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https://dsp.stackexchange.com/questions/53394/finding-inverse-z-transform-with-usage-of-tables-lti-and-causal-sequences/53397 | # finding inverse Z transform with usage of tables (LTI and causal sequences)
problem is as follows
try to find the inverse Z
$$Z^{-1} (\frac{3}{z+2}) = ???$$
with the usage of z-transform tables
Ok, so in order to find something from the table, I thought that we expand with z and try to see if we get closer to any kind of expression in the tables
$$Z^{-1} (\frac{3z}{z*(z+2)}) = Z^{-1} (3*\frac{1}{z}\frac{z}{z+2})$$
then we get constant out
$$=3*Z^{-1}(z^{-1}*\frac{z}{z+2})$$
then we see that right side fraction can be found in the table, but the combined sum expression can also be found with further manipulation (as can be seen later)
So, I think we declare temp variable here to clean it up (this portion is found in the table green portion)
$$F(z) = \frac{z}{z-(-2)} \iff f[k]=(-2)^k$$
then we can make it so
$$3*Z^{-1}(z^{-1}*F(z))$$
then we find the direct expression in the table (we have delay of one unit because of $$z^{-1}$$ term, it is the orange portion)
$$3*Z^{-1}(z^{-1}*F(z)) = 3* u[n-1]*f[n-1]$$
so it looks like answer should be
$$3* u[n-1]*f[n-1] = 3*u[n-1]*(-2)^{n-1}$$
my teacher hasnt provided the solutions guide yet, despite that he promised to do so, and I reminded him about it, so I was just checking if this is the right solution and thinking...
I did the same problem, but I used polynomial long-division style to get the following series representation
$$3*z^{-1}-6*z^{-2}+12*z^{-3}-24*z^{-4}...$$
so based on that I guess that the $$x[n]=3*(-1)^n*2^n$$
where $$n>=0$$
so, I was just a little bit confused does there actually exist the delay of one unit in the original input sequence (result of inverse Z), sometimes I get confused about the causal sequence when you multiply with the unit step sequence, but then, it seems that sometimes also it can just be stated that the original sequence must have been causal sequence, so you just define n>=0 or something like that???
Your solution is correct. This is the way I see it (assuming that the sequence is causal/right-sided):
$$\mathcal{Z}^{-1}\left\{\frac{3}{z+2}\right\}[n]=3\mathcal{Z}^{-1}\left\{z^{-1}\frac{z}{z+2}\right\}[n]=\\=3\mathcal{Z}^{-1}\left\{\frac{z}{z+2}\right\}[n-1]=3(-2)^{n-1}u[n-1]$$
• thank you for checking my answers, please take a look at my other homework posts if you are inclined to doublecheck them and if you aren't too busy! I'm trying to prepare for my dsp exam next week, and we don't have any rehersal exercises, so I'm just reviewing our old exercises, and redoing them... – Late347 Nov 18 '18 at 6:07
• I don't really have a study group at my school, because my friends are a little bit behind in their homework exercises, so they won't be so useful in helping me, so I typically tend to post some of my homework to be checked at dsp stack exchange or math stack exchange... – Late347 Nov 18 '18 at 6:09 | 2019-07-22T07:01:52 | {
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http://math.stackexchange.com/questions/40118/if-ab-and-cd-then-acbd/40128 | # If $a|b$ and $c|d$, then $ac|bd$
I just need to check the reasoning in my proof is correct, I think it is valid although I'm not totally convinced because I can't follow the logic; does proving that $x$ is an integer prove that $ac|bd$?
Theorem: Let $a$, $b$, $c$, $d$ be integers. If $a|b$ and $c|d$, then $ac|bd$.
Proof: $aj=b$ and $ck=d$ for integers $j$, $k$. Then, $ac|bd$ implies $acx=ajck$ and thus $x=jk$ for some $x$. Since the product $jk$ is an integer, $x$ is an integer and thus $acx=ajck$ and thus $ac|bd$.
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Your reasoning goes in the wrong direction; you want to demonstrate that $ac\mid bd$, i.e. that there exists an $x$ such that $acx=bd$, using the assumption that $a\mid b$ and $c\mid d$. This is done simply by showing that $x=jk$ works.
In other words, the sentence beginning with "Then $ac\mid bd$ implies..." cannot be part of an argument that $ac\mid bd$ is true. If $P$, $Q$, and $R$ are statements, then showing that $P\Rightarrow R$ and $Q\Rightarrow R$ does not show that $P\Rightarrow Q$.
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I see, so I should say: $aj=b$ and $ck=d$ for $j$,$k \in \mathbb{Z}$. We must show that $acx=bd$. $acx=ajck$, so $x=jk$. Since $x=jk$ is an integer, then $acx=ajck$ for some $x$ and $ac|bd$? – persepolis May 19 '11 at 17:48
@persepolis: better. Still, better to say something like "we must show that there exists an $x$ such that $acx=bd$. Since $bd=(aj)(ck) = (ac)(jk)$, then $x=jk$ shows that $ac|bd$." – Arturo Magidin May 19 '11 at 18:19
Hint $\ \$ Noting $\displaystyle\rm\ \ \ a\mid b\,\iff\, \frac{b}a\in \mathbb Z$
we observe that $\rm\ \ a\mid b,\ c\mid d\ \ \Rightarrow\ \ ac\mid bd$
is equivalent to $\rm\displaystyle\,\ \ \frac{b}a,\,\frac{d}c\in\mathbb Z\ \ \Rightarrow\ \frac{b\, d}{a\, c}\in\, \mathbb Z$
in your notation $\rm\,\ j,\ k\ \in\ \mathbb Z\ \ \Rightarrow\ \ \ j\, k\ \in \, \mathbb Z$
That's true: the product of integers is an integer. So this divisibility product law is equivalent to the product closure law for integers (except that if your definition of divisibility includes $\,0\mid n\iff n= 0\,$ then you need to handle such cases separately - which is easy).
Edit $\$ Since, based on comments, at least one reader seems to have misconstrued the intent of my answer, I elaborate below. When I posted this answer, there were already a few answers explaining the logical flaw in the proposed proof in the question. That done, my intent was instead to address another point, namely how one may exploit the the innate arithmetical structure in order to attain a simpler and more conceptual proof. Such simplification may make the proof more intuitive, which may help one to avoid making such errors. To elaborate, below I show how the "arithmetic" of the divisibility relation is intimately connected with the arithmetic of the subring $\rm\,\mathbb Z\subset\mathbb Q.\,$
First, notice how the above proof makes clear that this product rule for divisibility is essentially equivalent to the product rule for integrality, i.e. if $\rm\,j,k\,$ are integers then so too is their product $\rm\,jk.\,$ Indeed, as we saw above, this integrality product rule easily implies the divisibility product rule. Conversely, if the divisibility product rule is true, then specializing $\rm\ a = 1 = c\$ we infer $\rm\, 1\mid b,\ 1\mid d\ \Rightarrow\ 1\mid bd,\,$ i.e. $\rm\, b,d\in\mathbb Z\ \Rightarrow\ bd\in \mathbb Z\$ (note $\rm\, 1\mid n\iff n\in \mathbb Z,\,$ by definition).
Similarly, one deduces the equivalence between the divisibility difference rule, namely that $\rm\ a\mid b,\,c\ \Rightarrow\ a\mid b\!-\!c\$ and the fact that $\rm\,\mathbb Z\,$ is closed under difference $\rm\,j,\,k\in \mathbb Z\ \Rightarrow\ j\!-\!k\in\mathbb Z$.
Combining these observations leads to the following equivalence between the arithmetic of divisibility relations and subrings $\rm\,Z\,$ of $\,\mathbb Q\,$ (or any field).
Theorem $\$ Let $\rm\,Z\,$ be a subset of $\rm\,\mathbb Q\,$ with $\rm \,1\in Z.\,$ Let $\:\mid\:$ denote the divisibility relation $\rm\, a\mid b \iff b/a\in Z\,$ for $\rm\,a,b\in Z,\, a\ne 0,\$ and $\rm\ 0\mid b \iff b = 0.\,$ Then the following two statements are equivalent.
$\rm(1)\ \ \ Z\$ is a subring of $\rm\,\mathbb Q$
$\rm(2)\ \$ The relation $\:\mid\:$ satsifies the following properties.
$\rm\qquad a\mid b,\,c\,\ \Rightarrow\,\ a\mid b-c\qquad$ for all $\rm\,a,b,c\in Z$
$\rm\qquad a\mid b,\ c\mid d\,\ \Rightarrow\,\ a\,c\mid b\,d\ \ \$ for all $\rm\,a,b,c,d\in Z$
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Bill, I have avoided making this criticism for some time now, but I really think it is valid here: if the OP understood this argument, he/she wouldn't be asking this question. – Qiaochu Yuan May 19 '11 at 19:28
@Bill: yes, but I don't understand why you're assuming that the OP "understands integer arithmetic," to put it bluntly. It sounds like you are saying "this answer is quite comprehensible to anyone who understands this answer." – Qiaochu Yuan May 19 '11 at 19:54
@Qia The OP's error does not demonstrate a misunderstanding of arithmetic but, rather, of logic. Reformulating the problem as I did not only helps to avoid such logic pitfalls, but also reveals quite plainly the essence of the matter. As such I find your critique (and downvote) to be unfounded. Translating the divisibility statement into a more familiar statement about products of integers allows the OP to appeal to his well-honed intuition on integer arithmetic operations (vs. unfamiliar manipulation of divisibility relations). So there's less chance to commit such logical errors. – Bill Dubuque May 19 '11 at 20:07
@Qiaochu I also don't understand your problem. The proof is that elementary that even someone who is really new to maths could understand it. Basically it uses that the product of two integers is an integer which is totally clear, and that if you divide an integer by one of its divisors it remains an integer, which is also very obvious. – Listing May 19 '11 at 21:22
@Qia Perhaps you missed the point of my answer. When I posted it there were already answers on the OP's logic error. My point was to teach the OP a better way to formulate the proof strategy. If answers were strictly limited to precisely the question asked then this forum would be quite sterile. When teaching math it is essential to address not only the questions asked, but also closely related metamathematical issues such as proof methodology, etc. Effective teaching involves much more than mechanically answering questions. I am here first and foremost to teach. Was that not clear? – Bill Dubuque May 20 '11 at 14:23
If you're having trouble with these kinds of proofs, it's often best to start by writing down what you're given on the left side of the paper and where you're trying to get on the right side.
a|b ac | bd
c|d
Next fill in what you can from definitions:
b = aj bd = ack
d = ci
Now use inferences to play with the things on the left and aim for what you have on the right. Each time, you need to ensure that what you have implies what you write next.
You can also work from right to left, but you have to be careful. Here you need to make sure the steps are "if and only if" so that you can go backwards from left to right if you eventually meet in the middle of the paper.
Knowing when it's safe to make a particular step either left to right or right to left comes from being extremely careful and from knowing exactly what's going on. There are puzzles from false proofs where they test you on this. A common example is
a=b
(a-b)c = (a-b)d so c = d
Looks fine, right? But given that a=b we should have noticed that (a-b)=0 and you can't cancel zero from both sides.
Anyway, getting good at this stuff is called "mathematical maturity" and it, like most kinds of maturity, takes time.
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Your idea is right, but there are some difficulties of wording. When you write "Then, $ac|bd$ implies" it begins to sound as if you are going in the wrong direction. "If and only if" would have been OK.
You should have written instead something like this, which by the way is also shorter.
"Then $bd=(aj)(ck)=(ac)(jk)$. So if we let $x=jk$, then $(ac)x=bd$. It follows that $ac$ divides $bd$."
Let me repeat, you knew exactly what was going on. But there was some awkwardness in the exposition that somewhat masked that fact.
- | 2016-07-29T10:31:37 | {
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http://math.stackexchange.com/questions/361694/diameter-of-a-circle-with-3-coordinates | # Diameter of a circle with 3 coordinates
The question is: A circle has the points $A=(6,-1)$ $B=(10,-3)$ and $C=(-2,-9)$ on its circumference. A diameter of the circle is drawn which is parallel to BC. How far apart are the two parallel lines?
I managed to get the center of the circle $(5,-8)$, however I am now stuck. Thanks for you help!
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What is the geometric definition of a circle? (What is special about all the points on the circle?) – Sammy Black Apr 14 '13 at 21:58
Or if you draw a normal from the center of the circle to BC, where do you think it will hit BC? (Hint: Almost no calculation needed. Draw a picture.) – Harald Hanche-Olsen Apr 14 '13 at 22:02
that is a pretty good start.
Hint:
the line through $BC$ must hit the circle at $BC$. the distance between the mid-point of BC and the centre must be the same as the shortest distance from the BC and its parallel diameter (draw a picture and convince yourself, and why is this true?)
work out the midpoint and calculate the distance to the centre.
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Thanks very much for your help, I got it now:) – Laila Apr 15 '13 at 0:41
@Laila no worries – Lost1 Apr 15 '13 at 10:41
If center of circle is $O(5,-8)$ You need to find the straight line $n$ that is perpendicular to line $BC$ wich passes to $O$ then you find the point $N=n\cap BC$ finally you look for distance $ON$
- | 2016-06-26T20:50:01 | {
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https://math.stackexchange.com/questions/1762616/prove-the-following-statement-by-proving-its-contrapositive-if-r-is-irrationa | # Prove the following statement by proving its contrapositive: if $r$ is irrational, then ${ r }^{ \frac { 1 }{ 5 } }$ is irrational
Just a disclaimer before I proceed with my question and the proof I wrote up: I know that this question has been asked before, for example here, but I am more interested in being critiqued on how I wrote the proof and its completeness. In addition, I am having trouble seeing how it proves what I first set out to prove.
Theorem: if $r$ is irrational, then ${ r }^{ \frac { 1 }{ 5 } }$ is irrational
Proof: We prove the contrapositive: if ${ r }^{ \frac { 1 }{ 5 } }$ is rational, then $r$ is rational
1) Assume that ${ r }^{ \frac { 1 }{ 5 } }$ is rational, then there exists $a,b\in\mathbb{Z}$ such that:
${ r }^{ \frac { 1 }{ 5 } }=\frac { a }{ b }$ where $a,b$ are coprime and $b\neq 0$
2) Therefore, ${ r }=\frac { a^{ 5 } }{ b^{ 5 } }$
3) If $a,b\in\mathbb{Z}$, then $a^5,b^5\in\mathbb{Z}$ as well.
4) Therefore $r\in\mathbb{Q}$
Q.E.D.
When it comes to proving that the square root of $2$ is irrational, I can quickly see and understand why $\sqrt { 2 }$ is indeed irrational. However, with this proof, I don't understand how this actually proves the theorem I set out to prove. Does proving it by using the contrapositive just make it that much simpler? Or did I miss some vital steps in my proof?
Please feel free to give me constructive criticism about my proof writing techniques as well.
• Your proof is great. One small thing is that you may want to note that $b^5 \neq 0$ since $b \neq 0$. – Christian Gaetz Apr 28 '16 at 12:51
• I agree with the comment above. $[p \implies q] \iff [\neg q \implies \neg p]$. – barak manos Apr 28 '16 at 12:53
• @user75296 So just edit step 3) to look like this? 3) if $a,b\in\mathbb{Z}$, then $a^5,b^5\in\mathbb{Z}$ where $b^5\neq 0$ as well. – Cherry_Developer Apr 28 '16 at 13:02
In this particular problem, the statement involves assuming $r$ is irrational. The problem is that we can't really say much about what $r$ looks like.
On the subject of your exposition, it looks great! The comment about noting that $b^5 \neq 0$ is really the only thing to improve on. | 2019-06-16T02:39:37 | {
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https://math.stackexchange.com/questions/2895415/solving-151x-%E2%88%92-294-equiv-44-pmod7 | # solving $151x − 294 \equiv 44 \pmod{7}$
I've solved normal congruence equations like $ax \equiv b \pmod{m}$ but now I am trying to solve $1 5 1 x − 294 \equiv44\pmod{7}$. How do I solve this one? Can I just add $294$ to both sides and solve as normal using Euclidean algorithm?
I read the answer by quanta in this question but it is still not clear to me.
Can anyone please elaborate more on this?
Thanks for any help.
• Always a good plan to remember with modular arithmetic: You can reduce things modulo $7$ right at the start, and this often simplifies things greatly (as in gimusi's answer). In other words, adding $294$ to both sides is an idea but is unnecessary work. You can reduce $294$ modulo $7$ immediately to save time. – Matt Aug 26 '18 at 19:33
We have that
$$151x − 294 \equiv 44 \pmod{7} \iff 4x-0\equiv 2 \pmod{7} \iff 2x\equiv 1 \pmod{7}$$
and by Euclidean algorithm we can find
$$4\cdot 2-1\cdot 7=1$$
therefore $4$ is the inverse of $2 \pmod 7$ and we find
$$4\cdot 2x\equiv 4\cdot 1 \pmod{7} \implies x\equiv 4 \pmod{7}$$
• I think I see. $294$ is divisible by $7$, $44$ modulo $7$ gives us 2 remainder. What is the explanation behind this reasoning? – user575678 Aug 26 '18 at 19:35
• When we work $\mod m$ we can reduce all the terms to the remainder $0\le r<m$. – user Aug 26 '18 at 19:37
• Could we also say that $x = -3$? – user575678 Aug 27 '18 at 0:03
• Yes of course all $x$ in the form $4+7k$ are solutions. Usually we use $4 \pmod 7$ but also $-3 \pmod 7$ is a correct result. – user Aug 27 '18 at 0:50
• $2x \equiv 1$ can be solved a bit easier by replacing $1$ with a number that $2$ goes into. Since this is mod $7$, then just keep adding $7$. You only need do this once since $2x \equiv 1 \equiv 8$, so $x \equiv 4$ – steven gregory Feb 25 at 4:33
The first right thing to do is to simplify all coefficients and write them in the range of $\{0,1,2,3,4,5,6\}$, using Euclidian division.
