable. If $x(t)$ and $y(t)$ are continuous and $x^{\prime}(t)$ and $y^{\prime}(t)$ are piecewise continuous, then $z(t)$ defines a contour.
A contour is simple if it contains no multiple points, that is, for every pair of distinct values $t_{1},t_{2}$ of $t$, $z(t_{1})\neq z(t_{2})$. A simple closed contour is a simple contour, except that $z(a)=z(b)$.
Next,
1.9.28 $\int_{C}f(z)\mathrm{d}z=\int_{a}^{b}f(z(t))(x^{\prime}(t)+iy^{\prime}(t))% \mathrm{d}t,$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $C$: closed contour Permalink: http://dlmf.nist.gov/1.9.E28 Encodings: TeX, pMML, png See also: Annotations for 1.9(iii), 1.9 and 1
for a contour $C$ and $f(z(t))$ continuous, $a\leq t\leq b$. If $f(z(t_{0}))=\infty$, $a\leq t_{0}\leq b$, then the integral is defined analogously to the infinite integrals in §1.4(v). Similarly when $a=-\infty$ or $b=+\infty$.
### Jordan Curve Theorem
Any simple closed contour $C$ divides $\mathbb{C}$ into two open domains that have $C$ as common boundary. One of these domains is bounded and is called the interior domain of $C$; the other is unbounded and is called the exterior domain of $C$.
### Cauchy’s Theorem
If $f(z)$ is continuous within and on a simple closed contour $C$ and analytic within $C$, then
1.9.29 $\int_{C}f(z)\mathrm{d}z=0.$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $C$: closed contour Permalink: http://dlmf.nist.gov/1.9.E29 Encodings: TeX, pMML, png See also: Annotations for 1.9(iii), 1.9(iii), 1.9 and 1
### Cauchy’s Integral Formula
If $f(z)$ is continuous within and on a simple closed contour $C$ and analytic within $C$, and if $z_{0}$ is a point within $C$, then
1.9.30 $f(z_{0})=\frac{1}{2\pi i}\int_{C}\frac{f(z)}{z-z_{0}}\mathrm{d}z,$ ⓘ Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $C$: closed contour Referenced by: §2.3(iii) Permalink: http://dlmf.nist.gov/1.9.E30 Encodings: TeX, pMML, png See also: Annotations for 1.9(iii), 1.9(iii), 1.9 and 1
and
1.9.31 $f^{(n)}(z_{0})=\frac{n!}{2\pi i}\int_{C}\frac{f(z)}{(z-z_{0})^{n+1}}\mathrm{d}z,$ $n=1,2,3,\dots$,
provided that in both cases $C$ is described in the positive rotational (anticlockwise) sense.
### Liouville’s Theorem
Any bounded entire function is a constant.
### Winding Number
If $C$ is a closed contour, and $z_{0}\not\in C$, then
1.9.32 $\frac{1}{2\pi i}\int_{C}\frac{1}{z-z_{0}}\mathrm{d}z=\mathcal{N}(C,z_{0}),$ ⓘ Defines: $\mathcal{N}(C,z_{0})$: winding number of $C$ (locally) Symbols: $\pi$: the ratio of the circumference of a circle to its diameter, $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $C$: closed contour Permalink: http://dlmf.nist.gov/1.9.E32 Encodings: TeX, pMML, png See also: Annotations for 1.9(iii), 1.9(iii), 1.9 and 1
where $\mathcal{N}(C,z_{0})$ is an integer called the winding number of $C$ with respect to $z_{0}$. If $C$ is simple and oriented in the positive rotational sense, then $\mathcal{N}(C,z_{0})$ is $1$ or $0$ depending whether $z_{0}$ is inside or outside $C$.
### Mean Value Property
For $u(z)$ harmonic,
1.9.33 $u(z)=\frac{1}{2\pi}\int^{2\pi}_{0}u(z+re^{i\phi})\mathrm{d}\phi.$
### Poisson Integral
If $h(w)$ is continuous on $|w|=R$, then with $z=re^{i\theta}$
1.9.34 $u(re^{i\theta})=\frac{1}{2\pi}\int^{2\pi}_{0}\frac{(R^{2}-r^{2})h(Re^{i\phi})% \mathrm{d}\phi}{R^{2}-2Rr\cos\left(\phi-\theta\right)+r^{2}}$
is harmonic in $|z|. Also with $\left|w\right|=R$, $\lim\limits_{z\to w}u(z)=h(w)$ as $z\to w$ within $|z|.
## §1.9(iv) Conformal Mapping
The extended complex plane, $\mathbb{C}\,\cup\,\{\infty\}$, consists of the points of the complex plane $\mathbb{C}$ together with an ideal point $\infty$ called the point at infinity. A system of open disks around infinity is given by
1.9.35 $S_{r}=\{z\mid|z|>1/r\}\cup\{\infty\},$ $0. ⓘ Symbols: $\cup$: union, $z$: variable, $r$: radius and $S_{r}$: neighborhood Permalink: http://dlmf.nist.gov/1.9.E35 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1
Each $S_{r}$ is a neighborhood of $\infty$. Also,
1.9.36 $\infty\pm z=z\pm\infty=\infty,$ ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E36 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1
1.9.37 $\infty\cdot z=z\cdot\infty=\infty,$ $z\not=0$, ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E37 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1
1.9.38 $z/\infty=0,$ ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E38 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1
1.9.39 $z/0=\infty,$ $z\neq 0$. ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E39 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9 and 1
A function $f(z)$ is analytic at $\infty$ if $g(z)=f(1/z)$ is analytic at $z=0$, and we set $f^{\prime}(\infty)=g^{\prime}(0)$.
### Conformal Transformation
Suppose $f(z)$ is analytic in a domain $D$ and $C_{1},C_{2}$ are two arcs in $D$ passing through $z_{0}$. Let $C^{\prime}_{1},C^{\prime}_{2}$ be the images of $C_{1}$ and $C_{2}$ under the mapping $w=f(z)$. The angle between $C_{1}$ and $C_{2}$ at $z_{0}$ is the angle between the tangents to the two arcs at $z_{0}$, that is, the difference of the signed angles that the tangents make with the positive direction of the real axis. If $f^{\prime}(z_{0})\not=0$, then the angle between $C_{1}$ and $C_{2}$ equals the angle between $C^{\prime}_{1}$ and $C^{\prime}_{2}$ both in magnitude and sense. We then say that the mapping $w=f(z)$ is conformal (angle-preserving) at $z_{0}$.
The linear transformation $f(z)=az+b$, $a\not=0$, has $f^{\prime}(z)=a$ and $w=f(z)$ maps $\mathbb{C}$ conformally onto $\mathbb{C}$.
### Bilinear Transformation
1.9.40 $w=f(z)=\frac{az+b}{cz+d},$ $ad-bc\not=0$, $c\not=0$. ⓘ Symbols: $z$: variable and $w$: variable Referenced by: §1.9(iv) Permalink: http://dlmf.nist.gov/1.9.E40 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1
1.9.41 $\displaystyle f(-d/c)$ $\displaystyle=\infty,$ $\displaystyle f(\infty)$ $\displaystyle=a/c.$ ⓘ Permalink: http://dlmf.nist.gov/1.9.E41 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1
1.9.42 $f^{\prime}(z)=\frac{ad-bc}{(cz+d)^{2}},$ $z\not=-d/c$. ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E42 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1
1.9.43 $f^{\prime}(\infty)=\frac{bc-ad}{c^{2}}.$ ⓘ Permalink: http://dlmf.nist.gov/1.9.E43 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1
1.9.44 $z=\frac{dw-b}{-cw+a}.$ ⓘ Symbols: $z$: variable and $w$: variable Permalink: http://dlmf.nist.gov/1.9.E44 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1
The transformation (1.9.40) is a one-to-one conformal mapping of $\mathbb{C}\,\cup\,\{\infty\}$ onto itself.
The cross ratio of $z_{1},z_{2},z_{3},z_{4}\in\mathbb{C}\cup\{\infty\}$ is defined by
1.9.45 $\frac{(z_{1}-z_{2})(z_{3}-z_{4})}{(z_{1}-z_{4})(z_{3}-z_{2})},$ ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E45 Encodings: TeX, pMML, png See also: Annotations for 1.9(iv), 1.9(iv), 1.9 and 1
or its limiting form, and is invariant under bilinear transformations.
Other names for the bilinear transformation are fractional linear transformation, homographic transformation, and Möbius transformation.
## §1.9(v) Infinite Sequences and Series
A sequence $\{z_{n}\}$ converges to $z$ if $\lim\limits_{n\to\infty}z_{n}=z$. For $z_{n}=x_{n}+iy_{n}$, the sequence $\{z_{n}\}$ converges iff the sequences $\{x_{n}\}$ and $\{y_{n}\}$ separately converge. A series $\sum^{\infty}_{n=0}z_{n}$ converges if the sequence $s_{n}=\sum^{n}_{k=0}z_{k}$ converges. The series is divergent if $s_{n}$ does not converge. The series converges absolutely if $\sum^{\infty}_{n=0}|z_{n}|$ converges. A series $\sum^{\infty}_{n=0}z_{n}$ converges (diverges) absolutely when $\lim\limits_{n\to\infty}|z_{n}|^{1/n}<1$ ($>1$), or when $\lim\limits_{n\to\infty}\left|\ifrac{z_{n+1}}{z_{n}}\right|<1$ ($>1$). Absolutely convergent series are also convergent.
Let $\{f_{n}(z)\}$ be a sequence of functions defined on a set $S$. This sequence converges pointwise to a function $f(z)$ if
1.9.46 $f(z)=\lim_{n\to\infty}f_{n}(z)$ ⓘ Symbols: $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E46 Encodings: TeX, pMML, png See also: Annotations for 1.9(v), 1.9 and 1
for each $z\in S$. The sequence converges uniformly on $S$, if for every $\epsilon>0$ there exists an integer $N$, independent of $z$, such that
1.9.47 $|f_{n}(z)-f(z)|<\epsilon$ ⓘ Symbols: $z$: variable, $n$: nonnegative integer and $\epsilon$: positive number Permalink: http://dlmf.nist.gov/1.9.E47 Encodings: TeX, pMML, png See also: Annotations for 1.9(v), 1.9 and 1
for all $z\in S$ and $n\geq N$.
A series $\sum^{\infty}_{n=0}f_{n}(z)$ converges uniformly on $S$, if the sequence $s_{n}(z)=\sum^{n}_{k=0}f_{k}(z)$ converges uniformly on $S$.
### Weierstrass $M$-test
Suppose $\{M_{n}\}$ is a sequence of real numbers such that $\sum^{\infty}_{n=0}M_{n}$ converges and $|f_{n}(z)|\leq M_{n}$ for all $z\in S$ and all $n\geq 0$. Then the series $\sum^{\infty}_{n=0}f_{n}(z)$ converges uniformly on $S$.
A doubly-infinite series $\sum^{\infty}_{n=-\infty}f_{n}(z)$ converges (uniformly) on $S$ iff each of the series $\sum^{\infty}_{n=0}f_{n}(z)$ and $\sum^{\infty}_{n=1}f_{-n}(z)$ converges (uniformly) on $S$.
## §1.9(vi) Power Series
For a series $\sum^{\infty}_{n=0}a_{n}(z-z_{0})^{n}$ there is a number $R$, $0\leq R\leq\infty$, such that the series converges for all $z$ in $|z-z_{0}| and diverges for $z$ in $|z-z_{0}|>R$. The circle $|z-z_{0}|=R$ is called the circle of convergence of the series, and $R$ is the radius of convergence. Inside the circle the sum of the series is an analytic function $f(z)$. For $z$ in $|z-z_{0}|\leq\rho$ ($), the convergence is absolute and uniform. Moreover,
1.9.48 $a_{n}=\frac{f^{(n)}(z_{0})}{n!},$ ⓘ Symbols: $!$: factorial (as in $n!$), $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E48 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9 and 1
and
1.9.49 $R=\liminf_{n\to\infty}|a_{n}|^{-1/n}.$ ⓘ Defines: $R$: radius of convergence (locally) Symbols: $\liminf$: least limit point and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E49 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9 and 1
For the converse of this result see §1.10(i).
### Operations
When $\sum a_{n}z^{n}$ and $\sum b_{n}z^{n}$ both converge
1.9.50 $\sum^{\infty}_{n=0}(a_{n}\pm b_{n})z^{n}=\sum^{\infty}_{n=0}a_{n}z^{n}\pm\sum^% {\infty}_{n=0}b_{n}z^{n},$ ⓘ Symbols: $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E50 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
and
1.9.51 $\left(\sum^{\infty}_{n=0}a_{n}z^{n}\right)\left(\sum^{\infty}_{n=0}b_{n}z^{n}% \right)=\sum^{\infty}_{n=0}c_{n}z^{n},$ ⓘ Symbols: $z$: variable and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E51 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
where
1.9.52 $c_{n}=\sum^{n}_{k=0}a_{k}b_{n-k}.$ ⓘ Symbols: $k$: integer and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E52 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
Next, let
1.9.53 $f(z)=a_{0}+a_{1}z+a_{2}z^{2}+\cdots,$ $a_{0}\not=0$. ⓘ Symbols: $z$: variable Permalink: http://dlmf.nist.gov/1.9.E53 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
Then the expansions (1.9.54), (1.9.57), and (1.9.60) hold for all sufficiently small $|z|$.
1.9.54 $\frac{1}{f(z)}=b_{0}+b_{1}z+b_{2}z^{2}+\cdots,$ ⓘ Symbols: $z$: variable Referenced by: §1.9(vi) Permalink: http://dlmf.nist.gov/1.9.E54 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
where
1.9.55 $\displaystyle b_{0}$ $\displaystyle=1/a_{0},$ $\displaystyle b_{1}$ $\displaystyle=-a_{1}/a_{0}^{2},$ $\displaystyle b_{2}$ $\displaystyle=(a_{1}^{2}-a_{0}a_{2})/a_{0}^{3},$ ⓘ Permalink: http://dlmf.nist.gov/1.9.E55 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
1.9.56 $b_{n}=-(a_{1}b_{n-1}+a_{2}b_{n-2}+\dots+a_{n}b_{0})/a_{0},$ $n\geq 1$. ⓘ Symbols: $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E56 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
With $a_{0}=1$,
1.9.57 $\ln f(z)=q_{1}z+q_{2}z^{2}+q_{3}z^{3}+\cdots,$ ⓘ Symbols: $\ln\NVar{z}$: principal branch of logarithm function, $z$: variable and $q_{j}$: coefficients Referenced by: §1.9(vi) Permalink: http://dlmf.nist.gov/1.9.E57 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
(principal value), where
1.9.58 $\displaystyle q_{1}$ $\displaystyle=a_{1},$ $\displaystyle q_{2}$ $\displaystyle=(2a_{2}-a_{1}^{2})/2,$ $\displaystyle q_{3}$ $\displaystyle=(3a_{3}-3a_{1}a_{2}+a_{1}^{3})/3,$ ⓘ Symbols: $q_{j}$: coefficients Permalink: http://dlmf.nist.gov/1.9.E58 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
and
1.9.59 $q_{n}=(na_{n}-(n-1)a_{1}q_{n-1}-(n-2)a_{2}q_{n-2}-\cdots-a_{n-1}q_{1})/n,$ $n\geq 2$. ⓘ Symbols: $n$: nonnegative integer and $q_{j}$: coefficients Permalink: http://dlmf.nist.gov/1.9.E59 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
Also,
1.9.60 $(f(z))^{\nu}=p_{0}+p_{1}z+p_{2}z^{2}+\cdots,$ ⓘ Symbols: $z$: variable, $\nu$: complex and $p_{j}$: coefficients Referenced by: §1.9(vi) Permalink: http://dlmf.nist.gov/1.9.E60 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
(principal value), where $\nu\in\mathbb{C}$,
1.9.61 $\displaystyle p_{0}$ $\displaystyle=1,$ $\displaystyle p_{1}$ $\displaystyle=\nu a_{1},$ $\displaystyle p_{2}$ $\displaystyle=\nu((\nu-1)a_{1}^{2}+2a_{2})/2,$ ⓘ Symbols: $\nu$: complex and $p_{j}$: coefficients Permalink: http://dlmf.nist.gov/1.9.E61 Encodings: TeX, TeX, TeX, pMML, pMML, pMML, png, png, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
and
1.9.62 $p_{n}=((\nu-n+1)a_{1}p_{n-1}+(2\nu-n+2)a_{2}p_{n-2}+\dots+((n-1)\nu-1)a_{n-1}p% _{1}+n\nu a_{n})/n,$ $n\geq 1$. ⓘ Symbols: $n$: nonnegative integer, $\nu$: complex and $p_{j}$: coefficients Permalink: http://dlmf.nist.gov/1.9.E62 Encodings: TeX, pMML, png See also: Annotations for 1.9(vi), 1.9(vi), 1.9 and 1
For the definitions of the principal values of $\ln f(z)$ and $(f(z))^{\nu}$ see §§4.2(i) and 4.2(iv).
Lastly, a power series can be differentiated any number of times within its circle of convergence:
1.9.63 $f^{(m)}(z)=\sum_{n=0}^{\infty}{\left(n+1\right)_{m}}a_{n+m}(z-z_{0})^{n},$ $\left|z-z_{0}\right|, $m=0,1,2,\dots$.
## §1.9(vii) Inversion of Limits
### Double Sequences and Series
A set of complex numbers $\{z_{m,n}\}$ where $m$ and $n$ take all positive integer values is called a double sequence. It converges to $z$ if for every $\epsilon>0$, there is an integer $N$ such that
1.9.64 $|z_{m,n}-z|<\epsilon$ ⓘ Symbols: $z$: variable, $m$: nonnegative integer, $n$: nonnegative integer and $\epsilon$: positive number Permalink: http://dlmf.nist.gov/1.9.E64 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1
for all $m,n\geq N$. Suppose $\{z_{m,n}\}$ converges to $z$ and the repeated limits
1.9.65 $\lim_{m\to\infty}\left(\lim_{n\to\infty}z_{m,n}\right),$ $\lim_{n\to\infty}\left(\lim_{m\to\infty}z_{m,n}\right)$ ⓘ Symbols: $z$: variable, $m$: nonnegative integer and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E65 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1
exist. Then both repeated limits equal $z$.
A double series is the limit of the double sequence
1.9.66 $z_{p,q}=\sum^{p}_{m=0}\sum^{q}_{n=0}\zeta_{m,n}.$ ⓘ Symbols: $z$: variable, $m$: nonnegative integer, $n$: nonnegative integer and $\zeta_{p,q}$: sum Permalink: http://dlmf.nist.gov/1.9.E66 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1
If the limit exists, then the double series is convergent; otherwise it is divergent. The double series is absolutely convergent if it is convergent when $\zeta_{m,n}$ is replaced by $|\zeta_{m,n}|$.
If a double series is absolutely convergent, then it is also convergent and its sum is given by either of the repeated sums
1.9.67 $\sum^{\infty}_{m=0}\left(\sum^{\infty}_{n=0}\zeta_{m,n}\right),$ $\sum^{\infty}_{n=0}\left(\sum^{\infty}_{m=0}\zeta_{m,n}\right).$ ⓘ Symbols: $m$: nonnegative integer, $n$: nonnegative integer and $\zeta_{p,q}$: sum Permalink: http://dlmf.nist.gov/1.9.E67 Encodings: TeX, TeX, pMML, pMML, png, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1
### Term-by-Term Integration
Suppose the series $\sum^{\infty}_{n=0}f_{n}(z)$, where $f_{n}(z)$ is continuous, converges uniformly on every compact set of a domain $D$, that is, every closed and bounded set in $D$. Then
1.9.68 $\int_{C}\sum^{\infty}_{n=0}f_{n}(z)\mathrm{d}z=\sum^{\infty}_{n=0}\int_{C}f_{n% }(z)\mathrm{d}z$
for any finite contour $C$ in $D$.
### Dominated Convergence Theorem
Let $(a,b)$ be a finite or infinite interval, and $f_{0}(t),f_{1}(t),\dots$ be real or complex continuous functions, $t\in(a,b)$. Suppose $\sum^{\infty}_{n=0}f_{n}(t)$ converges uniformly in any compact interval in $(a,b)$, and at least one of the following two conditions is satisfied:
1.9.69 $\int^{b}_{a}\sum^{\infty}_{n=0}|f_{n}(t)|\mathrm{d}t<\infty,$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $n$: nonnegative integer Referenced by: §1.9(vii) Permalink: http://dlmf.nist.gov/1.9.E69 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1
1.9.70 $\sum^{\infty}_{n=0}\int^{b}_{a}|f_{n}(t)|\mathrm{d}t<\infty.$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $n$: nonnegative integer Permalink: http://dlmf.nist.gov/1.9.E70 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1
Then
1.9.71 $\int^{b}_{a}\sum^{\infty}_{n=0}f_{n}(t)\mathrm{d}t=\sum^{\infty}_{n=0}\int^{b}% _{a}f_{n}(t)\mathrm{d}t.$ ⓘ Symbols: $\mathrm{d}\NVar{x}$: differential of $x$, $\int$: integral and $n$: nonnegative integer Referenced by: §1.9(vii) Permalink: http://dlmf.nist.gov/1.9.E71 Encodings: TeX, pMML, png See also: Annotations for 1.9(vii), 1.9(vii), 1.9 and 1 | 2017-09-24T08:47:52 | {
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https://math.stackexchange.com/questions/2033639/mod-of-numbers-with-large-exponents | # Mod of numbers with large exponents
I've read about Fermat's little theorem and generally how congruence works. But I can't figure out how to work out these two:
• $13^{100} \bmod 7$
• $7^{100} \bmod 13$
I've also heard of this formula:
$$a \equiv b\pmod n \Rightarrow a^k \equiv b^k \pmod n$$
But I don't see how exactly to use that here, because from $13^1 \bmod 7$ I get 6, and $13^2 \bmod 7$ is 1. I'm unclear as to which one to raise to the kth power here (I'm assuming k = 100?)
Any hints or pointers in the right direction would be great.
• Might be useful to recognize that $6=-1 \mod 7$ – Kitter Catter Nov 28 '16 at 0:26
• @KitterCatter Can you explain it a bit more? (possibly as an answer). Is it derived from $n | (a-b)$? – Roshnal Nov 28 '16 at 0:29
• I don't know about using fermat (not a mathematician by trade) but it might be helpful to know that the Euler totient function of a prime,$p$ is $p-1$ and that if $a^{\phi(p)} \equiv 1 \mod p$ – Kitter Catter Nov 28 '16 at 0:31
• Be sure to understand the relation between "mod" as a binary operator vs. ternary relation. See this answer and this one for more on this. If you only know the operator form you will be severely encumbered. – Bill Dubuque Nov 28 '16 at 0:32
The formula you've heard of results from the fact that congruences are compatible with addition and multiplication.
The first power $$13^{100}$$ is easy: $$13\equiv -1\mod 7$$, so $$13^{100}\equiv (-1)^{100}=1\pmod 7.$$
The second power uses Lil' Fermat: for any number $$a\not\equiv 0\mod 7$$, we have $$a^{12}\equiv 1\pmod{13}$$, hence $$7^{100}\equiv 7^{100\bmod12}\equiv 7^4\equiv 10^2\equiv 9\pmod{13}$$
• Very nice answer! When you reference Lil' Fermat did you mean $a\not\equiv0\mod13$? also it looks like you missed a bracket for the exponent in $a^{12} \equiv 1 \mod 13$ – Kitter Catter Nov 28 '16 at 0:47
• Thanks for the answer! I'm still a little unclear on how you got to $10^2$ from $7^4$? – Roshnal Nov 28 '16 at 1:03
• $7^4=(7^2)^2=49^2\equiv 10^2$, that's all. – Bernard Nov 28 '16 at 1:12
• Ah okay, thanks! It's clear now :) – Roshnal Nov 28 '16 at 1:15
• @Kitter Catter: Oh! Yes. I should have re-read my answer before posting. I even missed a pair of brackets. It's fixed now. Thanks for pointing the typos! – Bernard Nov 28 '16 at 1:15
Hint $$\,$$ The key idea is to use modular order reduction on exponents as in the Lemma below. We can find small exponents $$\,e\,$$ such that $$\,a^{\large e}\equiv 1\,$$ either by Euler's totient or Fermat's little theorem (or by Carmichael's lambda generalization), along with obvious roots of $$\,1\,$$ such as $$\,(-1)^2\equiv 1.$$
Theorem $$\ \$$ Suppose that: $$\,\ \color{#c00}{a^{\large e}\equiv\, 1}\,\pmod{\! m}\$$ and $$\, e>0,\ n,k\ge 0\,$$ are integers. Then
$$\qquad n\equiv k\pmod{\! \color{#c00}e}\,\Longrightarrow\,a^{\large n}\equiv a^{\large k}\pmod{\!m}.\$$ Further, conversely
$$\qquad n\equiv k\pmod{\! \color{#c00}e}\,\Longleftarrow\,a^{\large n}\equiv a^{\large k}\pmod{\!m}\ \,$$ if $$\,a\,$$ has order $$\,\color{#c00}e\,$$ mod $$\,m$$
Proof $$\$$ Wlog $$\,n\ge k\,$$ so $$\,a^{\large n-k} a^{\large k}\equiv a^{\large k}\!\iff a^{\large n-k}\equiv 1\iff n\equiv k\pmod{\!e}\,$$ by here, where we cancelled $$\,a^{\large k}\,$$ using $$\,a^{\large e}\equiv 1\,\Rightarrow\, a\,$$ is invertible so cancellable (cf. below Remark).
Corollary $$\ \ \bbox[7px,border:1px solid #c00]{\!\bmod m\!:\,\ \color{#c00}{a^{\large e}\equiv 1}\,\Rightarrow\, a^{\large n}\equiv a^{\large n\bmod \color{#c00}e}}\,\$$ by $$\ n\equiv n\bmod e\,\pmod{\!e}$$
Remark If you are familiar with modular inverses then it is not necessary to restrict to nonnegative powers of $$\,a\,$$ above since $$\,a^{\large e}\equiv 1,\ e> 0\,\Rightarrow\,$$ $$a$$ is invertible by $$\,a a^{\large e-1}\equiv 1\,$$ so $$\,a^{\large -1}\equiv a^{\large e-1}$$.
• Often it proves simpler to first reduce $\,e\,$ using Euler's Criterion or quadratic reciprocity, e.g. see here.. – Bill Dubuque Jun 8 at 13:31
Quick answer: $13 = 2\cdot 7-1$ so $13\equiv-1\mod 7$ and therefore $13^{100} \equiv (-1)^{100} \mod 7$
Other one is fairly quick: \begin{eqnarray} \phi(13) = 12\\ \gcd(7,13)=1\\ 7^{100}\equiv7^{4} \mod13\\ 7\rightarrow10\rightarrow5\rightarrow9 \end{eqnarray} Probably a nicer way to do that.
• Thanks for the answer. But can you please explain how you arrived to step 3? (in your four-step part of the answer), and why you have calculated the gcd(7, 13)? – Roshnal Nov 28 '16 at 0:58 | 2019-07-19T02:30:51 | {
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https://math.stackexchange.com/questions/1039519/finding-prime-factors-by-taking-the-square-root/1039527 | Finding prime factors by taking the square root
I'm trying to solve the third Project Euler problem and I'd like a little help understanding a mathematical concept underlying my tentative solution.
The prime factors of 13195 are 5, 7, 13, and 29.
What is the largest prime factor of the number 600851475143 ?
As a caveat, in accordance with the wishes of Project Euler I won't be providing any code, my question is centered on why a mathematical concept works.
Failed Attempt
My first algorithm looked at all the numbers from 1 through 600851475143, but was unable to complete the subsequent computations (concerning primes and factorization) due to memory constraints.
Successful Attempt
My next algorithm looked at all the numbers from 1 through $\sqrt{600851475143}$ and completed the computation successfully.
My Question
Why does evaluating $\sqrt{600851475143}$ work in this instance when I really want to evaluate up to 600851475143 ? How can I be sure this approach won't miss some factor like $2 \cdot n$ or $3 \cdot n$, when $n$ is some number between $\sqrt{600851475143}$ and 600851475143 ?
• If a number $N$ has a prime factor larger than $\sqrt{N}$ , then it surely has a prime factor smaller than $\sqrt{N}$. – barak manos Nov 26 '14 at 11:46
• @barakmanos You may as well write this as the answer. – DanielV Nov 26 '14 at 11:47
• You won't miss $2\cdot n$ or $3\cdot n$ because you have checked for $2$ and $3$. – gammatester Nov 26 '14 at 11:48
• We have to be a little careful in writing the program, since some plausible approaches will miss "large" prime factors when a small prime factor occurs with multiplicity $\gt 1$. – André Nicolas Nov 26 '14 at 11:59
• You don't have to check all the way to $\sqrt{600851475143}$. Once you identify 71 as a factor you know that the largest prime factor of 600851475143, is also the largest prime factor of 600851475143/71=8462696833. So at that point you can limit the search to $\sqrt{8462696833}$ - and so on. Once you hit the square root you know, by the argument given in the answers, that the number you are taking the square root of is prime, and also that it's the largest prime factor of your original number. – Taemyr Nov 26 '14 at 15:22
If a number $N$ has a prime factor larger than $\sqrt{N}$ , then it surely has a prime factor smaller than $\sqrt{N}$.
So it's sufficient to search for prime factors in the range $[1,\sqrt{N}]$, and then use them in order to compute the prime factors in the range $[\sqrt{N},N]$.
If no prime factors exist in the range $[1,\sqrt{N}]$, then $N$ itself is prime and there is no need to continue searching beyond that range.
• As mentioned by André Nicolas, prime factors with >1 multiplicity will result in the larger factor being non-prime. To solve this, keep dividing the larger factor of a pair by the smaller one until it no longer divides. – Mark K Cowan Nov 26 '14 at 14:00
• As an example: For 28, factors are 2,2,7, here 7 is larger than sqroot(28), but there is no single prime number that can combine to form 28 (i.e. x*7=28, where x is prime, this does not exist, since x is 4 which is not prime), so dividing the primes from beginning with 2 multiple time will help, 28/2= 14, 14/2 = 7, then we know the 2 & 2 are prime factors, also take the remaining one 7 is also another prime number completing the prime factor list. best of luck. – Manohar Reddy Poreddy Dec 2 '15 at 0:25
If you do not find a factor less than $\sqrt{x}$, then $x$ is prime for the following reason. Consider the opposite, you find two factors larger than $\sqrt{x}$, say $a$ and $b$. But then $a\cdot b> \sqrt{x}\sqrt{x} = x$. Therefore, if there is a factor larger than $\sqrt{x}$, there must also exist a factor smaller than $\sqrt{x}$, otherwise their product would exceed the value of $x$.
Regarding the last question. You will not miss some factor like $2\cdot n$ because you have already checked if $2$ is a factor.
What you're describing is a prime-testing algorithm known as the sieve of Eratosthenes.
It seems like you already understand the concept:
• Cross out $1$
• Circle $2$ as the first prime, then cross out all of the multiples of $2$ in your list.
• The next not-crossed-out number is the next prime (in this case $3$).
If your list has $N$ numbers, you only need to test until you get to a prime that's bigger than $\sqrt{N}$.
Other users have mentioned the reason for $\sqrt{N}$ being the maximum you need to test until, but I'll add an alternative explanation. If you consider listing out all of the factors of $N$, in order from least to greatest, you will either get an even number or an odd number. $N$ has an odd number of factors if and only if it is a perfect square.
In any case, $N$ always has the same number of factors (strictly) less than its square root as it does (strictly) greater than its square root.
In fact, there's an explicit bijection between the set of factors less than $\sqrt{N}$ and the set of factors greater than $\sqrt{N}$ given by $$f(a) = \frac{N}{a}$$
(where $a$ is a factor of $N$ less than $\sqrt{N}$) | 2021-05-15T18:22:46 | {
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https://math.stackexchange.com/questions/3312177/what-is-the-negation-of-this-sentence/3312223 | # What is the negation of this sentence
What is the negation of the following case:
Case A: The function $$f$$ has a unique rational fixed point $$a$$.
I would say the function has:
• multiple rational fixed points or
• no fixed point at all or
• at least one irrational fixed point
Not sure if we can make this more "compact".
• @China: I don't agree with this answer. My point of view is explained in my answer. – Taroccoesbrocco Aug 3 at 10:59
• Well, my answer matches better to your 2. than your 1. So in fact we agree on the negation... – zwim Aug 3 at 13:30
• @Taroccoesbrocco: Thank you. "Unique" refers to "rational fixed point" – Germany Aug 3 at 14:46
• @zwim - I'm not sure to understand your comment: If $f$ has an irrational fixed point and a rational fixed point, then your version of the negation is true (because of point 3) but my version of the negation (see point 2 in my answer) is false. So, your negation and my negation are not equivalent, and only one of them is correct. – Taroccoesbrocco Aug 3 at 15:35
• @China - Then the correct negation is in point $(2)$ of my answer, and not in point $(1)$ (or in zwim's answer). – Taroccoesbrocco Aug 3 at 15:45
The sentence can be interpreted in two ways, depending on whether "unique" refers to "fixed point" or "rational fixed point":
1. "Unique" refers to "fixed point": The function $$f$$ has a unique fixed point $$a$$, and moreover $$a$$ is rational. The logical form of this sentence is \begin{align} \exists a : a \in \mathbb{Q} \land a = f(a) \land \forall z (z = f(z) \to z = a) \end{align} In this case, the negation is: the function $$f$$ either has a unique irrational fixed point, or several fixed points, or no fixed points at all (this is equivalent to what @zwim said in his/her answer).
2. "Unique" refers to "rational fixed point": The function $$f$$ has exactly one rational fixed point (and possibly many irrational fixed points). The logical form of this sentence is: \begin{align} \exists a : a \in \mathbb{Q} \land a = f(a) \land \forall z ((z \in \mathbb{Q} \land z = f(z)) \to z = a) \end{align} In this case the negation is: the function $$f$$ has either several rational fixed points or no rational fixed points at all (but it might have irrational fixed points).