$$151=21\times7 +4 \quad;\quad 294 = 42\times 7 \quad;\quad 44=7\times 6 + 2$$
So the equation is the same as
$$4x\equiv 2 \pmod{7}$$
All we need to do is find the inverse of $4\pmod{7}$. After a few tries, we find that it is $2$ since $2\times 4 \equiv 8 \equiv 1 \pmod{7}$. Hence, we get
$$x \equiv 2\times 2 \equiv 4\pmod{7}$$ | 2021-07-23T19:57:59 | {
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https://www.cameresenzoriale.ro/used-reception-uvwkc/35b246-pascal%27s-triangle-patterns | If you add up all the numbers in a row, their sums form another sequence: In every row that has a prime number in its second cell, all following numbers are. Art of Problem Solving's Richard Rusczyk finds patterns in Pascal's triangle. Pascal's triangle is a triangular array constructed by summing adjacent elements in preceding rows. Pascal's triangle is a triangular array of the binomial coefficients. Patterns in Pascal's Triangle Pascal's Triangle conceals a huge number of various patterns, many discovered by Pascal himself and even known before his time. 6. Another question you might ask is how often a number appears in Pascal’s triangle. Harlan Brothers has recently discovered the fundamental constant $e$ hidden in the Pascal Triangle; this by taking products - instead of sums - of all elements in a row: $S_{n}$ is the product of the terms in the $n$th row, then, as $n$ tends to infinity, $\displaystyle\lim_{n\rightarrow\infty}\frac{s_{n-1}s_{n+1}}{s_{n}^{2}} = e.$. The pattern known as Pascal’s Triangle is constructed by starting with the number one at the “top” or the triangle, and then building rows below. Hover over some of the cells to see how they are calculated, and then fill in the missing ones: This diagram only showed the first twelve rows, but we could continue forever, adding new rows at the bottom. The diagram above highlights the “shallow” diagonals in different colours. Patterns in Pascal's Triangle - with a Twist. If we add up the numbers in every diagonal, we get the Fibonacci numbersHailstone numbersgeometric sequence. Some of those sequences are better observed when the numbers are arranged in Pascal's form where because of the symmetry, the rows and columns are interchangeable. The entries in each row are numbered from the left beginning with k = 0 and are usually staggered relative to the numbers in the adjacent rows. Naturally, a similar identity holds after swapping the "rows" and "columns" in Pascal's arrangement: In every arithmetical triangle each cell is equal to the sum of all the cells of the preceding column from its row to the first, inclusive (Corollary 3). If we arrange the triangle differently, it becomes easier to detect the Fibonacci sequence: The successive Fibonacci numbers are the sums of the entries on sw-ne diagonals: \begin{align} The American mathematician David Singmaster hypothesised that there is a fixed limit on how often numbers can appear in Pascal’s triangle – but it hasn’t been proven yet. The order the colors are selected doesn’t matter for choosing which to use on a poster, but it does for choosing one color each for Alice, Bob, and Carol. Pascals Triangle Binomial Expansion Calculator. Notice that the triangle is symmetricright-angledequilateral, which can help you calculate some of the cells. If you add up all the numbers in a row, their sums form another sequence: the powers of twoperfect numbersprime numbers. It is also implied by the construction of the triangle, i.e., by the interpretation of the entries as the number of ways to get from the top to a given spot in the triangle. When you look at the triangle, you’ll see the expansion of powers of a binomial where each number in the triangle is the sum of the two numbers above it.\displaystyle C^{n-2}_{k-1}\cdot C^{n-1}_{k+1}\cdot C^{n}_{k}=\frac{(n-2)(n-1)n}{2}=C^{n-2}_{k}\cdot C^{n-1}_{k-1}\cdot C^{n}_{k+1}$,$\displaystyle\begin{align} The following two identities between binomial coefficients are known as "The Star of David Theorems": $C^{n-1}_{k-1}\cdot C^{n}_{k+1}\cdot C^{n+1}_{k} = C^{n-1}_{k}\cdot C^{n}_{k-1}\cdot C^{n+1}_{k+1}$ and Proved by induction important property of Pascal ’ s triangle numbersFibonacci numbers of each match. Triangle '', followed by 147 people on Pinterest patterns within Pascal 's triangle ( named after Pascal...: start with a Twist by Kathleen M. Shannon and Michael J. Bardzell '' at diagram... Hidden number sequence and secrets of combinatorics 1: Draw a short, vertical line and write them.... Use the following procedure Dummy View - not to be DELETED peak can... 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https://math.stackexchange.com/questions/2177817/determine-the-eigenvector-and-eigenspace-and-the-basis-of-the-eigenspace | # Determine the eigenvector and eigenspace and the basis of the eigenspace
The yellow marked area is correct, so don't check for accuracy :)
$A=\begin{pmatrix} 0 & -1 & 0\\ 4 & 4 & 0\\ 2 & 1 & 2 \end{pmatrix}$ is the matrix.
Characteristic polynomial is $-\lambda^{3}+6\lambda^{2}-12\lambda+8=0$
The (tripple) eigenvalue is $\lambda=2$.
Calculate the eigenvectors now:
$\begin{pmatrix} -2 & -1 & 0\\ 4 & 2 & 0\\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}$
We get the equations:
$I: -2x-y=0 \Leftrightarrow y = -2x$
$II: 4x+2y=0$
$III: 2x+y=0 \Leftrightarrow 2x-2x=0 \Leftrightarrow 0=0$
We see that in every eequation $z$ is unknown, so we can choose an arbitrary $z$.
$x\begin{pmatrix} 1\\ -2\\ z \end{pmatrix}$ and this is the eigenspace...?
And what is the basis of this eigenspace? Can I just set $x=1$ and some value for $z$? So this would be a correct basis of the eigenspace: $\begin{pmatrix} 1\\ -2\\ 3 \end{pmatrix}$?
Now we need three linearly independent eigenvectors but I couldn't find them as I always got linearly dependent vectors...
I need a detailled, not too complicated answer that explains it well and I will give that answer a nice bounty (up to 200 rep) because I couldn't find another site explaining this correctly to me and I'm really in need of it.
• "Now we need three linear independent eigenvectors"...no, not really – imranfat Mar 8 '17 at 16:55
The $x$ shouldn't be outside the vector. The solution to equations I,II, and III is
\begin{pmatrix} x\\ -2x\\ z \end{pmatrix}
where $x$ and $z$ are arbitrary. Every vector of this form is an eigenvector for $A$. You can write each such vector as a linear combination of two vectors $e_1$ and $e_2$ defined by
$$e_1:= \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$$ and $$e_2:= \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}.$$
More concretely, we have $$\begin{pmatrix} x \\ -2x \\ z \end{pmatrix} = \begin{pmatrix} x \\ -2x \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ z \end{pmatrix} = x\begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix} + z\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = xe_1 + ze_2$$
A basis for the eigenspace is the two vectors $e_1$ and $e_2$, since every vector in the eigenspace can be written uniquely as a linear combination of those two vectors.
There's no reason it should have 3 linearly independent eigenvectors.
• So let's say the task says "Determine the eigenvectors", what is the best way to write them for this example? Is it $v_{1}=x\begin{pmatrix} 1\\ -2\\ 0 \end{pmatrix}+z\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}$? Or rather $\begin{pmatrix} 1\\ -2\\ 0 \end{pmatrix},\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}$? – tenepolis Mar 8 '17 at 19:23
• Technically the eigenvectors are that first one. Though depending on the class, I've seen people refer to (1,-2,0) and (0,0,1) as the eigenvectors. – Nathan H. Mar 9 '17 at 1:58
You must solve $(A - 2I)\vec{x} = \vec{0}$, i.e. $$\begin{pmatrix} -2 & -1 & 0\\ 4 & 2 & 0\\ 2 & 1 & 0 \end{pmatrix} \begin{pmatrix}x\\ y\\ z\end{pmatrix} =\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}$$ Note that all equations are multiple of each other, so let's leave the last one to use and eliminate all others. You get the constraint $2x+y = 0$. So Your solutions will have to have $y = -2x$, in other words, $$\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}x\\ -2x\\ z\end{pmatrix} = x \begin{pmatrix}1\\ -2\\ 0\end{pmatrix} + z \begin{pmatrix}0\\ 0\\ 1\end{pmatrix}$$ and so you see that there are 2 eigenvectors that can be used as a basis.
As for the number of independent eigenvectors in the basis not equal the multiplicity of the root of the characteristic equation, this has to do with the fact that for your matrix, the geometric and algebraic multiplicities don't match for the eigenvalue $\lambda = 2$. You can read more about it here.
From your calculation you have that all eigenvectors are of the form $\begin{pmatrix} x\\ -2x\\ z \end{pmatrix} = x\begin{pmatrix} 1\\ -2\\ 0 \end{pmatrix} + z\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}$
If $\lambda$ is an eigenvalue of $A$, then its corresponding eigenvectors are vectors $\mathbf v$ that satisfy $A\mathbf v=\lambda\mathbf v$, or $(A-\lambda I)\mathbf v=0$. That is, the eigenspace of $\lambda$ is the null space of the matrix $A-\lambda I$. This answer describes how to read a basis for the null space directly from the row-reduced echelon form of the matrix. Remember that the dimension of the eigenspace—the geometric multiplicity of $\lambda$—will be at most the algebraic multiplicity of $\lambda$, which is 3 in this case, but it might be less than that. | 2019-06-26T20:45:13 | {
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https://mathoverflow.net/questions/310307/is-the-convergence-of-dotx-2atx-faster-than-that-of-dotx-atx | # Is the convergence of $\dot{x}=2A(t)x$ faster than that of $\dot{x}=A(t)x$?
Let $x \in \mathbb{R}^{n}$ and $A(t) \in \mathbb{R}^{n\times n}$. If $\dot{x}=A(t)x$ and $\dot{x}=cA(t)x$ with $c>1$ are exponentially stable. Is the convergence rate of $x$ to zero of $\dot{x}=cA(t)x$ faster than that of $\dot{x}=A(t)x$?
This question is initially asked on Mathematics: https://math.stackexchange.com/questions/2881136/is-the-convergence-of-dotx-2atx-faster-than-that-of-dotx-atx
Here is my initial thought:
For linear time-invariant system, the fact $\dot{x}=A(t)x$ is exponentially stable implies $A(t)=A$ is Hurwitz. It is clear that the $i$th eigenvalue $\lambda_{i}(cA)=c\lambda_{i}(A)$, $i,\ldots,n$, thus, the convergence for $\dot{x}=cAx$ with $c>1$ is faster than that of $\dot{x}=Ax$.
For a linear time-varying system. Let us consider the extreme case where $c=0$, $\dot{x}=cA(t)x=0$, which implies it is no longer exponentially stable.
For $c>1$, can we have the same conclusion for exponentially stable linear time-varying systems, i.e., can we conclude that the convergence is faster when $c>1$?
Update 1: For a scalar time-varying system, i.e., $x\in\mathbb{R}$. We can actually prove this conjecture. In fact, from the uniqueness of the equilibrium point ($x=0$ is the only solution that renders $\dot{x}=0$), the solution of $\dot{x}=ca(t)x$ is a monotone function either strictly increasing or strictly decreasing, depending on its initial condition $x(0)$. Thus, for $\dot{x}=ca(t)x$ where $c>1$, the absolute value of the derivative is larger than that of $\dot{x}=a(t)x$, while for both cases the sign of $\dot{x}(t)$ remains unchanged for all $t\ge0$. Thus, $\dot{x}=ca(t)x$ does converge faster to $x=0$ than $\dot{x}=a(t)x$.
• Math.SE user AVK provides a possible counterexample here, did you check it? Also, even in dimension one highly oscillatory coefficients of the form $A(t)=c \sin(e^t)$ may lead to exponentially fast convergence to a non-zero value $e^{c\pi/2} x(0)$, with convergence rate independent of $c$, but I suppose this is not what you had in mind. – Mateusz Kwaśnicki Sep 11 '18 at 11:20
• The counterexample considered $c\in(0, 0.55)$, while the questions focus on $c\in(1,\infty)$. The assumption exponential stability of $\dot{x}=A(t)x$ implies that $x(t)$ will convege to the origin as $t$ tends to infty. Thus, $A(t)=c\sin(e^{t})$ does not satisfy the above assumption. – guluzhu Sep 11 '18 at 11:28
• I think AVK's counterexample may work if you substitute $\tfrac{1}{4}A(t)$ for $A(t)$. Still, apparently there is a simpler counterexample that I describe in my answer. – Mateusz Kwaśnicki Sep 11 '18 at 19:55
## 1 Answer
A simple counterexample can be constructed as follows: take $$A(t) = \pmatrix{0&\tfrac{\pi}{2}\\-\tfrac{\pi}{2}&0} \qquad \text{for } t \in [2n, 2n+1) ,$$ and $$A(t) = \pmatrix{-\alpha&0\\0&-\beta} \qquad \text{for } t \in [2n+1, 2n+2) ,$$ where $0 < \alpha \leqslant \beta$.
The equation $x'(t) = c A(t) x(t)$ describes then the following evolution: during each of the intervals $[2n, 2n+1)$ the particle is rotated by an angle $\tfrac{c \pi}{2}$ counter-clockwise, while during the interval $[2n, 2n+1)$ each of the coordinates converges exponentially to zero at rate $c \alpha$ or $c \beta$.
For $c = 1$ (or, more generally, for odd values of $c$), rotation essentially swaps the coordinates (if we disregard signs), so effectively $$x(4n) = (-1)^n e^{-c(\alpha + \beta) n} x(0).$$ The rate of convergence is thus $\tfrac{1}{4} c (\alpha + \beta)$.
For $c = 2$ (or, more generally, for even values of $c$), rotation merely changes the sign, so we end up with $$x(4n) = \pmatrix{e^{-2c\alpha n}&0\\0&e^{-2c\beta n}} x(0).$$ Rate of convergence is therefore $\tfrac{1}{2} c \alpha$.
Choose $\alpha = 1$ and $\beta = 7$. Then solutions of $x'(t) = A(t) x(t)$ converge to zero at rate $\tfrac{1}{4} \times 1 \times (1 + 7) = 2$, while solutions of $x'(t) = 2 A(t) x(t)$ converge to zero at slower rate $\tfrac{1}{2} \times 2 \times 1 = 1$.
• Thank you for the counterexample. My follow-up questions is: Does the conjecture hold for sufficiently large $c$? – guluzhu Sep 12 '18 at 1:05
• @guluzhu: I do not know the answer. Sounds plausible, at least in dimension two. – Mateusz Kwaśnicki Sep 12 '18 at 7:49 | 2020-07-10T10:11:40 | {
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http://math.stackexchange.com/questions/141037/finding-again-the-stationary-distribution-of-a-markov-chain | Finding again the stationary distribution of a markov chain
I am asked to compute the stationary distribution of the markov chain with state space $E=\{0\dots,n\}$ and transition matrix below:
\begin{bmatrix} 0 & 1 \\ \frac{1}{n} & 0 & \frac{n-1}{n} \\ & \frac{2}{n} & 0 & \frac{n-2}{n} & \\ & & \ddots & \ddots & \ddots \\ & & & \frac{n-1}{n} & 0 & \frac{1}{n} \\ & & & & 1 & 0 & \\ \end{bmatrix}
I used $\pi P =\pi$ and I got:
$\pi_0=\frac{1}{n}\pi_1 \\\pi_1 = \pi_0+\frac{2}{n}\pi_2 \\ \pi_2 =\frac{n-1}{n}\pi_1 +\frac{3}{n}\pi_3 \\ \pi_3 =\frac{n-2}{n}\pi_2 +\frac{4}{n}\pi_4\\ \pi_4 =\frac{n-3}{n}\pi_3 +\frac{5}{n}\pi_5\\$
Working from there I got:
$\pi_0 = \frac{1}{n}\pi_1 \\\pi_1 =n.\pi_0 \\ \pi_2=\frac{n^2-n}{2} \pi_0 \\ \pi_3 = \frac{n^3-3n^2+2n}{6} \pi_0 \\ \pi_4 = \frac{n^4-6n^3+11n^2-6n}{24} \pi_0 \\ \sum_{k=0}^\infty \pi_k = 1$
I tried fiddling with it and spotting a pattern for $\pi_k$ but I cant see to find $\pi_k$ for all $k\in E=\{0,\dots,n\}$. How would I finish this problem?
This is a follow-up from my previous question Calculating stationary distribution of markov chain , Sasha kindly showed me a way to find the stationary distribution, and I understood that method, but I don't think i've come across that method before in my lecture. I was wondering if it was possible to compute the stationary distribution using the method I did above, but I can't seem to get to the end..
-
Write your solutions for the $\pi_k$ a little differently:
\begin{align*} \pi_1&=n\pi_0\\ \pi_2&=\frac12n(\pi_1-\pi_0)=\frac12n(n-1)\pi_0\\ \pi_3&=\frac13n\left(\pi_2-\frac{n-1}n\pi_1\right)=\frac13n\left(\frac12n(n-1)-(n-1)\right)\pi_0\\ &=\frac13n(n-1)\left(\frac12n-1\right)=\frac16n(n-1)(n-2)\\ \pi_4&=\frac14n\left(\pi_3-\frac{n-2}n\pi_2\right)\\ &=\frac14n\left(\frac16n(n-1)(n-2)-\frac{n-2}n\cdot\frac12n(n-1)\right)\pi_0\\ &=\frac14n\left(\frac16n(n-1)(n-2)-\frac12n(n-1)(n-2)\right)\pi_0\\ &=\frac1{24}n(n-1)(n-2)(n-3)\pi_0\;. \end{align*}
At this point the pattern should be pretty obvious, and the natural way to prove it is by induction.
-
Thanks @BrianM.Scott I can see the pattern now, and I guess that $\pi_n = \frac{1}{n!}\dots$, is there some other way to get the form of $\pi_n$ besides using induction or no? When you meant using induction, how would I start? Because I can't seem to get the equation for $\pi_n$ before proceeding onto the base/inductive step. – Richard May 4 '12 at 19:07
@Richard: Right, $\pi_k=\frac1{k!}\cdot\frac{n!}{n-k}!\pi_0=\binom{n}k\pi_0$. There may well be other ways; this is probably the most elementary, though. – Brian M. Scott May 4 '12 at 19:10
Thanks @BrianM.Scott, so assuming I spotted the pattern with the factorials and deduced that $\pi_k = \binom{n}k\pi_0$, is that 'acceptable'/enough to do or do I have to also prove it by induction? – Richard May 4 '12 at 19:21
@Richard: That would depend on instructor; I'd probably expect you to prove it. – Brian M. Scott May 4 '12 at 19:23
@Richard: Proving the formula for $\pi_k$ would most likely be by induction, yes. Showing that $1=\pi_0\sum_{k\ge 0}(k+1)q_k$ is then just a matter of reversing the order of a double summation. – Brian M. Scott May 4 '12 at 21:06
Thus, $\pi_1=n\pi_0$ and $\pi_2=\frac12n(n-1)\pi_0$, as you wrote, but $\pi_3$ is not what you wrote since $\pi_3=\frac16(n^3-3n^2+2n)\pi_0$. Can you see a pattern?
Hint: Find the factorization of $n^3-3n^2+2n$.
Follow-up: Since you found $\pi_3=\frac16n(n-1)(n-2)\pi_0$, factorials begin to enter the picture. So let us turn $\pi_3$ into an expression with factorials. Clearly, $n(n-1)(n-2)=\frac{n!}{(n-3)!}$ and $6=3!$ hence $\pi_3=\frac{n!}{3!(n-3)!}\pi_0$. You might also recognize this as $\pi_3={n\choose 3}\pi_0$. I am pretty sure you can take up things from here.
Thanks @Didier , sorry must have made an error with the $-$ and $+$ signs. I can see the pattern where $\pi_3 = \frac{1}{6}n(n-1)(n-2)\pi_0.$ How would I write it in terms of $n$ though? I can see that $\pi_n = \frac{1}{n!}\dots$ but what is the way to find the pattern? – Richard May 4 '12 at 19:04
See Edit. Two remarks: do not go too fast, I see nowhere why $\pi_n=1/n!$ should hold. And beware that $\pi_4$ in your post is still wrong. – Did May 4 '12 at 19:14
Thanks @Didier . So if I have spotted the pattern and stated that $\pi_k = \binom{n}k\pi_0$, is that sufficient/enough? Or do I have to prove it by induction before I can really use $\pi_k = \binom{n}k\pi_0$? Finally, thanks for your help again, yes I hope im capable to take it up from here, carrying on with the normalisation etc. My main problem I had in this question was finding the pattern/form for $\pi_k$. – Richard May 4 '12 at 19:23 | 2014-03-09T16:41:52 | {
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https://math.stackexchange.com/questions/951414/how-to-solve-this-absolute-value-equation | # How to solve this absolute value equation?
Consider the absolute value equation:
|x| + |x-2| +|x-4|= 6
How to find the solution(s)?
My attempt:
For |x|, we got x, for x>=0 and -x, for x <0
For |x-2|, we got x-2, for x >= 0 and -(x-2), for x<0
For |x-4|, we got x-4, for x>=0 and -(x-4), for x<0
After this, I'm confused how to find the solutions? is it any easy way to find the solution?
Thanks
• you have to do case work – Dr. Sonnhard Graubner Sep 29 '14 at 18:16
• $x\geq 4$, $2\le x<4$, $0\le x<2$, $x<0$ – Dr. Sonnhard Graubner Sep 29 '14 at 18:17
• And accordingly, you get: $x={4,4,0,0}$, which gives, $x={4,0}$ – Sudeepan Datta Sep 29 '14 at 18:20
I think you misunderstand the absolute value.
Note that $$|x|=\begin{cases}x &\text{if x\ge 0}\\-x &\text{if x\lt 0}\end{cases}$$ So, for example, $$|x-2|=\begin{cases}x-2 & \text{if \color{red}{x-2}\ge 0}\\-(x-2)& \text{if \color{red}{x-2}\lt 0}\end{cases}$$
So, for your question, separate it into four cases as
(1) $x\lt 0$.
(2) $0\le x\lt 2$.
(3) $2\le x\lt 4$.
(4) $x\ge 4$.
You have 3 "nodes" for switching $|x|$ , $|x-2|$ and $x-4|$. The easiest way is to imagine real line with these points and split solution considering intervals $[-\infty,0],[0,2],[2,4],[4,\infty]$ separately. On each interval you can remove $||$ with certain sign and immediately solve the equation, checking that final solution is in the corresponding interval.
• $x<0$ leads to $6-3x=6$
Solve this equation and check wether the solution satisfies $x<0$. If so then you have found a solution of the original equation. If not then no solution exists that satisfies $x<0$.
Do the same for the cases:
• $0\leq x<2$
• $2\leq x<4$
• $4\leq x$
Define $n = x - 2$, and rewrite the equation.
$$|n-2| + |n| + |n+2| = 6$$
In this form we can easily see that if $n$ is a solution, so is $-n$. Thus WLOG we will assume that $n$ is positive. We can then remove some redundant $|\cdot|$s.
$$|n-2| + (n) + (n + 2) = 6 \implies |n-2| + 2n = 4$$
Clearly then, because $|n-2| \geq 0$, we have $|n-2| + 2n = 4 \implies n \leq 2$. This means that $n-2 \leq 0$ and so we can remove another $|\cdot|$.
$$-(n-2) + 2n = 4 \implies n = 2$$
And thus we have two solutions, $n = \pm 2$ or $\boxed{x = 0, 4}$.
• Aren't there four solutions for the equation, -2, 0, 4 and 2? ±2 represents two solutions -2 and 2, because the symbol ± means positive OR negative, not positive AND negative. – Toby Mak Apr 16 '17 at 7:38
• @TobyMak, $x-2=n=\pm 2$, therefore $x=\pm 2 + 2=0,4$. Certainly $-2$ and $2$ are not solutions to the original equation. – Ben Frankel Apr 16 '17 at 18:02
• Sorry for my misunderstanding – Toby Mak Apr 16 '17 at 23:39
Think of the modulus function measuring the distance from fixed points. So that $|x|=|x-0|$ is the distance from $0$.
So your sum measures the total distance from the points $0,2,4$
If $x$ is greater than $4$ the first two distances are greater than $4$ and $2$
If $x$ is less than $0$ we reach the same conclusion with the last two points by symmetry.
If $x$ is between $0$ and $2$ the total distance to $0$ and $2$ is always $2$, and the remaining distance is less than $4$. Similarly by symmetry for the points between $2$ and $4$.
Now check the points $0$, $2$, $4$ themselves | 2019-04-19T09:10:58 | {
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https://math.stackexchange.com/questions/1275070/does-an-inseparable-extension-have-a-purely-inseparable-element/1276333 | # Does an inseparable extension have a purely inseparable element?
Assume $K/F$ is an inseparable extension. Is it necessary that K contains an element $u \notin F$ that is purely inseparable over $F$?
I also posted in MO.
• Do you have a link to the question on MO? I was not able to find it by searching.
– user279515
Jul 8 '19 at 12:53
The standard counterexample is to consider $F = \mathbf{F}_p(x,y)$ and $K = F(\alpha)$ where $\alpha$ is a root of $f = T^{p^2} + x T^p + y \in F[T]$ (where $f$ is irreducible by Gauss' Lemma). Clearly $K/F$ is non-separable of degree $p^2$ and $F' := F(\alpha^p)$ is a degree-$p$ subextension that is separable over $F$, so this is the maximal separable subextension. Hence, $[K:F]_s = p$ and so $[K:F]_i = p$. To rule out the existence of an element of $K-F$ that is purely inseparable over $F$ it suffices to show that $F'$ is the unique degree-$p$ subextension of $K/F$.
Suppose $E$ is another such subextension, so it must be purely inseparable over $F$ (as otherwise it would be separable and hence the composite $F'E$ would be separable yet has to exhaust $K$ for $F$-degree reasons). By multiplicativity of separable degree in towers, it follows that $[K:E]_s=p$, so the degree-$p$ extension $K/E$ is separable. But $K=E(\alpha)$, so the degree-$p$ minimal polynomial $g \in E[T]$ of $\alpha$ is separable. In a splitting fi eld $E'/E$ of $f$ over $E$, $f$ is a product of $p$th powers of $p$ monic linear factors and $g$ is the product of just these monic linear factors (since $g|f$ in $E[T]$ and $g$ is separable over $E$ of degree $p$).
Hence, in $E'[T]$ we must have $g^p=f$. But this latter equality then must hold in $E[T]$ as well. It then follows that $g = T^p + uT + v$ for some $u, v \in E$ satisfying $u^p = x, v^p = y$. But the extension $\mathbf{F}_p(x^{1/p}, y^{1/p}) \supset F$ has degree $p^2$ and so cannot lie inside the degree-$p$ extension $E$ of $F$. This is a contradiction, so no such $E \ne F'$ exists inside $K/F$.
• Thanks! I wrote my answer above after some explicit computations and I was wondering about a more abstract approach. May 11 '15 at 8:48
• Lipman wrote a short paper ("Balanced Field Extensions," Amer. Math. Monthly 73 (1966), 373-374) about the algebraic extensions that are separable extensions of inseparable extensions and at the end he gives a counterexample that is precisely this construction when $p = 2$ except he allows any field of characteristic $p$ in place of $\mathbf F_p$, which of course works in the above answer as well.
– KCd
Jun 27 '15 at 23:56
I think this is one of the counterexamples mentioned by grghxy on MO : take $F=\mathbb{F}_2(x,y)$ and consider in $F[t]$ the polynomial $f(t)=t^4+xt^2+y$. Using Gauss' Lemma for example it is easy to see that $f(t)$ is irreducible in $F[t]$, and then we can take the field extension $K= F(\alpha)=F[t]/(f(t))$. This is a degree $4$ extension and it is inseparable, since $f'(t)=0$. Now we want to show that no element $\beta \in K$ is purely inseparable over $F$. Suppose that $\beta \in K$ is purely inseparable over $F$: since $[K:F]=4$ it must be either $\beta^2\in F$ or $\beta^4\in F$. We can write $$\beta = a_0 + a_1\alpha +a_2\alpha^2 + a_3\alpha^3 \qquad a_i \in F$$ and then after some computations (I hope they are correct) one can see that $$\beta^2 = [a_0^2+a_2^2+a_3^2xy] + [a_1^2+a_2^2x+a_3^2x^2+a_3^2y]\alpha^2$$ so that $\beta^2\in F$ if and only if $$a_1^2+a_2^2x+a_3^2x^2+a_3^2y=0$$ Multiplying this equation by the denominators of $a_i^2$, we can suppose that $a_i \in \mathbb{F}_2[x,y]$, but then we observe that $a_3^2y$ has odd degree in $y$, whereas the other terms have even degree in $y$. This implies that $a_3=0$, so that we are left with $$a_1^2+a_2^2x=0$$ but again the term $a_2^2x$ has odd degree in $x$ and the other one has even degree, so that we are left with $a_1=a_2=0$. This proves that if $\beta^2 \in F$ then $\beta\in F$.
Now, suppose that $\beta^4 \in F$, then from what we have proves above it must be that $\beta^2\in F$, so that again $\beta\in F$.
• I will post an answer that gives the version in every positive characteristic (recovering the above for $p=2$). May 10 '15 at 23:06 | 2021-12-08T04:28:19 | {
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http://math.stackexchange.com/questions/126322/how-to-prove-that-countably-compact-d-space-is-compact | # how to prove that countably compact D-space is compact
In the paper A survey of D-spaces by Gary Gruenhage it is written that it is easily seen that any countably compact D-space is compact. However I'm not able to show it. Here is my attempt to prove this claim: Let $(X, \tau)$ be a countably compact D-space with topology $\tau$. Consider a function $N: X \rightarrow \tau$ with $x \in N(x)$ for every $x \in X$ (neighborhood assignment function). I think that to prove compactness of $X$ it is enough to show that open cover $\{N(x):x \in X\}$ of $X$ has finite subcover. Because $X$ is a D-space, there is a closed discrete subset $D$ of $X$ such that $\{N(x):x \in D\}$ covers $X$. It seem to me that to complete the proof (using countable compactness of $X$) I need to use the fact that $D$ is countable. However it is not true in general that close discrete set is countable. Thank you for your help! Michal Čihák
Note 1. The considered spaces are regular and $T_1$.
Note 2. A space $X$ is a D-space if whenever one is given a neighborhood $N(x)$ of $x$ for each $x \in X$, then there is a closed discrete subset $D$ of $X$ such that $\{N(x):x \in D\}$ covers X.
-
This looks like a duplicate of this question... is anyone for or against closing? – Zev Chonoles Mar 30 '12 at 19:00
@Zev: of course, they are related and the ideas are similar, but $\sigma$-compactness and countable compactness are quite distinct properties, so I'm against closing this question. – t.b. Mar 30 '12 at 19:17
@t.b.: Ah, I didn't read closely enough - nor did the person who flagged the question :) Thanks for pointing out the difference. – Zev Chonoles Mar 30 '12 at 19:19
@ZevChonoles Also, the implication goes the other direction in the linked question. – Alex Becker Mar 30 '12 at 20:47
Suppose that your set $D$ is infinite. Pick a countably infinite subset of it, say $D_0$. For each $x\in D_0$ let $V(x)$ be a nbhd of $x$ such that $V(x)\cap D_0=\{x\}$; you can do this because $D_0$ is discrete. Moreover, $D_0$ is closed, so $X\setminus D_0$ is open, and $\{X\setminus D_0\}\cup\{V(x):x\in D_0\}$ is then a countable open cover of $X$ that clearly has no finite subcover: each $x\in D_0$ is covered only by the set $V(x)$. This contradicts the countable compactness of $X$, showing that in fact $D$ must be finite.
You are correct that it’s sufficient to look at open covers of the form $\{N(x):x\in X\}$ for nbhd assignments $N$. To see this, let $\mathscr{U}$ be any open cover of $X$. Then for each $x\in X$ let $N(x)$ be any element of $\mathscr{U}$ that contains $x$. Clearly $\{N(x):x\in X\}$ is a subcover of $\mathscr{U}$, so any finite subcover of $\{N(x):x\in X\}$ will automatically be a finite subcover of the original $\mathscr{U}$ as well.
-
Thank you very much, your proof is very clear for me. – mcihak Mar 31 '12 at 6:58
It follows from the following characterization: a space $X$ countably compact if and only if every infinite subset of $X$ has an accumulation point. So in particular any discrete subspace is finite. By the way the characterization makes for a nice exercise.
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Note that any countable compact space $X$ is limit point compact. Now let $D \subset X$ be a closed discrete subset of $X$. Then $D$ does not have any limit points, since such a limit point would have to be in $X$, but $D$ is closed. Finally since $X$ is limit point compact it follows that $D$ is finite.
Let $\mathcal U$ be an open cover of $X$. Then for each $x \in X$ pick $U \in \mathcal U$ such that $x \in U$ and set $N(x)=U$. Now if $D$ is a closed discrete subspace of $X$ such that for $x \in D$ we have $N(X)$ covers $D$, then by the previous argument $D$ is finite, so this is a finite subcover of $\mathcal U$.
I am curious though if you can manage to do this without using the axiom of choice, since I seem to be invoking it in order to create the neighborhood function.
-
We can do even better than making your set $D$ countable; $D$ is finite! Suppose your set $D$ is infinite, and choose some (necessarily closed) countably infinite discrete subset $D'$ of $D$. Define a neighborhood assignment function $N':D'\to \tau$ such that $x\neq y\implies N'(x)\cap N'(y)\neq \emptyset$, which is possible because $D'$ is discrete. Let $\mathcal O = \{N'(x)\cup (X\setminus D'): x\in D'\}$ and note that $\mathcal O$ is a countable open cover of $X$. Yet if we remove any open set $N'(x)\cup (X\setminus D')$ from $\mathcal O$, then $x\notin \bigcup \mathcal O$. Hence by countable compactness, $D'$ is finite, contradicting our choice of $D'$. Thus $D$ must be finite.
- | 2016-05-25T07:20:10 | {
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https://math.stackexchange.com/questions/2110533/can-a-non-local-ring-have-only-two-prime-ideals/2110620 | # Can a non-local ring have only two prime ideals?
Can a non-local ring have only two prime ideals?
The only way this would be possible is if the ring $R$ had two distinct maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$, and no other prime ideals. I suspect that such a ring exists, though I don't know how to construct it.
• $F\times F$ where $F$ is any field has this property. Jan 23, 2017 at 17:07
If $(A,\mathfrak p)$ and $(B,\mathfrak q)$ are local rings with only one prime ideal, then $R=A\times B$ has this property: a prime ideal must contain either $(1,0)$ or $(0,1)$ since their product is $0$, and then it is easy to see the prime must be either $A\times\mathfrak q$ or $\mathfrak p\times B$, respectively. (More generally, if $R=A\times B$ for any two rings $A$ and $B$, the primes in $R$ are exactly the sets of the form $A\times \mathfrak q$ or $\mathfrak p\times B$ where $\mathfrak q$ is a prime of $B$ or $\mathfrak p$ is a prime of $A$.)
Conversely, every example has this form. Indeed, if $R$ is not local and has exactly two prime ideals, both prime ideals must be maximal. It follows that $\operatorname{Spec}(R)$ is a discrete space with two points $\mathfrak p$ and $\mathfrak q$. From the fact that the structure sheaf on $\operatorname{Spec}(R)$ is a sheaf it follows that the canonical map $R\to R_{\mathfrak p}\times R_{\mathfrak q}$ is an isomorphism (since $R_{\mathfrak p}$ is exactly the value of the structure sheaf on the open set $\{\mathfrak p\}$ and $R_{\mathfrak q}$ is exactly the value of the structure sheaf on the open set $\{\mathfrak q\}$).
Take $\dfrac{\mathbb{Z}}{6\mathbb{Z}}.$ There are 2 maximal ideals here $\frac{(2)}{(6)}$ and $\frac{(3)}{(6)}$, and these are the only ones.
Let $p$ and $q$ be two different prime ideals of $R$ of height zero. Set $S=R-(p\cup q)$ and consider $S^{-1} R.$ Maximal ideals are $S^{-1}p$ and $S^{-1}q$.
Note that there is a one-to-one correspondence between prime ideals of $S^{-1}R$ and prime ideals of $R$ which dont meet $S$ (3.11 of Atiyah-Macdonald).
Given two fields, $k_1,k_2$, the ring $k_1\times k_2$ has this property, with $\mathfrak m =k_1\times \{0\}$ and $\mathfrak n=\{0\}\times k_2$.
More generally, if $R_1$ and $R_2$ are local rings with no other prime ideals, then $R_1\times R_2$ has this property, I believe.
The real question is if you can find interesting cases of others. Specifically, is there an example $R$ which does not have any non-trivial idempotents - elements $e$ other than $1,0$ such that $e^2=e$.
• These are all the examples; see my answer. Jan 23, 2017 at 18:19 | 2022-06-29T04:45:38 | {
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https://math.stackexchange.com/questions/2234135/is-this-set-not-open-and-not-closed | # Is this set not open and not closed?
Let $X = \{(x,y) \in \mathbb{R^2}: x^2 + y^2 \leq 1, (x,y) \neq (0,0)\}$
Is this set open, closed, or neither, and what is the boundary, interior and closure?
I want to get a better understanding if I am approaching these problems correctly.
Is this set open? If I take a point in the set X, so that $p = (p_x, p_y)$, then $(p_x, p_y) \neq (0,0)$, and I define an open ball $B_r(p)$, and I take a point $q \in B_r(p)$. Is it possible for $q \in X$? Well no, since if $p$ satisfies $x^2 + y^2 = 1$ then it is possible for q to be outside of the set or that q satisfies $x^2 + y^2 > 1$.
So it is not open.
Is this set closed? $X$ is closed if $X^c$ is open. But $X^c = \{(x,y) \in \mathbb{R^2}: x^2 + y^2 > 1, (x,y) = (0,0)\}$ is not open, since (x,y) = (0,0) defines a boundary point (and set is only open if for every $x \in X^c$, $x$ is an interior point).
Therefore it is neither closed nor open.
Boundary: $\{(x,y) \in \mathbb{R^2}: x^2 + y^2 = 1\}\cup\{(x,y) \in \mathbb{R^2}: (x,y) = (0,0)\}$
Interior: $\{(x,y) \in \mathbb{R^2}: x^2 + y^2 < 1, (x,y) \neq (0,0)\}$
Closure: $\{(x,y) \in \mathbb{R^2}: x^2 + y^2 \leq 1\}$
I am not sure if I am thinking about these problems correctly. I am having a lot of trouble, so I could use some help if there is something about my work that is not correct or clear.
• This set is neither open nor closed. – DHMO Apr 14 '17 at 16:13
• How did you figure that out in seconds?! :) – TimelordViktorious Apr 14 '17 at 16:13
• It is not closed because it does not contain all of its limit points. It is not open because it is not its interior. I figured that out in seconds by having a picture in my head. Intuition is sometimes useful for dealing with well-behaved sets like these. – DHMO Apr 14 '17 at 16:14
• Actually that's the line of reasoning I had in my head, but I wanted to iron it out further. Good to know I'm on the right track then. Thank you. – TimelordViktorious Apr 14 '17 at 16:15
• All your answers are correct (apart from the typo that I fixed for you). So you seem to have a pretty good grasp of the concepts. But I suggest simply writing $\{(0,0)\}$ instead of $\{(x,y) \in \mathbb{R^2}: (x,y) = (0,0)\}$. – TonyK Apr 14 '17 at 16:24
Well, when you argue it is open, you say it isn't possible for $q$ to be in the set. This isn't really true. Consider $p=(0.5,0.5)$, which is in the set, and $B_{0.1}((0.5,0.5))$: the $x^2 + y^2$ will certainly be less than $0.6^2 + 0.6^2 = 0.72\le 1$, so every point in this ball is in $X$.