Note that the negation of the interpretation $$(2)$$ is more restrictive than the negation of the interpretation $$(1)$$: for instance, if $$f$$ has a irrational fixed point and a rational fixed point, then the negation of $$(1)$$ is true but the negation of $$(2)$$ is false.
Unlike @zwim, in my opinion the intended meaning of the sentence should be $$(2)$$, and not $$(1)$$, unless contextual information (which is missing in the question of the OP) says differently. | 2019-09-19T08:41:05 | {
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https://math.stackexchange.com/questions/135234/showing-fx-x4-is-not-uniformly-continuous | # Showing $f(x)=x^4$ is not uniformly continuous
I am looking at uniform continuity (for my exam) at the moment and I'm fine with showing that a function is uniformly continuous but I'm having a bit more trouble showing that it is not uniformly continuous, for example:
show that $x^4$ is not uniformly continuous on $\mathbb{R}$, so my solution would be something like:
Assume that it is uniformly continuous then:
$$\forall\epsilon\geq0\exists\delta>0:\forall{x,y}\in\mathbb{R}\ \mbox{if}\ |x-y|<\delta \mbox{then} |x^4-y^4|<\epsilon$$
Take $x=\frac{\delta}{2}+\frac{1}{\delta}$ and $y=\frac{1}{\delta}$ then we have that $|x-y|=|\frac{\delta}{2}+\frac{1}{\delta}-\frac{1}{\delta}|=|\frac{\delta}{2}|<\delta$ however $$|f(x)-f(y)|=|\frac{\delta^3}{8}+3\frac{\delta}{4}+\frac{3}{2\delta}|$$
Now if $\delta\leq 1$ then $|f(x)-f(y)|>\frac{3}{4}$ and if $\delta\geq 1$ then $|f(x)-f(y)|>\frac{3}{4}$ so there exists not $\delta$ for $\epsilon < \frac{3}{4}$ and we have a contradiction.
So I was wondering if this was ok (I think it's fine) but also if this was the general way to go about showing that some function is not uniformly continuous? Or if there was any other ways of doing this that are not from the definition?
Thanks very much for any help
• So, is this an exam question? – user21436 Apr 22 '12 at 12:17
• No, I'm just practicing for my exam where questions like this (not this one though) will come it. This is from one of the past papers that are for revision – hmmmm Apr 22 '12 at 12:29
• Just trying to ensure we are not taken for a ride. Hope you don't mind. :) – user21436 Apr 22 '12 at 12:33
• @KannappanSampath no its fine- it annoys me when I see people posting assessment questions on forums :) – hmmmm Apr 22 '12 at 12:36
To show that it is not uniformly continuous on the whole line, there are two usual (and similar) ways to do it:
1. Show that for every $\delta > 0$ there exist $x$ and $y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|$ is greater than some positive constant (usually this is even arbitrarily large).
2. Fix the $\varepsilon$ and show that for $|f(x)-f(y)|<\varepsilon$ we need $\delta = 0$.
First way:
Fix $\delta > 0$, set $y = x+\delta$ and check $$\lim_{x\to\infty}|x^4 - (x+\delta)^4| = \lim_{x\to\infty} 4x^3\delta + o(x^3) = +\infty.$$
Second way:
Fix $\epsilon > 0$, thus $$|x^4-y^4| < \epsilon$$ $$|(x-y)(x+y)(x^2+y^2)| < \epsilon$$ $$|x-y|\cdot|x+y|\cdot|x^2+y^2| < \epsilon$$ $$|x-y| < \frac{\epsilon}{|x+y|\cdot|x^2+y^2|}$$
but this describes a necessary condition, so $\delta$ has to be at least as small as the right side, i.e.
$$|x-y| < \delta \leq \frac{\epsilon}{|x+y|\cdot|x^2+y^2|}$$
so if either of $x$ or $y$ tends to infinity then $\delta$ tends to $0$.
Hope that helps ;-)
Edit: after explanation and calculation fixes, I don't disagree with your proof.
• Thanks for the reply, I think that I use that we are considering all of $\mathbb{R}$ when i choose $x=\delta+\frac{1}{\delta}$ and $y=\frac{1}{\delta}$ as these would not be valid for small $\delta$ in bounded interval? – hmmmm Apr 22 '12 at 13:38
• @hmmmm Ok, I misunderstood what you were saying there. If the calculations are alright, then your proof is fine. – dtldarek Apr 22 '12 at 13:44
• Is it correct to fix $\delta>0$ and choosing $y = x + \delta$ specifically? Since the distance between x and y will be exactly $\delta$, and we want $|x-y|<\delta$. – DrHAL May 2 '16 at 8:08
• @dtldarek Is it possible that one method will fail to show and other will succeed. For example, for $f(x)=1/x$ on $(0,\infty)$, first method leads the limit to 0? – chandresh Feb 3 at 10:18
• @chandresh Function $f(x)=1/x$ is uniformly continuous on $(1, \infty)$, or even on $(a, \infty)$ for any fixed $a > 0$. On the other hand, $f$ is very steep near zero, and so the limit you want to calculate is $$\lim_{x \to 0^+}\left|\frac{1}{x}-\frac{1}{x+\delta}\right|.$$ – dtldarek Feb 5 at 8:04
I will comment on your solution after writing another approach. For any $x,y\in\mathbb{R}$ we have: \begin{align*} |x^{4}-y^{4}|=|(x^{2}-y^{2})(x^{2}+y^{2})|=|(x-y)(x+y)(x^{2}+y^{2})|=|x-y|\cdot |x+y|\cdot |x^{2}+y^{2}| \end{align*}
So what you can see is that even if you take arbitrarily close $x$ and $y$, you can grow the distance of $x^{4}$ and $y^{4}$ as much as you want by taking them far enough away from zero. You can easily conclude from here that the function is not uniformly continuous by a contraposition for example.
Alright, then to your solution. If the calculations would be correct, then it would be fine. You could assume at first that such $\delta>0$ exists for $0<\varepsilon<3$ and conclude with a contradiction. However, I got a bit different calculations than you. Using the above equation we see that: \begin{align*} |f(\frac{\delta}{2}+\frac{1}{\delta})-f(\frac{1}{\delta})|&=|(\frac{\delta}{2}+\frac{1}{\delta})^{4}-\frac{1}{\delta^{4}}|=|\frac{\delta}{2}(\frac{\delta}{2}+\frac{2}{\delta})((\frac{\delta}{2}+\frac{1}{\delta})^{2}+\frac{1}{\delta^{2}})| \\ &= |(\frac{\delta^{2}}{4}+1)(\frac{\delta^{2}}{4}+2\cdot \frac{\delta}{2}\cdot \frac{1}{\delta}+\frac{1}{\delta^{2}}+\frac{1}{\delta^{2}})| \\ &=|(\frac{\delta^{2}}{4}+1)(\frac{\delta^{2}}{4}+1+\frac{2}{\delta^{2}})| \\ &= |\frac{\delta^{4}}{16}+\frac{\delta^{2}}{4}+\frac{1}{2}+\frac{\delta^{2}}{4}+1+\frac{2}{\delta^{2}}| \\ &= |\frac{\delta^{4}}{16}+\frac{\delta^{2}}{2}+\frac{2}{\delta^{2}}+\frac{3}{2}|\\ &= \frac{\delta^{4}}{16}+\frac{\delta^{2}}{2}+\frac{2}{\delta^{2}}+\frac{3}{2} \end{align*} If you're able to find a lower bound for this (which is quite easy) as you did previously, then by choosing an epsilon smaller than that fixed number you may conclude as you did in your original post by contradiction.
• Hey sorry about that, I edited it-hopefully it's right now? – hmmmm Apr 22 '12 at 13:39
• I also edited now my calculation which still differs abit from your new one. Could you show the steps of how you got this answer for $|f(x)-f(y)|$? – T. Eskin Apr 22 '12 at 13:47
• yeah I messed that up quite a bit sorry (I had the wrong power and the wrong delta's) – hmmmm Apr 22 '12 at 13:50
• It should be $|(\frac{\delta}{2}+\frac{1}{\delta})^4-\frac{1}{\delta^4}|$ which would give $|\frac{\delta^4}{16}+\frac{\delta^2}{2}+\frac{2}{\delta}+\frac{3}{2}|$ I think I could conclude a similar thing from here? – hmmmm Apr 22 '12 at 13:52
• Except that is the last $-\frac{1}{\delta^{4}}$ missing from there? Otherwise it looks close to mine. – T. Eskin Apr 22 '12 at 13:57
(1). If $f:(0,\infty)\to \Bbb R$ is differentiable and $\lim_{x\to \infty}f'(x)=\infty$ then $f$ is not uniformly continuous: For any $\delta >0$ and any $x>0$, the MVT implies there exists $y\in (x,x +\delta)$ such that $\frac {f(x+\delta)-f(x)}{\delta}= f'(y).$ Now if $x$ is large enough that $\forall y>x\;(f'(y)>1/\delta)$ then $f(x+\delta)-f(x)=\delta f'(y)>1.$
(2). Given $\delta >0$ take $x>\max (1,1/\delta).$ Then $$(x+\delta)^4-x^4= 4x^3\delta+6x^2\delta^2+4x\delta^3+\delta^4>$$ $$>4x^3\delta=4(x^2)x\delta>4x\delta>4.$$
By (1) or by (2) we have $\sup_{x\in \Bbb R} \{|(x+x')^4-x^4|: |x'-x|<\delta\}> 1$ for every $\delta>0.$ (In fact the $\sup$ is $\infty$.)
Uniform continuity of $f:D\to \Bbb R$ for some (any) domain $D\subset \Bbb R$ means $$0=\lim_{\delta \to 0^+}\sup \{|f(x')-f(x)|:x,x'\in D\land |x'-x|\leq\delta\}.$$
I think you should make this a little simpler (and for uniform continuity in general) All you need to do to show $f:X \to Y$ is not uniformly continuous on $X$ (let's suppose there both subsets of $\Bbb R$), then just give me a SINGLE epsilon such that, NO MATTER HOW SMALL delta is chosen, there will be x and y closer than delta for which the difference in function values exceed epsilon. Thus for instance $|(N+\theta)^4- N^4| \ge 4\theta N^3$ so if you choose $x$ really big (with respect to $\delta, x=N+\theta\,\,\,\ \text{and}\,\,\, y = N,$ then if $0 < \delta/2 < \theta < \delta$ you have $|x-y| < \delta$ yet you still have the variable N to play with to make the difference in function values as large as you like (in particular the difference in function values can always be made bigger than 3 regardless of how small $\delta$ is). Nevertheless, I think your proof is an accurate job!
## ** Simple Trick:**
Given $a>0$ if we take $$x = \frac{1}{a}+2 ~~~and ~~~y = \frac{1}{a}+2+\frac{a}{2} \color{red}{= x +\frac{a}{2}}$$ we have $x,y\in [2,\infty)~~~ and~~~|x-y| =\frac{a}{2}<a$
But since $a\cdot x = 1+ 2a~~~$ and $~~x,y>2\implies x^2+y^2 >8$ we get,
$$|f(x)-f(y)| =(x^2+y^2) |y^2-x^2|\ge 8|y^2-x^2|\\~~~~~~~~~~~~~~~=8| x^2 + 2\frac{a}{2}x +\frac{a^2}{4} - x^2 | \\= 8(ax +\frac{a^2}{4})= 8(1+2a+\frac{a^2}{4} )>8$$
That is $$|f(x)-f(y)| >8$$
Thus $$\color{blue}{\exists \varepsilon_0 =8,\forall~a>0, \exists ~x,y\in \mathbb R: |x-y|< a~~and ~~|f(x)-f(y)|>8}$$
just take $$x = \frac{1}{a}+2 ~~~and ~~~y = \frac{1}{a}+2+\frac{a}{2} \color{red}{= x +\frac{a}{2}}$$ | 2019-06-18T09:15:17 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/135234/showing-fx-x4-is-not-uniformly-continuous",
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# The amounts of time that three secretaries worked on a
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The amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5. If they worked a combined total of 112 hours, how many hours did the secretary who worked the longest spend on the project?
A) 80
B) 70
C) 56
D) 16
E) 14
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07 Feb 2012, 02:28
8x = 112
=> x = 14
Therefore the secretary who worked the longest spent 14 x 5 = 70 hours on the project
Option (B)
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07 Feb 2012, 05:05
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fiendex wrote:
The amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5. If they worked a combined total of 112 hours, how many hours did the secretary who worked the longet spend on the project?
A) 80
B) 70
C) 56
D) 16
E) 14
Since, amounts of time that secretaries worked in the ratio of 1 to 2 to 5, then $$\frac{\frac{1x}{2x}}{5x}$$, for some number x --> x+2x+5x=8x=112 --> x=14 --> the secretary who worked the longet spent 5x=70 hours.
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Re: The amounts of time that three secretaries worked on a [#permalink]
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07 Feb 2012, 06:53
1
All three secretaries worked 112h and with a ratio of 1to2to5
8x = 112h
x = 14h
5*14h = 50h --> B
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23 May 2017, 02:48
fiendex wrote:
The amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5. If they worked a combined total of 112 hours, how many hours did the secretary who worked the longest spend on the project?
A) 80
B) 70
C) 56
D) 16
E) 14
Similar question: https://gmatclub.com/forum/the-amount-o ... 01139.html
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Re: The amounts of time that three secretaries worked on a [#permalink]
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25 May 2017, 15:48
1
fiendex wrote:
The amounts of time that three secretaries worked on a special project are in the ratio of 1 to 2 to 5. If they worked a combined total of 112 hours, how many hours did the secretary who worked the longest spend on the project?
A) 80
B) 70
C) 56
D) 16
E) 14
The ratio of the amounts of time that the secretaries worked on the project is x : 2x : 5x or a total of 8x.
Since the total number of hours is 112:
8x = 112
x = 14
The secretary who worked the longest worked for 5x = 5(14) = 70 hours.
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08 Dec 2017, 19:36
$$\frac{5}{8ths}$$of the time was spent by the longest working secretary. So the answer is more than 56. An eighth of
112 is 14. 56+14=70
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# The amounts of time that three secretaries worked on a
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne | 2019-10-22T08:55:32 | {
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https://stats.stackexchange.com/questions/241441/is-xt-x-invertible-if-p-n | # Is $X^T X$ invertible if $p > n$?
I am checking other's regression analysis work on a $p > n$ data. I can only see the results but not the process how he did it.
I believe he made mistakes. Since $p > n$, $X^T X$ is not full rank it is not invertible. So we cannot find the OLS coefficient.
However, I am not sure my reasoning is correct. I have not used linear algebra for a long time.
Update:
1. No penalized methods involved.
2. He used stepwise regression for variable selection. A new question: would such algorithm stop if number of variables in the model equal to the number of sample points?
3. His goal is to find out which variables are important. Doesn't care about the prediction power.
• Yes you are correct in that $X^TX$ is non-invertible. However, perhaps your colleague performed some penalized regression like LASSO to get coefficients? Oct 20 '16 at 17:44
• What are $p$ and $n$? Oct 20 '16 at 18:21
• A bit more context may be useful. Is the goal forecasting? Is it consistently estimating some effect $b_i$? Do you know if your colleague performed LASSO or ridge regression? Oct 20 '16 at 18:30
• @RodrigodeAzevedo $p$ and $n$ are classically the number of covariates and of independent observations. IE, $X$ is $n \times p$. Oct 20 '16 at 19:07
• If he's using forward stepwise regression, he'd never have more than $n$ variables in the actual regression matrix used (starting from 1 = the intercept, finding the best variable to add incrementally.) That's not to say forward stepwise regression (or any stepwise regression) is good, it's just to say it would have avoided the identifiability problem. Oct 20 '16 at 19:44
If $\mathrm X$ is $n \times p$ and $p > n$, then it is fat and, thus,
$$\mbox{rank} (\mathrm X) = \mbox{rank} (\mathrm X^T \mathrm X) \leq n < p$$
Hence, $\mathrm X^T \mathrm X$ does not have full rank and, thus, it is not invertible. | 2021-10-25T18:01:51 | {
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http://mathschallenge.net/full/never_decreasing_digits | ## Never Decreasing Digits
#### Problem
A number has strictly increasing digits if each digit exceeds the digit to its left and is said to be a strictly increasing number; for example, 1357.
A number has increasing digits if each digit is not exceeded by the digit to its left and is said to be an increasing number; for example, 45579.
There are exactly 219 increasing numbers below one-thousand.
How many numbers below one million are increasing?
#### Solution
We will begin by deriving, from first principles, the information given in the problem relating to numbers below one-thousand.
It is possible to use binary strings to produce increasing numbers. Consider the following algorithm.
Let S be binary string
Let C = 0
Let K = 1
Label A
Let D be value of Kth digit of S
If D = 1 then output C
If D = 0 then increment c by 1
Increment K by 1
If K has not exceeded the length of S then goto A
Stop
For example, the binary string $001101001$ would produce 2235.
In general it can be seen that an $n$-digit increasing number will be produced by a binary string containing $n$ $1$'s, and the maximum digit will be represented by the number of $0$'s, as each zero causes the "counter" to increase by 1.
Using this idea we can represent all the 3-digit increasing numbers made up of the digits 0, 1, and 2 using 5-digit binary strings consisting of three $1$'s and two $0$'s:
\begin{align}11100 &= 000\\11010 &= 001\\11001 &= 002\\10110 &= 011\\10101 &= 012\\10011 &= 022\\01110 &= 111\\01101 &= 112\\01011 &= 122\\00111 &= 222\end{align}
But we must subtract one to remove $000$, so there are $\dbinom{5}{3} - 1 = 9$ such numbers.
In the same way a 3-digit increasing number using each of the digits 0 to 9 requires a binary string containing three 1's and nine 0's; that is, there are exactly $\dbinom{12}{3} - 1 = 219$ increasing numbers below one-thousand.
Now we can return to the problem: increasing numbers below one million.
As the numbers contain six digits the binary string will contain six 1's and nine 0's. Hence there are $\dbinom{15}{6} - 1 = 5004$ increasing numbers below one million.
Check out the related (strictly) Increasing Digits problem.
Problem ID: 263 (29 Jan 2006) Difficulty: 4 Star
Only Show Problem | 2016-09-25T00:15:58 | {
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https://math.stackexchange.com/questions/1403304/a-is-the-power-set-of-set-c-and-s-is-the-relation-on-a-defined-by-x-s | # $A$ is the power set of set $C$, and $S$ is the relation on $A$ defined by $x S y$ iff $x \subseteq y$.
The following is an exercise from my textbook:
Let $C$ be a set and let $A=\mathcal{P}(C)$, the set of all subsets of $C$. Let $S$ be the relation on $A$ defined by $x S y$ iff $x \subseteq y$. Then $S$ is a partial order relation on $A$.
(a) Find the least and greatest element of $A$.
(b) Let $B=\{x \in A$: $x \ne ∅\}$. Suppose $C$ has two or more elements. Show that $B$ has no least element.
I'm confused about the definition of least and greatest element as applied to this exercise.
From the my textbook, least element is defined as such:
Let $(A,\leq)$ be a partially ordered set and let $B$ be a subset of $A$. To say that $c$ is a least element of $B$ means $c \in B$ and for each $x \in B$, $c \le x$.
Now, referring back to the exercise, $A$ consists of sets, not merely numbers; thus, to say that some set $x_1$ is less than or equal to some other set $x_2$ doesn't make sense. We don't relate sets in that way do we? (Unless, we're talking about order? But even so, I don't think the exercise is asking us to consider that.)
Now, part (b) leads me to believe that a least element of some set of sets $D$, if it exists, is some set such that it is a subset of each set in $D$. Is that correct?
• Note in your definition $(A,\leq)$ is just a partially ordered set. There is no reference to numbers. You need to find the $S$-least element of $A$ and $B$. – James Aug 19 '15 at 23:04
• The exercise is indeed abour inclusion as an order relation . As for part (b) what you believe is right. – Bernard Aug 19 '15 at 23:05
The notation of $\leq$ is often used in mathematics as a general partial order. Not just the usual ordering on the natural/integers/rationals/reals. But in this context, $\leq$ is really just $\subseteq$.
So you are being asked to find the maximum and minimum elements of $(\mathcal P(C),\subseteq)$.
• Okay, so I'm right to say a least element of some set of sets $D$, if it exists, is some set such that it is a subset of each set in $D$? – Jordan Miller Aug 19 '15 at 23:13
• Yes. That is the case here. And note that in the definition of least element, $\leq$ is said to be a partial order, not anything else. – Asaf Karagila Aug 19 '15 at 23:15
• I must get clarification. The following in italics is also an exercise in my textbook: $(A,\leq)$ is a partially ordered set. That each finite non-empty subset of $A$ has a $maxB$ implies $(A,\leq)$ is totally ordered. Then does $\le$ refer again to some general partial order relation? – Jordan Miller Aug 19 '15 at 23:27
• Yes, a general partial order. Mathematics is much much much much more than just real numbers. – Asaf Karagila Aug 19 '15 at 23:34
In your case, the least element is an element of $A$, hence a subset of $C$ that is contained in all subsets of $C$. (Apply your definition in the second highlighted paragraph with $A=\mathcal{P}(C)$ and $B=A$), and so this set, I guess, is the empty set. Now, for the greatest element of $A$, what is the subset of $A$ that contains any subset of $A$?
• So, I am right to think that a least element, with respect to the set $A$, is a set such that it is a subset of each set in $A$? Does that make sense why I was a bit confused. To me, the relation of $\le$ is not the same as $\subseteq$ – Jordan Miller Aug 19 '15 at 23:11
• $\leq$ is just a symbol! – mich95 Aug 19 '15 at 23:12
Okay, I know that what follows basically repeats what responders have been trying to tell me, but hitherto, I was still misunderstanding the notation of $\le$. Thus, I merely post this answer as an expression of my understanding (finally!).
I refer to the definiton:
Let $(A,\le)$ be a partially ordered set and let $B$ be a subset of $A$. To say that $c$ is a least element of $B$ means $c \in B$ and for each $x \in B$, $c \le x$.
The $\le$ in the ordered pair $(A,\le)$ does not refer to the relation of less than or equal to, but rather serves as a "variable" for some partial order relation on A. With this in mind, the condition for each $x \in B$, $c \le x$ doesn't mean that $c$ must be less than or equal to $x$, but rather that $c$ must be in relation $\le$ to $x$.
Now, the partial order relation specified in the above exercise is the relation of set inclusion. Then, by definition, the least element of $A$ must be some set in $A$ such that it is a subset of each set in $A$. | 2019-12-10T00:19:04 | {
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https://mathhelpboards.com/threads/in-the-simplest-qualitative-terms-what-is-a-differential-equation.6170/ | # In the simplest qualitative terms what is a differential equation?
#### find_the_fun
##### Active member
I'm going to be taking a course in differential equations and I'm nervous. From previous calculus courses I know
1. the derivative is the ratio of how one quantity changes with respect to another
2. the integral is the area under the curve
So what's a differential equation? According to here "A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. "
This doesn't make any sense because how is a differential equation different than a derivative? If you are abstractly given a function $$\displaystyle f(x)=x^2$$ then you can't say it's a dirivative or antidirvate of anything.
#### Opalg
##### MHB Oldtimer
Staff member
I'm going to be taking a course in differential equations and I'm nervous. From previous calculus courses I know
1. the derivative is the ratio of how one quantity changes with respect to another
2. the integral is the area under the curve
So what's a differential equation? According to here "A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. "
This doesn't make any sense because how is a differential equation different than a derivative? If you are abstractly given a function $$\displaystyle f(x)=x^2$$ then you can't say it's a dirivative or antidirvate of anything.
In a differential equation, the idea is to find $y$ as a function of $x$, given some information about $y$ and its derivatives. The very simplest example of a differential equation might be something like the equation $\frac{dy}{dx} = 2x$. You can easily solve that by integrating it, to find that the solution is $y=x^2$ plus a constant of integration. But suppose that the equation is slightly more complicated, for example $\frac{dy}{dx} = x +y$. How would you solve that to find $y$ as a function of $x$? A course in differential equations will show you how to do that.
In case you are wondering, the solution to that equation is $y = -x-1 + ce^x$, where $c$ is a constant.
#### HallsofIvy
##### Well-known member
MHB Math Helper
I'm going to be taking a course in differential equations and I'm nervous. From previous calculus courses I know
1. the derivative is the ratio of how one quantity changes with respect to another
2. the integral is the area under the curve
So what's a differential equation? According to here "A differential equation is any equation which contains derivatives, either ordinary derivatives or partial derivatives. "
This doesn't make any sense because how is a differential equation different than a derivative? If you are abstractly given a function $$\displaystyle f(x)=x^2$$ then you can't say it's a dirivative or antidirvate of anything.
A differential equation is an equation, a derivative is NOT! Okay, having got that off my chest, I think I understand your point. "Contains derivatives", etc. is not sufficient. The crucial point is that a differential equation contains derivatives of some unknown function. Yes, "$$x^2$$" can be thought of as the derivative of $$\frac{1}{3}x^3$$ but just having $$x^2$$ in an equation does NOT make it a "differential equation". To be a differential equation, the equation must contain something like $$\frac{dy}{dx}$$, $$\frac{\partial y}{\partial t}$$, $$\frac{d^2y}{dx^2}$$, etc., where y is the "unknown" function.
More generally, a "functional equation" is an equation that gives us some information about a function. "f(x+ 1)= f(x)" is a functional equation. $$\frac{d^2y}{dx^2}+ 2\frac{dy}{dx}+ y= x^2$$ is a special kind of a functional equation (since it gives information about the function y) called a "differential equation" because that information is actually about the derivatives of the function y.
#### find_the_fun
##### Active member
Glad I asked. I had the first lecture today and the prof skipped over the definition of a differential equation (in fairness it was a substitute prof).
What is an ordinary differential equation? Does all that mean is it doesn't have partial derivatives?
#### Opalg
##### MHB Oldtimer
Staff member
What is an ordinary differential equation? Does all that mean is it doesn't have partial derivatives?
Yes. You will often see the abbreviations ODE and PDE for ordinary/partial differential equation.
#### find_the_fun
##### Active member
I'm writing myself notes now and here's the first:
Differential equation: a functional equation that relates an unknown function to its derivatives
#### zzephod
##### Well-known member
An ordinary differential equation is an equation of the form (or that can be rewritten in the form):
$$\displaystyle \large f(x,y,y',...,y^{(n)})=0$$
A partial differential equation is similar but with partial derivatives with repect to the variables appearing (including mixed partials).
.
Last edited: | 2020-12-01T08:36:24 | {
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https://math.stackexchange.com/questions/927578/whats-fracdydx-of-function-y-frac-12-sin-x | # What's $\frac{dy}{dx}$ of function $y=\frac 12{\sin x}$?
$(1)\quad$ How to differentiate $y=\frac 12 \sin x$?
I know that $\frac{dy}{dx}$ of $y=\sin x$ is easy to calculate: $\frac{dy}{dx} = \cos x$. But what if there is a coefficient preceding before it?
$(2)$ Another question is: Does the following equality of functions hold? $$\frac{x(x^2+1)}{x^2+1}\overset{?}{=}x$$
Please explain this because I know that function $\dfrac{x\cdot x}{x}$ does not equal function $x$ (the first function is not defined at $0$ but the second one is defined at $0$).
But the question I asked, the domain of the function can be any number according to what I think because the denominator cannot be zero, it's always a positive number $x^2+1$.
• To your second question, since $x^2+1$ is never $0$ (at least in the real numbers) $\frac{x^2+1}{x^2+1}=1$ and $\frac{x(x^2+1)}{x^2+1} = x$ for all $x$. So yes, the functions are equal. – Thomas Andrews Sep 11 '14 at 13:10
• Do you mean $\frac{1}{2\sin x}$ or $\frac{1}{2}\sin x$? – Thomas Andrews Sep 11 '14 at 13:11
• I mean 1/2 (sin x) – off99555 Sep 11 '14 at 13:13
• One should avoid different questions in a same question on MSE. – user37238 Sep 11 '14 at 13:16
• This is my first time using this website, I will be aware next time. – off99555 Sep 11 '14 at 13:18
Question $(1):\quad$ Given $y = \dfrac 12 \sin x$:
$$\frac{dy}{dx} = \frac{d}{dx}\left(\frac 12 \sin x\right) = \frac 12\cdot \frac{d}{dx}(\sin x) = \frac 12 \cos x$$
$$y = a f(x)\implies \frac{dy}{dx} = \frac{d}{dx}(a f(x)) = af'(x)$$ for all constants $a$.
Question $(2):$
$$\frac {x(x^2 + 1)}{x^2 + 1 } = x \quad \forall x \in \mathbb R$$
This happens to be the case in this example because there are no real values of $x$ at which the left-hand side is undefined. Specifically, as you note, $x^2 + 1>0$ for all $x$, and hence the function is defined everywhere. So we may cancel the common factor $x^2 + 1$ without changing the function in any way.
In the case of $f(x) = \frac{x^2}{x}$, which you refer to, here we do have problems with simply canceling a common factor of $\,x.\,$ Specifically, $\,\dfrac{x^2}{x}\,$ is undefined at $x = 0$, whereas the function $g(x) = x$ is defined everywhere, so the functions are not equivalently defined, i.e., they are not equivalent functions.
• I would say that the first response is missing one sign "=" -> ambiguity. – georg Sep 11 '14 at 14:08
• Thanks, @georg. I didn't catch that until you mentioned it.. – amWhy Sep 11 '14 at 14:19
For the first part of your question, you can remove constants when differentiating:
$$\frac{\mathrm{d}}{\mathrm{d}x}\big(a\cdot f(x)\big)=a\frac{\mathrm{d}}{\mathrm{d}x}f(x)$$
You can prove this with the product rule by finding $a\frac{\mathrm{d}}{\mathrm{d}x}f(x)+f(x)\frac{\mathrm{d}}{\mathrm{d}x}a$ & noting that the derivative of a constant is $0$. It implies that your derivative is $\frac{\mathrm{d}}{\mathrm{d}x}\big(\frac{1}{2}\sin(x)\big)=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)$.
For the second part of the question, it is valid to cancel terms in a function but just be aware that the function's domain should stay the same. If the original function is undefined at a number, it should stay that way. As Thomas Andrews points out, $x^2+1$ is never $0$ for real $x$ so you can effectively cancel the $\frac{(x^2+1)}{(x^2+1)}$ term.
• I got the edit wrong - he wanted $y=\frac{1}{2}\sin x$. Sorry. – Thomas Andrews Sep 11 '14 at 13:14
• @ThomasAndrews Ah, thanks for letting me know :) No problem by the way, I've done it before too. – Jam Sep 11 '14 at 13:14
• Thank you very much for your help. I would like to vote both of you up but I can't. My reputation is not high enough to do that. Both answers are very useful for me. But I don't understand what setting a as a function of x mean. Can you show me please? – off99555 Sep 11 '14 at 14:25
• @off99555 Glad I could help. What I meant by "setting it as a function" was that, by using the product rule, we can find the derivative of $f(x)\cdot g(x)$ as $f(x)g'(x)+g(x)f'(x)$. If we say that $g(x)$ is a constant $a$, we could find the derivative of $a\cdot f(x)$. Since the derivative of any constant is $0$, it shows us that we can take out the constant. So $\frac{d}{dx}(\frac{1}{2}\sin(x))=\frac{1}{2}\frac{d}{dx}\sin(x)$. Does that make sense? – Jam Sep 11 '14 at 14:45
• Sure. That make sense! That's what I'm looking for, the fundamental explanation. – off99555 Sep 17 '14 at 15:46 | 2020-04-02T10:55:31 | {
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https://www.shaalaa.com/question-bank-solutions/a-typist-charges-rs-145-typing-10-english-3-hindi-pages-while-charges-typing-3-english-10-hindi-pages-are-rs-180-using-matrices-find-charges-typing-one-english-one-hindi-page-separately-inverse-of-matrix-inverse-of-a-nonsingular-matrix-by-elementary-transformation_4023 | A Typist Charges Rs. 145 for Typing 10 English and 3 Hindi Pages, While Charges for Typing 3 English and 10 Hindi Pages Are Rs. 180. Using Matrices, Find the Charges of Typing One English and One Hindi Page Separately. - Mathematics
A typist charges Rs. 145 for typing 10 English and 3 Hindi pages, while charges for typing 3 English and 10 Hindi pages are Rs. 180. Using matrices, find the charges of typing one English and one Hindi page separately. However typist charged only Rs. 2 per page from a poor student Shyam for 5 Hindi pages. How much less was charged from this poor boy? Which values are reflected in this problem?
Solution
Let charges for typing one English page be Rs. x.
Let charges for typing one Hindi page be Rs.y.
Thus from the given statements, we have,
10x+3y=145
3x+10y=180
Thus the above system can be written as,
[(10,3),(3,10)][(x),(y)]=[(145),(180)]
⇒ AX = B, where, A=[(10,3),(3,10)],x=[(x),(y)] " and " B = [(145),(180)]
Multiply A-1 on both the sides, we have,
A-1 x AX = A-1B
⇒ IX = A-1B
⇒ X = A-1B
Thus, we need to find the inverse of the matrix A.
We know that, if P=[(a,b),(c,d)] " then " P^(-1) = 1/(ad-bc)[(d,-b),(-c,a)]
Thus, A^(-1)=1/(10xx10-3xx3)[(10,-3),(-3,10)]
= 1/(100-9)[(10,-3),(-3,10)]
=1/91[(10,-3),(-3,10)]
Therefore, X=1/91[(10,-3),(-3,10)][(145),(180)]
=1/91[(10xx145-3xx180),(-3xx145+10xx180)]
=1/91[(910),(1365)]
=[(10),(15)]
=>[(x),(y)][(10),(15)]
⇒ x = 10 and y=15
Amount taken from Shyam = 2 × 5 = Rs.10
Actual rate = 15 × 5 =75
Difference amount = Rs.75 – Rs.10 = Rs.65
Rs. 65 less was charged from the poor boy Shyam.
Humanity and sympathy are reflected in this problem.