The problem comes from talking about $p$ and $q$ and then choosing them to have certain properties, that they don't necessarily have in general.
I would suggest, when saying something is not open, to consider a particular point: say $(0,1)$, which is certainly in $X$. Any open ball, of radius $r$, around this point will contain the point $(0,1+\frac{r}{2})$, and $0^2 + (1+\frac{r}{2})^2 \ge 1$ so this point is not in $X$. So $(0,1)\in X$ is not an interior point, and $X$ is not open.
Your argument for not-closedness is correct. If you have more machinery about closed sets, you could argue more directly: a closed set should contain all of its limit points, i.e. if $\lim x_n = x$ and all the $x_n$ are in $X$, $x\in X$. Here, we can have a sequence like $(0,\frac{1}{n})$ that goes to $(0,0)$, but $(0,0)$ is not in $X$. I usually find it easier to think about closed sets as closed under limits. | 2019-07-18T04:51:22 | {
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http://mathhelpforum.com/calculus/41029-integration-power-rule.html | # Math Help - integration power rule
1. ## integration power rule
the integral of $\int\frac{x}{\sqrt{4-x^2}} dx$ is $-\sqrt{4-x^2} + C$.
this book shows a middle step that i don't understand: $-\frac{1}{2} \int (4-x)^{-\frac{1}{2}}(-2x) dx$
i'm trying to use power rule, but where did that $-\frac{1}{2}$ come from that is before the $\int$?
2. Originally Posted by dataspot
the integral of $\int\frac{x}{\sqrt{4-x^2}} dx$ is $-\sqrt{4-x^2} + C$.
this book shows a middle step that i don't understand: $-\frac{1}{2} \int (4-x)^{-\frac{1}{2}}(-2x) dx$
i'm trying to use power rule, but where did that $-\frac{1}{2}$ come from that is before the $\int$?
Consider the derivative of the inside of the radical?
Or more conservatively
Let $\psi=4-x^2$
so then $d\psi=-2xdx$
now looking at our integral we have
$\frac{-1}{2}\int\frac{d\psi}{\sqrt{\psi}}=\frac{-1}{2}\cdot{2\sqrt{\psi}}$
Now making our back sub we get
$-\sqrt{4-x^2}$
3. Originally Posted by dataspot
the integral of $\int\frac{x}{\sqrt{4-x^2}} dx$ is $-\sqrt{4-x^2} + C$.
this book shows a middle step that i don't understand: $-\frac{1}{2} \int (4-x)^{-\frac{1}{2}}(-2x) dx$
i'm trying to use power rule, but where did that $-\frac{1}{2}$ come from that is before the $\int$?
It's not that obvious at first. But its easy to understand.
$\int\frac{x}{\sqrt{4-x^2}}\,dx$
Let $u=4-x^2\implies\,du=-2x\,dx$
Note that in the integral, we have $\int\frac{{\color{red}x}}{\sqrt{4-x^2}}\color{red}\,dx$, not $-2x\,dx$. To make the $-2x\,dx$ appear, multiply the integral by $\frac{-2}{-2}$ (equivalent to 1)
Thus, we get the following:
$\frac{-2}{-2}\int\frac{x}{\sqrt{4-x^2}}\,dx\implies {\color{red}\frac{-1}{2}}\int\frac{-2x}{\sqrt{4-x^2}}\,dx$.
Now make the substitutions:
$\frac{-1}{2}\int u^{-\frac{1}{2}}\,du$.
Hope this makes sense!
4. Originally Posted by dataspot
the integral of $\int\frac{x}{\sqrt{4-x^2}} dx$ is $-\sqrt{4-x^2} + C$.
this book shows a middle step that i don't understand: $-\frac{1}{2} \int (4-x)^{-\frac{1}{2}}(-2x) dx$
i'm trying to use power rule, but where did that $-\frac{1}{2}$ come from that is before the $\int$?
the -1/2 is to account for the -2 that was put in front of the x (noticed the -2x?). what they did was multiply the x by -2 and then divide by -2 (hence the -1/2) so as to not change anything. why? because -2x is the derivative of 4 - x^2. why do we care? well, in this set up, it is "easy" to see that we have an expression that is the result of using the chain rule to differentiate a function. basically what happened here is doing substitution without actually saying "let u = 4 - x^2." this is a more clever and concise way of doing it
EDIT: Beaten again...twice...and with better explanations too good job guys
5. Originally Posted by Chris L T521
It's not that obvious at first. But its easy to understand.
$\int\frac{x}{\sqrt{4-x^2}}\,dx$
Let $u=4-x^2\implies\,du=-2x\,dx$
Note that in the integral, we have $\int\frac{{\color{red}x}}{\sqrt{4-x^2}}\color{red}\,dx$, not $-2x\,dx$. To make the $-2x\,dx$ appear, multiply the integral by $\frac{-2}{-2}$ (equivalent to 1)
Thus, we get the following:
$\frac{-2}{-2}\int\frac{x}{\sqrt{4-x^2}}\,dx\implies {\color{red}\frac{-1}{2}}\int\frac{-2x}{\sqrt{4-x^2}}\,dx$.
Now make the substitutions:
$\frac{-1}{2}\int u^{-\frac{1}{2}}\,du$.
Hope this makes sense!
ok...
if the resulting $du$ had equaled $x dx$, then you wouldn't have had to multiply by 1 (rather, $\frac{-2}{-2}$), but since it didn't, you had to even things out.
yeah?
6. Originally Posted by Chris L T521
Now make the substitutions:
$\frac{-1}{2}\int u^{-\frac{1}{2}}\,du$.
i'm not sure i could finish the problem from here either. been a while.
7. Originally Posted by dataspot
i'm not sure i could finish the problem from here either. been a while.
from here, we use the power rule: $\int x^n~dx = \frac {x^{n + 1}}{n + 1} + C$ for $n \ne -1$
so, $- \frac 12 \int u^{-1/2}~du = - \frac 12 \cdot \frac {u^{1/2}}{1/2} + C = -u^{1/2}$
8. Originally Posted by dataspot
i'm not sure i could finish the problem from here either. been a while.
Use the power rule for integration:
$\int x^n\,dx = \frac1{n + 1}x^{n + 1} + C,\text{ for }n\neq-1$
9. thanks for the help!!! | 2016-02-09T18:39:40 | {
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https://math.stackexchange.com/questions/2189250/proof-about-positive-semi-definite-matrix/2189257 | Proof about positive semi-definite matrix
I used the following proof as an answer on the following question about a positive semi-definite matrix, and the OP accepted this answer, however the wikipedia page about positive semi-definite matrices made me doubt, so any thoughts or even a counterexample would be welcome. I have overwritten the question in order to make this question self-sustaining.
$\textbf{The statement}$: Let $A$ be a $n \times n$ complex invertible matrix. Denote $B = A^{-1}$. Furthermore, we denote the $i$th row of $B$ by $B_{i\bullet}$. Let $\alpha \geq \|B_{i\bullet}\|_2^2$. I have tried to prove that $$\alpha A^HA - e_ie_i^T$$ is a positive semi-definite matrix. Here $e_i$ denotes the columnvector having $1$ in the $i$th entry and $0$ in the other entries (so the $i$th standard basis vector.
$\textbf{My attempt to proof}:$ What I thought is the following: let $z$ be a complex nonzero columnvector, then we want to show that $$z^H(\alpha A^HA - e_ie_i^T)z \geq 0.$$ This is equal with showing that $$\alpha (Az)^H(Az) - z^He_ie_i^Tz \geq 0.$$ We have that the $2$-norm $\|x\|_2^2 = \langle x, x \rangle$, where $\langle \cdot, \cdot \rangle$ denotes the inner product of vectors. Moreover, you can compute yourself that $z^He_ie_i^Tz = |z_i|^2$, where $z_i$ is the $i$th entry of the vector $z$.
Let us now focus on the term $\alpha(Az)^H(Az)$ for which we have the following inequality: $$\alpha(Az)^H(Az) \geq \|B_{i\bullet}\|_2^2 \|Az\|_2^2$$ because of the definition of $\alpha$. Let us now use the inequality of Cauchy-Schwarz, which states that for every $p$-norm we have that $|x^Hy| \leq \|x\|_p\|y\|_p$. Applying this for $p = 2$ and the fact that $f: \mathbb{R} \to \mathbb{R}: x \mapsto x^2$ is a strictly increasing function on $\mathbb{R}^+$, we have that $$\|B_{i\bullet}^H\|_2^2 \|Az\|_2^2 \geq |B_{i\bullet} Az|^2$$ (we can write $\|B_{i\bullet}^H\|_2^2 = \|B_{i\bullet}\|_2^2$ for a vector).
If we now note that $Az$ is a linear combination of the columns of $A$ with coefficients $z_j$ (the entries of the vector $z$) and we also have that $B_{i\bullet}a_j = \delta_{i,j}$ where $a_i$ denotes the $i$th column of the matrix $A$, we have that $$|B_{i\bullet} Az|^2 = |B_{i\bullet}a_iz_i|^2 = |z_i|^2.$$
As a result, we find that $$\alpha (Az)^H(Az) - z^He_ie_i^Tz \geq 0 \geq |z_i|^2 - |z_i|^2 = 0,$$ which concludes this proof.
This is what I thought should be a proof. However, I am not completely sure if this statement is even true (I started doubting when I read the wikipedia page about positive semi-definite matrices, which states that for real matrices $A^TA$ is positive semi-definite, but there is no such statement on this page regarding complex valued matrices).
$\textbf{Remark: }$ I guess it is unusual to ask a question about an answer I have given, but I am quite worried that I gave a wrong answer, which the OP accepted. Perhaps the most worrying part to me is that he asked this question regarding 'research' he is doing, so I am extremely worried that I might have given a (possible) wrong answer.
• This was the wikipedia page I consulted, but it only states that for real matrices $A$ we have that $A^TA$ is positive definite and since there appears to not be a similar statement for complex matrices, it seemed quite unintuitive to me that we coul make a matrix positive definite by substracting a zero matrix (with one diagonal entry being $1$) to make it positive semi-definite... But so you also think my proof is valid? – Student Mar 16 '17 at 10:59
• In your solution, you haven't used that $A^T A$ is pos. def., so I think your proof is fine. – Andreas Mar 16 '17 at 11:23
• @Student From that page: "For any matrix A, the matrix $A^*A$ is positive semidefinite" This includes complex matrices. – Exodd Mar 16 '17 at 11:25
• As for the definition of positive definiteness for complex matrices, indeed there is a stronger and a weaker definition (regarding only the real part of $z^* M z$) - the portion of the Wikipedia page starting at my link elaborates this nicely. But again, this doesn't affect your proof. – Andreas Mar 16 '17 at 11:26 | 2019-09-18T09:57:37 | {
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https://math.stackexchange.com/questions/1596392/determine-the-size-of-a-test-bank | # Determine the Size of a Test Bank
Suppose you have two people take an exam which is composed of 30 questions which are randomly chosen from a test bank of n questions.
Person A and Person B both take different randomly generated instances of the exam, and then compare the question sets they were given. Person B notices that 7 of their 30 questions were repeated from Person A's question set.
Is there anyway to deduce the likely total number of questions in the test bank given you know 7/30 of them were repeated in a second instance of the exam? Obviously you would not get an exact value, but could you determine a range of probabilities for each different size of the test bank? How you would you go about solving this?
Thank you!
• You have a population of $n$ questions, from which $K=30$ are tagged as successes (Questions of Person A) and you draw a sample of $m=30$ questions (Questions for Person B). Then the number of successes (same questions) is hypergeometrically distributed with the above parameters and hence the probability of $k=7$ successes (same questions) is equal to $$\dfrac{\dbinom{30}{7}\dbinom{n-30}{23}}{\dbinom{n}{30}}=\frac{30!^2(n-30)!^2}{n!7!23!^2(n-53)!}$$ So, you need to solve $$\max_n \frac{(n-30)!(n-30)!}{n!(n-53)!}$$ – Jimmy R. Jan 1 '16 at 20:41
• @Stef, it's probably enough to use Stirling's formula to approximate the factorials, perhaps something simplifies in the process. – vonbrand Jan 1 '16 at 20:44
• @GregoryGrant I cheated a little :) I read your comment (came second) so posted this only for comparison. – Jimmy R. Jan 1 '16 at 20:50
• @Stef Got it, nobody can say you're not honest :-) – Gregory Grant Jan 1 '16 at 20:51
• Use a numeric method – vonbrand Jan 1 '16 at 21:09
The minimum number in the pool must be $53$. Suppose there are $n$ in total.
So it's like if you had an urn with $n$ balls, $30$ are white and $n-30$ are red. Then you pull $30$ balls at random. You want to know how many of the balls you pulled are white. Or more specifically you want to know the probability that $7$ of the $30$ you pull are white.
Let $A$ be the number of white balls. Then $P(A=k)$ is hypergeometric and equal to
$\frac{{{30}\choose{k}}{ {n-30}\choose{30-k}}}{{n}\choose{30}}$
$\frac{{{30}\choose{7}}{{n-30}\choose{23}}}{{n}\choose{30}}$
This is the probability of an overlap of exactly $7$.
You now need to find the $n$ that maximizes that probability.
If you start plugging in numbers (using a calculator) starting at $n=53$ you'll probably see that it goes up and then soon starts to go back down. Choose the max before it starts going back down. Shouldn't be too much larger than 53. I'm guessing somewhere around 100.
• Thank you for this explanation! It was extremely clear and I really appreciate it! After a little WolframAlpha magic, I found 128 questions to be the maximum. So if I understand this correctly, the test bank is composed of anywhere between 53-128 questions, with 128 questions being the most probable? – ProfessorStealth Jan 1 '16 at 20:59
• @ProfessorStealth Well there could be more than $128$ questions, it's just that $128$ is the most likely. There could be $500$ questions, but then having $7$ come up in common starts to be quite unlikely when there are that many in the pool. – Gregory Grant Jan 1 '16 at 21:01
• @ProfessorStealth Oh and you're very welcome :-) – Gregory Grant Jan 1 '16 at 21:03
This can be solved by 'capture-recapture' or 'mark-recapture' methods of estimating population size. One person is 'capture' and the other is 'recapture'. The 'Chapman' estimator (see Wikipedia on 'mark recapture') in this case is $\hat N_C = (30 + 1)(30 + 1)/(7 + 1) -1 \approx 119.$ Based on a hypergeometric model, this estimator is nearly unbiased. The Wikipedia gives two methods for finding a corresponding confidence interval.
The older and simpler 'Lincoln-Peterson' estimator is simply $\hat N = 30^2/7 \approx 128.$ It gives an infinite value if there happen to be no repeated questions. Thus $E(\hat N)$ does not exist, and one cannot discuss the unbiassedness of this estimator.
Addendum: The comments and the answer by @GregoryGrant are using the Lincoln-Peterson estimator, which is the maximum likelihood estimator, based on knowledge that there are 7 coincidences. Here is some relevant R code and a figure:
N = 100:150
like = choose(30,7)*choose(N-30, 30-7)/choose(N, 30)
N[like==max(like)] # value of N that maximizes 'like'
## 128
plot(N, like, pch=20); abline(v=128, lty="dotted")
Note: Here is one method to get an analytic solution for the maximum: Let $f(N|7) = {30 \choose 7}{N-30 \choose 23}/{N \choose 30}.$ Then look at $f(N|7)/f(N-1|7),$ simplifying it with lots of cancellation. Then notice the behavior of the ratio.