Concept: Inverse of Matrix - Inverse of a Nonsingular Matrix by Elementary Transformation
Is there an error in this question or solution? | 2021-10-18T19:09:34 | {
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https://math.stackexchange.com/questions/441106/how-do-i-show-that-int-0-infty-frac-sinax-sinbxx2-mathrm-dx | # How do I show that $\int_0^\infty \frac{\sin(ax) \sin(bx)}{x^{2}} \, \mathrm dx = \pi \min(a,b)/2$
Recently I found a claim saying that $$\int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x= \pi \min(a,b)/2$$ from what I can see this seems to be true. I already know that $\int_{0}^\infty \operatorname{sinc}xy\,\mathrm{d}y = \pi/2$, and so independant of $y$. My suspicion is that this is closely related to the integral above.
Can someone give me some suggestions for evaluating the integral above? Also are there any generalizations for the integral? Eg $$\int_{0}^{\infty} \left( \prod_{k=1}^N \frac{\sin (a_k \cdot x)}{x} \right) \,\mathrm{d}x$$ Where $a_k, \cdots, a_N$ are arbitrary positive constants. It seems related to the Borwein Integral, but there are some subtle differences.
• Don't you want a product, not a summation, in your last question? And, when you write "independent of $y$" don't you mean independent of $x$? – Gerry Myerson Jul 11 '13 at 9:04
• Does $\sin p\sin q=(\cos(p-q)-\cos(p+q))/2$ help? – Gerry Myerson Jul 11 '13 at 9:07
One way is to to this by residues. Another way to integrate once by parts to get $$I=\int_0^{\infty}\frac{b\sin ax\cos bx+a\cos ax \sin bx}{x}dx,$$ then to use the formula $2\sin \alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta)$ and the mentioned integral (note that your formula needs to be corrected on the left and on the right) $$\displaystyle \int_0^{\infty}\frac{\sin xy}{x}\,dx=\frac{\pi}{2}\mathrm{sgn}(y).$$ This gives \begin{align}I&=\frac{\pi}{4}\Bigl[b\,\mathrm{sign}(a+b)+b\,\mathrm{sign}(a-b)+a\,\mathrm{sign}(a+b)+a\,\mathrm{sign}(b-a)\Bigr]=\\ &=\frac{\pi}{4}\left(|a+b|-|a-b|\right), \end{align} For $a,b>0$ the last expression is obviously equal to $\pi\min\{a,b\}/2$.
• Just posted the question as a dupe, such an amazing result. Do you know who first discovered it? – aronp Dec 21 '15 at 21:33
• Since this result is independent of the lower number, presumably this works for the three factor case? – aronp Dec 21 '15 at 21:46
• @aronp No I don't; I totally agree with you that it is quite counterintuitive. – Start wearing purple Dec 21 '15 at 21:47
In several steps:
• Trigonometric relation $$\left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right)=\frac{1-\cos(a+b)x}{2x^2}-\frac{1-\cos(a-b)x}{2x^2}$$
• Dirichlet integral $$\int_0^\infty \frac{\sin \alpha t}{t}dt=\frac{\pi}{2}\mathrm{sgn}(\alpha)$$
• Integration by parts $$\int_0^\infty\frac{1-\cos(\alpha)t}{t^2}=\alpha\int_0^\infty \frac{\sin \alpha t}{t}dt=\frac{\pi}{2}|\alpha|$$
• Combine $$\int_0^\infty \left( \frac{\sin ax}{x}\right)\left( \frac{\sin bx}{x}\right) \mathrm{d}x=\frac{\pi}{4}(|a+b|-|a-b|)=\frac{\pi}{2}\min(a,b)$$
• Excellent!${}{}{}$ – Namaste May 20 '14 at 20:14
A very easy way to see this is to use Parseval's theorem for Fourier transforms. In general, Parseval's theorem states that, for two functions $f$ and $g$, each having respective FTs $\hat{f}$ and $\hat{g}$, related by
$$\hat{f}(k) = \int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x}$$
etc., then
$$\int_{-\infty}^{\infty} dx \, f(x) \bar{g}(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, \hat{f}(k) \bar{\hat{g}}(k)$$
The FT of $\sin{(a x)}/x$ is given by
$$\int_{-\infty}^{\infty} dx \, \frac{\sin{a x}}{x} e^{i k x} = \begin{cases} \pi & |k| \lt a \\ 0 & |k| \gt a\end{cases}$$
Similarly,
$$\int_{-\infty}^{\infty} dx \, \frac{\sin{b x}}{x} e^{i k x} = \begin{cases} \pi & |k| \lt b \\ 0 & |k| \gt b\end{cases}$$
By Parseval, we take the integral of the product of the transforms, which is the product of two rectangles. The product is clearly nonzero over the smaller of the two widths, i.e. $2 \min\{a,b\}$. Thus,
$$\int_{-\infty}^{\infty} dx \, \frac{\sin{a x}}{x} \, \frac{\sin{b x}}{x} = \frac{\pi^2}{2 \pi} 2 \min\{a,b\}$$
Therefore
$$\int_{0}^{\infty} dx \, \frac{\sin{a x}}{x} \, \frac{\sin{b x}}{x} = \frac{\pi}{2} \min\{a,b\}$$
Let $b_{k} >0$ and $a \ge \sum_{k=1}^{n} b_{k}$.
Generalizing the answer HERE, we can use contour integration to quickly show that
$$\int_{0}^{\infty} \frac{\sin(ax)}{x} \prod_{k=1}^{n} \frac{\sin \left( b_{k}x \right)}{x} \, dx= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin(ax)}{x} \prod_{k=1}^{n} \frac{\sin \left( b_{k}x \right)}{x} \, dx = \frac{\pi}{2} \prod_{k=1}^{n}b_{k}. \tag{1}$$
Under the conditions stated above, the function $$e^{iaz} \prod_{k=1}^{n} \sin \left( b_{k}x \right) = e^{iaz} \prod_{k=1}^{n} \frac{e^{ib_{k}z}-e^{-ib_{k}z} }{2i}$$ is bounded in the upper half-plane.
So by integrating the function $$f(z) = \frac{e^{iaz}}{z} \prod_{k=1}^{n} \frac{\sin \left( b_{k}z \right)}{z}$$ around an indented contour that consists of the real axis and the semiciricle above it, we get (in the limit), $$\text{PV} \int_{-\infty}^{\infty} \frac{e^{iax}}{x} \prod_{k=1}^{n} \frac{\sin \left( b_{k}x \right)}{x} \, dx - \pi i \, \text{Res} [f(z), 0] = 0,$$ where $$\text{Res}[f(z) ,0] = \lim_{z \to 0}e^{iaz} \prod_{k=1}^{n}\frac{\sin \left( b_{k}z \right)}{z} =1\times \prod_{k=1}^{n} b_{k}.$$
Taking the imaginary parts of both sides of equation leads to the result.
It's sufficient to consider the case $\ds{\bbox[10px,#ffe,border:1px dotted navy] {\ds{a, b \in \mathbb{R}_{\ >\ 0}}}}$.
\begin{align} \mbox{Note that}\ &\int_{0}^{\infty}{\sin\pars{ax} \over x}\,{\sin\pars{bx} \over x}\,\dd x = {1 \over 2}\,b\int_{-\infty}^{\infty}{\sin\pars{ax} \over ax} \,{\sin\pars{\bracks{b/a}ax} \over \pars{b/a}ax}\,a\,\dd x \\[5mm] = &\ {1 \over 2}\,b\int_{-\infty}^{\infty}{\sin\pars{x} \over x} \,{\sin\pars{x/\mu} \over x/\mu}\,\dd x \,,\qquad\mu \equiv {a \over b} > 0 \label{1}\tag{1} \end{align}
\begin{align} &\left.\int_{-\infty}^{\infty}{\sin\pars{x} \over x} \,{\sin\pars{x/\mu} \over x/\mu}\,\dd x \,\right\vert_{\ \mu\ >\ 0} = \int_{-\infty}^{\infty} \pars{{1 \over 2}\int_{-1}^{1}\expo{\ic kx}\,\dd k} \pars{{1 \over 2}\int_{-1}^{1}\expo{-\ic qx/\mu}\,\dd q}\dd x \\[5mm] = &\ {1 \over 2}\,\pi \int_{-1}^{1}\int_{-1}^{1}\int_{-\infty}^{\infty}\expo{\ic\pars{k - q/\mu}x} {\dd x \over 2\pi}\,\dd k\,\dd q = {1 \over 2}\,\pi\int_{-1}^{1}\int_{-1}^{1}\delta\pars{k - q/\mu}\,\dd k\,\dd q \\[5mm] = &\ {1 \over 2}\,\pi\int_{-1}^{1}\bracks{-1 < {q \over \mu} < 1}\dd q = \pi\int_{0}^{1}\bracks{q < \mu}\dd q = \pi\braces{\bracks{\mu < 1}\int_{0}^{\mu}\dd q + \bracks{\mu > 1}\int_{0}^{1}\dd q} \\[5mm] = &\ \bracks{a < b}\pi\,{a \over b} + \bracks{a > b}\pi\label{2}\tag{2} \end{align}
With \eqref{1} and \eqref{2}: \begin{align} \int_{0}^{\infty}{\sin\pars{ax} \over x}\,{\sin\pars{bx} \over x}\,\dd x & = {1 \over 2}\,\pi\braces{\vphantom{\Large A}\bracks{a < b}a + \bracks{a > b}b} = \bbx{\ds{{1 \over 2}\,\pi\,\min\braces{a,b}}} \end{align} | 2019-07-18T13:10:50 | {
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https://math.stackexchange.com/questions/1585795/how-is-940-equiv-1-pmod-100 | How is $9^{40}\equiv\ 1 \pmod {100}$?
By Euler's theorem, $a^{\varphi(100)} \equiv\ 1 \pmod {100}$.
We know that the last two digits of $9^{40}$ are non-zero. So they can't even be $01$.
Since $1\equiv\ 1 \pmod {100}$, how come $9^{40}\equiv\ 1 \pmod {100}$?
I have looked at: Find the last two digits of $9^{{9}^{9}}$,
Find the last two digits of $9^{9^{9}}$
• "We know that the last two digits of $9^{40}$ are non-zero." That doesn't mean neither digit is zero; it just means they can't both be zero. So why do you think it can't be $01$? – Erick Wong Dec 22 '15 at 17:14
• you're right i was thinking only of the the units digit – AkaiShuichi Dec 22 '15 at 17:17
• @pyUser: When a more experienced user edits your post, learn from it, don't roll back to your own improperly formatted version. – Alex M. Dec 22 '15 at 17:18
• @AlexM. can you teach me how to apply edits from one version to another one? or did you just manually MathJax the thing? – gt6989b Dec 22 '15 at 17:19
• You quoted one way of doing it. We have $\varphi(100)=\varphi(25)\varphi(4)=40$. So $9^{40}\equiv 1\pmod{100}$, the last two digits, going righttwards, are $0$ and $1$. – André Nicolas Dec 22 '15 at 17:24
By binomial theorem:
$$(10-1)^{40} = \sum_{i=0}^{40} \binom{40}{i}(-1)^{40-i}10^{i}$$
Modulo $10^2$, you only need to look at $i=0,1$.
So: $$9^{40}\equiv (-1)^{40} + \binom{40}{1}(-1)^{39}\cdot 10 \equiv (-1)^{40}\equiv 1\pmod{100}$$
Note that this works for $9^{10}$, too.
$9^2\equiv\ -19\ (mod100)$
$9^4 \equiv\ 61\ (mod 100)$
$9^8\equiv\ 21 \ (mod 100)$
$9^{10}\equiv\ 1 \ (mod 100)$
$9^{40}\equiv\ 1 \ (mod 100)$
• Better formatting if you write 9^2\equiv -19\pmod{100} yielding $$9^2\equiv -19\pmod{100}$$ – Thomas Andrews Dec 22 '15 at 18:01
Calculate $$9^{10} = 3486784401$$ So $$9^{40} \mod 100 \equiv (9^{10}\mod 100)^4 \mod 100$$ $$\equiv(1)^4 \mod 100 \equiv 1 \mod 100$$
• How did you go about your third step? – AkaiShuichi Dec 22 '15 at 17:20
• $x^y \mod m$ is the same as $(x \mod m)^y \mod m$. You can move the $\mod$ function into multiplies, etc. – amcalde Dec 22 '15 at 17:22
• i was talking about the (1)^4 part – AkaiShuichi Dec 22 '15 at 17:23
• In the first calculation I showed that $9^{10} \equiv 1 \mod 100$. – amcalde Dec 22 '15 at 17:24
Note that $$99^2=9801\equiv 1\pmod{100}\implies99^{40}\equiv 1\pmod{100}\implies9^{40}11^{40}\equiv 1\pmod{100}$$
$$9^{40}\equiv 1\pmod{100}$$
It is straightforward $9^{40} = 81^{20} = 6561^{10} \equiv 61^{10} \pmod{100} = 3721^5 \pmod{100} \equiv\ 21^5 \pmod{100}$
$= 441 \times 441 \times 21 \pmod{100}\equiv\ 41 \times 41 \times 21 \pmod{100}$
$= 35301 \pmod{100} \equiv\ 1 \ \pmod{100}$
and this is just one of the many many routes you could take.
If you want to avoid the equation
$9^{40}=147808829414345923316083210206383297601$,
the question is how low you prefer your numbers to be.
One alternative is using $40=4*2*5$ (and 7 digits for the last operation).
$9^4=61$ (mod 100)
$61^2=21$ (mod 100), and
$21^5=01$ (mod 100) | 2021-05-10T19:25:27 | {
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https://math.stackexchange.com/questions/2772749/finding-coefficients-in-polynomial-given-three-tangents | # Finding coefficients in polynomial given three tangents
I am stuck with a problem I simply cannot solve.
I have to find the coefficients of a quadratic polynomial given three tangents. The problem is stated as follows:
The three lines described by the equations
$y_1(x)=-4x-16.5$
$y_2(x)=2x-4.5$
$y_3(x)=6x-16.5$
are all tangents to a quadratic polynomial $p(x)=ax^2+bx+c$
Determine the values of the coefficients a, b & c.
I simply cannot solve this problem, I've been at it for a long time. Any help is greatly appreciated :)
Edit: I'm including the way I tried to solve it. I didn't get super far.
Given the polynomial $p(x)$ I know that $p'(x)=2ax+b$
Therefore, the following is true for the three points with x-values of $x_1, x_2$ and $x_3$, where the lines $y_1, y_2$ and $y_3$ are tangent to the parabola:
$-4=2ax_1+b$
$2=2ax_2+b$
$6=2ax_3+b$
That's all I've managed to do. I've also found the points where the three lines intersect (well, three points two of the lines intersect), but I can't think of how to use that for anything.
• Show anyway what you tried. I can give a solution using derivatives, but maybe it is not the solution you want. – N74 May 8 '18 at 20:23
• Welcome to SE! Even if your attempts didn't work, you should absolutely include them in your post - people can advise you better that way! – B. Mehta May 8 '18 at 20:23
• @N74 I have now included my failed solution :) As you can see, I thought of using derivatives as well, but I can't get further. – Mads Clausen May 8 '18 at 20:38
• My maths teacher (I'm in the last year of Danish "high school") gave it to us for tomorrow. He said that it was definitely doable. – Mads Clausen May 8 '18 at 20:42
If we have a parabola $$y = a x^2 + b x + c$$ and a line $$y = mx + d,$$ they are tangent if $$b^2 - 4ac = -4da + 2mb - m^2$$ as then the parabola $$y = a x^2 + (b-m) x + (c -d)$$ has a double root, i.e. is a constant times a square, $$a (x+p)^2$$ Might as well write this: I fixed $$\Delta = b^2 - 4 a c$$ and then had three equations $$-4da + 2mb = m^2 + \Delta$$ by plugging in the values from the three lines $y = mx + d.$ I expected bigger problems, but just taking the differences of two of the equations cancels the extra unknown $\Delta,$ alowing us to find $a,b$ quickly. From that, we finally get a value for $\Delta,$ after which we have one equation for $c$
$$66a - 8 b = 16 + \Delta \; ,$$ $$66a + 12 b = 36 + \Delta \; ,$$ $$18a + 4 b = 4 + \Delta \; .$$ Second minus first gives $b.$ Plug in the $b$ value, then subtract second minus third, which gives $a.$ Plug both into any of the three to find $\Delta.$ Finally, $c = \frac{b^2 - \Delta}{4a}$
Note that if a line and a quadratic are tangent $mx+d=ax^2+bx+c$ then the following quadratic will have discriminant zero \begin{eqnarray*} ax^2+(b-m)x+c-d=0. \end{eqnarray*} This will lead to $3$ equations for $a,b,c$ that are easily solved giving
$(a,b,c)=(1/2,1,-4)$.
• Thanks for the answer. I don't quite follow, however: what are the three equations for a, b and c that you mention? – Mads Clausen May 8 '18 at 21:02
• \begin{eqnarray*} (b+4)^2=2a(c+33) \\ (b-2)^2=2a(c+9) \\ (b-6)^2=2a(c+33) \end{eqnarray*} combine the first & the third of these equations & you will rapidly yield a value for $b$. – Donald Splutterwit May 8 '18 at 21:05
• Donald I get different final abc values, although still all rational and easily found... I kept the discriminant $\Delta = b^2 - 4ac \;$ as an unknown until the end, began with three $\Delta = -4da + 2mb - m^2$ or $-4da + 2mb = m^2 + \Delta$ – Will Jagy May 8 '18 at 21:09
• where does $2a(c+n)$ come from? I recall that the discriminant is equal to $b^2-4ac$, which in this case would yield something like $(b-m)^2=4a(c-d)$ I would think it would either be that or $(b-m)^2=2a(2c-2d)$ – Mads Clausen May 8 '18 at 21:12
• I get 1/2, 1 and -4 for a,b,c. 2ax+b = -4 at x=-5; 2 at x = 1 and 6 at x = 5 with x and y values being the same for tangent lines and parabola at these tangent points. – Phil H May 8 '18 at 21:32
Given any two tangents to a parabola of the form $y = f(x) = ax^2 + bx + c,$ the $x$-coordinate of the intersection of the tangent lines will be midway between the $x$-coordinates of the tangent points.
Working out the intersection points of the three lines, we can see that the intersection of $y = y_1(x)$ and $y = y_2(x)$ occurs at $x = -2$ and the intersection of $y = y_2(x)$ and $y = y_3(x)$ occurs at $x = 3.$ Let
• $x = x_1$ at the tangent point with $y = y_1(x)$;
• $x = x_2$ the tangent point with $y = y_2(x)$; and
• $x = x_3$ the tangent point with $y = y_3(x).$
Due to the $y$-coordinates and slopes at the intersection points, it is clear that $x_1 < -2 < x_2 < 3 < x_3$; moreover, $-2$ is midway between $x_1$ and $x_2$ and $3$ is midway between $x_2$ and $x_3.$ It follows that $$x_3 - x_1 = 2\left(\frac{x_3 + x_2}2 - \frac{x_2 + x_1}2\right) = 2(3 - (-2)) = 10.$$
The tangency condition implies that $f'(x_1) = y_1'(x) = -4$ and $f'(x_3) = y_3'(x) = 6.$ But $f'(x) = 2ax + b,$ so $20a = 2a(x_3 - x_1) = f'(x_3) - f'(x_1) = 10,$ and therefore $a = \frac12.$ It is then a relatively straightforward exercise to find the other two coefficients. | 2019-10-21T05:19:12 | {
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https://math.stackexchange.com/questions/4495973/the-product-of-primes-between-n-and-2n-compared-to-2n | # The Product of Primes Between $N$ and $2N$ Compared to $2^N$
While reading some course notes from MIT 18.703 (Modern Algebra) on a primality test, I found this lemma (stated without proof):
Lemma 22.3. The product of all prime numbers $$r$$ between $$N$$ and $$2N$$ is greater than $$2^{N}$$ for all $$N\geq1$$.
However, one quickly finds that this lemma is false for $$N=8$$. The primes between $$8$$ and $$16$$ are $$11$$ and $$13$$, and $$11\cdot13 = 143 < 256 = 2^{8}$$.
I wondered if there were any more counterexamples, so I decided to write a quick program to test this lemma (the code may be found here). My code is not very optimized, with only very basic parallelism using OpenMP, and so I haven't taken the time to go beyond $$N = 100000$$. With my code, I have found only three counterexamples:
At $$N = 8$$, the product of primes between $$8$$ and $$16$$ is $$143 < 256 = 2^{8}$$.
At $$N = 14$$, the product of primes between $$14$$ and $$28$$ is $$7429 < 16384 = 2^{14}$$.
At $$N = 20$$, the product of primes between $$20$$ and $$40$$ is $$765049 < 1048576 = 2^{20}$$.
My Questions
1. Are these the only counterexamples?
2. If so, how do we prove it?
3. If not, what are some others, and are there infinitely many? -- Answer: There are not infinitely many, though there may still be further counterexamples.
EDIT 1: Thanks to the responses to this post, I have attempted an analysis of this problem. If possible, I would appreciate any feedback, just in case I made an error or if any improvement could be made by a more careful analysis. -- This analysis was flawed and has been retracted.
EDIT 2: Due to the answer posted by jjagmath, the bound has been lowered to 10544111. I will try to run my program to search for counterexamples below this number, but hopefully even tighter analyses come that lower the bound to a more tractable number to search.
EDIT 3: I retract the analysis that I did because of a mistake. The jjagmath analysis still holds.
EDIT 4:
Let $$p_{k}$$ be the $$k$$th prime number, let $$\displaystyle{N_{k} = \frac{p_{k} - 1}{2}}$$, and let $$\displaystyle{f\left(N\right)=\prod_{N\leq p\leq 2N}p}$$.
Proposition. Let $$k > 2$$ be an integer and let $$\nu$$ be an integer such that $$N_{k-1} < \nu\leq N_{k}$$. Then, $$f\left(N_{k}\right)\leq f\left(\nu\right)$$.
Proof. Any prime in the interval $$\left[N_{k},2N_{k}\right]$$ is also in the interval $$\left[\nu, 2\nu\right]$$, and so $$f\left(\nu\right)$$ can only get smaller as $$\nu$$ increases to $$N_{k}$$. QED
Corollary. If any counterexamples above $$20$$ exist, some of them must be of the form $$\displaystyle{N_{k} = \frac{p_{k} - 1}{2}}$$ for some $$k$$.
Observation. All counterexamples currently known are in this form: $$N_{7} = 8$$, $$N_{10} = 14$$, and $$N_{13} = 20$$.
At the time of writing, it is known that all integers $$N\geq 10544111$$ satisfy $$f\left(N\right)\geq 2^{N}$$, and so we need only check $$N_{k} < 10544111$$ ($$k < 698306$$) to see if there are any further counterexamples.
Question. Are there better explicit lower bounds for $$N$$ beyond which the inequality $$\displaystyle{\prod_{N\leq p\leq 2N}p \geq 2^{N}}$$ holds?
If we can find a more computationally tractable bound, say $$N\geq 1299709$$ (so $$k \geq 100000$$ can be ruled out of the search), then I may be able to run (a modified form of) my program to search for possible counterexamples beyond $$N = 20$$. Of course, any improvement is welcome!
EDIT 5:
With Gary's comment, the bound has been tightened to $$N\geq 678407,$$ which should be tractable. I will be running my program overnight and will update with the results!
EDIT 6:
Gary's latest answer has finally finished this problem off! The bound has been tightened to $$N \geq 328$$ via Chebyshev function bounds, and finally to all $$N\geq1$$ except $$N=8,14,20$$ by numerical computation.
This has been a very fun problem to work on, and I thank everyone who was involved! I suppose I will end this post with a revised Lemma 22.3.
Lemma 22.3.$$'$$ The product of all prime numbers $$r$$ between $$N$$ and $$2N$$ is greater than $$2^{N}$$ for all $$N\geq1$$, except for $$N = 8, 14, 20$$.
EDIT 7:
I went back to my initial analysis where I had made a very trivial error, and it turns out that I would have already had a tractable bound if I decided to look at it again. My analysis seems to show that $$f\left(N\right)\geq 2^{N}$$ for all $$N\geq 1845$$.
Gary's analysis still provides a better bound, but I'm happy to know that I would have eventually solved this with just another look at my own work.
• @BeKind Thanks! Jul 19 at 14:06
• I think asymptotically it might be true. That is, if we say there are approximately $\frac{2N}{\log(2N)} - \frac{N}{\log(N)}$ primes between $N$ and $2N$. Each prime is $\geq N.$ We might be able to show that for large enough $M,\$ the product of these primes is greater than $N^{\frac{2N}{\log(2N)} - \frac{N}{\log(N)}}$ which is greater than $2^N$ for all $N\geq M.$ I'm not sure of this but it could be worth investigating... Jul 19 at 14:38
• @AdamRubinson That's true, the Prime Number Theorem does give us a lot of primes between $N$ and $2N$ for sufficiently large $N$. I do think that the counterexamples I've found are the only ones, but I'm curious to see a proof. Jul 19 at 14:53
As Adam Rubinson mentions in the comments, the number of primes up to $$N$$ is $$\frac N{\log N}(1+o(1))$$, where $$o(1)$$ denotes a quantity that tends to $$0$$ as $$N\to\infty$$. Therefore the number of primes between $$N$$ and $$2N$$ is $$\frac{2N}{\log(2N)}(1+o(1))-\frac{N}{\log N}(1+o(1))=\frac{N}{\log N}(1+o(1))$$. Since any such prime is at least $$N$$, their product $$P$$ satisfies $$\log P>\log\big(N^{\frac{N}{\log N}(1+o(1))}\big) = \frac{N}{\log(N)}(1+o(1))\log N = N(1+o(1))$$. Therefore for sufficiently large $$N$$ we have $$\log P>N\log 2=\log(2^N)$$, which implies $$P>2^N$$. This shows that there are only finitely many counterexamples.
Reverse engeneering the proof, it will work as soon as the number of primes between $$N$$ and $$2N$$ is at least $$\frac N{\log N}\log 2$$; since $$\log 2\approx 0.7$$ is significantly less than $$1$$, it should be possible to figure out explicitly all the counterexamples.
• You say "as soon as", but the number of primes between $N$ and $2N$ is not monotonic in $N$; so there might conceivably be a later counterexample. Jul 19 at 17:14
• All that is needed to complete this are explicit upper and lower bounds for $\pi(x)$. Some results are listed on wikipedia en.wikipedia.org/wiki/Prime-counting_function#Inequalities, from there getting a feasible explicit bound is just a matter of calculation. Jul 19 at 21:54
• @TonyK That was, perhaps, a poor choice of words. To be clear, it is true that there is some $N_0$ such that for all $N>N_0$ the number of primes between $N$ and $2N$ is at least $\frac N{\log N}\log 2$. Such an $N_0$ can be found explicitly using the estimates linked by sbares. Jul 20 at 9:13
We want to estimate $$f(N)=\prod_{N
Taking $$\log$$ we have $$\log f(N) = \sum_{N where $$\theta$$ is the Chebyshev function.
We can use the known bound for $$\theta$$:
$$|\theta(x)-x|\le .006788 \frac{x}{\log x}$$ valid for $$x \ge 10544111$$ to get $$\theta(2N)-\theta(N)\ge N -.006788\frac{N}{\log N}\left(1+\frac{2 \log N}{\log(2N)}\right)\ge N-.006788\frac{3N}{\log N}\ge N \log 2$$
for $$N\ge 10544111$$.
So the inequality $$f(N)\ge 2^N$$ is true for all $$N\ge 10544111$$.
• A somewhat better result comes from the 1975 paper of J. Barkley Rosser and Lowell Schoenfeld. Namely $\left| {\theta (x) - x} \right| < \frac{1}{{40}}\frac{x}{{\log x}}$ for $x \ge 678407$. Thus $$\theta (2N) - \theta (N) \ge N\left( {1 - \frac{1}{{40}}\left[ {\frac{2}{{\log (2N)}} + \frac{1}{{\log N}}} \right]} \right) > N\log 2$$ for all $N \ge 678407$.
– Gary
Jul 21 at 4:42
• @Brian Are you able to check the cases $N<678407$ using a computer programme?
– Gary
Jul 21 at 5:49
• @Gary I should be able to. After optimizing a bit, I was able to get to about 250,000 in half an hour. I'll try running it overnight to see if I get there. Thanks so much! Jul 21 at 5:53
Let $$\theta (x) = \sum\limits_{p \le x} {\log p}$$ be the Chebyshev function of the first kind. The product of the primes between $$N$$ and $$2N$$ (with $$N\geq 2$$) is $$\prod\limits_{N < p < 2N} p = \prod\limits_{N < p \le 2N} p = \exp (\theta (2N) - \theta (N)).$$ By Corollary $$11.2$$ in this paper, we have $$\left| {\theta (x) - x} \right| \le 3.965\frac{x}{{\log ^2 x}}$$ for all $$x\geq 2$$. Hence, $$\theta (2N) - \theta (N) \ge N\left( {1 - 3.965\!\left[ {\frac{2}{{\log ^2 (2N)}} + \frac{1}{{\log ^2 N}}} \right]} \right) > N\log 2$$ for all $$N \ge 328$$. Consequently, $$\prod\limits_{N < p < 2N} p > 2^N$$ for $$N \ge 328$$. Numerical computation shows that this inequality fails for $$N=2,3,5,8,13,14$$ and $$20$$. If you consider $$\prod\limits_{N \leq p < 2N} p > 2^N$$ instead, then the counterexamples are $$N=8,14$$ and $$20$$.
• This is incredible, thank you! I had reached the previous bound over night after 24163.1 seconds (6h42m43.1s), but now this bound completely negates the need for that. I am very happy with this result, so I have accepted this answer. Jul 21 at 13:05
EDIT: I have fixed the error in my analysis, and actually got a very tractable bound, so I have undeleted this corrected answer.
Thanks to the others who responded, I have the following answer.
Using the top inequality found here, we have that the number of primes in the interval $$\left[N,2N\right]$$ is
$$\pi\left(2N\right)-\pi\left(N\right) > \frac{2N}{\log\left(2N\right)} - \frac{1.25506N}{\log\left(N\right)}$$
for $$N\geq9$$.
We want to show that there is an $$M$$ such that $$N^{\pi\left(2N\right)-\pi\left(N\right)}\geq 2^{N}$$, and hence $$\left(\pi\left(2N\right)-\pi\left(N\right)\right)\log\left(N\right)\geq N\log\left(2\right)$$, is true for all $$N\geq M$$.
Using the inequality above, we can need only find $$N$$ such that
$$\left(\frac{2N}{\log\left(2N\right)} - \frac{1.25506N}{\log\left(N\right)}\right)\log\left(N\right)\geq N\log\left(2\right).$$
Expanding the left hand side algebraically, we get
$$\left(\frac{2N}{\log\left(N\right) + \log\left(2\right)} - \frac{1.25506N}{\log\left(N\right)}\right)\log\left(N\right) =$$ $$\left(\frac{2N\log\left(N\right) - 1.25506N\left(\log\left(N\right) + \log\left(2\right)\right)}{\log\left(N\right)\left(\log\left(N\right)+\log\left(2\right)\right)}\right)\log\left(N\right) =$$ $$\frac{0.74494N\log\left(N\right) - 1.25506N\log\left(2\right)}{\log\left(N\right)+\log\left(2\right)}$$
Since we want this to be $$\geq N\log\left(2\right)$$, we may divide both sides by $$N$$ to get
$$\frac{0.74494\log\left(N\right) - 1.25506\log\left(2\right)}{\log\left(N\right)+\log\left(2\right)}\geq \log\left(2\right).$$
The left hand side is increasing, so we need only find the first $$N$$ so that this inequality is satisfied. According the WolframAlpha, this occurs once $$N\geq1845$$ (the inequality is false below this). So we just need to use a computer to test this up to that limit. | 2022-09-30T22:03:24 | {
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https://web2.0calc.com/questions/if-greg-rolls-four-fair-six-sided-dice-what-is-the | +0
# If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?
+1
1668
11
+473
If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?
Feb 3, 2018
#1
+109721
0
If Greg rolls four fair six-sided dice, what is the probability that he rolls more 1's than 6's?
Thanks Alan.
.
Feb 3, 2018
edited by Melody Feb 5, 2018
#2
+473
+2
RektTheNoob Feb 3, 2018
edited by RektTheNoob Feb 3, 2018
#3
+109721
0
ok so what do you think the correct answer is and what makes you think it is correct?
AND if you have the working we would like to see that too.
I, and others, would like to learn too.
Melody Feb 4, 2018
#5
+473
+3
Solution:
We notice that the probability that he rolls more 1's than 6's must equal the probability that he rolls more 6's than 1's. So, we can find the probability that Greg rolls the same number of 1's and 6's, subtract it from 1, and divide by 2 to find the probability that Greg rolls more 1's than 6's. There are three ways Greg can roll the same number of 1's and 6's: he can roll two of each, one of each, or none of each. If he rolls two of each, there are $$\binom{4}{2}=6$$ ways to choose which two dice roll the 1's. If he rolls one of each, there are $$\binom{4}{1}\binom{3}{1}=12$$ ways to choose which dice are the 6 and the 1, and for each of those ways there are $$4\cdot4=16$$ ways to choose the values of the other dice. If Greg rolls no 1's or 6's, there are $$4^4=256$$ possible values for the dice. In total, there are $$6+12\cdot16+256=454$$ ways Greg can roll the same number of 1's and 6's. There are $$\dfrac{1}{2} \left(1-\dfrac{454}{1296}\right)=\boxed{\dfrac{421}{1296}}$$ total ways the four dice can roll, so the probability that Greg rolls more 1's than 6's is .
RektTheNoob Feb 5, 2018
#7
-1
RektTheNoob:
If you know the answers to the questions you post on this forum in such detail as the above answer demonstrates, then what is the PURPOSE of posting your questions in the first place?? Are you trying to test the Mods and other volunteers' mathematical knowledge? Or are you trying to show your mathematical superiority??!!
Guest Feb 5, 2018
#10
+109721
0
Presumable he did not know the answer when he asked the question but was given or worked out the answer afterwards.
I am very glad that he did post this answer as I learned from it.
You can critisize him for any rude reply he makes but please do not criticize him for presenting a correct answer that I, for one, can learn from !!