• Based on the exact solution he found using the hypergeometric, your $129$ is remarkably close to the $128$ he found. I must say I'm impressed I'll have to look into this Lincoln-Peterson estimator. – Gregory Grant Jan 1 '16 at 21:04
• @GregoryGrant: If $a = b$ is the number on each test and 7 is the number of repeated questions, then intuitively $7/b \approx a/N$. Then $N \approx ab/7.$ Sort of a 'method of moments' approach. – BruceET Jan 1 '16 at 21:27
• Nice that also gives $128$ ($900/7$ right?). Obviously there are many ways to skin this cat. This was obviously a stimulating question for all. And no lack of civility. That's how it should be, heartwarming :-) – Gregory Grant Jan 1 '16 at 21:31
• @GregoryGrant: Wishing us all a Happy Snark-Free New Year. – BruceET Jan 1 '16 at 21:35
• Thank you, that would be wonderful :-) – Gregory Grant Jan 1 '16 at 21:38 | 2019-07-19T05:53:42 | {
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http://dnmmagazine.com/nc9vynz/0380ce-differential-equations-rate-of-change | # differential equations rate of change
It is one of the major calculus concepts apart from integrals. A simple illustration of this type of dependence is changes of the Gross Domestic Product (GDP) over time. Also, check: Solve Separable Differential Equations. In biology and economics, differential equations are used to model the behavior of complex systems. Another observer belives that the rate of increase of the the radius of the circle is proportional to $$\frac{1}{(t+1)(t+2)}$$ iv) Write down a new differential equation for this new situation. Kumarmaths.weebly.com 2 Past paper questions differential equations 1. Ordinary Differential Equations Well, maybe it's just proportional to population. derivative Or is it in another galaxy and we just can't get there yet? Required fields are marked *, Important Questions Class 12 Maths Chapter 9 Differential Equations, $$\frac{d^2y}{dx^2}~ + ~\frac{dy}{dx} ~-~ 6y$$, Frequently Asked Questions on Differential Equations. *Exercise 8. Free ordinary differential equations (ODE) calculator - solve ordinary differential equations (ODE) step-by-step This website uses cookies to ensure you get the best experience. simply outstanding We are learning about Ordinary Differential Equations here! It is therefore of interest to study first order differential equations in particular. Mathematics » Differential Calculus » Applications Of Differential Calculus. We solve it when we discover the function y (or set of functions y). The Solution Inside The Tank Is Kept Well Stirred And Flows Out Of The Tank At A Rate … dy 4. y’, y”…. For instance, if individuals only live for 2 weeks, that's around 50% of a month, and then δ = 1 / time to die = 1 / 0.5 = 2, which means that the outgoing rate for deaths per month ( δ P) will be greater than the number in the population ( 2 ∗ P ), which to me doesn't make sense: deaths can't be higher than P. the weight gets pulled down due to gravity. For this particular virus -- Hong Kong flu in New York City in the late 1960's -- hardly anyone was immune at the beginning of the epidemic, so almost everyone was susceptible. dx dy The response received a rating of "5/5" from the student who originally posted the question. Make a diagram, write the equations, and study the dynamics of the … Integrating factor technique is used when the differential equation is of the form dy/dx + p(x)y = q(x) where p and q are both the functions of x only. T0 is the temperature of the surrounding, dT/dt is the rate of cooling of the body. By using this website, you agree to our Cookie Policy. The rate of change in sales {eq}S {/eq} is the first derivative w.r.t time {eq}t {/eq}, i..e {eq}S' = \frac{dS}{dt} {/eq}. By constructing a sequence of successive … This rate of change is described by the gradient of the graph and can therefore be determined by calculating the derivative. So now that we got our notation, S is the distance, the derivative of S with respect to time … Money earns interest. Note: we haven't included "damping" (the slowing down of the bounces due to friction), which is a little more complicated, but you can play with it here (press play): Creating a differential equation is the first major step. Let us see some differential equation applications in real-time. The following example uses integration by parts to find the general solution. So the rate of change is proportional to the amount of the substance hence: dx x dt v Therefore: dx kx dt The negative is used to highlight decay. Differential equations describe relationships that involve quantities and their rates of change. It is widely used in various fields such as Physics, Chemistry, Biology, Economics and so on. The solution is said to be $\\dfrac{dP}{dt} = k\\sqrt{P}$, It is like travel: different kinds of transport have solved how to get to certain places. Dec 1, 2020 • 1h 30m . To solve this differential equation, we want to review the definition of the solution of such an equation. Rates of Change and Differential Equations When given the rate of change of a quantity and asked to find the quantity itself we need to integrate : If () t f dt dQ = then () dt t f Q ⌡ ⌠ = Example 3 Water is pouring into a container at a rate given by 2 5 t dt dV = where 3 cm V is the volume of water in the container after t … The rate of change of distance with respect to time. Is there a road so we can take a car? Derivatives are fundamental to the solution of problems in calculus and differential equations. Then those rabbits grow up and have babies too! Assuming a quantity grows proportionally to its size results in the general equation dy/dx=ky. Differential Equations can describe how populations change, how heat moves, how springs vibrate, how radioactive material decays and much more. Write the answer. Why do we use differential calculus? In our world things change, and describing how they change often ends up as a Differential Equation: The more rabbits we have the more baby rabbits we get. where P and Q are both functions of x and the first derivative of y. It has the ability to predict the world around us. Suppose further that the population’s rate of change is governed by the differential equation dP dt = f (P) where f (P) is the function graphed below. Separation of the variable is done when the differential equation can be written in the form of dy/dx = f(y)g(x) where f is the function of y only and g is the function of x only. The rate of change of a certain population is proportional to the square root of its size. Next we work out the Order and the Degree: The Order is the highest derivative (is it a first derivative? The differential equation giving the rate of change of the radius of the rain drop is? For example, the Single Spring simulation has two variables: the position of the block, x, and its velocity, v. Each of those variables has a differential equation … Differentiation Connected Rates of Change. Here, the differential equation contains a derivative that involves a variable (dependent variable, y) w.r.t another variable (independent variable, x). Homogeneous Differential Equations That is the fact that $$f'\left( x \right)$$ represents the rate of change of $$f\left( x \right)$$. , so is "First Order", This has a second derivative It contains only one independent variable and one or more of its derivative with respect to the variable. More formally a Linear Differential Equation is in the form: OK, we have classified our Differential Equation, the next step is solving. Past paper questions differential equations 1. Time Rates If a quantity x is a function of time t, the time rate of change of x is given by dx/dt. Then, given the rate equations and initial values for S, I, and R, we used Euler’s method to estimate the values at any time in the future. In differential calculus basics, you may have learned about differential equations, derivatives, and applications of derivatives. A differential equation expresses the rate of change of the current state as a function of the current state. Thanks in advance! For any given value, the derivative of the function is defined as the rate of change of functions with respect to … Substitute in the value of x. The function given is $$y$$ = $$e^{-3x}$$. Using the same initial conditions as before, find the the new value for the constant v) Hence solve the differential … It is Linear when the variable (and its derivatives) has no exponent or other function put on it. A simple illustration of this type of dependence is changes of the Gross Domestic Product (GDP) over time. dy Definition 5.7. Compare the SIR and SIRS dynamics for the parameters = 1=50, = 365=13, = 400 and assuming that, in the SIRS model, immunity lasts for 10 years. To do 4 min read. 3. y is the dependent variable. The different types of differential equations are: Anyone having basic knowledge of Differential equation can attend this clas. If the order of the equation is 2, then it is called a second-order, and so on. So this is going to be our speed. Solution for Give a differential equation for the rate of change of vectors. It is mainly used in fields such as physics, engineering, biology and so on. The code assumes there are 100 evenly spaced times between 0 and 10, the initial value of $$y$$ is 6, and the rate of change … then it falls back down, up and down, again and again. "Ordinary Differential Equations" (ODEs) have. The derivative of the function is given by dy/dx. etc): It has only the first derivative So mathematics shows us these two things behave the same. 180 CHAPTER 4. Differential Equations and Rate of Change are investigated. $$A$$ is the amount or quantity of chemical that is dissolved in the solution, usually with units of weight like kg. d2x modem theory of differential equations. We substitute the values of $$\frac{dy}{dx}, \frac{d^2y}{dx^2}$$ and $$y$$ in the differential equation given in the question, On left hand side we get, LHS = 9e-3x + (-3e-3x) – 6e-3x, = 9e-3x – 9e-3x = 0 (which is equal to RHS). We also provide differential equation solver to find the solutions for related problems. 4 CHAPTER 1 FIRST-ORDER DIFFERENTIAL EQUATIONS e−1 = e−λτ −1 =−λτ τ = 1/λ. The rate of change of ... \begin{equation*} \text{ rate of change of some quantity } = \text{ rate in } - \text{ rate out }\text{.} Rates of Change. dx2 6) The motion of waves or a pendulum can also be described using these equations. which outranks the Function and rate of change … 2. x is the independentvariable. History. awesome Some people use the word order when they mean degree! Is it near, so we can just walk? For the differential equation (2.2.1), we can find the solution easily with the known initial data. First-order differential equation is of the form y’+ P(x)y = Q(x). The solution is detailed and well presented. Let us imagine the growth rate r is 0.01 new rabbits per week for every current rabbit. Section 8.4 Modeling with Differential Equations. Express the rate of change of y wrt tin terms of the rate of change wrt to x. t 1 = 2 l n 10 l n 2 Illustration : The rate at which a substance cools in moving air is proportional to the difference between the temperatures of the substance and that of the air. Your email address will not be published. Now we again differentiate the above equation with respect to x. Finally, we complete our model by giving each differential equation an initial condition. I learned from here so much. Linear Differential Equations So we need to know what type of Differential Equation it is first. Syllabus Applications of Differentiation 4.2.1 use implicit differentiation to determine the gradient of curves whose equations are given in implicit form 4.2.2 examine related rates as instances of the chain rule: 4.2.3 apply the incremental formula to differential equations 4.2.4 solve simple first order differential equations of the form ; differential equations … Connected rates of change can be difficult if you don't break it down. 5. c is some constant. and so on, is the first order derivative of y, second order derivative of y, and so on. 0 Example 4 dy =4x-3 dx dy dy dx -=-X-dt dx dt =5(4x-3) =5[4x(-2)-3] =-55 A spherical metal ball is heated so that its radius is expanding at the rate of0.04 mm per second. DIFFERENTIAL EQUATIONS S, I, and R and their rates S′, I′, and R′. (b) Let h be the half-life, that is, the amount of time it takes for a quantity to decay to one-half of its original amount. Using the same initial conditions as before, find the the new value for the constant v) Hence solve the differential equation 3) They are used in the field of medical science for modelling cancer growth or the spread of disease in the body. Differential equations can be divided into several types namely. Differential Equations Most of the differential equation questions will require a number of integration techniques. T. Tweety. Differential Calculus and you are encouraged to log in or register, so that you can track your … The rate of change of the radiss r cms if a ball of ice is given by dr/dt = -.01r cm./mins. Differential Equations: Feb 20, 2011: Differential equations help , rate of change: Calculus: Jun 16, 2010: differential calculus rate of change problems: … The rate law or rate equation for a chemical reaction is an equation that links the initial or forward reaction rate with the concentrations or pressures of the reactants and constant parameters (normally rate coefficients and partial reaction orders). But we also need to solve it to discover how, for example, the spring bounces up and down over time. See how we write the equation for such a relationship. (a) Determine the differential equation describing the rate of change of glucose in the bloodstream with respect to time. Please help. For example, the Single Spring simulation has two variables: the position of the block, x, and its velocity, v. Each of those variables has a differential equation saying how that variable evolves over time. A differential equation contains derivatives which are either partial derivatives or ordinary derivatives. So this is going to be our speed. And how powerful mathematics is! dx. Generally, $\frac{dQ}{dt} = \text{rate in} – \text{rate out}$ Typically, the resulting differential equations are either separable or first-order linear DEs. But that is only true at a specific time, and doesn't include that the population is constantly increasing. The liquid entering the tank may or may not contain more of the substance dissolved in it. nice web "Partial Differential Equations" (PDEs) have two or more independent variables. Since λ = 1/τ,weget 1 2 r0 = r0e −λh 1 2 r0 = r0e −h/τ 1 2 = e −h/τ −ln2 =−h/τ. Differential Equation Contents. Differential equations describe relationships that involve quantities and their rates of change. The response received a rating of "5/5" from the student who originally posted the question. The differential equation for the mixing problem is generally centered on the change in the amount in solute per unit time. So it is a Third Order First Degree Ordinary Differential Equation. That short equation says "the rate of change of the population over time equals the growth rate times the population". The solution is detailed and well presented. The primary purpose of the differential equation is the study of solutions that satisfy the equations and the properties of the solutions. The order of the differential equation is the order of the highest order derivative present in the equation. So let us first classify the Differential Equation. The underlying logic that's just driven by the actual differential equation. But first: why? The rate of change of x with respect to y is expressed dx/dy. If the dependent variable has a constant rate of change: \begin{align} \frac{dy}{dt}=C\end{align} where $$C$$ is some constant, you can provide the differential equation in the f function and then calculate answers using this model with the code below. a simple model gives the rate of decrease of its … Watch Now. It can be represented in any order. The order of ordinary differential equations is defined as the order of the highest derivative that occurs in the equation. Sep 2008 631 2. It is a very useful to me. Differential equations help , rate of change Watch. Rates of Change; Example. Hi, I am from Bangladesh. View Answer. 2 k. B ... Form the differential equation of the family of circles touching the X-axis at the origin. We expressed the relation as a set of rate equations. So it is better to say the rate of change (at any instant) is the growth rate times the population at that instant: And that is a Differential Equation, because it has a function N(t) and its derivative. The rate of change of Note as well that in man… The population will grow faster and faster. But don't worry, it can be solved (using a special method called Separation of Variables) and results in: Where P is the Principal (the original loan), and e is Euler's Number. Verify that the function y = e-3x is a solution to the differential equation $$\frac{d^2y}{dx^2}~ + ~\frac{dy}{dx} ~-~ 6y$$ = $$0$$. Introduction to Time Rate of Change (Differential Equations 5) The interest can be calculated at fixed times, such as yearly, monthly, etc. Liquid leaving the tank will of course contain the substance dissolved in it. Rates of Change and Differential Equations: Filling and Leaking Water Tank: Differential Equations: Apr 20, 2013: differential equation from related rate of change. To gain a better understanding of this topic, register with BYJU’S- The Learning App and also watch interactive videos to learn with ease. Over the years wise people have worked out special methods to solve some types of Differential Equations. 5) They help economists in finding optimum investment strategies. If initially r =20cms, find the radius after 10mins. The higher-order differential equation is an equation that contains derivatives of an unknown function which can be either a partial or ordinary derivative. Share. Learn how to solve differential equation here. I don't understand how to do this problem: Write and solve the differential equation that models the verbal statement. In Mathematics, a differential equation is an equation with one or more derivatives of a function. Introducing a proportionality constant k, the above equation can be written as: Here, T is the temperature of the body and t is the time. I'm literally having trouble going about this question since there is no similar example to the following question in the book! The derivative represents a rate of change, and the differential equation describes a relationship between the quantity that is continuously varying with respect to the change in another quantity. Differential equations help , rate of change. The rate of change, with respect to time, of the population. Mohit Tyagi. Question: Write The Differential Equation, Do Not Evaluate, Represent The Rate Of Change Of Overall Rate Of The Sodium. The weight is pulled down by gravity, and we know from Newton's Second Law that force equals mass times acceleration: And acceleration is the second derivative of position with respect to time, so: The spring pulls it back up based on how stretched it is (k is the spring's stiffness, and x is how stretched it is): F = -kx, It has a function x(t), and it's second derivative The rate of change of population is proportional to its size. The degree is the exponent of the highest derivative. Section 5.2 First Order Differential Equations ¶ In many fields such as physics, biology or business, a relationship is often known or assumed between some unknown quantity and its rate of change, which does not involve any higher derivatives. But when it is compounded continuously then at any time the interest gets added in proportion to the current value of the loan (or investment). 4) Movement of electricity can also be described with the help of it. Jun 16, 2010 #1 A mathematician is selling goods at a car boot sale. The general definition of the ordinary differential equation is of the form: Given an F, a function os x and y and derivative of y, we have. To understand Differential equations, let us consider this simple example. Calculus. Let’s study about the order and degree of differential equation. This statement in terms of mathematics can be written as: This is the form of a linear differential equation. The solution to these DEs are already well-established. An ordinary differential equation Âcontains one independent variable and its derivatives. If the temperature of the air is 290K and the substance cools from 370K to 330K in 10 minutes, when will the temperature be 295K. A differential equation states how a rate of change (a "differential") in one variable is related to other variables. Find the general form of n-th order ODE is given as by a mass on spring! Family of circles touching the X-axis at the origin and leaving a holding tank contain... Mathematician is selling goods at a point volume be increasing when the population of a function of variables results the! Is proportional to population that involve quantities and their rates S′, I′, and so on initial.! ( e^ { -3x } \ ) by dr/dt = -.01r cm./mins we. Biology, economics and so on basic knowledge of differential equations of circles touching the X-axis at origin! State as a function at a rate problem, the derivative populations change, how heat,... ), where P is expressed in millions dy/dx does not count, as is. Quantities and their rates of change of x and the rate of change can utilized! Linear differential equations rate of change equation also used to model the behavior of complex systems can describe how change! And the rate of change being the difference between the rate of and... And study the dynamics of the graph and can therefore be determined by calculating derivative! Previous chapter 's just driven by the gradient of the economy starter Tweety ; start date Jun,. In these problems we will study questions related to rate change in which one or of! In it concepts apart from integrals is widely used in the mathematical modeling physical. Equation Âthat contains one or more independent variables be difficult if you do understand... These two things behave the same ( differential equations describe various exponential growths and.... 2 k. B... form the differential equation, we complete our model by giving each differential equation it a... Liquid leaving the tank is kept well Stirred and Flows out of available food } \ ) things in engineering! P is expressed in millions various differential equations rate of change of the highest order derivative of y, second order derivative present the. Governing differential equation which has degree equal to differential equations rate of change write the equations letÂ. Finally, we can just walk equations '' ( PDEs ) have example uses integration by parts could be.... ’ S study about the order of the population over time and one or more of its derivatives such... Physics, engineering, biology and economics, differential equations ( ifthey can be formulated as differential ''. Rate problem, the derivative of y equation applications in real-time used in the differential are... '' from the student who originally posted the question derivatives ) has no or... Is described by the actual differential equation says the rate of change of y this question since there no! Statement in terms of the derivatives temperature of the body dynamics of the current.. Medical science for modelling cancer growth or the spread of disease in the universe to the given equation! Related problems an application in the amount in solute per unit time Linear differential equation such... In this class we will start with a substance that is only true at car! The solutions we discover the function is a wonderful way to express something, but is hard to.! Jun 16, 2010 ; Tags change differential equations are very important in the universe some... Ce kt of dNdt as how much the population the ordinary differential equation is 2, then falls... 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Start new discussion reply, chemistry, biology, economics and so on, is the distance, the differential! Of time t, the more new rabbits per week problem: and... ( or set of rate equations very important in the engineering field finding... Equation for such a relationship for every current rabbit functions y ) electricity... Here some examples for different orders of the family of circles touching the X-axis at the origin study of that. Expressed dx/dy go on forever as they will soon run out of available.... The spread of disease in the engineering field for finding the relationship between various parts of the major concepts... Equations solution Guide to help you is like travel: different kinds of transport have how! Equation describing the rate of cooling of the solutions for related problems dNdt as how... Car boot sale our notation, S is the temperature of the highest order derivative present the! 6 ) the motion of waves or a pendulum can also be described these... The time rate of change wrt to x fixed times, such physics... Each differential equation is 2, then it is one of the population '' (... Awesome very very nice, find the radius after 10mins function which can be divided into several types namely growth! Population over time just ca n't get there yet some types of differential equations it a derivative! Unit time mathematics, the variable of integration is time t. 2 thought why a hot cup coffee! So on n't include that the population, the time rate of change a... ( 2.2.1 ), where P and Q are both functions of with. Then 1000×0.01 = 10 new rabbits per week for every current rabbit to be solved differential equations rate of change ) at... Species is described by the actual differential equation which has degree equal to 1 as physics, engineering,,! Change being the difference between the rate of change of a particular species is described the! Set of rate equations the total rate of change of the radiss cms. How springs vibrate, how springs vibrate, how springs vibrate, radioactive... An example of this is a Third order first degree ordinary differential equation says the rate of of... And can therefore be determined by calculating the derivative of y, second order present. The family of circles touching the X-axis at the origin a ) Determine the equation! ) they help economists in finding optimum investment strategies, find the radius after 10mins rate in! Have two or more independent variables well, but is hard to use x\ ) order degree... A spring describe various exponential growths and decays and r and their of... So mathematics shows us these two things behave the same derivative ) the body ’ S study the. A rating of 5/5 '' from the total rate of change wrt to.... An equation with respect to x ( and its derivatives ) has no exponent or other function put on.... ( e^ { -3x } \ ) 1 a mathematician is selling goods at a?. '' from the student who originally posted the question and does n't include that population. A simple illustration of this type of differential equations solution Guide to help you condition is m = kt... And the properties of the economy mathematics » differential Calculus » applications of differential says! Is selling goods at a rate … modem theory of differential equation Âcontains one variable... Investment over time Guide to help you mean degree Âthat contains one or more independent and! Radius is 3 mm out special methods to solve it to discover,... From integrals to be solved! ) true at a point spring bounces up and down, up down! Tank is kept well Stirred and Flows out of available food function P ( x ) include that population. There is no similar example to the solution of such an equation that relates function. Be utilized as an application in the book the difference between the rate of change dNdt is then 1000×0.01 10... Function given is \ ( y\ ) = \ ( e^ { -3x } \.. More examples here: an ordinary differential equation, we complete our model by giving each equation. | 2021-10-26T04:35:13 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=0f6jdnc8eanf2pk5uf3ta0o7l0&topic=1600.0 | ### Author Topic: Q7 TUT 0201 (Read 2647 times)
#### Victor Ivrii
• Elder Member
• Posts: 2569
• Karma: 0
##### Q7 TUT 0201
« on: November 30, 2018, 04:05:34 PM »
(a) Determine all critical points of the given system of equations.
(b) Find the corresponding linear system near each critical point.
(c) Find the eigenvalues of each linear system. What conclusions can you then draw about the nonlinear system?
(d) Draw a phase portrait of the nonlinear system to confirm your conclusions, or to extend them in those cases where the linear system does not provide definite information about the nonlinear system.
\left\{\begin{aligned} &\frac{dx}{dt} = 1 - xy, \\ &\frac{dy}{dt} = x - y^3. \end{aligned}\right.
Bonus: Computer generated picture
#### Yulin WANG
• Full Member
• Posts: 17
• Karma: 25
• MAT244H1 2018F
##### Re: Q7 TUT 0201
« Reply #1 on: November 30, 2018, 04:39:52 PM »
(a)
\left\{
\begin{array}{**lr**}
1-xy=0 & \\
x-y^{3}=0\\
\end{array}
\right.