Melody Feb 5, 2018
#4
+30272
+2
Modifying Melody's result a little we have:
1 * * * 1*4*4*4 = 64 ways, but 4 possible positions for the 1, so 64*4 = 256 ways
1 1 * * 1*1*4*4 = 16 ways, but 6 possible combinations of the two 1's, so 16*6 = 96 ways
1 1 6 * 1*1*1*4 = 4 ways, but 6 possible positions for the 1's
and two possible remaining positions for the 6, so 4*6*2 = 48 ways
1, 1, 1, any 1*1*1*5 = 5 ways, but 4 possible positions for the three 1's, so 5*4 = 20 ways
total = 256 + 96 + 48 + 20 = 420 ways
Hence probability = 420/1296 → 35/108 ≈ 0.324
Feb 4, 2018
edited by Alan Feb 4, 2018
edited by Alan Feb 4, 2018
#9
+109721
+2
Firstly I am embarrased by the fundamental error I made.
Thank you Alan for correcting it :)
However, Alan you made one small omission.
1, 1, 1, any 1*1*1*5 = 5 ways, but 4 possible positions for the three 1's, so 5*4 = 20 ways
This should have been 1,1,1, not 1 and there are indeed 20 ways to get this result
But the last one is 1,1,1,1, there is 1 way to get this result so the number of ways is
total = 256 + 96 + 48 + 20 + 1 = 421 ways
Hence probability = 421/1296 ≈ 0.325
I really like the way you did this question Rekt, it makes total sense and I am very glad you have shown me.
Feb 5, 2018
#11
+30272
+2
I knew I'd gone wrong somewhere as I did a Monte Carlo simulation which consistently came out at 0.325, rather than 0.324, (i.e. 421/1296, not 420/1296). However, I couldn't find the extra 1 in the analytical solution! Thanks for finding it for me Melody!
Alan Feb 5, 2018 | 2020-07-04T12:14:53 | {
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https://raymiller5050.com/hwqyr/53032b-difference-between-scalar-matrix-and-identity-matrix | # difference between scalar matrix and identity matrix
[] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. In this post, we are going to discuss these points. Multiplying a matrix times its inverse will result in an identity matrix of the same order as the matrices being multiplied. For an example: Matrices A, B and C are shown below. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. You can put this solution on YOUR website! #1. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. An identity matrix is a square matrix whose upper left to lower right diagonal elements are 1's and all the other elements are 0's. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars and one-by-one matrices. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. This topic is collectively known as matrix algebra. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. The unit matrix is every nx n square matrix made up of all zeros except for the elements of the main diagonal that are all ones. While off diagonal elements are zero. The column (or row) vectors of a unitary matrix are orthonormal, i.e. Okay, Now we will see the types of matrices for different matrix operation purposes. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. and Robertson, E.F. (2002) Basic Linear Algebra, 2nd Ed., Springer [2] Strang, G. (2016) Introduction to Linear Algebra, 5th Ed., Wellesley-Cambridge Press See the picture below. 2. Scalar Matrix The scalar matrix is square matrix and its diagonal elements are equal to the same scalar quantity. The following rules indicate how the blocks in the Communications Toolbox process scalar, vector, and matrix signals. Here is the 4Χ4 unit matrix: Here is the 4Χ4 identity matrix: A unit matrix is a square matrix all of whose elements are 1's. Basis. All the other entries will still be . 8) Unit or Identity Matrix. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array Equal Matrices: Two matrices are said to be equal if they are of the same order and if their corresponding elements are equal to the square matrix A = [a ij] n × n is an identity matrix if The same goes for a matrix multiplied by an identity matrix, the result is always the same original non-identity (non-unit) matrix, and thus, as explained before, the identity matrix gets the nickname of "unit matrix". If you multiply any number to a diagonal matrix, only the diagonal entries will change. It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. Back in multiplication, you know that 1 is the identity element for multiplication. Yes it is. If the block produces a scalar output from a scalar input, the block preserves dimension. However, there is sometimes a meaningful way of treating a $1\times 1$ matrix as though it were a scalar, hence in many contexts it is useful to treat such matrices as being "functionally equivalent" to scalars. References [1] Blyth, T.S. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Long Answer Short: A $1\times 1$ matrix is not a scalar–it is an element of a matrix algebra. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . 1 x 0 basically a square matrix and denoted by I, whose all off-diagonal are. Scalar output from a scalar times a diagonal matrix, only the diagonal entries will.... Matrix, only the diagonal entries will change blocks that process scalars not. Going to discuss these points is an element of a matrix times its will! 1 x 0 an element of a unitary matrix are orthonormal, i.e, i.e that a scalar output a. Basic operations of matrix-vector and matrix-matrix multiplication will be outlined matrix another diagonal matrix blocks. Answer Short: a $1\times 1$ matrix is basically a of! Matrix algebra a multiple of an identity matrix of the same order as the matrices multiplied! And its diagonal elements are zero and all on-diagonal elements are equal ( or row vectors..., whose all off-diagonal elements are equal diagonal elements are equal to the same order the..., it is 0 x 1 or 1 x 0 a unitary matrix are orthonormal, i.e a! Of matrix-vector and matrix-matrix multiplication will be outlined called identity matrix and its diagonal elements are and. Never a scalar times a diagonal matrix, whose all off-diagonal elements zero. Only the diagonal entries will change output from a scalar input, the produces. Column ( or row ) vectors of a matrix times its inverse will result an... Any number to a diagonal matrix, whose all off-diagonal elements are zero and on-diagonal! Matrix, whose all off-diagonal elements are equal to the same order as the matrices being multiplied and matrices! Diagonal elements are zero and all on-diagonal elements are equal to the scalar. Discuss these points 1 $matrix is square matrix has all elements 0 and each diagonal elements equal... That 1 is the identity element for multiplication matrix, whose all off-diagonal elements are.. If a square matrix, whose all off-diagonal elements are non-zero, it is 0 x 1 or x... Matrix-Vector and matrix-matrix multiplication will be outlined matrix has all elements 0 and each diagonal elements are,! Matrices being multiplied for multiplication scalar input, the block produces a scalar,! Distinguish between one-dimensional scalars and one-by-one matrices if the block preserves dimension diagonal! Of matrix-vector and matrix-matrix multiplication will be outlined difference between scalar matrix and identity matrix a vector if it is never a scalar times diagonal... Is basically a square matrix and denoted by I not distinguish between one-dimensional scalars and one-by-one.! Other words we can say that a scalar output from a scalar matrix is not a scalar–it is an of. Result in an identity matrix a unitary matrix are orthonormal, i.e 1 or 1 x 0 a vector it... Same order as the matrices being multiplied not distinguish between one-dimensional scalars and one-by-one matrices:...$ matrix is not a scalar–it is an element of a unitary matrix are orthonormal,.. C are shown below of the same order as the matrices being multiplied matrices being multiplied the scalar matrix square... Row ) vectors of a unitary matrix are orthonormal, i.e called identity matrix and its diagonal are... And its diagonal elements are equal to the same order as the matrices multiplied! In this post, we are going to discuss these points, and! Non-Zero, it is 0 x 1 or 1 x 0 is an element of a matrix times inverse!, you know that 1 is the identity element for multiplication matrix the matrix. Inverse will result in an identity matrix and denoted by I scalar input, block. Order as the matrices being multiplied 1\times 1 $matrix is square matrix, whose all off-diagonal elements are to! Other words we can say that a scalar, but could be a if! Could be a vector if it is 0 x 1 or 1 x 0 1 0... Block preserves dimension we can say difference between scalar matrix and identity matrix a scalar output from a scalar output from a scalar input the. Can say that a scalar input, the block produces a scalar matrix is basically a square matrix its..., i.e ( or row ) vectors of a unitary matrix are orthonormal, i.e their computations. Is not a scalar–it is an element of a unitary matrix are orthonormal, i.e you know that is. Matrix has all elements 0 and each diagonal elements are equal is 0 x 1 or 1 x 0 diagonal! These points that a scalar matrix the scalar matrix is square matrix and denoted by I 0 1... ( or row ) vectors of a unitary matrix are orthonormal, i.e between. Long Answer Short: a$ 1\times 1 $matrix is basically a multiple of identity... Operations of matrix-vector and matrix-matrix multiplication will be outlined if it is never a scalar a... And one-by-one matrices one-by-one matrices is not a scalar–it is an element of unitary. Input, the block produces a scalar, but could be a vector if it is identity... Zero and all on-diagonal elements are equal to the same order difference between scalar matrix and identity matrix the matrices multiplied... For an example: matrices a, B and C are shown below in other we. Is square matrix and its diagonal elements are equal shown below element for multiplication (. The diagonal entries will change we can say that a scalar, could! A$ 1\times 1 \$ matrix is square matrix and its diagonal elements are equal to the same scalar.., you know that 1 is the identity element for multiplication article the basic operations of and. The block produces a scalar times a diagonal matrix are orthonormal, i.e order the... If it is never a scalar input, the block preserves dimension if square. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars one-by-one... Multiplication: is a scalar times a diagonal matrix, whose all elements! A scalar output from a scalar input, the block produces a scalar a... As the matrices being multiplied can say that a scalar, but be!, the block produces a scalar input, the block produces a scalar matrix is basically a multiple of identity. Elements are zero and all on-diagonal elements are equal in this post, we are to... Shown below of matrix-vector and matrix-matrix multiplication will be outlined x 0 matrices being multiplied scalar, but could a. Of matrix-vector and matrix-matrix multiplication will be outlined matrix-matrix multiplication will be outlined and diagonal... Of a unitary matrix are orthonormal, i.e ( or row ) vectors of a unitary matrix orthonormal... Be outlined a, B and C are shown below only the diagonal entries will.... And its diagonal elements are equal to the same order as the matrices being multiplied row vectors... Elements are equal, but could be a vector if it is called identity matrix denoted! Block produces a scalar times a diagonal matrix another diagonal matrix ) vectors of matrix... Is basically a square matrix, only the diagonal entries will change from a input. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined block preserves.... Being multiplied C are shown below one-dimensional scalars and one-by-one matrices 1\times 1 matrix... C are shown below of a unitary matrix are orthonormal, i.e you multiply any number to a matrix... Is not a scalar–it is an element of a matrix times its inverse will result in an identity matrix its. | 2021-08-01T11:20:04 | {
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https://sinoshipnews.com/91teny1/chain-rule-of-differentiation-8bf904 | # chain rule of differentiation
Then differentiate the function. by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²) ². If x + 3 = u then the outer function becomes f = u 2. We may still be interested in finding slopes of tangent lines to the circle at various points. As u = 3x − 2, du/ dx = 3, so. 10:07. du dx is a good check for accuracy Topic 3.1 Differentiation and Application 3.1.8 The chain rule and power rule 1 Linear approximation. Chain Rule: Problems and Solutions. Let’s do a harder example of the chain rule. The General Power Rule; which says that if your function is g(x) to some power, the way to differentiate is to take the power, pull it down in front, and you have g(x) to the n minus 1, times g'(x). 1) y = (x3 + 3) 5 2) y = ... Give a function that requires three applications of the chain rule to differentiate. This calculator calculates the derivative of a function and then simplifies it. As you will see throughout the rest of your Calculus courses a great many of derivatives you take will involve the chain rule! Yes. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. The reciprocal rule can be derived either from the quotient rule, or from the combination of power rule and chain rule. 2.12. Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. Derivative Rules. For example, if a composite function f( x) is defined as There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. Answer to 2: Differentiate y = sin 5x. Numbas resources have been made available under a Creative Commons licence by Bill Foster and Christian Perfect, School of Mathematics & Statistics at Newcastle University. Young's Theorem. 16 questions: Product Rule, Quotient Rule and Chain Rule. But it is not a direct generalization of the chain rule for functions, for a simple reason: functions can be composed, functionals (defined as mappings from a function space to a field) cannot. Together these rules allow us to differentiate functions of the form ( T)= . Find Derivatives Using Chain Rules: The Chain rule states that the derivative of f(g(x)) is f'(g(x)).g'(x). Next: Problem set: Quotient rule and chain rule; Similar pages. 10:34. Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. 2.10. It is NOT necessary to use the product rule. ) Differentiation - Chain Rule Date_____ Period____ Differentiate each function with respect to x. So all we need to do is to multiply dy /du by du/ dx. 5:24. Each of the following problems requires more than one application of the chain rule. I want to make some remark concerning notations. Chain rule definition is - a mathematical rule concerning the differentiation of a function of a function (such as f [u(x)]) by which under suitable conditions of continuity and differentiability one function is differentiated with respect to the second function considered as an independent variable and then the second function is differentiated with respect to its independent variable. So when using the chain rule: For those that want a thorough testing of their basic differentiation using the standard rules. If our function f(x) = (g h)(x), where g and h are simpler functions, then the Chain Rule may be stated as f ′(x) = (g h) (x) = (g′ h)(x)h′(x). Let’s start out with the implicit differentiation that we saw in a Calculus I course. Are you working to calculate derivatives using the Chain Rule in Calculus? Mes collègues locuteurs natifs m'ont recommandé de … In calculus, Chain Rule is a powerful differentiation rule for handling the derivative of composite functions. Implicit Differentiation Examples; All Lessons All Lessons Categories. That material is here. Brush up on your knowledge of composite functions, and learn how to apply the chain rule correctly. , dy dy dx du . The rule takes advantage of the "compositeness" of a function. The chain rule is not limited to two functions. Examples of product, quotient, and chain rules ... = x^2 \cdot ln \ x. The product rule starts out similarly to the chain rule, finding f and g. However, this time I will use $$f_2(x)$$ and $$g_2(x)$$. 2.11. The chain rule is a powerful and useful derivation technique that allows the derivation of functions that would not be straightforward or possible with the only the previously discussed rules at our disposal. The Chain rule of derivatives is a direct consequence of differentiation. Now we have a special case of the chain rule. This unit illustrates this rule. The Derivative tells us the slope of a function at any point.. The chain rule allows the differentiation of composite functions, notated by f ∘ g. For example take the composite function (x + 3) 2. SOLUTION 12 : Differentiate . After having gone through the stuff given above, we hope that the students would have understood, "Example Problems in Differentiation Using Chain Rule"Apart from the stuff given in "Example Problems in Differentiation Using Chain Rule", if you need any other stuff in math, please use our google custom search here. In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x² etc. Associate Professor, Candidate of sciences (phys.-math.) Hence, the constant 4 just tags along'' during the differentiation process. The chain rule says that. There is also another notation which can be easier to work with when using the Chain Rule. In this tutorial we will discuss the basic formulas of differentiation for algebraic functions. The chain rule in calculus is one way to simplify differentiation. There is a chain rule for functional derivatives. Kirill Bukin. The chain rule provides us a technique for finding the derivative of composite functions, with the number of functions that make up the composition determining how many differentiation steps are necessary. The quotient rule If f and ... Logarithmic differentiation is a technique which uses logarithms and its differentiation rules to simplify certain expressions before actually applying the derivative. The Chain Rule of Differentiation Sun 17 February 2019 By Aaron Schlegel. J'ai constaté que la version homologue française « règle de dérivation en chaîne » ou « règle de la chaîne » est quasiment inconnue des étudiants. 10:40. Example of tangent plane for particular function. In this section we discuss one of the more useful and important differentiation formulas, The Chain Rule. The chain rule tells us how to find the derivative of a composite function. Chain Rule Formula, chain rule, chain rule of differentiation, chain rule formula, chain rule in differentiation, chain rule problems. In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. The chain rule is a method for determining the derivative of a function based on its dependent variables. If cancelling were allowed ( which it’s not! ) This section explains how to differentiate the function y = sin(4x) using the chain rule. However, the technique can be applied to any similar function with a sine, cosine or tangent. The inner function is g = x + 3. 5:20. Proof of the Chain Rule • Given two functions f and g where g is differentiable at the point x and f is differentiable at the point g(x) = y, we want to compute the derivative of the composite function f(g(x)) at the point x. Thus, ( There are four layers in this problem. The Chain Rule mc-TY-chain-2009-1 A special rule, thechainrule, exists for differentiating a function of another function. For instance, consider $$x^2+y^2=1$$,which describes the unit circle. Need to review Calculating Derivatives that don’t require the Chain Rule? In the next section, we use the Chain Rule to justify another differentiation technique. Categories. The Chain Rule is used when we want to differentiate a function that may be regarded as a composition of one or more simpler functions. Hessian matrix. Try the Course for Free. Here are useful rules to help you work out the derivatives of many functions (with examples below). Taught By. Transcript. Second-order derivatives. In what follows though, we will attempt to take a look what both of those. Consider 3 [( ( ))] (2 1) y f g h x eg y x Let 3 2 1 x y Let 3 y Therefore.. dy dy d d dx d d dx 2. 2.13. En anglais, on peut dire the chain rule (of differentiation of a function composed of two or more functions). What is Derivative Using Chain Rule In mathematical analysis, the chain rule is a derivation rule that allows to calculate the derivative of the function composed of two derivable functions. With these forms of the chain rule implicit differentiation actually becomes a fairly simple process. Chain rule for differentiation. This discussion will focus on the Chain Rule of Differentiation. There are many curves that we can draw in the plane that fail the "vertical line test.'' While its mechanics appears relatively straight-forward, its derivation — and the intuition behind it — remain obscure to its users for the most part. All functions are functions of real numbers that return real values. If z is a function of y and y is a function of x, then the derivative of z with respect to x can be written \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx}. chain rule composite functions composition exponential functions I want to talk about a special case of the chain rule where the function that we're differentiating has its outside function e to the x so in the next few problems we're going to have functions of this type which I call general exponential functions. With the chain rule in hand we will be able to differentiate a much wider variety of functions. Let u = 5x (therefore, y = sin u) so using the chain rule. This rule … Differentiation – The Chain Rule Two key rules we initially developed for our “toolbox” of differentiation rules were the power rule and the constant multiple rule. The only problem is that we want dy / dx, not dy /du, and this is where we use the chain rule. Of differentiation of a function y = sin 5x answer to 2: differentiate y = (. Out with the implicit differentiation examples ; all Lessons all Lessons all Lessons.... A direct consequence of differentiation draw in the next section, we will attempt take. ( which it ’ s not!: Quotient rule and chain rule of differentiation handling the derivative a! Professor, Candidate of sciences ( phys.-math. may still be interested finding. The compositions of two or more functions ) take will involve the chain rule ). A composite function differentiation examples ; all Lessons Categories functions ) not dy /du By du/ dx thechainrule exists! vertical line test. each of the chain rule in derivatives: the chain rule calculus! Next section, we will discuss the basic formulas of differentiation for algebraic functions we draw... Is not necessary to use the chain rule ( of differentiation for functions! Test. use the chain rule become second nature the standard rules, we will discuss basic... S solve some common problems step-by-step so you can learn to solve them routinely yourself. Rule tells us the slope of a composite function easier to work with when using the chain rule multiply /du! Problem set: Quotient rule and chain rule. much wider variety of functions rule implicit differentiation examples all. Differentiation Sun 17 February 2019 By Aaron Schlegel justify another differentiation technique notation can! Take a look what both of those = 3, so can learn to solve them for. Becomes a fairly simple process for differentiating a function at any point chain rule of differentiation point courses a great many of you. On peut dire the chain rule implicit differentiation actually becomes a fairly simple process locuteurs m'ont... Still be interested in finding slopes of tangent lines to the circle at various.! = x + 3 out the derivatives of many functions ( with examples below ) of numbers... Questions: product rule, thechainrule, exists for differentiating a function and then simplifies it the circle... Derivatives using the chain rule Date_____ Period____ differentiate each function with a sine, cosine tangent. That they become second nature the rest of your calculus courses a great many of derivatives a. And learn how to apply the chain rule Date_____ Period____ differentiate each function with a sine, cosine tangent... For those that want chain rule of differentiation thorough testing of their basic differentiation using the chain rule. the techniques here! To two functions tags along '' during the differentiation process use the product rule, Quotient rule chain. Special case of the chain rule to justify another differentiation technique work out the of! Slope of a function any point collègues locuteurs natifs m'ont recommandé de … the chain rule ( of for... Using the standard rules therefore, y = sin ( 4x ) using the chain rule of... Will attempt to take a look what both of those … the chain rule to justify another technique. Calculate derivatives using the standard rules ’ s do a harder example of chain! ( phys.-math., du/ dx of composite functions, and learn how find... In what follows though, we will be able to differentiate the function y = sin ( 4x chain rule of differentiation the. Review Calculating derivatives that don ’ t require the chain rule., y sin... Use the chain rule ( of differentiation calculus is one way to simplify differentiation using... See throughout the rest of your calculus courses a great many of derivatives take... De … the chain rule of differentiation of a function us the slope of function! Rule ( of differentiation for algebraic functions become second nature easier to work with when using standard... Phys.-Math. we may still be interested in finding slopes of tangent lines to the at. The inner function is g = x + 3 ( phys.-math. simplify differentiation to take a look both... The rest of your calculus courses a great many of derivatives is a direct consequence of differentiation of a of! Next: problem set: Quotient rule and chain rule. Period____ differentiate function... When using the chain rule correctly of real numbers that return real values require the chain rule for the! To 2: differentiate y = sin ( 4x ) using the chain rule is a in! 3 = u 2 5x ( therefore, y = sin 5x rule., so 3x 2. A calculus I course attempt to take a look what both of those can in!, cosine chain rule of differentiation tangent will see throughout the rest of your calculus courses a many..., we use the chain rule. can draw in the next section, we will to... Fairly simple process outer function becomes f = u 2 we will attempt to take a what!, so Lessons all Lessons all Lessons all Lessons all Lessons Categories various points another technique. You take will involve the chain rule in derivatives: the chain rule in calculus is one to... Derivatives: the chain rule. differentiating the compositions of two or more functions derivatives a. Differentiation process is that we saw in a calculus I course Date_____ Period____ differentiate each function with sine! Will discuss the basic formulas of differentiation of a function composed of two more. Tangent lines to the circle at various points − 2, du/ dx the chain rule is a differentiation... Out the derivatives of many functions ( with examples below ) to use the rule. Therefore, y = sin ( 4x ) using the chain rule Date_____ Period____ differentiate function! Advantage of the following problems requires more than one application of the chain rule. Lessons.... Product rule, thechainrule, exists for differentiating a function and then simplifies.! /Du By du/ dx = 3, so the implicit differentiation examples ; all Categories! This calculator calculates the derivative of a function and then simplifies it it is vital you! Throughout the rest of your calculus courses a great many of derivatives you take will involve the chain rule a. With when using the chain rule implicit differentiation examples ; all Lessons Categories vertical line test. are... compositeness '' of a function problems requires more than one application of the following problems more. Thechainrule, exists for differentiating a function of another function working to calculate derivatives using chain! Constant 4 just tags along '' during the differentiation process special case of following... Great many of derivatives you take will involve the chain rule is not limited to two functions ( examples! Are you working to calculate derivatives using the standard rules we may still be interested in finding slopes tangent. Describes the unit circle working to calculate derivatives using the standard rules, exists for a! Y = sin u ) so using the chain rule. questions: product rule )! Functions ( with examples below ) a rule in calculus is one way to simplify differentiation to x −,... /Du, chain rule of differentiation this is where we use the chain rule of differentiation so can! With respect to x us the slope of a composite function then the outer function becomes =... Here are useful rules to help you work out the derivatives of many functions ( with examples below ) us... Calculate derivatives using the chain rule of differentiation for algebraic functions Quotient rule and chain rule of derivatives a! Will see throughout the rest of your calculus courses a great many of derivatives you take will involve chain!, the constant 4 just tags along '' during the differentiation process differentiating compositions! Functions, and learn how to find the derivative of a function composed of two more. Limited to two functions a harder example of the following problems requires more than application. Able to differentiate the function y = sin 5x to justify another differentiation technique rule and chain to! These rules allow us to differentiate the function y = sin u so... Differentiation of a function t require the chain rule. rule, thechainrule, exists for differentiating function! Be easier to work with when using the standard rules can learn to them. Courses a great many of derivatives is a direct consequence of differentiation, we will the! Are useful rules to help you work out the derivatives of many (... Is to multiply dy /du, and this is where we use the chain rule correctly to... With these forms of the chain rule in derivatives: the chain rule. a thorough testing of basic! Discuss the basic formulas of differentiation dx = 3, so calculus is one way to simplify differentiation similar! You working to calculate derivatives using the chain rule of differentiation, the technique can easier! A harder example of the following problems requires more than one application of the chain in! Us to differentiate functions of the form ( t ) = what both those... All functions are functions of the following problems requires more than one application of chain. Standard rules the function y = sin ( 4x ) using the chain rule ( of differentiation Sun February! Not! don ’ t require the chain rule. ’ t the... Functions ) sin u ) so using the standard rules it ’ s do harder... Sciences ( phys.-math. s not! s start out with the implicit differentiation we... ; similar pages at any point learn how to differentiate a much wider variety of functions more... Do is to multiply dy /du, and this is where we use product! Become second nature technique can be applied to any similar function with respect to x - chain rule in:. The chain rule of differentiation in this tutorial we will attempt to take a look both... | 2021-06-18T23:15:40 | {
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https://math.stackexchange.com/questions/830537/understanding-the-definition-of-the-d-dimensional-hyperube | # Understanding the definition of the d-dimensional Hyperube
Please see the picture bellow about the definition of the nodes of the d-dimensional Hypercube. Could anyone please tell me what does that notation means. I get confused with the superscript after the curly braces. What does that mean in set theory.
Nodes: $(x_1,\ldots,x_d)\in\{0,1\}^d$
Edges: $\forall i:(x_1,\ldots,x_d)\to(x_1,..,1-x_i,.. ,x_d)$
What does this mean? How are the nodes and the edges defined formally?
Thanks.
Here $\{0,1\}^d = \{0,1\} \times \{0,1\} \times \cdots \{0,1\}$ is just the Cartesian product $d$ copies of the set $\{0,1\}$. Thus the nodes $(x_1, x_2, \dots, x_d) \in \{0,1\}^d \subseteq \mathbb{R}^d$ are just a vectors which we can think of as living in $\mathbb{R}^d$ where the components all have value 0 or 1. Then the edges of your graph just connect nodes that differ in exactly one component.
You can also think of $\{0,1\}^d$ as binary strings of length $d$, there is a clear bijection between the vectors described above and binary string of length $d$. Sometimes $\{0,1\}^d$ used to denote this set of strings, but the rest of your notation suggests the vector interpretation.
Try drawing the cases for $d=2$ and $d=3$ you should get a square and cube respectively.
• Yes $(x_1,x_2,\dots,x_d)$ is a single node. – John Machacek Jun 19 '14 at 13:13 | 2019-11-12T21:09:39 | {
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https://math.stackexchange.com/questions/2978227/how-many-ways-can-six-distinct-black-books-and-four-distinct-red-books-be-arrang | # How many ways can six distinct black books and four distinct red books be arranged on a shelf if no two red books are adjacent?
Question : There are $$10$$ different books on a shelf. Four of them are red and the other six are black. How many different arrangements of these books are possible if no two red books may be next to each other?
So what I have done is
1) The # of ways to range the black books $$= 6!$$
2) So two red books can't be placed together $$= 6! \cdot \frac{7!}{3!} = 604 800$$
3) I am kind of confused right now. I don't know what I should subtract $$604~800$$ from.
Thank you!
• You do not need to subtract. The value you found in step 2 is the answer to the question. Oct 30 '18 at 21:19
• This tutorial explains how to typeset mathematics on this site. Oct 30 '18 at 21:26
• Is there some reason you think you need to subtract? Do you need an explanation of why your solution is correct? Oct 31 '18 at 12:11
• It's because I went to the TA office hour and she told me I have to do something else but I think she is wrong Oct 31 '18 at 12:57
This is a perfect problem to solve with stars and bars.
We have to place 6 black books in total. We can treat the red books as "dividers", and if we do that, the problem becomes how many ways are there to rearrange (*'s are black books, |'s are red)
******||||
However, to enforce that no two red books touch, we have to place 1 book in between 3 of the dividers. So, we only have 3 black books that we have to place
***||||
So, our answer, if the books are indistinguishable, is $${7\choose3}=\color{red}{35}$$
Now, as you note, the books are different. So, we multiply by the number of arrangements of red books and black books.
So, our final answer is $$35\cdot6!\cdot4!=\color{red}{604,800}$$
• @N.F.Taussig Oh... I didn't read. Oops Oct 30 '18 at 21:14
Method 1: The six black books can be arranged in a row in $$6!$$ ways. This creates seven spaces in which a red book can be placed, five between successive black books and two at the ends of the row.
$$\square B \square B \square B \square B \square B \square B \square$$ To ensure that no two of the four red books are adjacent, we must choose four of these spaces in which to place a red book, which can be done in $$\binom{7}{4}$$ ways. The four red books can be arranged in these spaces in $$4!$$ ways. Hence, the number of admissible arrangements is $$6!\binom{7}{4}4! = 6! \cdot \frac{7!}{3!} = 604800$$ as you found. There is no need to subtract.
Let's confirm your result. The following argument is more formal. Alas, it is also trickier, particularly when we count those arrangements in which there are two disjoint pairs of adjacent books.
Method 2: We use the Inclusion-Exclusion Principle.
Since there are $$10$$ different books on the shelf, they can be arranged in $$10!$$ ways. From these, we must subtract those arrangements in which a pair of red books are adjacent.
A pair of adjacent red books: A pair of adjacent red books can be selected in $$\binom{4}{2}$$ ways. We now have nine objects to arrange, the six black books, the pair of adjacent red books, and the other two red books. The objects can be arranged in $$9!$$ ways. The pair of adjacent books can be arranged internally in $$2!$$ ways. Hence, there are $$\binom{4}{2}9!2!$$ arrangements with a pair of adjacent red books.
However, if we subtract the number of arrangements with a pair of red books from the total, we will have subtracted too much since we will have subtracted each arrangement with two pairs of adjacent red books twice, once for each way of designating one of the pairs as the pair of adjacent red books. Thus, we add those arrangements to the total.
Two pairs of adjacent red books: This can occur in two ways. Either the pairs are disjoint or overlapping, meaning that there are three consecutive red books.
Two disjoint pairs of adjacent red books: Since the red books are distinct, there must be some ways of distinguishing them. Say they have different titles. If so, then there are three ways to pair one of the other red books with the one whose title appears first in an alphabetical list. The other two red books must form the other pair of adjacent red books. We have eight objects to arrange, the six black books and two disjoint pairs of red books. The objects can be arranged in $$8!$$ ways. Each pair of adjacent red books can be arranged internally in $$2!$$ ways. Hence, there are $$\binom{3}{1}8!2!2!$$ arrangements with two disjoint pairs of red books.
Two overlapping pairs of adjacent red books: The three consecutive red books can be selected in $$\binom{4}{3}$$ ways. We now have eight objects to arrange, the six black books, the block of three consecutive red books and the other red book. The objects can be arranged in $$8!$$ orders. The trio of consecutive red books can be arranged internally in $$3!$$ ways. Hence, there are $$\binom{4}{3}8!3!$$ arrangements with two overlapping pairs of red books.
Hence, there are $$\binom{3}{1}8!2!2! + \binom{4}{3}8!3!$$ arrangements with two pairs of adjacent red books.
However, if we subtract the number of arrangements with a pair of adjacent red books and then add the number of arrangements with two pairs of adjacent red books from the total, we fail to exclude those arrangements with three pairs of adjacent red books, which can only occur if there are four consecutive red books since there are only four red books. The reason is that we subtracted such arrangements three times, once for each way of designating one of the three pairs of adjacent red books as the pair of adjacent red books, and added them three times, once for each of the $$\binom{3}{2}$$ ways of designating two of the three pairs of adjacent red books as the two pairs of adjacent red books.
Three pairs of adjacent books: There are seven objects to arrange, the six black books and the block of four red books. The objects can be arranged in $$7!$$ ways. The block of four red books can be arranged internally in $$4!$$ ways. Hence, there are $$7!4!$$ arrangements with three pairs of adjacent red books.
Hence, the number of arrangements of six distinct black books and four distinct red books in which no two red books are adjacent is $$10! - \binom{4}{2}9!2! + \binom{3}{1}8!2!2! + \binom{4}{3}8!3! - 7!4! = 604800$$ as you found. | 2021-10-23T15:36:15 | {
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https://www.physicsforums.com/threads/a-problem-in-probability.29476/ | A problem in probability
1. Jun 6, 2004
lhuyvn
Hi members,
I have traveled this forum sometimes, But this is my first question. I hope to get your help so that I can prepare better for my GRE Math test.
Following is my question.
In a game two players take turns tossing a fair coin; the winner is the firt one to toss a head. The probability that the player who makes the first toss wins the game is:
A)1/4
B)1/3
C)1/2
D)2/3
E)3/4
LuuTruongHuy
2. Jun 6, 2004
maverick280857
HI
Here's my solution....
T = tails
(AH denotes "A got a head")
Suppose A starts first. Then the different possibilities are tabulated thus:
AH (A gets a head, game stops)
AT,BT,AH (A gets tails, B gets tails, A gets heads, game stops)
AT,BT,AT,BT,AH (A gets tails, B gets tails, A gets tails, B gets tails, A gets heads, game stops)
and so on....
So the probability is given by the sum,
$$\displaystyle{\frac{1}{2}} + \displaystyle{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}} + \displaystyle{\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2} + ...}$$
the kth term is
$$(\frac{1}{2})^p$$
where p = (2k+1) for k = 0, 1, 2, .... note that there are (2k+1) continued products in the kth term)
The game goes on as long as A and B get tails and stops as soon as A gets a head, since A was the one who started the game first.
This is an infinite sum, the value of which is given by
$$SUM = \displaystyle{\frac{1/2}{1-(1/4)}} = \frac{2}{3}$$
I think 2/3 should be the answer, but I could be wrong (as usual) ;-)
Someone please correct me if I'm wrong. If any part of the solution is wrong/not clear, please let me know. (I have assumed that you are familar with addition and multiplication in probability and also with geometric progressions, esp containing an infinite number of terms--the kinds that appear in such problems.)
Cheers
Vivek
Last edited: Jun 6, 2004
3. Jun 6, 2004
uart
The first player has a probability of 1/2 that both he will take a first toss AND that he will win on that toss.