\left\{
\begin{array}{**lr**}
xy=1 & \\
x=y^{3}\\
\end{array}
\right.
\left\{
\begin{array}{**lr**}
x=1 & \\
y=1\\
\end{array}
\right.
or
\left\{
\begin{array}{**lr**}
x=-1 & \\
y=-1\\
\end{array}
\right.
Therefore, the critical points are (1,1) and (-1,-1)
(b)
The Jacobian matrix of the vector field is:
\begin{align*}
J &= \begin{bmatrix}
-y & -x \\
1 & -3y^{2}
\end{bmatrix}\\
~\\
J(1,1) &= \begin{bmatrix}
-1 & -1 \\
1 & -3
\end{bmatrix}\\
~\\
J(-1,-1) &= \begin{bmatrix}
1 & 1 \\
1 & -3
\end{bmatrix}
\end{align*}
(c)
\begin{align*}
For (1,1), let A&= \begin{bmatrix}
-1 & -1 \\
1 & -3
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
-1-\lambda & -1 \\
1 & -3-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda+3)(\lambda+1)+1=0\\
~\\
\lambda_{1} &= \lambda_{2} = -2 \\
~\\
Then \ the \ system \ has \ a \ stable \ improper \ node \ at \ (1,1) \\
~\\
For (-1,-1), let A&= \begin{bmatrix}
1 & 1 \\
1 & -3
\end{bmatrix}\\
~\\
A-\lambda I &= \begin{bmatrix}
1-\lambda & 1 \\
1 & -3-\lambda
\end{bmatrix}\\
~\\
det(A-\lambda I) &=(\lambda+3)(\lambda-1)-1=0\\
~\\
\lambda = -1 \pm \sqrt{5} \\
~\\
Then \ the \ system \ has \ a \ unstable \ saddle \ point \ at \ (1,1) \\
\end{align*}
(d) In the attachment.
« Last Edit: November 30, 2018, 05:25:52 PM by Yulin WANG »
#### Zhuojing Yu
• Newbie
• Posts: 3
• Karma: 2
##### Re: Q7 TUT 0201
« Reply #2 on: November 30, 2018, 06:29:30 PM »
I think when (1,1), it is node or spiral point, not IN(improper node).
#### Jingze Wang
• Full Member
• Posts: 30
• Karma: 25
##### Re: Q7 TUT 0201
« Reply #3 on: November 30, 2018, 08:31:23 PM »
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted
#### Yulin WANG
• Full Member
• Posts: 17
• Karma: 25
• MAT244H1 2018F
##### Re: Q7 TUT 0201
« Reply #4 on: November 30, 2018, 11:13:57 PM »
I think the phase portrait that Yulin drew is correct, here is computer generated picture since no one posted
Thanks for submitting the computer-generated phase portrait!!!
BTW, how do u plot the phase portrait on a computer? | 2022-01-18T05:31:22 | {
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https://math.stackexchange.com/questions/1259412/please-help-collecting-examples-of-finite-infinite-rings-satisfying-different-co | 0) Every nonzero element of a finite ring is either a zero divisor or a unit. This is proved in Every nonzero element in a finite ring is either a unit or a zero divisor
1) If a ring R satisfies the condition that "every nonzero element is either a zero divisor or a unit", must R be finite? If not, can you please give at least two non-isomorphic counterexamples?
2) If a ring R satisfies the condition that "every nonzero element is a unit", will R be finite or infinite? If both cases are possible, can you please give at least two non-isomorphic examples in both cases?
3) Does there exist finite/infinite rings such that "every nonzero nonidentity element is a zero divisor"? If both answers are yes, can you please give at least two non-isomorphic examples in both cases?
Edit: 4) If a ring R has an element that is neither a unit nor a zero divisor, then R must be infinite. Now will R be countable or uncountable? Can you please give examples (especially if both cases are possible)?------(Ok I know $\mathbb{Z}$ is a countable example. Does there exist uncountable examples?)
P.S. I am self-learning undergraduate level mathematics. Sorry if the question is trivial or stupid.
Edit: After reading the answers, I find that doing textbook excises is still not enough to be proficient in this subject. I need to learn to think in more various ways.
• For 1, just start by looking through a few of the rings you know for counterexamples. You should find one or two very soon. For 3, a stupid (but fun!) counterexample would be for example if the set of non-unit, non-zero elements were empty... – Circonflexe Apr 30 '15 at 17:01
• Ok for 1), $\mathbb{Z}$ and $\mathbb{R}$ are non-isomorphic examples. – willhuang00 Apr 30 '15 at 17:27
• counterexample for 1: $\mathbb Z$. 3: every non-unit is in maximal ideal. – user 1 Apr 30 '15 at 17:28
• @willhuang00 $\Bbb{Z}$ is no-good for 1, because it has no zero divisors, but the only units are $\pm 1$. – A.P. Apr 30 '15 at 17:28
• Yes, you can wonder. I fear that many people (myself included), especially among the oldest users, have been bittered by the too many people resorting to MSE as a homework-solving mechanical Turk. I'm curious, though: if you did know that $2 \times 2$ matrices over $\Bbb{R}$ are a counterexample for 1), why didn't you try to see if $3 \times 3$ (or even $n \times n$) matrices worked, too? – A.P. Apr 30 '15 at 18:13
1) This is true for all fields, like $\mathbb{R}$, $\mathbb{C}$, $\mathbb{Q}$ and $\mathbb{Z}/2\mathbb{Z}$ and all finite-dimensional algebras over these fields. Finite dimensional algebras covers matrix rings, and lots more besides.
2) This is exactly the condition that defines division rings, and the zero ring which is not a division ring. A division ring with a commutative product is called a field and I gave plenty of examples of fields above.
3) $(\mathbb{Z}/2\mathbb{Z})$ and $(\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z})$ and $(\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z})$ etc. I think that these and the zero ring are the only finite rings that meet this condition.
Response to edit
4) The ring of polynomials over the real numbers. Also the ring of continuous function from the real numbers to the real numbers; the function $x \rightarrow x$ is not a unit and not a zero divisor.
• Technically the class of cardinals doesn't count since a ring must be a SET with binary operations... – Nishant Apr 30 '15 at 23:18
• @Nishant I added this – jkabrg Apr 30 '15 at 23:22
• More problematically, no cardinal besides $0$ has an additive inverse. – Nishant Apr 30 '15 at 23:27
The rings $\mathbb{Q}$, $\mathbb{R}$,$\mathbb{C}$ (these are also fields) and $\mathbb{H}$ (the quaternions) are infinite examples of 2).
For 1) and 3) you can consider the subrings of $M(2,\mathbb{R})$ formed by the matrices of the form: $$D=\begin{bmatrix}a&0\\0&b\end{bmatrix} \qquad T=\begin{bmatrix}a&c\\0&b\end{bmatrix} \qquad U=\begin{bmatrix}0&c\\0&0\end{bmatrix}$$ The matrices $D$ and $T$ form two non isomorphic rings, The matrices $U$ are a ring without unity that is not isomorphic to $D$-ring and $T$-ring.
You can see that matrices of the form $\begin{bmatrix}0&0\\0&b\end{bmatrix}$ are zero divisors in $D$-ring and in the $T$-ring and you can easely see that there are invertible elements in these rings.
The ring of matrices $U$ has no invertible elements, all its elements are zero divisors.
Finally, You can construct similar exemples in $M(n;\mathbb{K})$ for any field $\mathbb{K}$, and all are not isomorphic.
1/2) Every field has this property; as part of the definition of a field every non-zero element is a unit. $\mathbb{Q}[x]/(p_n(x))$ where $p_n(x)$ is an irreducible polynomial of degree $n$ gives an infinite non-isomorphic family. As an example of an infinite ring where nonzero elements are zero divisors or units and where both occur take $n \times n$ matrices over some infinite field ($\mathbb{Q},\mathbb{R},\mathbb{C}$ for example).*
3) Check out http://en.wikipedia.org/wiki/Boolean_ring. These have the property that all elements $x$ are idempotent meaning $x^2=x$. But then $x(x-1)=0$ so every element besides $1$ is a zero divisor.
• * to explain this example, take some matrix $x \not \in \{0,1\}$. Either it's invertible (a unit) or has a nonzero kernel. If there's a nonzero kernel pick some other matrix which which maps a basis vector into the kernel of $x$ and sends all other basis vectors to 0. Then the product of this with $x$ will be 0. – rVitale Apr 30 '15 at 17:32
• Even (arguably) more easily: the ring of $n \times n$ matrices on a field $K$ is a vector space on $K$, thus it has property 1 because it is a product of fields. All products of fields have property 1 because multiplication is defined component-wise, so the $0$ divisors are exactly the elements with $0$ in at least one component, while every other element has an inverse because each of its components has one. – A.P. Apr 30 '15 at 17:45
• As a remark for the OP our examples are not isomorphic as rings. Matrix multiplication in the sense of composition of linear maps and component-wise multiplication give different ring structures. – rVitale Apr 30 '15 at 20:12
Define a ring R such that R has underlying group structure Z/(nZ), where n is an arbitrary integer greater or equal to 2 but define multiplication as xy=0 for all x,y in R. Then R is a ring and R satisfies part (3). | 2019-10-22T23:34:13 | {
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http://bootmath.com/there-are-at-least-three-mutually-non-isomorphic-rings-with-4-elements.html | # There are at least three mutually non-isomorphic rings with $4$ elements?
Is the following statement is true?
There are at least three mutually non-isomorphic rings with $4$ elements.
I have no idea or counterexample at the moment. Please help. So far I know about that a group of order $4$ is abelian and there are two non isomorphic groups of order $4$ like $K_4(non cyclic)$ and $\mathbb Z_4(cyclic)$.
#### Solutions Collecting From Web of "There are at least three mutually non-isomorphic rings with $4$ elements?"
By ring, I always mean unital ring. Each of the following rings has four elements:
$R_1 = \mathbb{Z}/4~, ~R_2 = \mathbb{F}_2 \times \mathbb{F}_2 = \mathbb{F}_2[x]/(x^2+x)~, ~R_3 = \mathbb{F}_4=\mathbb{F}_2[x]/(x^2+x+1)~, ~R_4 = \mathbb{F}_2[x]/(x^2)$
They are non-isomorphic because only $R_3$ is a field, only $R_1$ has characteristic $\neq 2$, and only $R_2,R_3$ are reduced.
Conversely, let $R$ be a ring with four elements. If $a \in R \setminus \{0,1\}$, then the centralizer of $a$ is a subgroup of $(R,+)$ with at least three elements $0,1,a$, so by Lagrange also the fourth element has to commute with $a$. Thus, $R$ is commutative. If $R$ is reduced, then it is a finite product of local artinian reduced rings, i.e. fields, so that $R \cong R_2$ or $R \cong R_3$. If $R$ is not reduced, there is some $a \in R \setminus \{0\}$ such that $a^2=0$. Since $0,1,a,a+1$ are pairwise distinct, these are the elements of $R$. If $2=0$, then we get an injective homomorphism $\mathbb{F}_2[x]/(x^2) \to R, x \mapsto a$. Since both sides have four elements, it is an isomorphism. If $2 \neq 0$, the characteristic has to be $4$, i.e. we get an embedding $\mathbb{Z}/4 \to R$, which again has to be an isomorphism.
Of course, this classification can also be obtained by more elementary methods. For other orders, see:
The smallest non-commutative ring has $8$ elements and is given by $\begin{pmatrix} \mathbb{F}_2 & \mathbb{F}_2 \\ 0 & \mathbb{F}_2 \end{pmatrix} \subseteq M_2(\mathbb{F}_2)$.
Consider the ring $R = \Bbb{Z}/2\Bbb{Z}[i]$. Alternatively $R$ can be constructed as a quotient
$$R \cong \Bbb{Z}[x]/(2,x^2+1).$$
As a ring $R$ is not isomorphic to either $S = \Bbb{Z}/2\Bbb{Z} \times \Bbb{Z}/2\Bbb{Z}$ or $T = \Bbb{Z}/4\Bbb{Z}$.
Edit: Perhaps I should add why the ring $R$ is not isomorphic to either $S$ or $T$. Firstly by counting orders of elements $R$ cannot be isomorphic to $T$; $T$ has an element of order 4 while $R$ does not. So now the penultimate question is why is $R$ is not isomorphic to $S$? As groups they are certainly isomorphic but as rings they can’t be. The reason is because the presence of $i$ means that the multiplication in $\Bbb{Z}/2\Bbb{Z}[i]$ is not the same as the multiplication in $S$ which is the usual one coming from the product ring structure.
In view of this we see that $R$ has non-trivial nilpotent elements, $(1+i)^2 = 1 – 2i +i^2 = 1 – 1 = 0$ while obviously $\text{nilrad}$ $S = 0$. Thus $R \not\cong S$.
These can be exhaustively enumerated using alg. To enumerate the number of non-isomorphic rings of orders $1,2,\ldots,8$, we enter:
./alg theories/ring.th --size 1-8 --count
which outputs:
# Theory ring
Constant 0.
Unary ~.
Binary + *.
Axiom plus_commutative: x + y = y + x.
Axiom plus_associative: (x + y) + z = x + (y + z).
Axiom zero_neutral_left: 0 + x = x.
Axiom zero_neutral_right: x + 0 = x.
Axiom negative_inverse: x + ~ x = 0.
Axiom negative_inverse: ~ x + x = 0.
Axiom zero_inverse: ~ 0 = 0.
Axiom inverse_involution: ~ (~ x) = x.
Axiom mult_associative: (x * y) * z = x * (y * z).
Axiom distrutivity_right: (x + y) * z = x * z + y * z.
Axiom distributivity_left: x * (y + z) = x * y + x * z.
size | count
-----|------
1 | 1
2 | 2
3 | 2
4 | 11
5 | 2
6 | 4
7 | 2
8 | 52
Check the numbers [2, 2, 11, 2, 4, 2, 52](http://oeis.org/search?q=2,2,11,2,4,2,52) on-line at oeis.org
If we want it to print out the addition and multiplication tables, we can remove --count from the command line:
./alg theories/ring.th --size 4
If we want to work with unital rings, we can use:
./alg theories/unital_ring.th --size 1-8 --count
which gives:
# Theory unital_ring
Theory unital_ring.
Constant 0 1.
Unary ~.
Binary + *.
Axiom plus_commutative: x + y = y + x.
Axiom plus_associative: (x + y) + z = x + (y + z).
Axiom zero_neutral_left: 0 + x = x.
Axiom negative_inverse: x + ~ x = 0.
Axiom mult_associative: (x * y) * z = x * (y * z).
Axiom one_unit_left: 1 * x = x.
Axiom one_unit_right: x * 1 = x.
Axiom distrutivity_right: (x + y) * z = x * z + y * z.
Axiom distributivity_left: x * (y + z) = x * y + x * z.
# Consequences of axioms that make alg run faster:
Axiom zero_neutral_right: x + 0 = x.
Axiom negative_inverse: ~ x + x = 0.
Axiom zero_inverse: ~ 0 = 0.
Axiom inverse_involution: ~ (~ x) = x.
Axiom mult_zero_left: 0 * x = 0.
Axiom mult_zero_right: x * 0 = 0.
size | count
-----|------
1 | 0
2 | 1
3 | 1
4 | 4
5 | 1
6 | 1
7 | 1
8 | 11
Check the numbers [1, 1, 4, 1, 1, 1, 11](http://oeis.org/search?q=1,1,4,1,1,1,11) on-line at oeis.org | 2018-06-25T06:05:31 | {
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http://mathhelpforum.com/algebra/133721-square-numbers-problem-solving.html | # Thread: Square numbers problem solving
1. ## Square numbers problem solving
Hello MHF,
Which of the following cannot be the last digit of the sum of the squares of seven consecutive numbers?
A: 3
B:5
C:6
D:7
E:8
I have no idea where to begin or what to do, i tried using the squares of 1-7
but obviously that wouldn't give me an answer (hopeless thing to do)
Any help appreciated.
2. Wow. I don't know any fancy way to do this but I would just go up the list until I was able to rule out four of the five numbers. Start with 4 + 9 + 16 +25 + 36 + 49 + 64 then go from there.
3. Originally Posted by 99.95
Hello MHF,
Which of the following cannot be the last digit of the sum of the squares of seven consecutive numbers?
A: 3
B:5
C:6
D:7
E:8
I have no idea where to begin or what to do, i tried using the squares of 1-7
but obviously that wouldn't give me an answer (hopeless thing to do)
Any help appreciated.
I can only offer a trial and error method:
Consider the last digit of a number and the last digit of it's square:
$\displaystyle \begin{array}{c|c}\text{last digit of number}& \text{last digit of square}\\ \hline\\0&0\\9 \ or\ 1&1 \\8\ or\ 2&4\\7\ or\ 3&9\\6\ or\ 4&6\\5&5\\ \hline\end{array}$
If you add the squares of numbers with end digits 0 to 6 you'll get 31 that means the end digit 1;
if you add the squares of numbers with end digits 1 to 7 you'll get 40 that means the end digit 0;
if you add the squares of numbers with end digits 2 to 8 you'll get 43 that means the end digit 3;
if you add the squares of numbers with end digits 3 to 9 you'll get 40 that means the end digit 0;
and so on...
Compare your results with the given answers (and then pick the right one!)
If I didn't make a mistake the answer is D.
4. Hello, 99.95!
Which of the following cannot be the last digit
of the sum of the squares of seven consecutive numbers?
. . $\displaystyle (A)\;3\qquad (B)\;5 \qquad (C)\;6\qquad (D)\;7 \qquad (E)\;8$
Note the last digit of all squares.
. . . . $\displaystyle \begin{array}{c} 0^2 \to 0 \\ 1^2 \to 1 \\ 2^2 \to 4 \\ 3^2 \to 9 \\ 4^2 \to 6 \end{array}\qquad \begin{array}{c}5^2 \to 5 \\ 6^2 \to 6 \\ 7^2 \to 9 \\ 8^2 \to 4 \\ 9^2 \to 1 \end{array}$
We see that squares cannot end in 2, 3, 7, or 8.
Let the 7 consecutive numbers be: .$\displaystyle x-3,\;x-2,\;x-1,\;x,\;x+1,\;x+2,\;x+3$
Then we have:
. . $\displaystyle \begin{array}{ccc} (x-3)^2 &=& x^2 - 6x + 9 \\ (x-2)^2 &=& x^2-4x+4 \\ (x-1)^2 &=& x^2-2x + 1 \\ x^2 &=& x^2\qquad\qquad \\ (x+1)^2 &=& x^2 + 2x + 1 \\ (x+2)^2 &=& x^2 + 4x + 4 \\ (x+3)^2 &=& x^2 + 6x + 9 \\ \hline \text{Sum} &=& 7x^2 + 28\end{array}$
Suppose the sum $\displaystyle 7(x^2+4)$ ends in 7.
. . Then $\displaystyle x^2+4$ must end in 1.
. . And $\displaystyle x^2$ must end in 7.
But no square ends in 7.
Answer: .$\displaystyle (D)\;7$
5. Originally Posted by 99.95
Hello MHF,
Which of the following cannot be the last digit of the sum of the squares of seven consecutive numbers?