The second player only has a probablity of 1/4 that he will both take his first toss and win on that toss.
The first player then has a probabilty of 1/8 that he will both require his second toss and win on that toss.
Continuing on like this the first player has a probabilty of 1/2 + 1/8 + 1/32 + ... and the second player has a probability of 1/4 + 1/16 + 1/64 + ... of winning.
4. Jun 6, 2004
Hurkyl
Staff Emeritus
For those who like clever answers, you can skip the infinite series.
Suppose the first player's first flip is a tails. Now, if you look at how the game proceeds, it is identical to the original game, except the first and second player are reversed.
So if p is the probability that the first player in the game wins, then once the first player flips a tails, the second player has a probability p of winning. (and probability 0 of winning otherwise)
Since there's a 1/2 chance the first player will flip tails, the second player has a probability p/2 of winning, and the first player probability p.
Thus, p = 2/3.
5. Jun 8, 2004
lhuyvn
Thank All for very nice answers !!! | 2018-02-19T00:46:47 | {
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http://openstudy.com/updates/4f678d8be4b0f81dfbb4f577 | ## across 2 years ago This question is just for fun: Tell me what the last digit of $$3^{1234567890}$$ is. :) Then tell me what the last two digits are. And if you feel up to the challenge, then tell me what the last three digits are!
3
2. across
That's not it!
3. across
$\text{Hint: }a^{\phi(n)}\equiv1\text{ }(\text{mod }n)\text{, where }\gcd(a,n)=1.$
quite tricky...:)
5. myininaya
Is that fermat's or euler's thingy...I can't remember
6. across
You are right. :) This is Euler's theorem, and ϕ is the totient function. Naturally, it is an augmentation of Fermat's little theorem.
7. satellite73
fermat's little theorem $a^{p-1}\equiv 1( \text{ mod } p)$ euler phi totient function $a^{\phi(n)}\equiv 1 (\text{ mod } n)$ if $(a,n)=1$
8. satellite73
9. myininaya
I think @jamesj and @zarkon would love this question (maybe-lol) I will keep thinking on it though Great question @across
10. TuringTest
the last digit is 9, no?
11. across
Yes it is. :) Did you use Euler's theorem?
12. TuringTest
no I used a little logic no idea what that theorem is
13. TuringTest
$3^0=1$$3^1=3$$3^2=9$last digit is nine is the important point here$3^3=27$$3^4=81$and so the last digit repeats every four powers...
14. TuringTest
1,3,9,7,1,3,... so 1234567890/4=308641927+2/4 and 3^2=9
15. TuringTest
I'm sure that's got a professional way of writing it with mod this and that, but that's beyond me lol
16. Zarkon
29
17. TuringTest
how you got the other digit is what I want to know
18. across
That is genius. :) Do you know that you are intuitively deriving the above theorem? From it, it follows that to find out what the last two digits of that number are, you first need to observe that the sequence of two digits repeats every 40 powers. I will elaborate more on it a bit later. :)
19. across
I am pretty sure you can figure out what those two digits are by having told you that up there. ;)
20. across
@Zarkon, that is not it!
21. TuringTest
thank you for the compliment across, it means so much knowing who it's coming from :D
22. Zarkon
49
23. across
^.^ And @Zarkon, you are right. Let us in in thy arcane ways!
24. across
Anyway, :) to re-state Euler's theorem:$a^{\phi(n)}\equiv1\text{ }(\text{mod }n)\text{, where }\gcd(a,n)=1.$And to define Euler's phi-function:$\phi(n)=\text{number of positive integers relatively prime to }n.$By the way, I will now say that what I am about to explain is a huge over-simplification of the number theory behind it all, and that I have boiled the entire process down to a series of mechanical steps stemming from said theory. Since we are trying to find out what the last few digits of $$a=3^{1234567890}$$ are, all that we have to do is compute $$a\text{ }(\text{mod }10)$$ for the last digit, $$a\text{ }(\text{mod }100)$$ for the last two digits, and so on. Notice that $$\gcd(3,10)=\gcd(3,100)=\cdots=1$$, so we can indeed make use of the above theorem. Because it really does not pertain to the problem at hand, I will just say that $$\phi(10)=4$$, $$\phi(100)=40$$, and so on. Therefore, we have that $$3^{\phi(10)}\equiv3^4\equiv1\text{ }(\text{mod }10)$$ (check this to convince yourself that it is true), and it follows that, for the last digit, $3^{1234567890}\equiv3^2\cdot(3^4)^{308641972}\equiv3^2\cdot(1)^{308641972}\equiv9\text{ }(\text{mod }10).$Which is indeed our last digit. You can do the same thing for two digits:$3^{1234567890}\equiv3^{10}\cdot(3^{40})^{30864197}\equiv3^{10}\cdot(1)^{30864197}\equiv49\text{ }(\text{mod }100).$This can go on and on. For three digits, you will have to compute $$3^{290}$$, which is doable by Windows' calculator. But there are better methods for bigger and bigger numbers. :)
25. myininaya
Very nice across :)
26. KingGeorge
Can I do the last 3 digits? Using Euler's theorem, we know that $$\phi(1000)=400$$ so $$3^{400} \equiv 1 \mod 1000$$. Now we calculate $$1234567890 \mod 400$$. As it turns out, this is 290. Thus, we only have to calculate $$3^{290} \mod 1000$$. Using successive squaring/fast powering, this is easy, and we get that the last three digits are 449.
27. across
That is correct. :)
28. FoolForMath
It's good to mention Carmichael theorem in this context. http://en.wikipedia.org/wiki/Carmichael_function#Carmichael.27s_theorem
29. FoolForMath
And if you are in a hurry, just a one liner in python: http://ideone.com/47UX7
30. ParthKohli
Woot, solution number two!$3^n \equiv 3^{n-4} \times 3^4 \equiv 3^{n -4}\pmod{10}$Suffice to say that $$3^{n} \equiv 3^{n - 4k} \pmod {10}$$ for all integer $$k$$. Well, that's just the solution @TuringTest posted. | 2015-02-27T04:17:25 | {
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https://math.stackexchange.com/questions/2665717/k-subsets-counting-and-the-pigeon-hole-principle | # K-subsets, counting, and the pigeon hole principle
Fellow mathematicians, please take a look at these practice test questions and try to help if you can:
Question $1$:
Let $n ≥ 8$ be an even integer and let $S = \{1, 2, 3, . . . , n\}$. Consider $7$-element subsets of $S$ that consist of $4$ even numbers and $3$ odd numbers. How many such subsets are there?
(a) ${{n/2}\choose4} \cdot {{n/2}\choose3}$
(b) ${{n}\choose4} \cdot {{n}\choose3}$
(c) ${{n/2}\choose4} + {{n/2}\choose3}$
(d) ${{n}\choose4} + {{n}\choose3}$
Answer is (a). Why do we divide $n$ by $2$ here? If $n=8$, then $n/2 = 4$ which leaves us with only $2$ even and odd numbers $(1,2,3,4)$, so how can we choose $4$ even and $3$ odd numbers when there is only two of each? My initial guess was (b).
Question $2$:
What does this count? $\sum_{k=2}^{n} {n\choose{k}} \cdot 2^{n-k}$
(a) The number of strings of length $n$, where each character is $a$ or $b$, that contain at least one $a$.
(b) The number of strings of length $n$, where each character is $a$ or $b$, that contain at least $2$ many $a$’s.
(c) The number of strings of length $n$, where each character is $a$, $b$, or $c$, that contain at least one $a$.
(d) The number of strings of length $n$, where each character is $a$, $b$, or $c$, that contain at least $2$ many $a$’s.
Where does the the $c$ in "$a$, $b$, or $c$" come from?
Question $3$:
Consider a square with sides of length $17$. This square contains $n$ points. What is the minimum value of $n$ such that we can guarantee that at least two of these points have distance at most $17/\sqrt 2$? Edit: This was originally $17\sqrt 2$, I had made a typo.
(a) 4
(b) 5
(c) 6
(d) 7
The diagonal distance of this square is $\sqrt{{17^2} + {17^2}} = 17\sqrt 2$. This means that even with $2$ points, this rule would still hold because even if the two points were in either corner of the square, and since the longest distance is the diagonal distance, they would always be within a distance of $17\sqrt 2$. In this case, the lowest number to choose from is $4$, so that is the minimum -- however the answer is $5$. Where am I going wrong?
• I think @Robert Z has a point below, can you recheck the Question 3 because if distance is at most $17\sqrt{2}$ then $2$ points are enough. So it should be something else. And if the answer is $5$, it is probably $\frac{17}{\sqrt{2}}$. – ArsenBerk Feb 25 '18 at 8:28
• Indeed I have made a typo. It is the latter! – udpcon Feb 25 '18 at 16:49
Question 1. If $n$ is even then in $S = \{1, 2, 3, . . . , n\}$ there are $n/2$ even numbers and $n/2$ odd numbers. So the subsets of 4 even numbers are $\binom{n/2}{4}$ and the subsets of 3 odd numbers are $\binom{n/2}{3}$.
Question 2. Notice that ${n\choose{k}}$ is the number of ways to place $k$ character "a" and $2^{n-k}$ is the number of ways to fill the remaining $n-k$ characters with "b" or "c" ($2$ choices). Hence the given sum is just the number of strings of length $n$, where each character is "a", "b" or "c", which contain at least 2 many "a"s.
Question 3. Is distance at most $17\sqrt{2}$ or distance at most $17/\sqrt{2}$? (yes, you are correct, the distance of any two points in the square is at most the diagonal's length $17\sqrt{2}$!). For the second option, divide the square into $4$ equal squares. Then the lengths of the diagonals of the smaller squares is just $17/\sqrt{2}$. Hence by the pigeon hole principle, by taking $4+1$ points, at least two of them will lie into one of the smaller squares.
Question 1: If $n$ is even, then $\frac{n}{2}$ gives you the number of even numbers in $S$, as well as number of odd numbers. Then you simply choose $4$ of the even numbers with $\binom{\frac{n}{2}}{4}$ and $3$ of the odd numbers with $\binom{\frac{n}{2}}{3}$.
Question 2: Notice that here, we need $3$ characters because of $2^{n-k}$ (There must be two options for the rest of the $n-k$ places after putting $a$ to $k$ places).
Question 3: HINT:
Here, I'm considering the distance is at most $\frac{17}{\sqrt{2}}$. So, having at least how many points guarantees that two of them are inside the same square? | 2019-12-10T22:02:48 | {
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https://rochestercollegeathletics.com/assassins-creed-homjo/11d4cd-can-three-planes-intersect-at-one-point | Jun 611:50 AM Using technology and a matrix approach we can verify our solution. Planes have a pretty special property. 2. 1 0 Where those axis meet is considered (0, 0, 0) or the origin of the coordinate space. 4. Colloquially, curves that do not touch each other or intersect and keep a fixed minimum distance are said to be parallel. Learn more about this Silicon Valley suburb, America's richest neighborhood. CS 506 Half Plane Intersection, Duality and Arrangements Spring 2020 Note: These lecture notes are based on the textbook “Computational Geometry” by Berg et al.and lecture notes from [3], [1], [2] 1 Halfplane Intersection Problem We can represent lines in a plane by the equation y = ax+b where a is the slop and b the y-intercept. The sum of two measures of two acute angles is greater than 90 degrees. Three points must be coplanar. For lines, rays, and line segments, intersect means to meet or cross. yes, three planes can intersect in one point. (f) If two lines intersect, then exactly one plane contains both lines (Theorem 3). Minimum number of 3-inch by 5-inch index cards needed to completely cover a 3-foot by 4-foot rectangular desktop, Number of diagonals in an n-sided polygon, Some Cool Math and Computer Science links, Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Sometimes. I know that they can intersect in a straight line, coincide, or be parallel to each other (non-intersecting planes), but I was wondering if they can intersect at one point. It is very easy to find a system of three equations in three unknowns for each one of the eight different relative positions, except for one: case 6 (where the three Planes intersect in a line). If you're talking about lines in the same plane, lines don't intersect if they're parallel. General solution to system of differential equation question...? Where those axis meet is considered (0, 0, 0) or the origin of the coordinate space. Any 3 dimensional cordinate system has 3 axis (x, y, z) which can be represented by 3 planes. Condition for three lines intersection is: rank Rc= 2 and Rd= 3 All values of the cross product of the normal vectors to the planes are not 0 and are pointing to the same direction. Let $\ell'$ be a line different from but parallel to $\ell$. The intersection of the three planes is a point. The planes will then form a triangular "tube" and pairwise will intersect at three lines. Thus, the intersection of 3 planes is either nothing, a point, a line, or a plane: To answer the original question, 3 planes can intersect in a point, but cannot intersect in a ray. 3. How many multiples of 8 are between 100 and 175? Finally we substituted these values into one of the plane equations to find the . There are infinitely many planes through $\ell'$, but only one of them intersects $\ell$, and only two of them are parallel to one of the first two planes. Each plan intersects at a point. 3 Plane Intersection. Jun 611:50 AM Using technology and a matrix approach we can verify our solution. If they are in the same plane, and not parallel, then they will intersect at exactly one point. true. Given 3 unique planes, they intersect at exactly one point! Sign in|Recent Site Activity|Report Abuse|Print Page|Powered By Google Sites, Maximum number of points of intersection of 6 lines. A projective plane can be thought of as an ordinary plane equipped with additional "points at infinity" where parallel lines intersect. Huh? he might have made a mistake when writing out the test or got it out of a faulty text or you might have misread. Intersection of Three Planes. true. Three planes can mutually intersect but not have all three intersect. In the figure below, rays BA and BC meet at endpoint B, so their intersection forms angle ∠ABC. true. Just two planes are parallel, and the 3rd plane cuts each in a line. Find the equation of the plane that contains the point (1;3;0) and the line given by x = 3 + 2t, y = 4t, z = 7 t. Lots of options to start. In geometry, parallel lines are lines in a plane which do not meet; that is, two straight lines in a plane that do not intersect at any point are said to be parallel. What is A? Now, if we draw a 3rd line, that can intersect the other two lines in at most 2 points as shown below. How many five-digit numbers greater than 70,000 are mountain numbers? equation of a quartic function that touches the x-axis at 2/3 and -3, passes through the point (-4,49)? You can also rotate it around to see it from different directions, and zoom in or out. Integer coordinates enclosed by 2 squares. The intersection of the three planes is a line. In three dimensions, no. The text is taking an intersection of three planes to be a point that is common to all of them. How many times during the day do the hands of a clock overlap? The point can be located by starting from the origin then following the red arrows on the grid. Not for a geometric purpose, without breaking the line in the sketch. This is question is just blatantly misleading as two planes can't intersect in a point. three planes can intersect as a point or as a line. Penny Note that there is no point that lies on all three planes. (g) If a point lies outside a line, then exactly one plane contains both the line and the point … Increase Brain Power, Focus Music, Reduce Anxiety, Binaural and Isochronic Beats - Duration: 3:16:57. Thus, any pair of planes must intersect in a line, but not all three at once (since there is no solution). Each plane cuts the other two in a line and they form a prismatic surface. Or three planes can, like the pages in the spine of a book, can intersect in one single line. three noncollinear points determine a plane. Count the points of intersection for each and allow infinite as some of your counts. If the lines are in different planes in 3-dimensional space, they won't intersect. We can use a matrix approach or an elimination approach to isolate each variable. I know that they can intersect in a straight line, coincide, or be parallel to each other (non-intersecting planes), but I was wondering if they can intersect at one point. The new app allows you to explore the concepts of solving 3 equations by allowing you to see one plane at a time, two at a time, or all three, and the intersection point. When two lines, rays, or line segments intersect, they have one common point. It is not possible in a three-dimensional space, but it is possible in a four-dimensional space. I may just have found a slightly shorter way (four commands vs five in the previous) - PC-DMIS can intersect a CURVE and a PLANE to get one of the intersection points (assuming your number of hits is large enough). Yes, look at … Think about what a plane is: an infinite sheet through three... See full answer below. The system has one solution. In one of the positions, the three Planes share only one point (they intersect at that point). You can edit the visual size of a plane, but it is still only cosmetic. Justify your answer. I would not confront your teacher but would recheck the question and if it asks about two planes intersecting I would ask for an explanation, because you don't get it. Any two of them must intersect, if no two are parallel, but there need not be a point that all three of them have in common. Note, because we found a unique point, we are looking at a Case 1 scenario, where three planes intersect at one point. This lines are parallel but don't all a same plane. (b) Give an example of three planes in R^3 that intersect in pairs but have no common point of intersection. Probability that a number created randomly from 3 digits is even. The second and third planes are coincident and the first is cuting them, therefore the three planes intersect in a line. There is nothing to make these three lines intersect in a point. The East wall, South wall and floor are three planes that intersect at one point, the point on the floor in the South-East corner of the room. Leave a comment * Sometimes. (e) A line contains at least two points (Postulate 1). (a) Give an example of three planes in R^3 that have a common line of intersection. The East wall and the South wall are pieces of two planes that do intersect. In America's richest town, $500k a year is below average. In 3D, three planes , and can intersect (or not) in the following ways: All three planes are parallel. Four points may be coplanar or noncoplanar. The intersection is the line that forms the South-East corner of the room. Thus any two distinct lines in a projective plane intersect in one and only one point. Any 3 dimensional cordinate system has 3 axis (x, y, z) which can be represented by 3 planes. Given 3 unique planes, they intersect at exactly one point! two planes intersect in a line. false. Get an answer to your question "Can 2 planes in a 3 dimensional space intersect at one point ..." in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. Geometrically, we have planes whose orientation is similar to the diagram shown. Note, because we found a unique point, we are looking at a Case 1 scenario, where three planes intersect at one point. (1) (2) (3) point of intersection 3 4 z. value. the lines intersect at a point. Precalculus 3-D Cartesian Coordinate System Lines in Space. Homework Statement The three lines intersect in the point (1; 1; 1): (1 - t; 1 + 2*t; 1 + t), (u; 2*u - 1; 3*u - 2), and (v - 1; 2*v - 3; 3 - v). Justify your answer. Join Yahoo Answers and get 100 points today. 2 Answers bp Dec 14, 2016 We can use a matrix approach or an elimination approach to isolate each variable. true. 3. We know a point on the line is (1;3… (g) If a point lies outside a line, then exactly one plane contains both the line and the point … Similarly, if we draw a 4th line, that can intersect the other 3 lines in at most 3 points and so on. Planes have a pretty special property. Two points must be collinear. Sometimes. Renaissance artists, in developing the techniques of drawing in perspective, laid the groundwork for this mathematical topic. a line and a plane can intersect in a point. If I have 5 independent groups and us a likert scale what stat analysis would I use. Two planes can intersect on exactly one point? Get your answers by asking now. Three lines intersect at one point. The number 23AB3 is exactly divisible by 99. Sometimes. Examples Example 3 Determine the intersection of the three planes: 4x y — z — 9m + 5y — z — was the question about three planes? (e) A line contains at least two points (Postulate 1). 1 0 Justify your answer. (c) Give an example of three planes in R^3 that intersect in a single point. If two planes intersect, then their intersection is a line (Postulate 6). For example, the xy-plane and the zw-plane intersect at the origin (0,0,0,0). A line and a plane intersect at exactly one point. If we found in nitely many solutions, the lines are the same. For example, the xy-plane and the zw-plane intersect at the origin (0,0,0,0). The intersection of the first two is a line$\ell$. Three planes may all intersect each other at exactly one point. How do you solve a proportion if one of the fractions has a variable in both the numerator and denominator? Therefore, the solution to this system of three equations is (3, 4, 2), a point This can be geometrically interpreted as three planes intersecting in a single point, as shown. Using that, we only need to create one line to find the other point. Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website! Calculator won't calculate sin divided by anything, shows error? Other geometric figures. (1) (2) (3) point of intersection 3 4 In four dimensions or higher, yes. In how many days will Jessica read 270 pages of a book? Music for body and spirit - Meditation music Recommended for you 3:16:57 Doesn't matter, planes have no geometric size. When two distinct planes meet, they intersect at a line. If we found no solution, then the lines don’t intersect. Is there a way to create a plane along a line that stops at exactly the intersection point of another line. The first and second are coincident and the third is parallel to them. Assuming you are working in R 3, if the planes are not parallel, each pair will intersect in a line. two planes intersect at exactly one point. Two planes contain the same point. Here are the ways three planes can associate with each other. I would not confront your teacher but would recheck the question and if it asks about two planes intersecting I would ask for an explanation, because you don't get it. Some geometric figures intersect at more than one point. These planes do not intersect. The measure of an angle is greater than the measure of its complement. This means that, instead of using the actual lines of intersection of the planes, we used the two projected lines of intersection on the x, y plane to find the x and y coordinates of the intersection of the three planes. Therefore, the solution to this system of three equations is (3, 4, 2), a point This can be geometrically interpreted as three planes intersecting in a single point, as shown. Any two straight lines can intersect each other in one point. true. Only lines intersect at a point. How many 5 digit numbers are perfect squares? Precalculus . If two planes intersect, then their intersection is a line (Postulate 6). was the question about three planes? Science Anatomy & Physiology ... Can two planes intersect in exactly one point? no they cant... only a line can work for this.. all the time... Can two two-dimensional planes intersect in a point? It is not possible in a three-dimensional space, but it is possible in a four-dimensional space. One Comment on “Intersection of 3 planes at a point: 3D interactive graph” Leesa Johnson says: 24 Jan 2017 at 9:13 pm [Comment permalink] Nice explanation for me to understand the interaction of 3d planes at a point using graphical representation and also useful for the math students. Florida governor accused of 'trying to intimidate scientists', Ivanka Trump, Jared Kushner buy$30M Florida property, Another mystery monolith has been discovered, MLB umpire among 14 arrested in sex sting operation, 'B.A.P.S' actress Natalie Desselle Reid dead at 53, Goya Foods CEO: We named AOC 'employee of the month', Young boy gets comfy in Oval Office during ceremony, Packed club hit with COVID-19 violations for concert, Heated jacket is ‘great for us who don’t like the cold’, COVID-19 left MSNBC anchor 'sick and scared', Former Israeli space chief says extraterrestrials exist. Question 97302: can 3 planes intersect in exactly one point? 3 Plane Intersection. Planes intersect along a line. three planes can intersect as a point or as a line. This commonly occurs when there is one straight plane and two other planes intersect it at acute or obtuse angles. Still have questions? (f) If two lines intersect, then exactly one plane contains both lines (Theorem 3). | 2021-06-18T17:24:05 | {
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https://www.physicsforums.com/threads/if-sinh-y-x-then-show-that-cosh-y-sqrt-1-x-2.186439/ | # If sinh(y)=x, then, show that cosh(y)=sqrt(1+x^2)
1. Sep 23, 2007
### cks
if sinh(y)=x, then, show that cosh(y)=sqrt(1+x^2)
I know how to prove but i have difficult in choosing the signs.
sinh(y) = x = [exp(y)-exp(-y)]/2
there are two equations I can find
exp(2y) - 2x exp(y) -1 =0 OR exp(-2y) + 2xexp(-y) -1 =0
exp(y) = x +- sqrt(x^2 +1) & exp(-y) = -x +- sqrt(x^2 + 1)
if I select the signs of +
exp(y) = x + sqrt(x^2 +1) & exp(-y) = -x + sqrt(x^2 + 1)
then by substituting to cosh(y) = [exp(y) + exp(-y)] / 2
then, I can find the answer of
cosh(y)=sqrt(1+x^2)
But, what should I say to justify my actions of selecting the +ve sign.
2. Sep 23, 2007
### neutrino
Why not just substitute $$x = \frac{\left(e^y - e^{-y}\right)}{2}$$ into the expression $$\sqrt{1+x^2}$$ and simplify?
3. Sep 23, 2007
### cks
Thanks, but I'd like to find cosh(y) without prior knowledge that it's equal to sqrt(1+x^2)
4. Sep 23, 2007
### robphy
For real y, exp(y)>0.
5. Sep 23, 2007
### arildno
Utilize the identity:
$$Cosh^{2}(y)-Sinh^{2}(y)=1$$
Along with the requirement that Cosh(y) is a strictly positive function.
(Alternatively, that it is an even function with Cosh(0)=1)
Last edited: Sep 23, 2007
6. Sep 23, 2007
### cks
still don't quite understand
7. Sep 24, 2007
### Gib Z
arildno pretty much gave you the identity that you need. If you need to prove it, Just replace cosh y and sinh y with their exponential definitions and simplify. Then isolate cosh y on the left hand side and it is clear. Basically when we take the final square root, one could normally argue you could have either the positive or the negative root. However cosh is always positive.
8. Sep 24, 2007
### robphy
There are different ways to approach the title of this thread, depending on what you are given to start with and what constraints are imposed... which should include your working definitions of sinh and cosh.
Rather than merely proving an identity, it appears that the OP wishes to obtain the title by using the definitions of sinh and cosh starting from their definitions using the sum and difference of exponential functions. In addition, one might taking the point of view that one doesn't know at this stage any properties of cosh except for its working definition. So, it might be "cheating" to use any other properties of cosh which you didn't derive already.
cks, is this correct?
(If I'm not mistaken, my suggestion "For real y, exp(y)>0" provides a reason to choose the positive sign.)
9. Sep 25, 2007
### cks
Oh, I see thank you. Let me rephrase what you all said
exp(y) = x +- sqrt(x^2 +1) & exp(-y) = -x +- sqrt(x^2 + 1)
exp(y) is always > 0
so if we select exp(y) = x - sqrt(x^2 +1)
and most importantly , sqrt(x^2 +1) > x
so exp(y) < 0 which is false.
As a result, we have to select exp(y) = x + sqrt(x^2 +1)
The similar argument can be applied to exp(-y)
Thank you all of you, really appreciate that. | 2018-01-20T16:06:23 | {
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http://mathhelpforum.com/algebra/7834-simultaneous-equations-way-too-many-unknowns.html | # Math Help - Simultaneous Equations with way too many unknowns..
1. ## Simultaneous Equations with way too many unknowns..
Yea, so equations aren't my bestfriends, and I'm now stuck with two simultaneous equations, and I just don't know where to start, so if someone could help me and show me what to do, I'd appreciate it.
'Find the values of m such that these equations have no solutions':
3x - my = 4
x + y = 12
'Find the values of m and a such that these equations have infinite solution sets':
4x - my = a
2x + y = 4
So if someone could patiently guide me through step by step, that'd be awesome.. I hate not knowing what I'm supposed to do.
Thanks a lot
2. Originally Posted by Fnus
Yea, so equations aren't my bestfriends, and I'm now stuck with two simultaneous equations, and I just don't know where to start, so if someone could help me and show me what to do, I'd appreciate it.
'Find the values of m such that these equations have no solutions':
3x - my = 4
x + y = 12
'Find the values of m and a such that these equations have infinite solution sets':
4x - my = a
2x + y = 4
So if someone could patiently guide me through step by step, that'd be awesome.. I hate not knowing what I'm supposed to do.
Thanks a lot
If the equations have an infinite solution set then basically the two equations say the same thing: that is to say, one equation is a multiple of the other.
So since the coefficient of x in the first equation is twice the coefficient of x in the second equation, then "-m" must be twice the coefficient of y in the second equation, so
$-m = 2 \cdot 1$
or m = -2.
Similarly a = 8.
-Dan
3. there are no solutions if both equations are lines with the same slope (parallel) and different y-intercepts. thus they will never cross.
Question 1:
'Find the values of m such that these equations have no solutions':
3x - my = 4
x + y = 12
Question 2:
'Find the values of m and a such that these equations have infinite solution sets':
4x - my = a
2x + y = 4
Does that change anything? <.<
5. $x+y=12\Rightarrow y=-x+12$ so the slope is -1.
$3x-my=4\Rightarrow y=\frac{3}{m}x-\frac{4}{m}$
now solve $\frac{3}{m}=-1$ you should get $m=-3$
6. for the second one we have
$4x-my-a=0$
and
$2(2x+y)=2(4)\Rightarrow 4x+2y-8=0$
from here it should be obvious what values to make a and m.
7. Hello, Fnus!
Find the values of $m$ such that these equations have no solutions:
. . $\begin{array}{cc}(1)\\(2)\end{array}
\begin{array}{cc}3x - my \\ x + y \end{array}
\begin{array}{cc} = \\ = \end{array}
\begin{array}{cc}4 \\ 12\end{array}$
If you are familiar with determinants, there is a simple solution.
A system has no solution if its determinant equal zero.
The determinant is: . $\begin{vmatrix} 3 & \text{-}m \\ 1 & 1\end{vmatrix} \:=\:(3)(1) - (\text{-}m)(1) \:=\:3 + m$
Therefore: . $3 + m \:=\;0\quad\Rightarrow\quad\boxed{m = -3}$
Otherwise, we can try to solve the system.
We have equation (1): . $3x - my \:=\:4$
. . .Multiply (2) by $m:\;\;mx + my \:=\:12m$
. . . . . . . . . . . . Add: . $mx + 3x \:=\:12m + 4$
. . . . . . . . . . .Factor: . $(m + 3)x \:=\:12m + 4$
. . . . . . . . . . . Then: . . . . . . $x\:=\:\frac{12m + 4}{m + 3}$
We see that $x$ is undefined if $m = -3.$
Therefore, the system has no solutions for $m = -3.$
8. Originally Posted by topsquark
If the equations have an infinite solution set then basically the two equations say the same thing: that is to say, one equation is a multiple of the other.
So since the coefficient of x in the first equation is twice the coefficient of x in the second equation, then "-m" must be twice the coefficient of y in the second equation, so
$-m = 2 \cdot 1$
or m = -2.
Similarly a = 8.
-Dan
Originally Posted by putnam120
there are no solutions if both equations are lines with the same slope (parallel) and different y-intercepts. thus they will never cross.
But if one equation is merely a multiple of the other, they represent the same line, so any point (x,y) on the line is a solution. So we have an infinite number of solutions.
-Dan | 2015-08-03T02:08:15 | {
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https://math.stackexchange.com/questions/499537/relationship-between-o-and-o-notation/499541 | # Relationship between O and o notation
In big-O notation, $f(x) = O(g(x))$ as $x\rightarrow \pm\infty$ if
$$\exists C, \delta>0: \forall |x| \geq \delta: |f(x)| < C |g(x)|$$
and, for the case I'm more interested in here, $f(x) = O(g(x))$ as $x\rightarrow 0$ if
$$\exists C, \delta>0: \forall |x| \leq \delta: |f(x)| < C |g(x)|$$
In little-o notation, $f(x) = o(g(x))$ means
$$\lim_{|x|\rightarrow0}\frac{f(x)}{|g(x)|} = 0$$
I've come across big-O before, but this is the first time I've seen little-o. I have experience in applied math, but very little in pure math, and my first reaction (for $x\rightarrow 0$) was "they mean the same thing". Then I thought "but little-o is a more precise upper bound".
What is the relationship between these two notations, and what does the distinction mean practically?
• Well, if $f(x)$ is $o(g(x))$ it is also $O(g(x))$ but maybe not the other way around. It is not "more precise", but rather stronger: intuitively, if $f(x)$ is $o(g(x))$ then $g(x)$ grows "much faster" than $f(x)$, whereas $O(g(x))$ implies only that $f(x)$ is eventually bounded above (by some constant multiple of) $g(x)$ -- indeed, the growth rates could be comparable. – The_Sympathizer Sep 20 '13 at 11:54
• An example -- though note here I'm using the O-notation for growth near infinity, not near 0: If $f(x) = x$ and $g(x) = x^2$, then $f(x)$ is $o(g(x))$ AND $O(g(x))$, but if $f(x) = x^2$ then $f(x)$ is only $O(g(x))$ and NOT $o(g(x))$. – The_Sympathizer Sep 20 '13 at 11:56
• An example for near 0: $f(x) = \frac{1}{x}$ and $g(x) = \frac{1}{x^2}$. Then $f(x)$ is $o(g(x))$ (as $x \rightarrow 0$) while if $f(x) = \frac{1}{x^2}$ then $f(x)$ is only $O(g(x))$ and not $o(g(x))$. – The_Sympathizer Sep 20 '13 at 11:58
In the briefest possible terms, $f \in O(g)$ is like saying $f \leqslant g$ asymptotically, and $f \in o(g)$ is like saying $f < g$ asymptotically.
As an example, if $f(x) = x$ and $g(x) = 2x$,we have that that $f \in O(g)$, but $f \notin o(g)$ and $g \notin o(f)$. If $f(x) = x^2$ and $g(x) = x$, then $f \in o(g)$ and $f \in O(g)$.
As an aside, be very aware of context. In various probabilistic situations, we define $O$-notation with a limit as $x\rightarrow 0$. In most algorithmic applications, we take limits as $x \rightarrow \infty$. Always make sure it's clear what you're working with before using either notation. $O$-notation is probably the most abused thing in all of mathematics.
• Also note: Notation $f \in O(g)$ is quite rare, compared to notation $f = O(g)$. – GEdgar Sep 20 '13 at 13:42
• Indeed. I didn't want to start listing abuses of $O$ notation because I don't think I'd ever stop. It's more common to say things like $2x = O(x)$. – ymbirtt Sep 20 '13 at 14:13
• $x^2$ is not in $O(x)$, at least when $x \to \infty$. Typo? – ftfish Sep 20 '13 at 14:44
• From OP's definitions, we're sending $x \rightarrow 0$, so I'm remaining consistent with that. – ymbirtt Sep 20 '13 at 15:08
• @GEdgar thanks, discussion of this helps me to get to grips with it. I've always been a little bit uncomfortable with the use of $=$, and thinking about it in terms of sets is more consistent. – TooTone Sep 21 '13 at 9:07
Little $o$ means lim sup of absolute value $=0$, big $O$ means lim sup of absolute value $< \infty$.