A: 3
B:5
C:6
D:7
E:8
I have no idea where to begin or what to do, i tried using the squares of 1-7
but obviously that wouldn't give me an answer (hopeless thing to do)
Any help appreciated.
Here is another approach. Let
$\displaystyle s(n) = 1^2 + 2^2 + 3^2 + \dots + n^2$
$\displaystyle = (1/6)\; n (n+1) (2n+1)$
(a well-known identity).
Then the sum of the squares of the 7 consecutive integers ending in n is
$\displaystyle s(n) - s(n-7) = (1/6) \; [ n (n+1) (2n+1) - (n-7)(n-6)(2n-13)] = 7 \; (n^2 - 6n + 13)$
Evaluating $\displaystyle 7\; (n^2 - 6n + 13)$ for n = 0, 1, 2, ..., 9 modulo 10 will then reveal the possibilities for the last digit of the sum.
Soroban posted his answer while I was composing this-- essentially the same approach? Or maybe not.[/edit]
6. Thanks guys both methods (although similiar) are definately useful and yes the answer is "D" (7). May I ask how you guys go about solving these, is it just some logical approach you take?
Thanks again
By the way, does modulo = remainder?
Eg: 10/4= 2 r 2 or 2 mod (2) ?
7. Yes, "Modulo 10" means divide by 10 and keep the remainder, i.e., the last digit.
As for how I do it, I just have lots of experience solving problems, and I am sure Soroban has more experience than I. | 2018-05-23T01:42:18 | {
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https://math.stackexchange.com/questions/3107702/a-finite-state-space-markov-chain-has-no-null-recurrent-states | # A finite state space Markov chain has no null-recurrent states
I'm fairly new to Markov chains. At the moment, I'm trying to understand why a finite state space Markov chain cannot have any null-recurrent states.
Searching on math.SE, I found this answer, which argues as follows:
Consider a communicating class $$C$$ containing a null, recurrent state. Then $$p_{ij}(n)\to 0$$ for every $$i,j\in C$$, but since this class $$C$$ is closed this gives the contradiction $$1=\lim_n \sum_j p_{ij}(n)=0.$$
My question is: How can we show that $$p_{ij}(n) \to 0$$ for all $$i, j$$ in $$C$$?
I suppose it's sufficient to show that, if $$k$$ is the null-recurrent state, then $$p_{kk}(n) \to 0$$. Indeed, for any $$i, j \in C$$, there exist $$m_1$$ and $$m_2$$ such that $$p_{ki}(m_1) > 0$$ and $$p_{jk}(m_2) > 0$$, so if $$p_{kk}(n) \to 0$$, then $$p_{ij}(n) \leq \frac{p_{kk}(m_1 + n + m_2) }{ p_{ki}(m_1) p_{jk}(m_2) }\to 0.$$
But then, how do I show that $$p_{kk}(n) \to 0$$? The fact that $$k$$ is null-recurrent tells us that $$\sum_{n = 1}^\infty n f_{kk}(n) = \infty$$, where $$f_{kk}(n)$$ is the probability of the first return from $$k$$ to $$k$$ taking place after $$n$$ steps. I don't see the link between $$\sum_{n = 1}^\infty n f_{kk}(n)$$ being infinite and $$p_{kk}(n)$$ tending to zero.
Edit: I thought of using the fact that $$P_{kk}(s) = 1 + F_{kk}(s) P_{kk}(s), \ \ \ \ -1 < s \leq 1,$$ where $$P_{kk}(s) := \sum_{n=0}^\infty p_{kk}(n) s^n, \ \ \ \ \ \ \ F_{kk}(s) := \sum_{n=1}^\infty f_{kk}(n)s^n$$ See page 12 of these notes. If you rearrange, and differentiate, you get $$F_{kk}'(s) = - \frac{P_{kk}'(s)}{(P_{kk}(s))^2}.$$ If $$\sum_{n = 1}^\infty n f_{kk}(n) = \infty$$, then $$F_{kk}'(1)$$ must be infinite. But then I'm not sure what to do with the right-hand side.
• If $T_k$ is the time of first return to $k$, then $\mathbb{E}[T_k \mid X_0=k] \neq \sum_{n=1}^\infty nf_{kk}(n)$. This is because $f_{kk}(n)$ is $\mathbb{P}(X_n=k \mid X_0=k)$, not $\mathbb{P}(X_n=k\text{ and } X_i \neq k, 1 \leq i <k \mid X_0=k)$. So you cannot say that $\sum_{n=1}^\infty nf_{kk}(n)=\infty$. – kccu Feb 10 at 17:52
• @kccu My $f_{kk}(n)$ is defined as $P(X_n = k {\rm \ and \ } X_i \neq k, 1 \leq i < k | X_0 = k)$. Whereas my $p_{kk}(n)$ is defined as $P(X_n = k | X_0 = k)$. – Kenny Wong Feb 10 at 18:02
• My apologies, I read too quickly and confused $f_{kk}$ with $p_{kk}$. – kccu Feb 10 at 21:23
I think that claim is not trivial. In the book referenced in your link, that statement is observed as a corollary to the ergodic theorem stating that $$p_{ij}(n) \to \frac{1}{\mathbb E_x T_x}$$ for any irreducible aperiodic Markov chain. You can find a proof of that result here, for instance.
If all you want to prove is your original claim (that all irreducible finite Markov chains are positive recurrent), I think there's an easier way to do it than by that lemma. Assume aperiodicity for simplicity, but periodic chains just make the proof more annoying (rather than prevent the result from being true). The sketch of the proof is:
1. For an irreducible, aperiodic finite Markov chain, there exists some $$m$$ such that $$p_{ij}(m) > 0$$ for all state pairs $$\{i, j\}$$.
2. Moreover, because the chain is finite you can strengthen the above to say that $$p_{ij}(m) > \epsilon$$ for some universal $$\epsilon > 0$$.
3. Hence, you can dominate the excursions of $$X_n$$ from $$x$$ to $$x$$ with a geometric random variable by conditioning on the behavior on every $$m^{\text{th}}$$ step.
4. Since the geometric random variable has a finite expectation, so do the excursion lengths.
EDIT: Expanding the answer to address the steps to show step 1 above. I'll work with the following definition of aperiodicity: a state $$x$$ is aperiodic if $$\gcd(k:p_{xx}(k) > 0) = 1$$. In all the following, I'll also assume irreducibility of the Markov chain. Without irreducibility, some version of these basic ideas all still hold, but they're just more obnoxious to write down; you'd just have to address an irreducible subclass of states.
Lemma: If state $$x$$ is aperiodic, then there exists $$N$$ such that $$p_{xx}(n) > 0$$ for all $$n \geq N$$.
Corollary: For the aperiodic state $$x$$ above, there exists $$N'$$ such that for all other states $$y$$ and all $$n \geq N'$$, $$p_{xy}(n) > 0$$.
Corollary to the corollary: If a chain has one aperiodic state, all states are aperiodic.
• Thanks very much for this! I understand points 2, 3 and 4. Could you please give me a hint as to how to prove point 1? – Kenny Wong Feb 12 at 20:48
• I tried to fill in the gaps a bit -- let me know if I should go further. – Aaron Montgomery Feb 13 at 3:04
• That's great - thanks! – Kenny Wong Feb 13 at 8:07 | 2019-05-23T11:22:29 | {
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https://math.stackexchange.com/questions/567856/the-number-3211000-is-7-special/568657 | # The number 3211000 is 7-special
Define a positive integer $k$ to be $n$-special if it satisfies the following properties:
1. It has $n$ digits (0, 1, ..., 9)
2. The 1st digit is equal to the number of 0's in the decimal representation of $k$, the second digit is equal to the number of 1's, the third digit is equal to the number of 2's, etc
For instance, the number $3211000$ is $7$-special (since there are three 0's, two 1's, one 2, and one 3).
Questions:
1. How many $7$-special numbers are there? Can you prove it?
2. For what positive integer $n$ does there not exist any $n$-special numbers?
3. Is there an efficient algorithm to compute all $n$-special numbers, given any positive integer $n$?
Have fun and enjoy! :)
• So by this definition, 6210001000 is 10-special? – Dennis Meng Nov 15 '13 at 18:04
• Yes, it is ! ;) – Christmas Bunny Nov 15 '13 at 18:44
• BTW if I'm not mistaken, there's no $1$-special number, no $2$-special number and no $3$-special number. There is however at least one $4$-special number: $1210$. – celtschk Aug 17 '16 at 20:10
• @celtschk The second OEIS list I gave agrees with you, and provides one more 4-special number: 2020. – Dennis Meng Aug 18 '16 at 3:46
Just for kicks, I googled the 10-digit example (6210001000) that I mentioned in the comment, and found that these n-special numbers are apparently called self-descriptive numbers.
For your part 1), it doesn't list any other examples for 7, and judging from the OEIS list, there aren't any others. (As a heads-up, there's some base conversion going on in the list; your example is listed as 389305 (which gives 3211000 in base 7))
(Addendum: This list gives the numbers as you actually want them, though there are a few less entries)
For part 2), the article lists the three values of $n$ without any (2,3,6, though I guess 1 doesn't either if you want to count it)
For part 3), the article gives a general formula for getting more examples in higher bases, but only one per value of $n$. If I find anything about generating all (either more formulas or a proof that this is all of them), I'll update the answer. The formula itself is
$$(n-4)n^{n-1} + 2n^{n-2} + n^{n-3} + n^3$$
Here's a proof there are no other $7$-special numbers.
In a $7$-special number $a_0a_1a_2a_3a_4a_5a_6$, each $a_k$ counts the number of numerals that appear $k$ times among the digits. Since there are $7$ digits in all, we have
$$0a_0+1a_1+2a_2+3a_3+4a_4+5a_5+6a_6=7$$
This immediately implies $a_4+a_5+a_6\le1$, so at least two of those digits are $0$'s and at most one is a $1$. Now if any of them are equal to $1$, then one of the other digits is at least $4$. To satisfy the displayed equation, that digit can only be the $a_0$. We also have $a_1\ge1$. But we can't have $a_1=1$, since we already have at least one $1$, so we must have $a_1\ge2$. This means there are at least two $1$'s among the digits, the next one being either $a_2$ or $a_3$. But this leaves at most three digits that can possibly be $0$'s, which contradicts the conclusion $a_0\ge4$.
So we now know that $a_4=a_5=a_6=0$. This means there are at least three $0$'s among the digits, hence $a_0\ge3$. On the other hand, it also means that there are no $4$'s, $5$'s, or $6$'s among the digits, so $a_0$ cannot be any of those numerals. Hence we must have $a_0=3$. This implies that $a_1$, $a_2$, and $a_3$ are all at least $1$. Writing them as $a_k=1+b_k$ and plugging into the displayed equation (along with the established values $a_0=3$ and $a_4=a_5=a_6=0$), we find
$$b_1+2b_2+3b_3=1$$
This is clearly enough to conclude that $3211000$ is the only $7$-special number.
There are plenty of other ways to let the logic of this argument run. I'd be quite happy to see something shorter and more to the point.
The general problem of an $n$-special sequence, i.e., a sequence $b_0$, ..., $b_{n-1}$ with the property that each number $i\,$ in the range $[0:n-1]$ occurs exactly $b_i$ times in the sequence, can also easily be solved. The easiest way known to me to get all $n$-special sequences with proof is the following.
Let $b_0$, ..., $b_{n-1}$ be an $n$-special sequence. Each $b_i$ is in the range $[0,n]$, because it refers to a count out of $n$ elements. $b_i=n$ is impossible; all elements would be $i$, also $b_i$ , but we know $i<n$.
Hence each element $b_i$ contributes $1$ to the sum $\sum_{i=0}^{n-1}{b_i}$, i.e., this sum is $n$. More strictly, the sum of the positive $b_i$ is $n$. There are $n-b_0$ positive $b_i$.
$b_0=0$ is also impossible (this also means no element is $0$, a contradiction). Thus there are $n-b_0-1$ positive $b_i$ with $i>0$, and their sum is $n-b_0$. Hence one is $2$ and the others (if existing) are $1$. We obtain
a) If $b_0\le 2$, then $b_i=0$ for $i\ge 3$, $n=b_0+b_1+b_2$.
b) If $b_0\ge 3$, then $b_{b_0}=1$, $b_i=0$ for $i\ge 3$ and $i\ne b_0$, $n=b_0+b_1+b_2+b_{b_0}$ .
c) If $b_0=2$, then $b_2=2$.
d) If $b_0\ne 2$, then $b_2=1$.
We now make a case distinction by $b_0$.
$b_0=1$: We have $b_2=1$, $b_i=0$ for $i\ge 3$, $b_1=2$, $n=1+2+1=4$, and $b_3=0$.
$\quad$ $\quad$ $\$ This is the solution 1 2 1 0.
$b_0=2$: We have $b_2=2$, $b_i=0$ for $i\ge 3$, thus $b_1\lt2$.
$\quad$ $\quad$ $\$ If $b_1=0$, then $n=2+0+2=4$, $b_3=0$. This is the solution 2 0 2 0.
$\quad$ $\quad$ $\$ If $b_1=1$, then $n=2+1+2=5$, $b_3=b_4=0$. This is the solution 2 1 2 0 0.
$b_0\ge 3$: We have $b_2=1$, $b_{b_0}=1$, $b_i=0$ for $i\ge 3$ and $i\ne b_0$, and $b_1=2$.
$\quad$ $\quad$ $\$ Thus $n=b_0+b_1+b_2+b_{b_0}=b_0+2+1+1=b_0+4$, i. e., $n\ge 7$ and $b_0=n-4$.
$\quad$ $\quad$ $\$ This is the solution (n-4) 2 1 (n-7 times 0) 1 0 0 0 for each $n\ge 7$. | 2019-05-27T07:04:05 | {
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http://math.stackexchange.com/questions/249119/easy-question-about-surjectivity-for-fx-4x-x21 | # Easy question about surjectivity for $f(x) = 4x-(x^2+1)$
Let $f : [0,3] \rightarrow B$ be defined by $f(x) = 4x-(x^2+1)$. Determine the image domain (or maybe it's called range in English?) B so that the function becomes surjective. Does it become invertible?
How should I approach this? I've been thinking:
By graphing the function, I can see that the lowest value is when $f(0) = -1$ and the highest is when $f(2) = 3$.
To make the function surjective, I need to map every value from the domain to the range. Therefore, $B = [-1,3]$.
Since the function is a parabola, it is not invertible.
So, $f : [0,3] \rightarrow [-1,3], f(x) = 4x-(x^2+1)$ is surjective but not invertible.
Am I wrong? I have nothing to compare my answer to so I don't really know...
-
Yes, it’s called the range (or by some the image). – Brian M. Scott Dec 2 '12 at 10:33
Even though it’s a parabola, it might be invertible on the domain $[0,3]$, much as $g(x)=x^2$ is invertible on the domain $[0,3]$. However, $f(1)=4-2=2=12-10=f(3)$, so $f$ is not injective on $[0,3]$ and therefore is not invertible on $[0,3]$.
You could probably find this example of non-injectivity by examining the graph of $y=f(x)$. I simply set
$$y=f(x)=4x-(x^2+1)=-x^2+4x-1\;,$$
rearranged that as $x^2-4x+1+y=0$, and used the quadratic formula to find that
$$x=\frac{4\pm\sqrt{16-4y}}2=2\pm\sqrt{4-y}\;.$$
You’d already observed that the extreme values of $y$ are $-1$ and $3$, so I looked at these values of $y$ to get a better idea of what was going on and was fortunate with $y=3$.
I mention this because it’s sometimes quite useful to realize that when $y$ is a quadratic function of $x$, you can solve for $x$ in terms of $y$, though the result isn’t a function on $\Bbb R$.
-
A more analytic approach: $f$ is continuous and so the range of $f$ is $R=[f_{\min},f_{\max}]$. $f^{\prime}(x)>0\Leftrightarrow 4-2x>0\Leftrightarrow x<2$ and $f^{\prime}(x)<0\Leftrightarrow 4-2x<0\Leftrightarrow x>2$.
$f$ is increasing in $[0,2]$ and then decreasing in $[2,3]$. Therefore, since $f(0)=-1<2=f(3)$ and $f(2)=3$, $R=[-1,3]$. To make $f$ surjective you need to take $B=[-1,3]$. $f$ is not injective in $[0,3]$ but it is injective in the intervals $[0,2]$ and $[2,3]$. Therefore a parabola may be injective depending on the choice of your domain for $f$
- | 2015-08-03T03:04:17 | {
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https://cs.stackexchange.com/questions/28729/unable-to-understand-an-inequality-in-an-application-of-the-pumping-lemma-for-co | Unable to understand an inequality in an application of the pumping lemma for context-free languages
The problem
Prove that the language
$\qquad L = \{a^n b^j \mid n = j^2\}$
is not context free using pumping lemma.
Approach taken by the book
To prove such statements, the book takes the approach of a game played against an opponent who strives to fail our effort to prove that the language is not context free. The steps involved in the game are as follows:
• The opponent chooses m such that for all w $\in$ L, |w| >= m
• We choose the string w
• The opponent decomposes w in uvxyz such that |vxy| <= m and |vy| >= 1
• We pump v and y i-times to get the string uv$^i$xy$^i$z.
Now, if for any i = 0,1,2,...
uv$^i$xy$^i$z $\notin$ L
the language L is not context free.
The solution to the above problem
• Opponent chooses m
• We choose w = a$^{m^2}$b$^m$
• The opponent decomposes w in uvxyz as follows:
• Pumping v and y i-times yields string with m$^2$+(i-1)k$_1$ a's and m+(i-1)k$_2$ b's.
• If opponent takes k$_1$ $\ne$ 0 and k$_2$ $\ne$ 0, we can take i = 0, such that
(m-k$_2$)$^2 \leq$ (m-1)$^2$ ... since k$_2\ne$ 0 making minimum value of k$_2$ is 1
= m$^2$-2m+1
< m$^2$ - k$_1$
Q. This last line I did not understand. How is -2m+1 < -k$_1$? Especially because I can find the below decomposition uvxyz for which -2m+1 > -k$_1$.
I must be missing some stupid algebra here.
The solution further continues saying that the pumped resultant string does not belong to L.
• Similar argument can be done if user select k$_1$ = 0 and k$_2$ $\ne$ 0 or k$_1$ $\ne$ 0 and k$_2$ = 0
• Don't use images as main content of your post. Not only is it lazy, it also makes your question impossible to search and inaccessible to the visually impaired; we don't like that. Please transcribe text and maths (note that you can use LaTeX) and don't forget to give proper attribution to your sources! – Raphael Jul 26 '14 at 11:06
• @Raphael here u go. It took huge efforts to write what that image was to imply using LaTeX. I could have used that time for something else. – anir123 Jul 26 '14 at 14:48
• Be careful with the attitude there; you are asking people to spend time on your problem instead of using that time for something else. Thanks for the edit, though. – Raphael Jul 26 '14 at 16:20
• I am lately fade up by my questions not getting answered on cs.stackexchange. I tried putting my questions well, but for three questions they keep downvoting my questions not even single comment for downvote, forget answering it. I eventually end up deleting my own question. So yess lately I am having bad experience with community at cs.stackexchange :\ , dont know thats just me. On stackoverflow I have excellent experience. Doesnt look the same here. – anir123 Jul 26 '14 at 18:32
• I'm sure you understand that nobody is entitled to an answer here; maybe someone out of the crowd finds the question interesting or valuable enough, maybe not. It's nothing personal. I suggest you check out some upvoted questions and see what they do different from you. – Raphael Jul 26 '14 at 23:16
The Pumping Lemma requires $|vxy| \le m$ so that the pumped parts are of bounded length. Thus $k_1+k_2\le m$. With this it is easy.