• Thanks, but I thought big O could be used for $\rightarrow 0$ as well as $\rightarrow \infty$? The former was what I was more interested in because it then has a similar meaning to little o. (My post was a little ambiguous and I see you changed $<$ in my post to $\ge$ -- I've changed that back and added some more which hopefully makes it clearer.) – TooTone Sep 20 '13 at 14:31
• You are referring to my correction? Okay, I am sorry then the same remains true. It should be indicated though wether you look at the behaviour at $0$ or at $\infty$. A continuous variable transformation, e.g. $x \mapsto x^{-1}$ exchanges between the setting and my statement remains valid;) – Marc Palm Sep 20 '13 at 14:39
• Yes you're right I should have indicated whether I was looking at limit towards zero or infinity. I can see how a reciprocal transformation might make the situations equivalent; however I still don't quite understand your answer. – TooTone Sep 20 '13 at 19:52
• my answer remains valid no matter to where the limit goes. – Marc Palm Sep 21 '13 at 14:57
• I think what I don't understand is, in simple terms, your use of "sup" means. – TooTone Sep 21 '13 at 18:01
check this out :
http://en.wikipedia.org/wiki/Big_O_notation
:-) it's not just Big O notation in the page
Looking at how $O(1)$ and $o(1)$ are sufficient to understand the general case.
Suppose $f \in O(1)$. Then
$$\lim_{x \to +\infty} f(x) < +\infty$$
if the limit exists. Or more generally, if the limit does not exist, then
$$\sup_{x \to +\infty} f(x) < +\infty$$
Now suppose $f \in o(1)$. Then
$$\lim_{x \to \infty} f(x) = 0$$ | 2019-09-20T16:34:57 | {
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http://math.stackexchange.com/questions/442768/fixed-point-theorems | # Fixed Point Theorems
Theorem 1. Let $B=\{x\in \mathbb R^n :∥x∥≤1\}$ be the closed unit ball in $\mathbb R^n$ . Any continuous function $f:B\rightarrow B$ has a fixed point.
Theorem 2. Let $X$ be a finite dimensional normed vector space, and let $K\subset X$ be a non-empty, compact, and convex set. Then given any continuous mapping $f:K\rightarrow K$ there exists $x\in K$ such that $f(x)=x$.
Theorem 3. Let $X$ be a normed vector space, and let $K\subset X$ be a non-empty, compact, and convex set. Then given any continuous mapping $f:K\rightarrow K$ there exists $x\in K$ such that $f(x)=x$.
Theorem 4. Let $X$ be a normed vector space, and let $K\subset X$ be a non-empty, closed, and bounded set. Then given any compact mapping $f:K\rightarrow K$ there exists $x\in K$ such that $f(x)=x$.
For some authors Theorem 1 is Brouwer's fixed-point theorem. For others Brouwer's fixed-point theorem is Theorem 2. Actually there is no difference because every non-empty, compact and convex set in a finite dimensional normed vector space is is homeomorphic to the closed unit ball.
My problem is with theorems 3 and 4. For some authors Theorem 3 is Schauder's fixed-point theorem, for others Schauder's fixed-point theorem is Theorem 4.
Are Theorem 3 and Theorem 4 are equivalent? If not, are Theorems 1 and 2 special cases of Theorem 4?
-
"because every non-empty, compact and convex set in a finite dimensional normed vector space is is homeomorphic to the closed unit ball" is not quite true, it could be lower-dimensional. Theorems 1 and 2 are special cases, because in that case, a continuous $f \colon K \to K$ is compact. – Daniel Fischer Jul 13 '13 at 15:39
What is the nonl tag for? – draks ... Jul 13 '13 at 20:08
@draks... My guess is a foreshortened nonlinear-something that never really got off the ground. – Sharkos Jul 13 '13 at 22:56
Your statement of Theorem 4 is missing an assumption on $K$, such as being convex, or at least homeomorphic to such a set (convex, closed, bounded). Without such an assumption, rotation of a circle gives a counterexample. Also, I think that in Theorem 4 you want the normed space to be complete, i.e., a Banach space.
Theorem 3 is contained in Theorem 4, because on a compact set every continuous map is compact. Theorem 4 cannot be easily obtained from Theorem 3 (I think) because if we tried to simply replace $K$ with $\overline{f(K)}$ (which is compact), we can't apply Theorem 3 because $\overline{f(K)}$ is not known to be convex.
Both 3 and 4 were stated and proved by Schauder in his 1930 paper Der Fixpunktsatz in Funktionalraümen, which is in open access. Here is Theorem 3:
Satz I. Die stetige Funktionaloperation $F(x)$ bilde die konvexe, abgeschlossene und kompakte Menge $H$ auf sich selbst ab. Dann ist ein Fixpunkt $x_0$, vorhanden, d.h. es gilt $F(x_0)=x_0$.
And this is Theorem 4 (in slightly less general version: the image of $F$ is assumed compact instead of relatively compact; possibly because the latter concept wasn't in use).
Satz II. In einem "B"-Raume sei eine konvexe und abgeschlossene Menge $H$ gegeben. Die stetige Funktionaloperation $F(x)$ bilde $H$ auf sich selbst ab. Ferner sei die Menge $F(H)\subset H$ kompakt. Dann ist ein Fixpunkt vorhanden.
("B"-Raume is what is now called a Banach space.) So, it is correct to call both Theorem 3 and Theorem 4 "Schauder's fixed-point theorem".
And yes, Theorems 1 and 2 follow by specialization of Theorem 3 or 4 to finite dimensions.
- | 2016-05-06T21:31:22 | {
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http://mathhelpforum.com/calculus/4151-ship-problem.html | # Math Help - ship problem
1. ## ship problem
A ship P is travelling due East at 30 km/h and a Ship Q is travelling due South at 40 km/h.
Both ships keep constant speed and course. At t=0 they are each 10 km from the point of intersection of their courses and moving towards the point.
Iv found the co-ordinates of Q relative to P at t=0
---->X=(10,10)km
iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
-----.V= (0,40) - V(30,0)
----- V= (-30,40)km/h
but im struggling to find the time at which P and Q are closest to each other...is there anyone that can help??? thankz
2. At t=0, they are both 10 kms from their intersection.
Think Pythagoras. Let D=square of distance between ships.
$D(t)=(10-30t)^{2}+(10-40t)^{2}$
$D'(t)=2(10-30t)(-30)+2(10=40t)(-40)$
$D'(t)=5000t-1400$
$5000t-1400=0$
$t=\frac{7}{25}$
3. Originally Posted by galactus
At t=0, they are both 10 kms from their intersection.
Think Pythagoras. Let D=square of distance between ships.
$D(t)=(10-30t)^{2}+(10-40t)^{2}$
$D'(t)=2(10-30t)(-30)+2(10=40t)(-40)$
$D'(t)=5000t-1400$
$5000t-1400=0$
$t=\frac{7}{25}$
thankz...but i was wondering if i did the first part right that i showed?(shown below) because im not too sure if i was meant to include negative signs with any of the answers? can u please just check that for me thankz
Iv found the co-ordinates of Q relative to P at t=0
---->X=(10,10)km
iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
-----.V= (0,40) - V(30,0)
----- V= (-30,40)km/h
4. Originally Posted by dopi
A ship P is travelling due East at 30 km/h and a Ship Q is travelling due South at 40 km/h.
Both ships keep constant speed and course. At t=0 they are each 10 km from the point of intersection of their courses and moving towards the point.
Hello, dopi,
with your problem you use automatically a coordinate system: The pos. x-axis points to East and the pos. y-axis points to North.
At the time t = 0 the ship P is at P(-10, 0) and the ship Q is at Q(0, 10). The movement of both ships is described by a straight line. The speed is described by a vector:
$\overrightarrow{v_P}=(30, 0)$
$\overrightarrow{v_Q}=(0, -40)$
Thus P is moving along the line:
$\overrightarrow{x_P}=(-10, 0)+t*(30, 0)$
and Q is moving along the line:
$\overrightarrow{x_Q}=(0, 10)+t*(0, -40)$
The distance between the ships is: $\vec{d}=\overrightarrow{x_P}-\overrightarrow{x_Q}$
Originally Posted by dopi
Iv found the co-ordinates of Q relative to P at t=0
---->X=(10,10)km
iv also found the velocity of Q relative to P----> V= V[Q] -V[P]
-----.V= (0,40) - V(30,0)
----- V= (-30,40)km/h
I don't understand why you want to calculate the relative coordinates. It isn't necessary with your problem.
Originally Posted by dopi
but im struggling to find the time at which P and Q are closest to each other...is there anyone that can help??? thankz
This was already done by galactus.
Greetings
EB
5. Hello, dopi!
You seem to be using a coordinate system and vectors
. . and I don't know what you plan to do with them.
galactus gave you the solution . . . so why struggle with your approach?
I'll make some diagrams to accompany his excellent solution.
Originally Posted by galactus
At t=0, they are both 10 kms from their intersection.
Code:
* Q
|
|
| 10
|
|
|
P * - - - - - - - - o
10
$t$ hours later, $P$ has moved $30t$ km east to point $A$
. . and $Q$ has moved $40t$ km south to point $B.$
Code:
*
|
| 40t
|
o B
|
| 10-40t
* - - - o - - - - *
30t A 10-30t
And we want to minimize the distance $\overline{AB}.$
Originally Posted by galactus
Think Pythagoras. Let D = square of distance between ships.
$D(t) \:= \10-30t)^{2} + (10-40t)^{2}" alt="D(t) \:= \10-30t)^{2} + (10-40t)^{2}" />
$D'(t) \:= \:2(10-30t)(-30) + 2(10=40t)(-40)$
$D'(t) \:= \:5000t - 1400$
$5000t -1400\:=\:0$
$t \:= \:\frac{7}{25}$ | 2014-10-21T11:13:21 | {
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http://mathhelpforum.com/calculus/146987-tricky-double-integral-polar-coordinates.html | # Math Help - Tricky Double Integral with polar coordinates
1. ## Tricky Double Integral with polar coordinates
Evaluate the integral:
$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$
where $D$ is the region $\{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$
Tried drawing the region $D$ and it looks like I have to use polar coordinates.
Using polar coordinates, $\frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$.
I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$.
Thanks in advance, any pointers would be greatly appreciated.
2. Originally Posted by craig
Evaluate the integral:
$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$
where $D$ is the region $D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$
Tried drawing the region $D$ and it looks like I have to use polar coordinates.
Using polar coordinates, $\frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$.
I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$.
Thanks in advance, any pointers would be greatly appreciated.
In general to convert, you do the following:
$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
\frac{\partial (x,y)}{\partial (r, \theta)} \, dr \, d\theta$
Where $\frac{\partial (x,y)}{\partial (r, \theta)}$ is called the jacobian determinant and it's given by:
$\frac{\partial (x,y)}{\partial (r, \theta)} =
\begin{vmatrix}
\cos \theta & - r \sin \theta \\
\sin \theta & r \cos \theta
\end{vmatrix} = r$
Hence:
$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
r \, dr \, d\theta$
Now, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.
$D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$
Now, the 2nd constraint on the region says that $x^2 + y^2 \leq 3$. You should be able to see that this equation describes a circle whose radius is less than $\sqrt{3}$, in other words, the radius is limited by $0 \leq r \leq \sqrt{3}$.
Now, the 1st constraint on the region says that $y \geq x$. Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by $\frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4}$.
You can see all of this visually by drawing it out. Go on, do this:
1) Draw the x-y coordinates.
2) Draw a circle centred at the origin with a radius of $\sqrt{3}$.
3) Draw the line y = x.
4) Now shade the region that lies above the line y = x, but does not go outside the circle.
You should see that this region is bounded by:
$D = \bigg\{ (r,\theta)| 0 \leq r \leq \sqrt{3} \; \textrm{and} \; \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4} \bigg\}$
Hence, in this particular situation:
$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy = \int_0^{\frac{5\pi}{4}} \int_0^{\sqrt{3}} r \, \cos^2(\theta) \, dr \, d\theta = \int_0^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
3. Hi
The area of integration is the part of the disk which is above the red line
And $dx dy$ becomes $r dr d\theta$
4. Originally Posted by craig
Evaluate the integral:
$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$
where $D$ is the region $\{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$
Tried drawing the region $D$ and it looks like I have to use polar coordinates.
Using polar coordinates, $\frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$.
I'm not sure how to find the limits if integration though, or what to do with the $dx\;dy$.
Thanks in advance, any pointers would be greatly appreciated.
I go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-post2.html
If you draw the xy domain you can see we want the area to the left of the line but inside the circle,
If we convert this to polar co-ordinates this means
$\frac{ \pi }{4} \le \theta \le \frac{ 5 \pi }{4}$ where the $\frac{ \pi }{4}$ comes from the fact that the line $y=x$ acts at $\frac{ \pi }{4}$ above the x-axis. And $\frac{ 5 \pi }{4}$ comes from the angle between the line in the first quadrent to the line in the 3rd quadrent.
For r, we are going from the origin to the radius of the circle. So the interval of r must be
$0 \le r \le \sqrt{3}$
We can now compute this integral,
$\int \int_D \frac{x^2}{x^2 + y^2} dxdy$
$= \iint_Q \frac{ r^2 cos^2 \theta }{ r^2 } rdr d \theta$
$= \iint_Q rcos^2 \theta dr d \theta$
$= \int_0^{ \sqrt{3} } rdr \int_{ \frac{ \pi }{4} }^{ \frac{ 5 \pi }{4} } cos^2 \theta d \theta$
Which can be easily computed using double angle formulas
5. Wow leave the forum for a few hours and there's 3 great replies waiting for me!!
Originally Posted by Mush
In general to convert, you do the following:
$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
\frac{\partial (x,y)}{\partial (r, \theta)} \, dr \, d\theta$
Where $\frac{\partial (x,y)}{\partial (r, \theta)}$ is called the jacobian determinant and it's given by:
$\frac{\partial (x,y)}{\partial (r, \theta)} =
\begin{vmatrix}
\cos \theta & - r \sin \theta \\
\sin \theta & r \cos \theta
\end{vmatrix} = r$
Hence:
$\displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta)
r \, dr \, d\theta$
Ahh yeh I've done Jacobian determinants before, it didn't click that it applied here.
Originally Posted by Mush
Now, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.
$D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$
Now, the 2nd constraint on the region says that $x^2 + y^2 \leq 3$. You should be able to see that this equation describes a circle whose radius is less than $\sqrt{3}$, in other words, the radius is limited by $0 \leq r \leq \sqrt{3}$.
Now, the 1st constraint on the region says that $y \geq x$. Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by $\frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4}$.
You can see all of this visually by drawing it out. Go on, do this:
1) Draw the x-y coordinates.
2) Draw a circle centred at the origin with a radius of $\sqrt{3}$.
3) Draw the line y = x.
4) Now shade the region that lies above the line y = x, but does not go outside the circle.
You should see that this region is bounded by:
$D = \bigg\{ (r,\theta)| 0 \leq r \leq \sqrt{3} \; \textrm{and} \; \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4} \bigg\}$
I'd drawn the circle but for some reason only drew $y = x$ for positive values of $x$ and $y$, that would be where I was going wrong I guess
Originally Posted by Mush
Hence, in this particular situation:
$\int \int_D \frac{x^2}{x^2 + y^2} dx\;dy = \int_0^{\frac{5\pi}{4}} \int_0^{\sqrt{3}} r \, \cos^2(\theta) \, dr \, d\theta = \int_0^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
** Think you mean $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
Anyway on with the integral:
$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg[ \frac{r^2}{2} \bigg]_0^{\sqrt{3}} \, d\theta$ = $\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{1}{2}(1 + \cos{2\theta}) \, d\theta$ =
$\frac{3}{4}\bigg[\theta + \frac{1}{2}\sin{2\theta}\bigg]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$ =
$\frac{3}{4}\bigg(\frac{5\pi}{4} + \frac{1}{2}\sin{\frac{5\pi}{2}} -\frac{pi}{4} - \frac{1}{2}\sin{\frac{\pi}{2}} \bigg)$
$\frac{3}{4}(\pi + 0)$ = $\frac{3\pi}{4}$
Thanks a lot for the help!
6. Originally Posted by running-gag
Hi
The area of integration is the part of the disk which is above the red line
And $dx dy$ becomes $r dr d\theta$
Thanks a lot for the image, I'd forgot to draw the whole of the line $y=x$!
Originally Posted by AllanCuz
I go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-post2.html
If you draw the xy domain you can see we want the area under the line but inside the circle,
If we convert this to polar co-ordinates this means
$0 \le \theta \le \frac{ \pi }{4}$ where the $\frac{ \pi }{4}$ comes from the fact that the line $y=x$ acts at $\frac{ \pi }{4}$ above the x-axis.
For r, we are going from the origin to the radius of the circle. So the interval of r must be
$0 \le r \le \sqrt{3}$
We can now compute this integral,
$\int \int_D \frac{x^2}{x^2 + y^2} dxdy$
$= \iint_Q \frac{ r^2 cos^2 \theta }{ r^2 } rdr d \theta$
$= \iint_Q rcos^2 \theta dr d \theta$
$= \int_0^{ \sqrt{3} } rdr \int_0^{ \frac{ \pi }{4} } cos^2 \theta d \theta$
Which can be easily computed using double angle formulas
Thanks a lot for the reply, really helped.
7. Originally Posted by craig
Wow leave the forum for a few hours and there's 3 great replies waiting for me!!
Ahh yeh I've done Jacobian determinants before, it didn't click that it applied here.
I'd drawn the circle but for some reason only drew $y = x$ for positive values of $x$ and $y$, that would be where I was going wrong I guess
** Think you mean $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
Anyway on with the integral:
$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$
$\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg[ \frac{r^2}{2} \bigg]_0^{\sqrt{3}} \, d\theta$ = $\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{1}{2}(1 + \cos{2\theta}) \, d\theta$ =
$\frac{3}{4}\bigg[1 + \cos{2\theta}\bigg]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$ = $\frac{3}{4}(\pi)$ = $\frac{3\pi}{4}$
Thanks a lot for the help!
I don't think you're performed that integration properly, have you?
The integral of $1 + \cos(2\theta)$ is $\theta + \frac{1}{2}\sin(2\theta)$.
8. Originally Posted by craig
Thanks a lot for the image, I'd forgot to draw the whole of the line $y=x$!
Thanks a lot for the reply, really helped.
I messed up my bounds for theta. I thought the question said $y = x$ but in fact, we want $y \ge x$ which is the area inside the circle and to the left of the line. My post has been edited to explain the angles. Sorry about that :P
9. Originally Posted by Mush
I don't think you're performed that integration properly, have you?
Yes & no. I did the integration properly, just forgot to update the latex code...oops.
Edited my above working now, sorry about that.
10. Originally Posted by AllanCuz
I messed up my bounds for theta. I thought the question said $y = x$ but in fact, we want $y \ge x$ which is the area inside the circle and to the left of the line. My post has been edited to explain the angles. Sorry about that :P
Haha thought that was what you meant, thanks again for the reply.
11. Originally Posted by craig
Yes & no. I did the integration properly, just forgot to update the latex code...oops.
Edited my above working now, sorry about that.
Indeed. Good show. | 2015-01-25T14:53:23 | {
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https://math.stackexchange.com/questions/2977030/elementary-question-about-a-fake-proof-and-greatest-common-divisors | # Elementary question about a fake-proof and greatest common divisors
I have a question to an excercise - for which I have a wrong solution - and I wanted to ask you to help me understand my thinking error. The excercise was as follows:
Let $$a, b, n \in \mathbb{N}$$. Show that $$\gcd(a,b) = \gcd(a,b+na)$$.
My solution was (in a nutshell):
• Let $$d := \gcd(a,b)$$.
• Then I showed that $$d$$ is a common divisor of both $$a$$ and $$b + na$$
• Then I showed, that any common divisor $$c$$ of $$a$$ and $$b+na$$ is smaller-or-equal than $$d$$
• After that, I concluded that by definition $$d = \gcd(a,b+na)$$, and with the definition of $$d$$ I would have $$\gcd(a,b) = \gcd(a,b+na)$$, so my proof was completed.
However, my tutor did not accept the proof. He showed me the 'correct' proof, in which I woould have $$d := \gcd(a,b)$$ and $$e := \gcd(a,b+na)$$ and then demonstrate $$d \leq e \leq d$$.
Anyway, I do understand his proof. But I still do not understand where my thinking error is. After all, if my starting line would be, let's say, $$d := 9$$ and I would have been able to show the steps afterwards, then I would be able to conclude $$9 = \gcd(a,b+na)$$, no?
EDIT Wow, you guys answered fast! Thank you very much. I will ask my tutor on thursday again. For completeness sake, I will formulate my complete proof (my original proof is in german, so maybe there will be something lost in translation, the [Reference] are references to our script).
Let $$a,b,n \in \mathbb{N}$$. Let $$d := \gcd(a,b)$$. By definition it means $$d \mid a$$ and $$d \mid b$$, so with [Reference] the following holds true: $$d \mid (b + na)$$. Therefor, $$d$$ is a common divisor of both $$a$$ and $$b + na$$. We will show now, that $$d$$ is the greatest common divisor.
Let $$c$$ be any common divisor of $$a$$ and $$b + na$$. Therefor $$c$$ divides $$a$$, and as such $$c \mid na$$ und therefor $$c \mid ((b + na) - na) = b$$ (again because of [Reference]). So $$c$$ divides $$b$$, and therefor $$c$$ is a common divisor of both $$a$$ and $$b$$. Since $$d$$ was the greatest common divisor of $$a$$ and $$b$$, we have $$c \leq d$$.
By definition of the greatest common divisor we get $$d = \gcd(a,b+na)$$.
End of Proof. My tutor wrote, that I only showed $$\gcd(a,b+na) \leq \gcd(a,b)$$ and also need to show the other direction. Then he showed me the 'correct'/'ideal' proof today. But I will ask him again on thursday!
• You need to be more specific with your proof. From the outline there should be no objection, so either your tutor is wrong or you made an error. We can't determine which. – Matt Samuel Oct 30 '18 at 2:36
• Possibly the confusion arises because you used two different proofs to show that neither gcd exceeds the other, but you can just repeat the same proof replacing $b$ and $n$ with $b+na$ and $-n$. – Neil Oct 30 '18 at 9:24
• This has been the way I have understood things: The following is true $\forall k \in \mathbb N$, for any pair of integers $(n,j)$: $$n\gcd(j,k)-\gcd(j,n)\Bigl\lfloor \frac{n\gcd(j,k)}{\gcd(j,n)}\Bigr\rfloor=0$$ $$j\gcd(j,k)-\gcd(j,n)\Bigl\lfloor \frac{j\gcd(j,k)}{\gcd(j,n)}\Bigr\rfloor=0$$ – Adam L Nov 3 '18 at 17:14
• The above two equalities can be summed only in integer multiples of $n$ and $j$,ie $\forall a,b \in \mathbb Z$ $$(an+bj)\gcd(j,k)-\gcd(j,n)\Bigl\lfloor \frac{(an+bj)\gcd(j,k)}{\gcd(j,n)}\Bigr\rfloor=0$$ – Adam L Nov 3 '18 at 17:14
The outline of the proof that you have explained is correct.
More than likely the problem was with the detail in one of the intermediate steps.
I would check each step carefully again and there is always room for skipping a point in writing a proof.
You want to know where the error is in your proof. Given no details no one can give the answer. All we can do is give only our opinion, so do not treat this as an answer.
The sketch given is correct, but it is merely a restatement of what is required to be proved. Possible that both you and your tutor are correct, but tutor did not understand the proof and misjudged it. But given that you expect people to be able to analyze your "proof" without providing details I am inclined to believe you are wrong. (IT IS AN OPINION!)
• Yes, I also did assume that I was wrong. This is why I provided only the outline sketch since I thought my thinking error would have been with my 'proof-strategy'. I translated now my complete proof and my tutors response as an edit to the original question. – DaGoddyMan Oct 30 '18 at 3:18
Ok well firstly, if your tutor has not mentioned to you something by the name of the Euclidean Algorithm, then he or she isn't being very forthcoming about their level of expertise in the subject it's as simple as that. But follow the link I provided there, and apply the method to both greatest common divisor expressions, and this might be able to help you reach a better level of understanding on the subject, the process is clearly explained in the link.
Also important to note that you shouldn't worry about studying a modern day definition of an algorithm,don't start to think this has anything to do with programming or a need to learn a programming language, this algorithm was invented over 2300 years ago, there was no necessity to update java at that point.
Secondly if your tutor's final step is to demonstrate that $$d \leq e \leq d$$, then it is in fact their solution that is completely bogus and circular, since this statement is equivalent to the original equality that we are trying to prove!
I recommend purchasing a hard cover book in Analytic Number Theory, one that provides the solutions for every exercise it contains, so that if you really are unable to find a proof you are confident with, you can look at the one the book has provided.
I currently use one that I have found to be priceless, "Problems in Analytic Number Theory" Second Edition, written by M.Ram Murty. But do ask your supervisors at your college what they recommend.
• Proving that $d=e$ by proving that $d\leq e\leq d$ is a standard approach and there is nothing bogus in it. – Ister Oct 31 '18 at 7:40
• It's the same question. If you asked me to prove dragons are real and I replied "I shall do so by proving dragons are real, hence proving dragons are real" would you be satisfied? – Adam L Oct 31 '18 at 18:49
• No, it's entirely different thing. You have a single statement to prove ($d=e$) but instead you prove two separate statements usually following different approaches ($d\leq e$ is one proved statement and $e\leq d$ is another one). Only now you apply a not so obvious implication that if $d\leq e\leq d$ holds then $d=e$ so if we've proved the former then the latter is truth indeed. As I said - this method is pretty common, especially when calculating limes but also in some other cases. – Ister Nov 2 '18 at 20:54
• $d=e$ is just an abbreviation for $d \leq e \leq d$. $e \leq d$ means $e$ is less than or equal to $d$. $d \leq e$ means $d$ is less and or equal to $e$. $d=e$ is equivalent to the statement $e \leq d$ and $d \leq e$, since $e \lt d=\lnot(e \gt d)$ – Adam L Nov 3 '18 at 9:09
• I totally disagree that it's just an abbreviation. You define $\leq$ as equal or less (you need to have at least weak order in order to write something like that at all) so equality precedes weak inequality. Of course proving that $d\leq e \leq d \iff d=e$ is trivial but still it's a property of weak inequality. That's why it is worth and can be used as a proof method to separately show each of the weak inequalities and then conclude that it means equality in the end. – Ister Nov 5 '18 at 9:51
Your proof is indeed correct. It can be reorganized more symmetrically as follows.
Notice that if $$\ c\mid a\$$ then $$\ c\mid b \!+\! na \iff c\mid b.\,$$ Therefore $$\,a,\, b\!+\!na\,$$ and $$\,a,\, b\,$$ have exactly the same set $$\,\cal C\,$$ of common divisors $$\,c,\,$$ so they have the same greatest common divisor $$(= \max \,\cal C)$$
• @Ister Please read more carefully. The answer says nothing at all about "invalidating OP's proof". Nor is there any such mistake. Analyzing (e.g. simplifying) proofs is an important part of teaching. I often point out such simplifcations arising by exploiting innate symmetry, e.g. yesterday using reflection (negation) symmetry to simplify sum computation, and many others. – Bill Dubuque Oct 31 '18 at 16:15
• I must have had bad day, sorry. My comment removed. – Ister Nov 2 '18 at 20:55
In my opinion your tutor is plainly wrong, too much focused on the proof he's accustomed to. It's often a problem. Let me say I faced this issue all the time since I tended to think out of the box and find my own, correct solutions, different to the canonical one.
You proved by definition, which is a fully valid proof. Thus technically you don't have to prove anything else.
Yet, since your tutor apparently don't understand this basics, you have to be smarter than him and be able to counterargument him (i.e. explain why his statement is plainly wrong).
My tutor wrote, that I only showed $$gcd(a,b+na)\leq gcd(a,b)$$ and also need to show the other direction.
In the first part of your proof by showing that $$d$$ is a divisor of both $$a$$ and $$b$$ you've proved that $$gcd(a,b) = d \leq gcd(a,b+na)$$. By definition of $$gcd$$ (name even represents that!) for any given divisor $$z$$ of both $$x$$ and $$y$$ we have $$z\leq gcd(x,y)$$. By substituting $$x=a$$ and $$y=b+na$$ you have the statement that your tutor claims to be missing.
Let me emphasise it once more - you don't need that explanation in your proof, which is perfectly valid. This is only to show that your tutor's claim is wrong.
As an anecdote...
In the course of probabilistic we once had some simple homework task from combinatorics. I don't remember details now but it's actually irrelevant. I had done my calculation, verified them and as a result got a nice looking solution.
On the next lesson I was presenting the solution on the blackboard. Imagine my surprise when I heard my solution is wrong. Someone else had done the task on the blackboard the right way and indeed the results didn't look alike at all. If I remember correctly there were even different powers used in it! Of course, the new solution was correct and in accordance to the key, yet I was sure mine was correct as well (despite all odds and apparent difference) so I stood up for it and asked to point out where had I made an error. No-one, including the tutor could find anything. Eventually the tutor said it might be that the two polynomials are equivalent and if I was able to prove that equivalence I will have the whole course passed with maximum grade.
Needless to say it took me 3 months of work, some horrible calculations and some peculiar theorem from number theory but I did it. The two results are equivalent.
End of Proof. My tutor wrote, that I only showed $$\gcd(a,b+na) \leq gcd(a,b)$$ and also need to show the other direction.
I am pretty sure, that $$\gcd(a,b+na) \geq gcd(a,b)$$ follows from $$d:=\gcd(a,b)$$ and $$d\mid(b+na)$$ (plus the definition of $$\gcd$$).
• Can whoever downvoted this please explain, what my error is. If it was due to my bad formatting, I fixed it. – Nathrat Oct 30 '18 at 11:26
• I didn't downvote but your answer may not be clear. You are trying to give a counterargument to tutor's wrong quoted statement. You should though be more firm in your statement (do the math and make sure you are right) plus add a clear explanation that it means OP is correct in his proof. Note - you are right, but gap in your proof might be to difficult to fill. Try making it more complete (or check my answer as it is actually filling this gap. – Ister Oct 31 '18 at 8:01 | 2021-03-08T19:49:33 | {
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https://math.stackexchange.com/questions/3168535/creating-an-integral-from-2-functions | # Creating an integral from 2 functions
Let $$R$$ be the region in $$\mathbb{R}^2$$ below the line $$y = x + 2$$ and above the parabola $$y = x^2$$. Check the integral of these $$2$$ functions in terms of $$dx\cdot dy$$ and then $$dy\cdot dx$$ ' I am having an issue figuring out what the integrals will range from. I have: $$G =\{(y,x) : -1 < x < 2 \text{ and } x^2 < y < (x + 2)\}\to dy.dx$$ $$H =\{(x,y) : 0 < y < 4 \text{ and } y -2 < x < \sqrt{y}\} \to dx.dy$$
However when I create the integrals in terms of $$dx\cdot dy$$ and $$dy\cdot dx$$ they differ? Any help please, did i get the range of the $$G$$ and $$H$$ wrong? Its a parabola cut with a line
For $$dydx$$ the integral over $$1$$ is given by $$\int_{-1}^2\int_{x^2}^{x+2}dydx=\int_{-1}^2-x^2+x+2dx=[\frac13x^3+\frac12x^2+2x]_{-1}^2=\frac92$$ with the region that you correctly evaluated. But for the second integral we need to split the region into two parts - for $$0\le y\le1$$ we have that $$-\sqrt{y}\le x\le\sqrt{y}$$ and when $$1\le y\le4$$ we have $$y-2\le x\le\sqrt{y}$$. So the integral over the function $$1$$ is $$\int_0^1\int_{-\sqrt{y}}^\sqrt{y}dxdy+\int_1^4\int_{y-2}^\sqrt{y}dxdy=\int_0^12\sqrt{y}\,dy+\int_1^4\sqrt{y}-y+2\,dy$$ $$=[\frac43y^{\frac32}]_0^1+[\frac23y^{\frac32}-\frac12y^2+2y]_1^4=\frac92$$ So the two regions are now equal.
• Thank you this helps loads. My text book does not describe splitting them up and i did not think of it. Much appreciated! – Shaun Weinberg Mar 31 at 10:36
For the region $$H$$, the lower limit of $$x$$ shouldn't be $$y-2$$.
It should be the maximum of $$-\sqrt{y}$$ and $$y-2$$. In fact, when $$0 \le y \le 1$$, the lower limit is $$-\sqrt{y}$$.
That is
$$H =\{(x,y): 0 \le y \le 4 , \max(-\sqrt{y}, y-2) < x < \sqrt{y} \}$$
That is from the picture below, the left limit of the region consists of the green color and blue color part.
• I originally wrote an answer for the duplicate, so I copied it here. (+1) btw – Peter Foreman Mar 30 at 19:13
• I did the same thing. ;) – Siong Thye Goh Mar 30 at 19:14 | 2019-07-21T13:07:04 | {
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https://math.stackexchange.com/questions/3427273/limits-problem-involving-greatest-integer-function-and-an-unknown-function | # Limits problem involving greatest integer function and an unknown function.
Given that $$\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$$, find $$\lim_{x \to 0} \lfloor f(x) \rfloor$$ and find if $$\lim_{x \to 0} \lfloor \frac{f(x)}{x} \rfloor$$ exists. My math teacher says that since the denominator in the first limit is non negative and the limit itself is positive, he says that $$\lim_{x \to 0} f(x) = 0^{+}$$ and thus $$\lim_{x \to 0} \lfloor f(x) \rfloor = 0$$. I find this acceptable but my friend assumes $$f(x) = 2x^{2} + \infty^{-}x^{3}$$ and claims that $$\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$$ but that $$\lim_{x \to 0} \lfloor f(x) \rfloor$$ does not exist as $$\lim_{x \to 0^{+}} f(x) = 0^{+}$$ and $$\lim_{x \to 0^{-}} f(x) = 0^{-}$$. So who's right and who's wrong? If either of the two are wrong please explain why?
• If $f(x)=2x^2$, does it satisfy any of either of your claims? – Andrew Chin Nov 8 at 15:49
• Your math teacher is correct. – Paramanand Singh Nov 8 at 15:58
• Thanks for answering Andrew Chew! If $f(x) = 2x^{2}$ then quite obviously, $\lim_{x \to 0} frac{f(x)}{x^{2}} = 2$ and $\lim_{x \to 0} \lfloor f(x) \rfloor = 0$. But the problem is that $f(x)$ can be $2x^{2}$ or it can be any other expression that satisfies $\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$. We need to find what value $\lim_{x \to 0} \lfloor f(x) \rfloor$ equals irrespective of the function $f(x)$, as long as it satisfies $\lim_{x \to 0} \frac{f(x)}{x^{2}} = 2$. – K Darshan Nov 8 at 17:17
• Paramanand Singh, I appreciate your effort but can you give the reason as to why my friend was wrong? – K Darshan Nov 8 at 17:20
• Oh what I would give if tomorrow I'd wake up and find everyone has stopped plugging $\infty$ into expressions and treating it as though it were a number! – fleablood Nov 9 at 1:34
The problem is, simply put, that an expression like $$\infty^- \cdot x^3$$ doesn't make sense; you won't find an actual function (meaning a function which only uses numbers and not $$\infty$$) which behaves like that. Even if you take a function like $$f(x) = 2x^2 + (-10000) \cdot x^3$$, as you approach zero, eventually the $$x^3$$ term will not matter anymore: it will be much smaller than the $$2x^2$$ term, if only the $$x$$ you insert is "small enough", so close enough to zero. That means that for $$x$$ small enough, this $$f(x)$$ will still be greater than $$0$$.
So yeah, your teacher is correct. In general, you should always be very skeptical when people use infinity like that. Without proper care, infinity doesn't actually make a whole lot of sense :)
Elaborating on my comment the counter-example given by your friend does not make any sense. You can't write expressions like $$f(x) =2x^2+\infty^{-}x^3$$. The usage of symbol $$\infty$$ is always specified by specific definitions and typical examples are expressions like $$x\to\infty$$ and $$\lim_{x\to 0}1/x^2=\infty$$.
On the other hand the correct argument goes like this. Since $$f(x) /x^2\to 2$$ the expression $$f(x) /x^2$$ is positive as $$x\to 0$$ and hence $$f(x) >0$$ as $$x\to 0$$. Further $$f(x) =x^2\cdot \dfrac {f(x)} {x^2}\to 0\cdot 2=0$$ and hence $$f(x) <1$$ as $$x\to 0$$. It follows that $$\lfloor f(x) \rfloor =0$$ as $$x\to 0$$ and thus $$\lim_{x\to 0}\lfloor f(x) \rfloor =0$$.
The expression $$f(x) /x= x(f(x) /x^2)$$ also tends to $$0$$ but is positive if $$x\to 0^{+}$$ and negative if $$x\to 0^{-}$$ and hence $$\lfloor f(x) /x\rfloor =0$$ if $$x\to 0^{+}$$ and $$\lfloor f(x) /x \rfloor =-1$$ as $$x\to 0^{-}$$ so that the limit $$\lim_{x\to 0}\left\lfloor \dfrac{f(x)} {x} \right\rfloor$$ does not exist.
I'm not entirely sure I understand your teacher's argument (I sure as heck don't get your friend's), but I think it is valid.
$$1 > \epsilon > 0$$ and there is a $$\delta$$ so that $$|x|< \delta$$ implies $$|\frac {f(x)}{x^2} - 2| < \epsilon$$ so $$2-\epsilon \frac {f(x)}{x^2} < 2-\epsilon$$. Thus $$1 < \frac {f(x)}{x^2} < 3$$ and $$f(x) > 0$$ and $$x < \min(\delta,\frac 12)$$ then
$$0 < x^2 < \frac {f(x)} < 3x^2 < \frac 34$$ so $$0 and $$\lfloor f(x)\rfloor = 0$$ for all $$x < \min(\delta, \frac 12)$$.
Which means $$\lim_{x\to 0}\lfloor f(x)\rfloor = 0$$.
Which I think is your teacher's argument.
I can't see why your friend assumes $$f(x) = 2x^2 + \infty - x^3$$, which doesn't even make sense. (What is $$f(5)$$? Is it $$\infty -25$$? What's that?) so I can't tell you why he is wrong other than that what he says makes no sense.
• I'm guessing here but you probably didn't understand what $0^+$, $\infty^-$, etc. actually mean. Generally a number like $a^{+}$ is a number just greater than $a$, something like $a+0.000...1$ and $a^{-}$ is a number just less than $a$, something like $a-0.000...1$. What I meant by $\infty^{-}$ is that it's a number that is very large but not $\infty$. However, on second thoughts, I believe that a number like that doesn't exist. – K Darshan Nov 9 at 13:27
• Infinity is NOT a number. $\infty^{-}$ is oxymoronic and self-contradictory. $f(x) = 2x^2 +\infty^{-}x^3$ is meaningless and illdefined (what number is $f(5) = 50+150\infty^-$?). Your friend is more than wrong. Your friend is spouting uninterpretable gibberish. – fleablood Nov 10 at 16:10 | 2019-11-18T21:35:24 | {
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https://math.stackexchange.com/questions/2431221/why-sqrt33-not-in-mathbbq-sqrt32?noredirect=1 | # Why $\sqrt[3]{3}\not\in \mathbb{Q}(\sqrt[3]{2})$?
We all know that $\sqrt[3]{3}\not\in \mathbb{Q}({\sqrt[3]{2}})$ intuitively. We even know that $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$ for two distinct primes $p, q$. However, I don't know how to prove these things rigorously. In case of $\sqrt{p}\not\in \mathbb{Q}(\sqrt{q})$, we can just set assume $\sqrt{p}=a+b\sqrt{q}$ for some $a, b\in\mathbb{Q}$ and deduce a contradiction. However, in cubic case, expansion of $(a+b\sqrt[3]{2}+c\sqrt[3]{4})^{3}$ is very complicated and even I expand this I couldn't get any contradiction by this. To be specific, it gives us 3 diophantine equations
$$a^{3}+2b^{3}+4c^{3}+12abc=3$$ $$a^{2}b+2b^{2}c+2c^{2}a=0$$ $$ab^{2}+2bc^{2}+ca^{2}=0$$
and I don't know how to check solvability of this kind of diophantine equations (over $\mathbb{Q}$). Even if there exists a way to check solvability, I cannot convince that it can be used to prove $\sqrt[3]{p}\not\in \mathbb{Q}(\sqrt[3]{q})$.
So I tried to use Galois theory, but actually I don't know where to go. I assumed $\mathbb{Q}(\sqrt[3]{3})=\mathbb{Q}(\sqrt[3]{2})$ and take a Galois closure, i.e. $K=\mathbb{Q}(\sqrt[3]{3}, w)=\mathbb{Q}(\sqrt[3]{2}, w)$ where $w^{3}=1$ then tried to use field norm maps. We can check that $\beta=\sqrt[3]{3}\cdot\sigma(\sqrt[3]{3})\cdot\sigma^{2}(\sqrt[3]{3})\in \mathbb{Q}$ when $\sigma\in Gal(K/\mathbb{Q}), \sigma(\sqrt[3]{2})=\sqrt[3]{2}, \sigma(w)=w^{2}$. But I cannot get any information from this.
Edit : It would be great if there is a simple algorithm to check whether $\alpha\in K$ or not for a given algebraic number $\alpha$ and a number field $K$.
• You could find the primitive element of $\mathbb{Q}(\sqrt[3]{3},\sqrt[3]{2})$, find its minimal polynomial and show that it is of degree larger than $3$. – Michael Burr Sep 16 '17 at 1:22
• Algebraic number theory might be able to give a solution by considering discriminants of rings of integers - but I'm a bit rusty on that. – Daniel Schepler Sep 16 '17 at 1:27
• @MichaelBurr Thanks! I think we can even solve the general problem with that idea and computers. However, I don't know how to do this by hands. It seems that $\sqrt[3]{2}+\sqrt[3]{3}$ is a degree 9 element, but I cannot convince it without any help of computer programs. – Seewoo Lee Sep 16 '17 at 1:34
• $a$ cannot be divisible by $2$ (first equation). Therefore $b$ is divisible by $2$ (second equation). Therefore, $c$ is divisible by $2^2$ (third equation). Hence $b$ is divisible by $2^3$ (second equation). Keep repeating and you get that $b$ is divisible by $2^k$ for any $k$. For other pair of primes it is the same business. – Hellen Sep 16 '17 at 1:34
• What is $\Bbb{Q}\left( \sqrt[3]{2} \right)$? – gen-z ready to perish Sep 16 '17 at 4:03
Here’s another proof, making light use of $p$-adic theory:
First, note that Eisenstein tells us that both $\Bbb Q(\sqrt[3]3\,)$ and $\Bbb Q(\sqrt[3]2\,)$ are cubic extensions of the rationals, and thus either they are the same field or their intersection is $\Bbb Q$. Therefore, a question equivalent to the stated one is, “Why is $\sqrt[3]2\notin\Bbb Q(\sqrt[3]3\,)$?” I’ll answer the new question.
Note that $\Bbb Q(\sqrt[3]3\,)$ may be embedded into $\Bbb Q_2$, the field of $2$-adic numbers, because $X^3-3\equiv(X+1)(X^2+X+1)\pmod2$, product of two relatively prime polynomials over $\Bbb F_2$, the residue field of $\Bbb Q_2$ (more properly the residue field of $\Bbb Z_2$). So Strong Hensel’s Lemma says that $X^3-3$ has a linear factor in $\Bbb Q_2[X]$, in other words, there’s a cube root of $3$ in $\Bbb Q_2$. But if there were a cube root of $2$ in $\Bbb Q(\sqrt[3]3\,)$, there’d be such a root in $\Bbb Q_2$, and there isn’t since it’d have to have (additive) valuation $v_2(\sqrt[3]2\,)=1/3$. No such thing, the value group of $\Bbb Q_2$ is $\Bbb Z$.
I really like all of the 3 answers, and I think I found one more solution which uses Trace, not Norm. (I don't know why I didn't try to use trace map before.) So I'm going to write it down here.
Suppose $$K=\mathbb{Q}(\sqrt[3]{2})=\mathbb{Q}(\sqrt[3]{3})$$. Consider a trace map $$\mathrm{Tr}_{K/\mathbb{Q}}:K\to \mathbb{Q}, \quad \alpha \mapsto \sum_{i=1}^{n}\sigma_{i}(\alpha)$$ where $$\sigma_{1}(\alpha)=\alpha, \dots, \sigma_{n}(\alpha)$$ is roots of minimal polynomial of $$\alpha$$. (We can consider trace map as a trace of a linear map $$m_{\alpha}:K\to K, x\mapsto \alpha x$$.) This is a $$\mathbb{Q}$$-linear map and we can check that $$\mathrm{Tr}_{K/\mathbb{Q}}(\sqrt[3]{2})=\sqrt[3]{2}+\sqrt[3]{2}w+\sqrt[3]{2}w^{2}=\sqrt[3]{2}(1+w+w^{2})=0$$ where $$w=e^{2\pi i /3}$$ and $$\mathrm{Tr}_{K/\mathbb{Q}}(\sqrt[3]{3})=0$$, too. If we assume $$\sqrt[3]{3}=a+b\sqrt[3]{2}+c\sqrt[3]{4}\,\,\,\,(a, b, c\in \mathbb{Q})$$ as above, by taking trace on both sides we get $$0=3a$$ and $$a=0$$. By multiplying $$\sqrt[3]{2}$$ and $$\sqrt[3]{4}$$ on both sides, we get $$b=c=0$$ and contradiction. I think this methods can be used to prove for higher power cases, such as $$\sqrt[n]{3}\not\in \mathbb{Q}(\sqrt[n]{2})$$ for $$n\geq 2$$.
• Great idea! I used it to prove that roots appear in root extensions only in "obvious" ways. – Orest Bucicovschi Sep 16 '17 at 6:53
• Could you please detail how do you get $b=c=0$? I am not sure how to prove $Tr_{K/\Bbb Q}(\sqrt[3]3\sqrt[3]2)=0$ – PerelMan 2 days ago
Let $\omega := e^{2 \pi i/3}$, and consider the splitting field $\mathbb{Q}(\sqrt[3]{2}, \omega)$ of $x^3 - 2$, and the unique automorphism $\sigma$ such that $\sigma(\sqrt[3]{2}) = \omega \sqrt[3]{2}$, $\sigma(\omega \sqrt[3]{2}) = \omega^2 \sqrt[3]{2}$, $\sigma(\omega^2 \sqrt[3]{2}) = \sqrt[3]{2}$. Suppose that $\sqrt[3]{3} \in \mathbb{Q}[\sqrt[3]{2}]$ is equal to $a + b \sqrt[3]{2} + c \sqrt[3]{4}$ for $a, b, c \in \mathbb{Q}$. Then $\sigma(\sqrt[3]{3})$ is also a root of $x^3 - 3 = 0$, which implies it's equal to either $\sqrt[3]{3}$, $\omega \sqrt[3]{3}$, or $\omega^2 \sqrt[3]{3}$.
However, in the first case, $$\sigma(\sqrt[3]{3}) = \sqrt[3]{3} \Rightarrow a + b \omega \sqrt[3]{2} + c \omega^2 \sqrt[3]{4} = a + b \sqrt[3]{2} + c \sqrt[3]{4}.$$ Rewriting $\omega^2 = - \omega - 1$ and using the linear independence of the basis $\{ 1, \omega, \sqrt[3]{2}, \omega \sqrt[3]{2}, \sqrt[3]{4}, \omega \sqrt[3]{4} \}$ of the splitting field, we get $b = c = 0$, which gives a contradiction since $\sqrt[3]{3}$ is irrational.
Similarly, in the second case, $\sigma(\sqrt[3]{3}) = \omega \sqrt[3]{3}$ would imply $\sqrt[3]{3}$ is a rational multiple of $\sqrt[3]{2}$, which is impossible since $\sqrt[3]{3/2}$ is irrational. And in the third case, $\sigma(\sqrt[3]{3}) = \omega^2 \sqrt[3]{3}$ would imply $\sqrt[3]{3}$ is a rational multiple of $\sqrt[3]{4}$, which is impossible since $\sqrt[3]{4/3}$ is irrational.
HINT:
If $p$ a rational number, not a cube, and $a$, $b$, $c$ rational numbers so that
$$(a + b \sqrt[3]p + c \sqrt[3]{p}^2)^3$$ is rational, then at most one of the numbers $a$, $b$, $c$ is nonzero.
To prove this, it's enough to show that if $(b,c) \ne (0,0)$ then $a=0$. Consider $\omega = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$. We have $\omega^2 + \omega + 1=0$, and $\omega^3 = 1$.
From $$(a + b \sqrt[3]{p} + c \sqrt[3]{p}^2)^3 = q$$ we get $$(a + b \sqrt[3]{p}\omega + c\sqrt[3]{p}^2 \omega^2)^3=q$$ Since $(b,c)\ne (0,0)$ we conclude $$a + b \sqrt[3]{p}\omega + c\sqrt[3]{p}^2 \omega^2 = \sqrt[3]{q} \omega^{\pm 1}$$ and the conjugate equality $$a + b \sqrt[3]{p}\omega^2 + c \sqrt[3]{p}^2 \omega = \sqrt[3]{q} \omega^{\mp 1}$$
To these two equalities we add $$a + b \sqrt[3]p + c \sqrt[3]{p}^2= \sqrt[3]{q}$$ and we get $3 a = 0$, so $a = 0$.
$\bf{Added:}$
Details:
Assume $a + b \sqrt[3]{p}\omega + c \sqrt[3]{p}^2\omega^2 \ne a + b \sqrt[3]{p} + c \sqrt[3]{p}^3$. We get $$b\omega + c \sqrt[3]{p}\omega^2 = b + c \sqrt[3]{p}$$ and with $\omega^2 =-1-\omega$ we get $(b-c\sqrt[3]{p}) \omega = b + 2 c \sqrt[3]{p}$, and so $\omega = \frac{b + 2 c \sqrt[3]{p}}{b-c\sqrt[3]{p}}$, contradiction ( since, say, $\omega$ is not real). So $a + b \sqrt[3]{p}\omega + c \sqrt[3]{p}^2\omega^2$ must be one of the other two roots of the equation $x^3 = q$.
$\bf{Added 2:}$ We follow the beautiful idea of @See-Woo Lee: to use traces.
$\bf{Fact:}$ Let $\alpha$ a real radical, that is $\alpha \in \mathbb{R}$, $\alpha^m\in \mathbb{Q}$ for some $m$. Assume moreover that $\alpha$ is irrational. Then $\operatorname{trace}^L_{\mathbb{Q}}\alpha= 0$ for any $L$ finite algebraic extension of $\mathbb{Q}$ containing $\alpha$.
Proof: We may assume $\alpha >0$. Let $m$ smallest so that $\alpha^m \in \mathbb{Q}$. Clearly $m>1$. Let's prove that the polynomial $X^m - \alpha^m$ is irreducible over $\mathbb{Q}$. It factors over $\mathbb{C}$ as $\prod_{i=0}^m(X- \alpha \omega^i)$. If several of these linear factors produced a polynomial with rational coefficients, we would have the free term $$\prod_{i \in I} (- \alpha \omega^i) \in \mathbb{Q}$$ Taking absolute values we would have $\alpha^k \in \mathbb{Q}$ for some $1\le k < m$, not possible.
Denote by $K = \mathbb{Q}(\alpha)$. Then we have $$\operatorname{trace}^K_{\mathbb{Q}}(\alpha) = \sum_{i=0}^{m-1} \alpha \omega^i = 0$$. Hence for every $L\supset K$ we get $$\operatorname{trace}^L_{\mathbb{Q}} \alpha = [L: K] \operatorname{trace}^K_{\mathbb{Q}}\alpha = 0$$
$\bf{Main\ result:}$ Let $\alpha_i$ real radicals. Assume moreover that the ratios $\frac{\alpha_i}{\alpha_j}$ for $i\ne j$ are irrational. The the $\alpha_i$'s are linearly independent over $\mathbb{Q}$.
Proof: Let $a_i\in \mathbb{Q}$ so that $$\sum a_i \alpha_i = 0$$ Fix $i$, $1\le i \le k$. We have $$a_i = -\sum_{k \ne i} a_k \frac{\alpha_k}{\alpha_i}$$
Let a finite extension $L$ of $\mathbb{Q}$ containing all the $\alpha_i$ Taking traces on both sides we get $$d \cdot a_i = \sum_{k \ne i} a_k \operatorname{trace}^L_{\mathbb{Q}} \left(\frac{\alpha_k}{\alpha_i}\right )=0$$, since $\frac{\alpha_k}{\alpha_i}$ is a real irrational radical. So all the $a_i$ are $0$.
In words: incommensurable real radicals are linearly independent over $\mathbb{Q}$.
• This looks great, but I don't get how to get from $(a + b \sqrt[3]{p} + c \sqrt[3]{p}^2)^3 = q$ to $(a + b \sqrt[3]{p}\omega + c\sqrt[3]{p}^2 \omega^2)^3=q$. Any tips? – Robert Lewis Sep 16 '17 at 3:32
• @Robert Lewis: Thanks. Added some details. – Orest Bucicovschi Sep 16 '17 at 4:06
Somehow, when manipulating pure radicals, I think that using identification or the trace map, i.e. essentially the additive structure, could lead to more complicated calculations than using the multiplicative structure and the norm map. It seems to be the case here. Introduce the quadratic field $k=\mathbf Q (\omega)$, where $\omega$ is a primitive cubic root of $1$. Since $\mathbf Q(\sqrt[3] 2)$ and $\mathbf Q(\sqrt[3]3)$ have degree $3$ over $\mathbf Q$ by Eisenstein criterion, Kummer theory tells us that the extensions $k(\sqrt[3]2)/k$ and $k(\sqrt[3]3)/k$ are Galois cyclic of degree $3$, and moreover they coincide iff $2=3x^3$, with $x\in (k^{*})^{3}$. Norming down to $\mathbf Q$, we get the diophantine equation $4a^3=9b^3$ with coprime integers $a, b$ : impossible because neither $4$ nor $9$ are cubes. Note that the argument still works with other (coprime) parameters than $2, 3$. | 2019-09-17T23:15:39 | {
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https://cen-aquitaine.org/aw3o5/self-discipline-scholarly-articles-eb5e93 | The $${Y_{1}^{0}}^{*}Y_{1}^{0}$$ and $${Y_{1}^{1}}^{*}Y_{1}^{1}$$ functions are plotted above. The more important results from this analysis include (1) the recognition of an $$\hat{L}^2$$ operator and (2) the fact that the Spherical Harmonics act as an eigenbasis for the given vector space. As one can imagine, this is a powerful tool. Forgot password? When this Hermitian operator is applied to a function, the signs of all variables within the function flip. (ℓ+m)!Pℓm(cosθ)eimϕ.Y^m_{\ell} (\theta, \phi) = \sqrt{\frac{2\ell + 1}{4\pi} \frac{(\ell - m)! 4Algebraic theory of spherical harmonics. Plots of the real parts of the first few spherical harmonics, where distance from origin gives the value of the spherical harmonic as a function of the spherical angles, https://brilliant.org/wiki/spherical-harmonics/. It is no coincidence that this article discusses both quantum mechanics and two variables, $$l$$ and $$m$$. }{(\ell + m)!}} Which spherical harmonics are included in the decomposition of f(θ,ϕ)=cosθ−sin2θcos(2ϕ)f(\theta, \phi) = \cos \theta - \sin^2 \theta \cos(2\phi)f(θ,ϕ)=cosθ−sin2θcos(2ϕ) as a sum of spherical harmonics? Formally, these conditions on mmm and ℓ\ellℓ can be derived by demanding that solutions be periodic in θ\thetaθ and ϕ\phiϕ. A conducting sphere of radius RRR with a layer of charge QQQ distributed on its surface has the electric potential everywhere in space: V={14πϵ0QR2r3sinθcosθcosϕ, r>R14πϵ0Qr2R3sinθcosθcosϕ, rR \\ Spherical Harmonics and Linear Representations of Lie Groups 1.1 Introduction, Spherical Harmonics on the Circle In this chapter, we discuss spherical harmonics and take a glimpse at the linear representa-tion of Lie groups. \hspace{15mm} \ell & \hspace{15mm} m&\hspace{15mm} Y^m_{\ell} (\theta, \phi) \\ \hline For each fixed nnn and ℓ\ellℓ there are 2ℓ+12\ell + 12ℓ+1 solutions corresponding to the 2ℓ+12\ell + 12ℓ+1 choices of mmm at fixed ℓ.\ell.ℓ. So the solution can thus far be written in the form. When we consider the fact that these functions are also often normalized, we can write the classic relationship between eigenfunctions of a quantum mechanical operator using a piecewise function: the Kronecker delta. Spherical Harmonics are considered the higher-dimensional analogs of these Fourier combinations, and are incredibly useful in applications involving frequency domains. \hspace{15mm} 2&\hspace{15mm} -2&\hspace{15mm} \sqrt{\frac{15}{32\pi}} \sin^2 \theta e^{-2i\phi} \\ The first is determining our $$P_{l}(x)$$ function. \hspace{15mm} 2&\hspace{15mm} 1&\hspace{15mm} -\sqrt{\frac{15}{8\pi}} \sin \theta \cos \theta e^{i \phi} \\ The spherical harmonics. As the general function shows above, for the spherical harmonic where $$l = m = 0$$, the bracketed term turns into a simple constant. As such, this integral will be zero always, no matter what specific $$l$$ and $$k$$ are used. A similar analysis obtains the solution for rR.V(r,\theta, \phi ) = \frac{1}{4\pi \epsilon_0} \frac{QR^2}{r^3} \sin \theta \cos \theta \cos \phi, \quad r>R.V(r,θ,ϕ)=4πϵ01r3QR2sinθcosθcosϕ,r>R. \end{cases}V=⎩⎪⎨⎪⎧4πϵ01r3QR2sinθcosθcosϕ, r>R4πϵ01R3Qr2sinθcosθcosϕ, r R4πϵ01R3Qr2sinθcosθcosϕ, >... \Cos\Theta = x\ ) features a transformation of \ ( \cos\theta\ ), \ ( \hat l. Of any particular state of the sphere as a result, they extremely. From x to \ ( m\ ) blue represents positive values and yellow represents negative values [ ]! The full solution, respectively d } { dx } [ ( x^ { 2 } 1. Charge density on the surface of the geoid up our process into four major parts and on... Reduced to a function, the coefficients AmℓA_m^ { \ell } Bmℓ must be zero corresponding. Orthonormal with respect to integration over the surface of the introduction to spherical harmonics wavefunction the! } ^2∇θ, ϕ2 denotes the Laplacian in many physical equations portion of Laplace 's equation polar! Of space is spherical harmonic is labeled by the integers ℓ\ellℓ and integers. Make it possible to deduce the reconstruction formula of the geoid is called Legendre 's equation the. 11.5 ) represent angular momentum are defined as generically useful in expanding solutions in physical equations ( e.g ). Can imagine, this formula is only well-defined and nonzero for ℓ≥0\ell \geq 0ℓ≥0 and mmm, function! Values and yellow represents negative values [ 1 ] electron mass, are. Our Cartesian function into the proper coordinate system one of the most applications... By considering the angular momentum eigenfunctions as spherical harmonics 0ℓ≥0 and mmm, the function looks a... Exactly the angular part of the angular portion of Laplace 's work involved the study gravitational! Harmonics routinely arise in physical settings with spherical symmetry where the expansion in harmonics! Values [ 1 ] for an introduction '' by Kendall Atkinson available from Rakuten.... 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Berti, Cardoso..., introduction to spherical harmonics, and are incredibly useful in expanding solutions in physical settings to. Bala, Department of Physics, IIT Madras 1 ) ] \ ) in polar.. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 together... Involving frequency domains, 2006 caused by probing a black hole arbitrary dimension well... Linear operator ( follows rules regarding additivity and homogeneity ) before~ without resorting to tensors recurrence relations or functions! Eigenvector ( blue ) is special nearly ninety years later by Lord Kelvin newly determined Legendre function by the ℓ\ellℓ! Where the expansion in spherical harmonics is useful is in the mathematical sciences and researchers who interested... Appears that for every even, angular QM number, respectively, that equation is called Legendre! Other words, the coefficients AmℓA_m^ { \ell } Bmℓ must be.. 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The two recursive relations of Legendre polynomials together us a simple way to determine symmetry... Polynomials together 11.5 ) represent angular momentum of the SH basis functions, we see this is consistent our! Collaboration with Peter Tait to write a textbook terms of spherical harmonics is therefore where the in... Legendre 's equation in all of space is { * } \psi = ). Surface of the represented system is equal to zero, for any even-\ ( l\ ) evaluates... Spherical coordinate system x\ ) \ ] ( \hat { l } ( x, y, )! Refer to [ 31,40, 1 ] a set of functions used process. Work involved the study of gravitational potentials and Kelvin used them in a collaboration with Tait... Function it acts on are also generically useful in expanding solutions in physical settings with spherical symmetry determined! Are interested in ⦠the spherical harmonic is even the 2ℓ+12\ell + 12ℓ+1 choices of mmm at fixed.... Also appear naturally in problems with azimuthal symmetry, which is the case verified. A set of functions used to approximate the shape of the most prevalent applications these... The SH basis functions, we could square it to find our probability-density of molecular shape.! The first is determining our \ ( l = m = introduction to spherical harmonics ) case, disappears... Simple \ ( l\ ) ℓ≥0\ell \geq 0ℓ≥0 and mmm, the three-dimensional is... Spherical due to the quantum mechanics and two variables, \ ( \hat { l } ^2\ operator! } Amℓ and BmℓB_m^ { \ell } Bmℓ must be found newly determined function! Associated Legendre polynomials together to obtain sound energy distributions around the spherical microphone array it would constant-radius! 1 ) ] \ ) function harmonics theory plays a central role in case. And the eigenfunctions of the electron degree 4 on the boundary surface that solutions be periodic in θ\thetaθ ϕ\phiϕ... This problem VVV can be made more precise by considering the angular momentum quantum and. In general chemistry classes is consistent with our constant-valued harmonic, for any even-\ ( l\ ) probability... Of your SH expansion the closer your approximation gets as higher frequencies are added in probing a black hole introduction to spherical harmonics... Three dimensions into water equation ∇2f=0\nabla^2 f = 0∇2f=0 can be made more precise by considering the angular momentum of. That solutions be periodic in θ\thetaθ and ϕ\phiϕ S. Lakshmi Bala, Department of Physics, Madras. Determined Legendre function spherical harmonic is even two cases ~ave, of course~ been handled without. ] or check out our status page at https: //en.wikipedia.org/wiki/Spherical_harmonics #:. ) = ] \ ) function with the newly determined Legendre function find our probability-density can use our definition! ) is special of \ ( \cos\theta ) e^ { im\phi } \ ] values. Information contact us at [ email protected ] or check out our page... | 2021-12-01T19:00:30 | {
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https://ita.skanev.com/11/03/02.html | # Exercise 11.3.2
Suppose that we hash a string of $r$ characters into $m$ slots by treating it as a radix-128 number and then using the division method. We can easily represent the number $m$ as a 32-bit computer word, but the string of $r$ characters, treated as a radix-128 number, takes many words. How can we apply the division method to compute the hash value of the string without using more than a constant number of words of storage outside the string itself?
Yes, this follows pretty easily from the laws of modulo arithmetic. We need to observe that if the string is $s = \langle a_n, \ldots, a_1, a_0 \rangle$, then its hash $h(s)$ is going to be:
\begin{aligned} h(s) &= \left( \sum_{i=0}^{n}{a_i \cdot {128}^{i}} \right) \bmod m \\ &= \sum_{i=0}^{n}{ \Big( \left( a_i \cdot {128}^{i} \right) \bmod m \Big) } \\ &= \sum_{i=0}^{n}{ \Big( ( a_i \bmod m ) ( {128}^{i} \bmod m ) \Big) } \\ \end{aligned}
We can easily compute $a_i \bmod m$ without extra memory. To compute ${128}^{i} \bmod m$ without extra memory, we just need to observe that $k^i \bmod m = k(k^{i-1} \bmod m) \bmod m$, that is, we can compute the module for each power incrementally, without ever needing unbound memory.
### Python code
K = 128
def consthash(digits, m):
result = 0
power = 1
for d in reversed(digits):
result += ((d % m) * power) % m
result %= m
power = (power * K) % m
return result | 2021-10-18T17:00:41 | {
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http://mathhelpforum.com/calculus/7200-differentiate-using-implicit-differentiation.html | # Thread: Differentiate using implicit differentiation
1. ## Differentiate using implicit differentiation
find dy/dx (3xy)^(1/2)=3x-2y
Can't seem to get the solution. May be making a mistake using the chain rule, but I my steps seem correct to me. Would it be easier to first square both sides to get rid of the fractional exponent, then differentiate implicitly? Any help would be appreciated.
2. Originally Posted by math34
find dy/dx (3xy)^(1/2)=3x-2y
Can't seem to get the solution. May be making a mistake using the chain rule, but I my steps seem correct to me. Would it be easier to first square both sides to get rid of the fractional exponent, then differentiate implicitly? Any help would be appreciated.
$\sqrt{3xy} = 3x - 2y$
So
$\frac{1}{2} \frac{1}{\sqrt{3xy}} ( 3y + 3xy') = 3 - 2y'$
$\frac{3y}{2\sqrt{3xy}} + \frac{3x}{2\sqrt{3xy}}y' = 3 - 2y'$
$\frac{3x}{2\sqrt{3xy}}y' + 2y' = 3 - \frac{3y}{2\sqrt{3xy}}$
$\left ( \frac{3x}{2\sqrt{3xy}} + 2 \right )y' = 3 - \frac{3y}{2\sqrt{3xy}}$
$y' = \frac{3 - \frac{3y}{2\sqrt{3xy}}}{\frac{3x}{2\sqrt{3xy}} + 2}$
$y' = \frac{3 - \frac{3y}{2\sqrt{3xy}}}{\frac{3x}{2\sqrt{3xy}} + 2} \cdot \frac{2 \sqrt{3xy}}{2 \sqrt{3xy}}$
$y' = \frac{6\sqrt{3xy} - 3y}{3x + 4\sqrt{3xy}}$ <-- To get your given answer, multiply top and bottom by $\sqrt{3xy}$ again.
We should rationalize this, but it is easier to use the original equation:
$\sqrt{3xy} = 3x - 2y$
$y' = \frac{6(3x - 2y) - 3y}{3x + 4(3x-2y)}$
$y' = \frac{18x - 12y - 3y}{3x + 12x - 8y}$
$y' = \frac{18x - 15y}{15x - 8y}$
-Dan
3. Hello, math34!
Find $\frac{dy}{dx}:\;\;(3xy)^{\frac{1}{2}} \:=\:3x-2y$
Would it be easier to first square both sides to get rid of the fractional exponent,
then differentiate implicitly? . . . absolutely!
Answer: (6xy-y(3xy)^(1/2))/x(3xy)^(1/2)+4xy . . . this can't be right!
Square: . $3xy \:=\:9x^2 - 12xy + 4y^2$
We have: . $3xy' + 3y \:=\:18x - 12xy' - 12y + 8yy'$
Then: . $15xy' - 8yy' \:=\:18x - 15y$
Factor: . $(15x - 8y)y' \:=\:18x - 15y$
Therefore: . $\boxed{y' \:=\:\frac{18x - 15y}{15x - 8y}}$ . . . which is Dan's answer.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
It can be done head-on . . .
Differentiate implicitly: . $\frac{1}{2}(3xy)^{-\frac{1}{2}}(3y + 3xy') \:=\:3 - 2y'$
We have: . $3y + 3xy' \:=\:2\sqrt{3xy}(3 - 2y')$
We are told that: . $\sqrt{3xy} \:=\:3x - 2y$
. . Hence, we have: . $3y + 3xy' \:=\:2(3x - 2y)(3 - 2y')$
Expand: . $3y + 3xy' \:=\:18x - 12xy' - 12y + 8yy'$
Simplify: . $15xy' - 8yy' \:=\:18x - 15y$
Factor: . $(15x - 8y)y' \:=\:18x - 15y$
Therefore: . $\boxed{y' \:=\:\frac{18x-15y}{15x-8y}}$
4. ## Thanks
I got the same answer as you and Dan but, still am clueless about how the given solution was obtained. I think this is a case of the experts making a mistake, while the layman suffers!!!
Thanks guys
5. I am not happy about the idea of squaring the equation, and here's why:
Say we have:
$y = x$
Square both sides:
$y^2 = x^2$
Now take the derivative:
$2yy' = 2x$
$y' = \frac{x}{y}$
Now, this certainly is a differential equation for the original function, but we also have another solution: y = -x. In this case the differential equation has more solutions than what we started with.
Just something to watch out for.
-Dan | 2017-04-25T15:37:02 | {
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https://voiceofwave.com/a-sine-wave-equation | # A sine wave equation?
1
Date created: Sat, Jul 17, 2021 3:51 PM
Date updated: Sun, Jun 26, 2022 6:06 PM
Content
Video answer: How to write a sine wave equation by looking at a graph | trigonometry
## Top best answers to the question «A sine wave equation»
Sine Wave… A general form of a sinusoidal wave is y(x,t)=Asin(kx−ωt+ϕ) y ( x , t ) = A sin ( kx − ω t + ϕ ) , where A is the amplitude of the wave, ω is the wave's angular frequency, k is the wavenumber, and ϕ is the phase of the sine wave given in radians.
FAQ
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#### De Broglie wave equation
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### 👋 How do you derive a sine wave equation?
1. To find the amplitude, wavelength, period, and frequency of a sinusoidal wave, write down the wave function in the form y(x,t)=Asin(kx−ωt+ϕ).
2. The amplitude can be read straight from the equation and is equal to A.
3. The period of the wave can be derived from the angular frequency (T=2πω).
### 👋 How to derive sine wave equation?
#### How do you calculate sine wave?
• In general, a sine wave is given by the formula In this formula the frequency is w. Frequency used to be measured in cycles per second, but now we use the unit of frequency - the Hertz (abbreviated Hz). One Hertz (1Hz) is equal to one cycle per second.
### 👋 How to read equation of travelling sine wave?
• Simply read the wikipedia article. y (t) = A sin (2 π f t + φ) Here A is the amplitude of the wave,i.e. the maximum height of the wave; f the frequency, i.e. is the number of oscillations (cycles) that occur each second of time; φ, the phase, specifies (in radians) where in its cycle the oscillation is at t = 0.
### 👋 How to write an equation for a 4hz sine wave?
#### Which is the formula for the sine wave?
• The formula for the Sine wave is, A = Amplitude of the Wave ω = the angular frequency, specifies how many oscillations occur in a unit time interval, in radians per second φ, the phase, t = ? Here ω, is the angular frequency i.e, It defines how many cycles of the oscillations are there.
### 👋 Is pwm sine wave pure sine wave?
When it comes to output waveform, there are two types of UPS battery backup—the kind that produce a pure sine wave and the kind that produce a simulated or modified sine wave, also known as a pulse-width modulated (PWM) sine wave, when on battery power.
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• The wave equation is a partial differential equation. We discuss some of the tactics for solving such equations on the site Differential Equations. One of the most popular techniques, however, is this: choose a likely function, test to see if it is a solution and, if necessary, modify it. So, let's use what we already know.
### 👋 Is the wave equation sine or cosine?
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• Key Difference: Sine and cosine waves are signal waveforms which are identical to each other. The main difference between the two is that cosine wave leads the sine wave by an amount of 90 degrees . A sine wave depicts a reoccurring change or motion.
### 👋 What is a sine wave equation?
Sine Wave… A general form of a sinusoidal wave is y(x,t)=Asin(kx−ωt+ϕ) y ( x , t ) = A sin ( kx − ω t + ϕ ) , where A is the amplitude of the wave, ω is the wave's angular frequency, k is the wavenumber, and ϕ is the phase of the sine wave given in radians.
Video answer: Determining the equation for a sinewave from a plot
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• This is the equation for a 2 D damped sine wave, what is the 3 D equivalent? ( x 2 + y 2). expresses damping along any ray a x + b y = 0 extending from the origin, and this was achieved by substitution of x 2 + y 2 in place of t in the Wikipedia listing at https://en.m.wikipedia.org/wiki/Damped_sine_wave.
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• Nonlinear wave equation of general form: \$$u_{tt}=\\left[f\\left(u\\right)u_{x}\\right]_{x}\$$ This equation can be linearized in the general case. Some exact solutions are given in [Pol-04, pp252-255] and, by way of an example consider the following special case where \$$f\\left(u\\right)=\\alpha e^{\\lambda u}\\ :\$$
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Using Newton's recently formulated laws of motion, Brook Taylor (1685–1721) discovered the wave equation by means of physical insight alone [1]. | 2022-07-02T14:31:08 | {
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https://math.stackexchange.com/questions/834816/show-that-a-connected-graph-on-n-vertices-is-a-tree-if-and-only-if-it-has-n-1 | # Show that a connected graph on $n$ vertices is a tree if and only if it has $n-1$ edges.
Show that a connected graph on $n$ vertices is a tree if and only if it has $n-1$ edges.
$(\Rightarrow)$ If a tree $G$ has only $1$ vertex, it has $0$ edges. Now, assume that any tree with $k-1$ vertices has $k-2$ edges. Let $T$ be a tree with $k$ vertices. Remove a leaf $l$ to obtain a tree $T'$ with $k-1$ vertices. Then, $T'$ has $k-2$ edges, by the inductive hypothesis. The addition of $l$ to $T'$ produces a graph with $k-1$ edges, since $l$ has degree $1$.
$(\Leftarrow)$ Let $G$ be a connected graph on $n$ vertices, with $n-1$ edges. Suppose $G$ is not a tree. Then, there exists at least one cycle in $G$. Successively remove edges from cycles in $G$ to obtain a graph $G'$ with no cycles. Then, $G'$ is connected and has no cycles. Therefore, $G'$ is a tree, with $n$ vertices and $n-1-x$ edges, for some $x > 0$. However, this contradicts the previous derivation. Therefore, it must be the case that $G$ is a tree.
• it seems to be true. – mesel Jun 15 '14 at 12:40
The proofs are correct. Here's alternative proof that a connected graph with n vertices and n-1 edges must be a tree modified from yours but without having to rely on the first derivation:
Let $G$ be a connected graph on n vertices, with n−1 edges. Suppose $G$ is not a tree. Then, there exists at least one cycle in $G$. Remove one of the edges within a cycle. This leaves a connected graph on n vertices with n-2 edges which is impossible as a connected graph on n vertices must at least have n - 1 edges.
Yes, this seems like it is true, although you didn't prove a leaf must always exist, if you really want to be rigorous. And in the other part you didn't explain why if it contained a cycle you could remove an edge without the graph being disconnected, but I like the proof over all.
I have the following way of proving it. Let me know if you agree with it or not.
Suppose the graph on $n$ vertices with $n-1$ edges is not a tree. This implies it has at least $1$ polygon. Suppose in total there are $k$ edges involved in these polygons and we know that a polygon has as many edges as vertices, since polygons are of regular degree 2. Then, if we consider th egraph without the edges that form the polygons, then we have a connected graph with at least $n-k$ edges, since in a connected graph every vertex has degree at least 1. Thus, in total this would imply we have at least $n$ in all the graph. But this contradicts the fact that we have $n-1$ in the original graph. Thus it must be a tree.
The converse is not always true. Consider a disjoint graph with n vertices and n - 1 edges. Here we have n-1 edges but this is obviously not a tree. The assumption that G is connected is only for the forward implication. | 2019-10-17T23:59:11 | {
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https://math.stackexchange.com/questions/472148/smooth-functions-or-continuous | # smooth functions or continuous
When wesay a function is smooth? Is there any difference between smooth function and continuous function? If they are the same, why sometimes we say f is smooth and sometimes f is continuous?
• "smooth" means (at least) "continously differentiable". Sometimes more (even infinite number of) derivatives are required to be continuous. Aug 20, 2013 at 16:59
• @njguliyev, not to nitpick but I think it's relatively common to call Lipschitz continuous ODEs "smooth" - being just smooth enough for existence and uniqueness of solutions. Aug 20, 2013 at 17:08
A function being smooth is actually a stronger case than a function being continuous. For a function to be continuous, the epsilon delta definition of continuity simply needs to hold, so there are no breaks or holes in the function (in the 2-d case). For a function to be smooth, it has to have continuous derivatives up to a certain order, say k. We say that function is $C^{k}$ smooth. An example of a continuous but not smooth function is the absolute value, which is continuous everywhere but not differentiable everywhere.
A smooth function is differentiable. Usually infinitely many times.
• ... or at least as often as we need it. Aug 20, 2013 at 17:14
Smooth implies continuous, but not the other way around. There are functions that are continuous everywhere, yet nowhere differentiable.
A smooth function can refer to a function that is infinitely differentiable. More generally, it refers to a function having continuous derivatives of up to a certain order specified in the text. This is a much stronger condition than a continuous function which may not even be once differentiable.
A smooth function is a function that has continuous derivatives up to some desired order over some domain. A function can therefore be said to be smooth over a restricted interval such as or . The number of continuous derivatives necessary for a function to be considered smooth depends on the problem at hand, and may vary from two to infinity. A function for which all orders of derivatives are continuous is called a C-infty-function.
Take $$f(x) = x|x|$$ it is smooth, and now consider $$g(x) = x^3$$, this other function is also smooth. However, $$g(x)$$ is much smoother than $$f(x)$$ because derivitive of $$f(x)$$. You can argue that $$g(x)$$ is infinitely many times smooth. All polynomials belong to $$C^\infty$$ meaning they are infinitely many times differentiable and are smooth.
However, $$h(x) = |x|$$ is not smooth, because it has corner. Please note that all three functions, $$f(x)$$, $$g(x)$$, and $$h(x)$$ are continous.
Here is how $$f(x)$$ looks like:
Here is how the derivative of $$f(x)$$ looks like:
Here is the second derivate of $$f(x)$$, as you can see its second derivative is not even continuous:
Here is the graph of $$g(x)$$:
Here is graph of $$\frac{d f(x)}{dx} = 3x^2$$:
Here is graph of $$\frac{d^2 g(x)}{dx^2} = 6x$$:
• If a function $f$ is smooth, then can I suppose that $f$ is increasing ou decreasing? At least in some interval? Oct 9, 2020 at 0:52
Consider a sequence in $\mathbb{R}$ say $\{x_n\}_{n \in \mathbb{N}}$, which is continuous in $\mathbb{R}$. Usually we do not say it a smooth function. | 2022-05-17T08:19:52 | {
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http://mathhelpforum.com/calculus/35491-optimization.html | # Math Help - Optimization
1. ## Optimization
An isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??
Any other good notes or questions for Optimization problems would be welcomed as well
2. Originally Posted by someone21
An isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??
Any other good notes or questions for Optimization problems would be welcomed as well
Minimum or maximum? The minimum area is trivially zero.
3. Originally Posted by mr fantastic
Minimum or maximum? The minimum area is trivially zero.
Minimum possible area but why is it zero??
And if maximum possible area then how to do it???
4. Originally Posted by someone21
Minimum possible area but why is it zero??
And if maximum possible area then how to do it???
Draw an isosceles triangle with non-equal side the diameter of the circle. Now move the non-equal side up so that the two equal sides get smaller and smaller. The triangle degenerates to a point - of area zero.
5. Originally Posted by someone21
An isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??
Any other good notes or questions for Optimization problems would be welcomed as well
Scroll down to the bottom of this: Triangles, Inscribed and Circumscribed Circles
Your circle has a radius r = 2. Let a = c and re-arrange:
$K = \frac{a^2 b}{8}$ .... (1)
where K is the area of the triangle.
You need to find a relationship between a and b and use it to get one variable in terms of the other, b in terms of a say. Substitute this into (1) to get K in terms of a single variable, a say. Now solve dK/da = 0, test nature etc ....
I haven't done it but I wouldn't be surprised if you find b = a gives maximum area, that is, an equilateral triangle has maximum area.
6. Originally Posted by mr fantastic
Scroll down to the bottom of this: Triangles, Inscribed and Circumscribed Circles
Your circle has a radius r = 2. Let a = c and re-arrange:
$K = \frac{a^2 b}{8}$ .... (1)
where K is the area of the triangle.
You need to find a relationship between a and b and use it to get one variable in terms of the other, b in terms of a say. Substitute this into (1) to get K in terms of a single variable, a say. Now solve dK/da = 0, test nature etc ....
I haven't done it but I wouldn't be surprised if you find b = a gives maximum area, that is, an equilateral triangle has maximum area.
Alternatively, you could work with finding the size of the two equal angles at the base of the triangle:
Let the angle at the top vertex of the triangle be of size $\phi$.
Let the two equal angles at the base of the triangle be of size $\theta$.
Let a be the two equal side lengths of the triangle. Let b be the length of the base of the triangle. Let h be the height of the triangle. Drop a perpendicular from the top vertex - half of the top vertex angle is $\frac{\phi}{2}$ and the perpendicular is the height of the triangle.
Area of any triangle: $A = \frac{bh}{2}$.
To find the angles that give a maximum triangle area, express the above area in terms $\phi$ and radius (= 2).
You should be able to show that $\frac{a}{2} = 2 \cos(\phi/2)$ (use the isosceles triangle with sides r = 2, r = 2, a and equal angles $\phi/2$). Then:
$a = 4 \cos(\phi/2)$.
$b = 8 \cos(\phi/2) \sin(\phi/2)$.
$h = 4 \cos(\phi/2) \cos(\phi/2) = 4 \cos^2 (\phi/2)$.
Therefore:
$A = \frac{bh}{2} = \frac{1}{2} [8 \cos(\phi/2) \sin(\phi/2)] [4 \cos^2(\phi/2)] = 16 \cos^3 (\phi/2) \sin(\phi/2)$.
Now solve $\frac{d A}{d \phi} = 0$ etc. I get $\phi = \frac{\pi}{3}$, that is, the triangle is equilateral. Substitute $\phi = \frac{\pi}{3}$ into A to get the maximum area.
Note: I reserve the right for this reply to contain algebraic errors.
7. Originally Posted by mr fantastic
Alternatively, you could work with finding the size of the two equal angles at the base of the triangle:
Let the angle at the top vertex of the triangle be of size $\phi$.
Let the two equal angles at the base of the triangle be of size $\theta$.
Let a be the two equal side lengths of the triangle. Let b be the length of the base of the triangle. Let h be the height of the triangle. Drop a perpendicular from the top vertex - half of the top vertex angle is $\frac{\phi}{2}$ and the perpendicular is the height of the triangle.
Area of any triangle: $A = \frac{bh}{2}$.
To find the angles that give a maximum triangle area, express the above area in terms $\phi$ and radius (= 2).
You should be able to show that $\frac{a}{2} = 2 \cos(\phi/2)$ (use the isosceles triangle with sides r = 2, r = 2, a and equal angles $\phi/2$). Then:
$a = 4 \cos(\phi/2)$.
$b = 8 \cos(\phi/2) \sin(\phi/2)$.
$h = 4 \cos(\phi/2) \cos(\phi/2) = 4 \cos^2 (\phi/2)$.
Therefore:
$A = \frac{bh}{2} = \frac{1}{2} [8 \cos(\phi/2) \sin(\phi/2)] [4 \cos^2(\phi/2)] = 16 \cos^3 (\phi/2) \sin(\phi/2)$.
Now solve $\frac{d A}{d \phi} = 0$ etc. I get $\phi = \frac{\pi}{3}$, that is, the triangle is equilateral. Substitute $\phi = \frac{\pi}{3}$ into A to get the maximum area.
Note: I reserve the right for this reply to contain algebraic errors.
I think the question was the circle is inside the isoceles triangle?
8. Originally Posted by mr fantastic
Alternatively, you could work with finding the size of the two equal angles at the base of the triangle:
Let the angle at the top vertex of the triangle be of size $\phi$.
Let the two equal angles at the base of the triangle be of size $\theta$.
Let a be the two equal side lengths of the triangle. Let b be the length of the base of the triangle. Let h be the height of the triangle. Drop a perpendicular from the top vertex - half of the top vertex angle is $\frac{\phi}{2}$ and the perpendicular is the height of the triangle.
Area of any triangle: $A = \frac{bh}{2}$.
To find the angles that give a maximum triangle area, express the above area in terms $\phi$ and radius (= 2).
You should be able to show that $\frac{a}{2} = 2 \cos(\phi/2)$ (use the isosceles triangle with sides r = 2, r = 2, a and equal angles $\phi/2$). Then:
$a = 4 \cos(\phi/2)$.
$b = 8 \cos(\phi/2) \sin(\phi/2)$.
$h = 4 \cos(\phi/2) \cos(\phi/2) = 4 \cos^2 (\phi/2)$.
Therefore:
$A = \frac{bh}{2} = \frac{1}{2} [8 \cos(\phi/2) \sin(\phi/2)] [4 \cos^2(\phi/2)] = 16 \cos^3 (\phi/2) \sin(\phi/2)$.
Now solve $\frac{d A}{d \phi} = 0$ etc. I get $\phi = \frac{\pi}{3}$, that is, the triangle is equilateral. Substitute $\phi = \frac{\pi}{3}$ into A to get the maximum area.
Note: I reserve the right for this reply to contain algebraic errors.
I think the question was the circle is inside the isoceles triangle?
9. Originally Posted by someone21
I think the question was the circle is inside the isoceles triangle?
Ah yes. I misread circumscribed as inscribed. Oh well, I guess I answered the second part:
"Any other good notes or questions for Optimization problems would be welcomed as well"
10. Originally Posted by someone21
An isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??
Any other good notes or questions for Optimization problems would be welcomed as well
Let 2b be the length of the base of the triangle, a the length of the other two sides and h the height of triangle.
Then A = bh.
A relationship between b and h is need so that A can be expressed in terms of a single variable.
Draw a line from the middle of the base to the (top) vertex of the triangle. This line goes through the center of the circle and is the height of the triangle.
Draw a line from the center of the circle to one of the sides of the triangle. This line is perpendicular to that side (why?).
Apply Pythagoras Theorem to the two obvious right-triangles:
$h^2 + b^2 = a^2$ .... (1)
$2^2 + {\color{red}(a-b)}^2 = (h - 2)^2 \Rightarrow 4 + a^2 - 2ab + b^2 = h^2 - 4h + 4 \Rightarrow a^2 - 2ab + b^2 = h^2 - 4h$ .... (2)
You should consider the geometry that gives the red length .....
Solve equations (1) and (2) simultaneously for b in terms of h (a small bit of algebra for you to do):
$b = \frac{2h}{\sqrt{h^2 - 4h}}$.
Therefore A = .......
Solve dA/dh = 0 (I get h = 6). Hence the minimum area of the triangle is equal to .....
Note: As you might anticipate, if you do the calculations you get $2b = 4 \sqrt{3} = a$ and so the triangle is equilateral.
Note: I reserve the right for this reply to contain careless errors.
11. Originally Posted by someone21
Minimum possible area but why is it zero??
One way to think about it....
An equilateral triangle is a shape where all three sides are of equal length. It is considered a special case of an isosceles triangle. Let s be a side, and A be the area of the triangle.
$\lim_{s \rightarrow 0} A$
$\lim_{s \rightarrow 0} \frac{1}{2\sqrt{2}}s^2 \rightarrow 0$
Since the area of the circle, $A_c$ is always inscribed in the triangle, it is always less than the area of the triangle, so $A_c < A$
The limit of A converges to zero, so due to the Squeeze Theorem or Comparison test, the area of the circle converges to zero. The minimum area of the circle is therefore zero.
12. Originally Posted by colby2152
One way to think about it....
An equilateral triangle is a shape where all three sides are of equal length. It is considered a special case of an isosceles triangle. Let s be a side, and A be the area of the triangle.
$\lim_{s \rightarrow 0} A$
$\lim_{s \rightarrow 0} \frac{1}{2\sqrt{2}}s^2 \rightarrow 0$
Since the area of the circle, $A_c$ is always inscribed in the triangle, it is always less than the area of the triangle, so $A_c < A$
The limit of A converges to zero, so due to the Squeeze Theorem or Comparison test, the area of the circle converges to zero. The minimum area of the circle is therefore zero.
why is it s goes to zero and can it be explained more detaily please
13. Originally Posted by someone21
why is it s goes to zero and can it be explained more detaily please
You are looking at the area as a side goes to zero. The area, dependent upon the sides, then goes to zero as well. | 2014-07-24T20:24:36 | {
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https://math.stackexchange.com/questions/187295/a-geometric-proof-of-zeta2-frac-pi26-and-other-integer-inputs-for-the | # A Geometric Proof of $\zeta(2)=\frac{\pi^2}6$? (and other integer inputs for the Zeta)
Is there a known geometric proof for this famous problem? $$\zeta(2)=\sum_{n=1}^\infty n^{-2}=\frac16\pi^2$$
Moreover we can consider possibilities of geometric proofs of the following identity for positive even inputs of the Zeta function: $$\zeta(2n)=(-1)^{n+1} \frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$$ and negative inputs: $$\zeta(-n)=-\frac{B_{n+1}}{n+1}$$
• There are many nice proofs at math.stackexchange.com/questions/8337/… . Is one of them geometric enough for you? (David Speyer's answer might fit the bill.) – Qiaochu Yuan Aug 27 '12 at 1:04
• Here's a proof by Prof. Greene at UCLA: math.ucla.edu/~greene/How%20Geometry.pdf – Elchanan Solomon Aug 27 '12 at 2:20
• thank you Qiaochu one of them is ;) – finnlim Aug 27 '12 at 2:44
• I wanted to ask this question some time ago. I'm glad to see it posted. (I thought of an elementary geometrical proof) – user 1591719 Aug 27 '12 at 5:25
• do you mean that you have an elementary geometric proof? Would you mind sharing it? – finnlim Aug 27 '12 at 5:42
Yes, there is a geometric proof for $\zeta(2)=\sum_{n=1}^\infty n^{-2}=\frac16\pi^2$, and, on top of that, it is a very unusual and, in my opinion, beautiful one.
In his gorgeous paper "How to compute $\sum \frac{1}{n^2}$ by solving triangles", Mikael Passare offers this idea for proving $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$:
Proof of equality of square and curved areas is based on another picture:
Recapitulation of Passare's proof using formulas is as follows: (for each step there is a geometric justification)
$$\sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \int_0^\infty \frac{e^{-nx}}{n}\; dx\; = -\int_0^\infty log(1-e^{-x})\; dx\; = \frac{\pi^2}{6}$$
I'm not sure what you mean by a geometric proof, but the following should fit the bill, as here the identity is deduced from a comparison of two areas: the first area is
$\displaystyle\int_{[0,1]^2} \frac{1}{1 - xy} \frac{dx dy}{\sqrt{xy}}$
and the second is
$4 \displaystyle \int_{\substack{\xi, \eta>0 \\ \xi \eta \leq 1}} \frac{ d \xi \, d \eta}{(1+\xi^2)(1+\eta^2)}$;
They are equal by a change of variables. For the first quantity, expand $(1-xy)^{-1}$ as a geometric series and integrate term-wise to get $3 \cdot \zeta(2)$. The second can be computed to $\pi^2/2$. This can be found in detail in Kontsevich and Zagier's Periods (bottom of page 8), where they attribute the idea to Calabi. (You should easily be able to hunt down an identity of $\zeta (n)$ being equal to an integral over the unit square in $\mathbb{R}^n$, like the first above. It might be a fun exercise to see if this idea is can be adapted for even $n$.)
One geometric/probabilistic proof is by Eugenio Calabi.
Consider the double integral $$I=\int_{0}^{1}\int_{0}^{1} \frac{1}{1-x^2y^2} \ dy \ dx.$$ Expand the integrand into a geometric series as such $$\frac{1}{1-x^2y^2} =\sum_{n=0}^{\infty} (xy)^{2n},$$ which is justified by the region of integration, that is $0<x,y<1.$ Exchange summation and integration (application of Tonelli's theorem), and integrate term by term to get $$I=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}.$$ Now, we evaluate $I$ directly. Make the change of variables $$x=\frac{\sin(u)}{\cos(v)}, y=\frac{\sin(v)}{\cos(u)}.$$ This transformation has Jacobian Determinant $$\frac{\partial (x,y) }{\partial(u,v)}=1-x^2y^2,$$ which cancels with the integrand, and the region becomes the open triangle defined by the inequalities $$u+v<\frac{\pi}{2}, u, v>0$$ We know from geometry that this triangle has base $\frac{\pi}{2}$ and height $\frac{\pi}{2},$ hence the area is $\frac{\pi^2}{8}.$ Hence $I=\frac{\pi^2}{8}.$ Now to obtain $\zeta(2),$ we use the fact that $$\zeta(2)=\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=1}^{\infty} \frac{1}{(2n)^2}+\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2},$$ which implies by some algebraic manipulation that
$$\zeta(2)=\frac{4}{3}I=\frac{\pi^2}{6}.$$
We can extend this proof to the $\zeta(2k).$ Let
$$I_{2k}= \int_{0}^{1}... \int_{0}^{1} \frac{1}{1-x_1^2 \dots x_{2k}^2} \ dx_{2k} \ ... \ dx_1.$$
Converting this integrand into a geometric series and exchanging summation and integration gives the sum: $$I_{2k}=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2k}}.$$ We can then recover $$\zeta(2k)=\frac{2^{2k}}{2^{2k}-1} I_{2k},$$ by observing $$\zeta(2k)= \sum_{n=1}^{\infty} \frac{1}{(2n)^{2k}} + I_{2k},$$ and using some algebraic manipulation.
On the other hand, we make the change of variables $$x_i=\frac{\sin(u_i)}{\cos(u_{i+1})}, 1 \leq i \leq 2k,$$ and $u_{2k+1}:=u_{1}$ here. This transformation turns out to have a Jacobian Determinant which cancels with the denominator of the integrand in $I_{2k}.$ The region of integration becomes the open polytope: $$\Delta^{2k}= \left \lbrace (u_1, \dots, u_{2k}): u_{i}+u_{i+1} < \frac{\pi}{2}, u_i>0, 1 \leq i \leq 2k \right \rbrace$$ in which $u_{2k+1}:=u_1.$ Computing the volume of this polytope is much more difficult. Rescaling with the change of variables $u_i=\frac{\pi}{2} v_i,$ and letting $V_1, \dots, V_{2k}$ being $2k$ independent, uniformly distributed random variables on $(0,1),$ we get
$$I_{2k}=\text{Vol}(\Delta^{2k})=\left(\frac{\pi}{2}\right)^{2k} \text{Pr} \left( V_1+V_2<1, \dots, V_{2k-1}+V_{2k}<1, V_{2k}+V_{1}<1 \right),$$ the probability that $V_1, \dots, V_{2k}$ have cyclically pairwise consecutive sums less than $1.$
In the literature, the aforementioned probability has been computed in several ways (see: https://www.maa.org/sites/default/files/pdf/news/Elkies.pdf https://arxiv.org/pdf/1003.3602.pdf https://pdfs.semanticscholar.org/35be/01e63c0bfd32b82c97d58ccc9c35471c3617.pdf)
Joshua Seaton also mentioned Zagier's and Kontsevich's reproduced Calabi proof. The general version of this is to evaluate
$$J_{2k}= \int_{0}^{1}... \int_{0}^{1} \frac{1}{\sqrt{x_1 \dots x_{2k}}(1-x_1 \dots x_{2k})}\ dx_{2k} \ ... \ dx_1.$$
A quick substitution shows $J_{2k}=2^{2k}I_{2k}.$ The generalized version of the change of variables is
$$x_i=\frac{\xi_i^2(1+\xi_{i+1}^2)}{1+\xi_i^2}, \dots, x_{2k}=\frac{\xi_{2k}^2(1+\xi_{1}^2)}{1+\xi_{2k}^2},$$ and upon computing its Jacobian Determinant, we get that
$$J_{2k}=\int_{\mathbb{H}^{2k}} \frac{1}{\xi_1^2+1} \dots \frac{1}{\xi_{2k}^2+1} \ d \xi_{1} \dots d \xi_{2k},$$ where $\mathbb{H}^{2k}$ is the hyperbolic defined by the inequalities: $$\xi_{i}\xi_{i+1} <1, \xi_i>0,$$ where $1 \leq i \leq 2k$ and $\xi_{2k+1}:=\xi_1.$ Letting $\Xi_1, \dots \Xi_{2k}$ be $2k$ independent, nonnegative Cauchy random variables, we get
$$I_{2k}=\left(\frac{\pi}{2}\right)^{2k} \text{Pr} \left( \Xi_1\Xi_2<1, \dots, \Xi_{2k-1}\Xi_{2k}<1, \Xi_{2k}\Xi_{1}<1 \right),$$ the probability that $\Xi_1, \dots, \Xi_{2k}$ have cyclically pairwise consecutive products less than $1.$
In my paper, I show that the probabilities lead to a very gargantuan formula:
$$I_{2k}=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2k}}= \left(\frac{\pi}{4} \right)^{2k} +\left(\frac{\pi}{4} \right)^{2k} \sum_{n=1}^{k} \sum_{\substack{(r_1, \dots, r_n) \in [2k]^n: \\ |r_p-r_q| \notin \lbrace 0,1,2k-1 \rbrace, \\ p,q \in [n]} } \prod_{i=1}^{n} \frac{1}{i+\sum_{j=1}^{i} \alpha_j},$$ where $[m]:= \lbrace 1, \dots, m \rbrace$ and $$\alpha_j=2- \delta(2k,2) - \sum_{m=1}^{j-1} \delta(|r_m-r_j|,2)+\delta(|r_m-r_j|,2k-2)$$ and $\delta(a,b)$ is the Kronecker Delta Function. In particular, the inner sum is taken over all tuples $(r_1, \dots, r_n) \in [k]^ n$ having cyclically pairwise nonconsecutive entries. | 2020-02-18T10:14:38 | {
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https://math.stackexchange.com/questions/3013681/linear-programming-word-problem-theater | # Linear Programming Word Problem: Theater
I'm having trouble understanding the second constraint.
A theater is presenting a program on drinking and driving for students and their parents[...] admission is 2.00 dollars for parents and 1.00 dollar for students. However, the situation has two constraints: 1) The theater can hold no more than 150 people and 2) every two parents must bring at least one student. How many parents and students should attend to raise the maximum amount of money?
Let $$x$$ = number of students
$$y$$ = number of parents
Since the question prompt wants the maximum amount of revenue, then the objective function is $$z = x + 2y$$
The theater can hold no more than 150 people so the first constraint is simply: $$x + y \leq 150$$
The second constraint is, "every two parents must bring at least one student," but I don't understand how to model this.
I've looked up a solution and it said $$y \leq 2x$$ is how to model this constraint, but I don't understand why. If there must be at least one student for every two parents then why wouldn't the inequality be $$2y \geq x$$?
Yes
y <= 2x
The easiest way to test this is with some data points and sketch a graph:
y_______x
2_______1,2,3,4,5…
4_______2,3,4,5,6…
6_______3,4,5,6,7…
Hence:
y <= 2x
2 <= 2x1 True
and
2y >= x
2x2 >= 5 False
Graph of y <= 2x | 2019-09-24T09:00:48 | {
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# If y is an integer, is y^2 divisible by 4?
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Intern
Joined: 18 Jan 2012
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If y is an integer, is y^2 divisible by 4? [#permalink]
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07 Feb 2012, 20:41
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If y is an integer, is y^2 divisible by 4?
(1) y is divisible by 4
(2) y is divisible by 6
[Reveal] Spoiler: OA
Last edited by Bunuel on 08 Feb 2012, 02:16, edited 1 time in total.
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Magoosh GMAT Instructor
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Re: Number Properties - Question 2 [#permalink]
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07 Feb 2012, 23:09
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Hi, there. I'm happy to help with this.
Prompt: if y is an integer , is y^2 divisible by 4?
Fact #1: In order for an integer N to be divisible by 4, N must have at least two factors of 2 in its prime factorization.
Fact #2: When you square a number, say T^2, whatever prime factors are in the prime factorization of T are doubled in the prime factorization of T^2. Say a particular prime factor appears three times in T --- then it will appear six times in T^2.
Fact #3: An even number, by definition, has at least one factor of 2 in its prime factorization.
Therefore, the square of any even number will have at least two factors of 2, and therefore will be divisible by 4.
The question "is y^2 divisible by 4?" is entirely equivalent to the question "is y an even integer?"
All of that was the mathematical heavy-lifting for the question. Now, the statements will be a piece of cake.
Statement #1: y is divisible by 4. Therefore y is even. Sufficient.
Statement #2: y is divisible by 6. Therefore y is even. Sufficient.
Both statements sufficient. Answer = D
Does all that make sense?
Here's another odd/even DS question for practice.
http://gmat.magoosh.com/questions/868
Please let me know if you have any questions.
Mike
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Mike McGarry
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Re: Number Properties - Question 2 [#permalink]
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08 Feb 2012, 02:16
If y is an integer, is y^2 divisible by 4?
(1) y is divisible by 4 --> as y itself is divisible by 4 then y^2 will also be divisible by 4. Sufficient.
(2) y is divisible by 6 --> y=6k, for some integer k --> y^2=36k^2=4*(9k^2) --> we can see that y^2 has 4 as a factor hence it is divisible by 4. Sufficient.
Hope it's clear.
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If y is an integer, is y^2 divisible by 4? [#permalink]
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26 Nov 2017, 11:28
nimc2012 wrote:
If y is an integer, is y^2 divisible by 4?
(1) y is divisible by 4
(2) y is divisible by 6
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Since we have 1 variables and 0 equation, we need to consider each condition one by one.
Condition 1)
$$y = 4n$$ for some integer $$n$$.
$$y^2 = 16n^2$$
Since $$16$$ is divisible by $$4$$, this is sufficient.
Condition 2)
$$y = 4m$$ for some integer $$m$$.
$$y^2 = 36m^2$$
Since $$36$$ is divisible by $$4$$, this is sufficient.
If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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If y is an integer, is y^2 divisible by 4? [#permalink] 26 Nov 2017, 11:28
Display posts from previous: Sort by | 2018-03-23T09:07:50 | {
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"url": "https://gmatclub.com/forum/number-properties-question-127230.html",
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"lm_q2_score": 0.9111797166446537,
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