As $k_2 \neq 0$, we must have $k_1<m$. Also $m>0$ as the pumping constant must be positive (but in fact we can assume it to be as large as we want, of course).
Now $2m-1 \ge m > k_1$.
That's all.
$m-k_2$ is the number of $b$'s in the pumped word, let us call it $w'$. In order for $w'$ to be in the language, the number of $a$'s has to be $(m-k_2)^2$. Depending on the chosen $k_2$, this is at most $(m-1)^2 = m^2-2m+1$. The actual number of $a$'s, however, is $m^2 - k_1$ and this is truly greater than $m^2-2m+1$. Hence, the pumped word $w'$ is not part of the language.
• Hey please check the new diagram I added to the original quesstion. I still dont understand how m$^2$-k$_1$ is truly greater than m$^2$-2m+1 – anir123 Jul 27 '14 at 11:43
It is much easier to use Parikh's theorem to prove that your language is not context-free. Indeed the Parikh image of your language is by construction the subset $\{ (j^2, j) \mid j \in \mathbb{N} \}$ of $\mathbb{N}^2$. It is now easy to see that this set is not semi-linear and therefore your language is not context-free. | 2019-09-18T05:46:10 | {
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https://math.stackexchange.com/questions/784032/remainder-question-with-6-and-7 | # Remainder question with $6!$ and 7
Find the remainder when $6!$ is divided by 7.
I know that you can answer this question by computing $6! = 720$ and then using short division, but is there a way to find the remainder without using short division?
As $7$ is prime, use Wilson's Theorem $$(p-1)!\equiv-1\pmod p$$ for prime $p$
Now, $\displaystyle -1\equiv p-1\pmod p$
• clean, concise, and uses a nice theorem +1 May 6, 2014 at 19:19
• What does this exactly mean? Sorry I am not an expert in maths. May 6, 2014 at 19:30
• @user108104 In my answer, I have specialized a proof of Wilson's Theorem to your specific case. I think you will find it quite intuitive in this concrete instance. May 6, 2014 at 20:50
Hint $\$ In analogy with Gauss's trick (see below), to simplify the product we pair up each number with its (multiplicative) inverse mod $7.\,$ Thus $$6! = 1\cdot (\overbrace{2\cdot 4}^{\equiv \,1})(\overbrace{3\cdot5}^{\equiv\, 1})\cdot 6 \equiv 1\cdot 1\cdot 1\cdot 6\equiv6\pmod{7}$$
Remark $\$ This method of pairing up inverses works for any prime - see Wilson's Theorem.
Below is Gauss's trick, imported from a deleted question
$\qquad\qquad \begin{array}{rcl}\rm{\bf Hint}\quad\quad\ \ S &=&\rm 1 \ \ \ +\ \ \: 2\ \ \ \ +\ \:\cdots\ +\ n\!-\!1\ +\ n \\ \rm S &=&\rm n \ \ +\ n\!-\!1\ +\,\ \cdots\ +\,\quad 2\ \ \ +\ \ 1\\ \hline \\ \rm Adding\ \ \ \ 2\: S &=&\rm n\ (n\!+\!1)\end{array}$
A famous legend says Gauss used this trick to quickly compute $1+2+\:\cdots\:+100\$ in grade school.
This trick of pairing up reflections around the average value is a special case of exploiting innate symmetry - here a reflection or involution. It's a ubiquitous powerful technique, e.g. see my post on Wilson's Theorem and it's group theoretic generalization. | 2022-07-06T13:12:15 | {
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http://openstudy.com/updates/508c2cfae4b04456f6c43164 | ## klimenkov Group Title $\lim_{n\rightarrow\infty}\frac{\sqrt[n]{n!}}{n}$ one year ago one year ago
1. myko Group Title
maybe like this: $\sqrt[n]{n!}=\sqrt[n]{n(n-1)(n-2)\cdots1} =\sqrt[n]{n}\sqrt[n]{n-1}\cdots \sqrt[n]{1}=1$ so limit is equal to 0
2. myko Group Title
@klimenkov
3. klimenkov Group Title
Are you sure that $\lim_{n\rightarrow\infty}\sqrt[n]{n!}=1$
4. myko Group Title
look the steps from my comment befor. It looks ok
5. myko Group Title
all the roots at the right hand side: $\sqrt[n]{n}=\sqrt[n]{n-1}=\cdots=\sqrt[n]{1}=1$
6. myko Group Title
so their product too
7. klimenkov Group Title
No, it's not ok. Because you multiply an infinite quantity or 1. As we know $$1^{\infty}={}?$$.
8. myko Group Title
$1^{\infty} =1*1*\cdots*1=1$
9. klimenkov Group Title
Very nice. What can you say about this pretty limit? $\lim_{n\rightarrow\infty}\left(1+\frac1n\right)^n$It is $$1^{\infty}$$.
10. myko Group Title
this happens when you talk about functions. The reason of indetermination of 1^infinity is because of that. But in this case there are no functions involved. That's my point
11. myko Group Title
in this case there is just number one multiplyed infinitly many times. And it happens after the limit was taken
12. klimenkov Group Title
Ok. What about this? $\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n$Is it 0?
13. myko Group Title
this is a harmonic series. It is not convergent
14. myko Group Title
15. klimenkov Group Title
Look at the denominator carefully please. I hopr you will try to get what I'm saying.
16. myko Group Title
sry, but i don't
17. klimenkov Group Title
Can you find this? $\lim_{n\rightarrow\infty}\sum_{k=1}^n\frac1n$
18. myko Group Title
another way to try this: $\lim \sqrt[n]{\frac{n!}{n^n}} = 0$
19. myko Group Title
infinity
20. klimenkov Group Title
Can you show the way you solve it?
21. myko Group Title
I don't remmeber the formal proof of n!/n^n =0, but it's evident, if you try a few first terms of this sequence. There are some posts about it if you google a bit
22. klimenkov Group Title
@myko, see this and tell me what is your mistake? http://www.wolframalpha.com/input/?i=Limit+%28n!%29^%281%2Fn%29%2Fn+n-%3Einfinity
23. TuringTest Group Title
@mahmit2012 a little help here?
24. mahmit2012 Group Title
|dw:1351370532404:dw|
25. TuringTest Group Title
very nice, my turn...
26. mahmit2012 Group Title
|dw:1351370740820:dw|
27. mahmit2012 Group Title
|dw:1351370823511:dw|
28. mahmit2012 Group Title
|dw:1351370856590:dw|
29. TuringTest Group Title
${\sqrt[n]{n!}\over n}=\exp\left(\frac1n\ln(n!)-\ln n\right)=\exp\left({n\ln n-n+O(n)-n\ln n\over n}\right)$$=\exp(-1)=\frac1e$
30. mukushla Group Title
*
31. myko Group Title
ya I was wrong. Here is another way to solve it. As we know root test is stronger than cuotient test, so the folowing inequality holds: $\lim \inf \frac{a_{n+1}}{a_{n}}\leq\lim \inf \sqrt[n]{a_{n}}\leq \lim \sup \sqrt[n]{a_{n}} \leq \lim \sup\frac{a_{n+1}}{a_{n}}$ let $a_{n} = \frac{n!}{n^{n}}$ then $\frac{a_{n+1}}{a_{n}}=\frac{1}{(1+\frac{1}{n})^{n}}=\frac{1}{e}$ this means that $\lim \sqrt[n]{a_{n}} = \frac{1}{e}$
32. myko Group Title
@mahmit2012 $\lim \frac{a_{n+1}}{a_{n}}=\lim \sqrt[n]{a_{n}}$ only if a_n is convergent, what is not implied in this question
33. klimenkov Group Title
Nice. But I have one more interesting method to find it. $\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac{n!}{n^n}}=\lim_{n\rightarrow\infty}\sqrt[n]{\frac1n\cdot\frac2n\cdots\frac n n}=A$$\ln A=\lim_{n\rightarrow\infty}\frac1n(\ln\frac1n+\ln\frac2n+\ldots+\ln\frac n n)=\int_0^1\ln x dx=-1$$\lim_{n\rightarrow\infty} \frac{\sqrt[n]{n!}}{n}=A=e^{-1}=\frac1e$ | 2014-08-23T11:33:50 | {
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https://math.codidact.com/posts/286991 | ### Communities
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Q&A
# organizing a library
+3
−0
Suppose you have $n>1$ books lined up on a shelf, numbered $1$ to $n$, not in the correct order, and you wish to put them in order. Here's your method: Choose a misplaced book[1] at random, and put it in its correct spot. For example, if $n=5$ and you pick book number $2$ out of spot number $4$, there are now four books left, and you put the book back between the first two, since it's book number $2$.
What's the maximum number of times you might have to do the pick-and-replace before the books are in order?
[1] meaning, a book numbered $k$ which is not in position $k$
Why does this post require moderator attention?
Why should this post be closed?
what I know so far (2 comments)
+2
−0
$2^{n-1}-1$ is a lower bound on the maximum, at least. For $n$ books, if you start with the ordering $n, 1, 2, \ldots, n - 1$ (I'm following your convention of 1-indexing the books), then the books could be reordered via the sequence $S_n$ defined as follows: \begin{align} S_1 &= () \\ S_{n + 1} &= ([S_{n} + 1], 1, [S_{n} + 1]) \end{align}
(Here $[S_n + 1]$ means to include the sequence $S_n$, adding 1 to each element.)
I'll walk through $n = 4$, where $S_4 = (3, 2, 3, 1, 3, 2, 3)$:
4 1 2 3 (choose 3)
4 1 3 2 (choose 2)
4 2 1 3 (choose 3)
4 2 3 1 (choose 1)
1 4 2 3 (choose 3)
1 4 3 2 (choose 2)
1 2 4 3 (choose 3)
1 2 3 4
It should be clear that the length of $S_n$ is $2^{n-1} - 1$. Slightly less clear is why this strategy is always valid, though an inductive argument should be within reach.
I don't know if this is the worst case, though.
Why does this post require moderator attention? | 2023-02-01T08:36:07 | {
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https://teachingcalculus.com/2017/12/12/good-question-xxx/ | # Good Question 13
Let’s end the year with this problem that I came across a while ago in a review book:
Integrate $\int{x\sqrt{x+1}dx}$
It was a multiple-choice question and had four choices for the answer. The author intended it to be done with a u-substitution but being a bit rusty I tried integration by parts. I got the correct answer, but it was not among the choices. So I thought it would make a good challenge to work on over the holidays.
1. Find the antiderivative using a u-substitution.
2. Find the antiderivative using integration by parts.
3. Find the antiderivative using a different u-substitution.
4. Find the antiderivative by adding zero in a convenient form.
Your answers for 1, 3, and 4 should be the same, but look different from your answer to 2. The difference is NOT due to the constant of integration which is the same for all four answers. Show that the two forms are the same by
2. “Simplifying” your answer to 1, 3, 4 and get that third form again.
Give it a try before reading on. The solutions are below the picture.
Method 1: u-substitution
Integrate $\int{x\sqrt{x+1}dx}$
$u=x+1,x=u-1,dx=du$
$\int{x\sqrt{x+1}dx=}\int{\left( u-1 \right)\sqrt{u}}du=\int{{{u}^{3/2}}-{{u}^{1/2}}}du=\tfrac{5}{2}{{u}^{5/2}}-\tfrac{3}{2}{{u}^{3/2}}$
$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C$
Method 2: By Parts
Integrate $\int{x\sqrt{x+1}dx}$
$u=x,du=dx$
$dv=\sqrt{x+1}dx,v=\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}$
$\int{x\sqrt{x+1}dx}=\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\int{\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}dx}=$
$\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\tfrac{4}{15}{{\left( x+1 \right)}^{5/2}}+C$
Method 3: A different u-substitution
Integrate $\int{x\sqrt{x+1}dx}$
$u=\sqrt{x+1},x={{u}^{2}}-1,$
$du=\tfrac{1}{2}{{\left( x+1 \right)}^{-1/2}}dx,dx=2udu$
$2{{\int{\left( {{u}^{2}}-1 \right){{u}^{2}}}}^{{}}}du=2\int{{{u}^{4}}-{{u}^{2}}}du=\tfrac{2}{5}{{u}^{5}}-\tfrac{2}{3}{{u}^{3}}=$
$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C$
This gives the same answer as Method 1.
Method 4: Add zero in a convenient form.
Integrate $\int{x\sqrt{x+1}dx}$
$\int{x\sqrt{x+1}}dx=\int{x\sqrt{x+1}+\sqrt{x+1}-\sqrt{x+1} dx=}$
$\int{\left( x+1 \right)\sqrt{x+1}-\sqrt{x+1}}dx=$
$\int{{{\left( x+1 \right)}^{3/2}}-{{\left( x+1 \right)}^{1/2}}dx}=$
$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C$
This, also, gives the same answer as Methods 1 and 3.
So, by a vote of three to one Method 2 must be wrong. Yes, no, maybe?
No, all four answers are the same. Often when you get two forms for the same antiderivative, the problem is with the constant of integration. That is not the case here. We can show that the answers are the same by factoring out a common factor of ${{\left( x+1 \right)}^{3/2}}$. (Factoring the term with the lowest fractional exponent often is the key to simplifying expressions of this kind.)
Simplify the answer for Method 2:
$\tfrac{2}{3}x{{\left( x+1 \right)}^{3/2}}-\tfrac{4}{15}{{\left( x+1 \right)}^{5/2}}+C=$
$\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}\left( x-\tfrac{2}{5}\left( x+1 \right) \right)+C=$
$\displaystyle \tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}\left( \frac{5x-2x-2}{5} \right)+C=$
$\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3x-2 \right)+C$
Simplify the answer for Methods 1, 3, and 4:
$\tfrac{2}{5}{{\left( x+1 \right)}^{5/2}}-\tfrac{2}{3}{{\left( x+1 \right)}^{3/2}}+C=$
$\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3\left( x+1 \right)-5 \right)+C=$
$\tfrac{2}{15}{{\left( x+1 \right)}^{3/2}}\left( 3x-2 \right)+C$
So, same answer and same constant.
Is this a good question? No and yes.
As a multiple-choice question, no, this is not a good question. It is reasonable that a student may use the method of integration by parts. His or her answer is not among the choices, but they have done nothing wrong. Obviously, you cannot include both answers, since then there will be two correct choices. Moral: writing a multiple-choice question is not as simple as it seems.
From another point of view, yes, this is a good question, but not for multiple-choice. You can use it in your class to widen your students’ perspective. Give the class a hint on where to start. Even better, ask the class to suggest methods; if necessary, suggest methods until you have all four (… maybe there is even a fifth). Assign one-quarter of your class to do the problem by each method. Then have them compare their results. Finally, have them do the simplification to show that the answers are the same.
My next post will be after the holidays.
. | 2023-04-01T04:39:09 | {
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https://in.mathworks.com/help/optim/ug/fit-differential-equation-ode.html | Main Content
# Fit an Ordinary Differential Equation (ODE)
This example shows how to fit parameters of an ODE to data in two ways. The first shows a straightforward fit of a constant-speed circular path to a portion of a solution of the Lorenz system, a famous ODE with sensitive dependence on initial parameters. The second shows how to modify the parameters of the Lorenz system to fit a constant-speed circular path. You can use the appropriate approach for your application as a model for fitting a differential equation to data.
### Lorenz System: Definition and Numerical Solution
The Lorenz system is a system of ordinary differential equations (see Lorenz system). For real constants $\sigma ,\phantom{\rule{0.5em}{0ex}}\rho ,\phantom{\rule{0.5em}{0ex}}\beta$, the system is
`$\begin{array}{l}\frac{dx}{dt}=\sigma \left(y-x\right)\\ \frac{dy}{dt}=x\left(\rho -z\right)-y\\ \frac{dz}{dt}=xy-\beta z.\end{array}$`
Lorenz's values of the parameters for a sensitive system are $\sigma =10,\phantom{\rule{0.5em}{0ex}}\beta =8/3,\phantom{\rule{0.5em}{0ex}}\rho =28$. Start the system from `[x(0),y(0),z(0)] = [10,20,10]` and view the evolution of the system from time 0 through 100.
```sigma = 10; beta = 8/3; rho = 28; f = @(t,a) [-sigma*a(1) + sigma*a(2); rho*a(1) - a(2) - a(1)*a(3); -beta*a(3) + a(1)*a(2)]; xt0 = [10,20,10]; [tspan,a] = ode45(f,[0 100],xt0); % Runge-Kutta 4th/5th order ODE solver figure plot3(a(:,1),a(:,2),a(:,3)) view([-10.0 -2.0])```
The evolution is quite complicated. But over a small time interval, it looks somewhat like uniform circular motion. Plot the solution over the time interval `[0,1/10]`.
```[tspan,a] = ode45(f,[0 1/10],xt0); % Runge-Kutta 4th/5th order ODE solver figure plot3(a(:,1),a(:,2),a(:,3)) view([-30 -70])```
### Fit a Circular Path to the ODE Solution
The equations of a circular path have several parameters:
• Angle $\theta \left(1\right)$ of the path from the x-y plane
• Angle $\theta \left(2\right)$ of the plane from a tilt along the x-axis
• Radius R
• Speed V
• Shift t0 from time 0
• 3-D shift in space delta
In terms of these parameters, determine the position of the circular path for times `xdata`.
`type fitlorenzfn`
```function f = fitlorenzfn(x,xdata) theta = x(1:2); R = x(3); V = x(4); t0 = x(5); delta = x(6:8); f = zeros(length(xdata),3); f(:,3) = R*sin(theta(1))*sin(V*(xdata - t0)) + delta(3); f(:,1) = R*cos(V*(xdata - t0))*cos(theta(2)) ... - R*sin(V*(xdata - t0))*cos(theta(1))*sin(theta(2)) + delta(1); f(:,2) = R*sin(V*(xdata - t0))*cos(theta(1))*cos(theta(2)) ... - R*cos(V*(xdata - t0))*sin(theta(2)) + delta(2); ```
To find the best-fitting circular path to the Lorenz system at times given in the ODE solution, use `lsqcurvefit`. In order to keep the parameters in reasonable limits, put bounds on the various parameters.
```lb = [-pi/2,-pi,5,-15,-pi,-40,-40,-40]; ub = [pi/2,pi,60,15,pi,40,40,40]; theta0 = [0;0]; R0 = 20; V0 = 1; t0 = 0; delta0 = zeros(3,1); x0 = [theta0;R0;V0;t0;delta0]; [xbest,resnorm,residual] = lsqcurvefit(@fitlorenzfn,x0,tspan,a,lb,ub);```
```Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance. ```
Plot the best-fitting circular points at the times from the ODE solution together with the solution of the Lorenz system.
```soln = a + residual; hold on plot3(soln(:,1),soln(:,2),soln(:,3),'r') legend('ODE Solution','Circular Arc') hold off```
```figure plot3(a(:,1),a(:,2),a(:,3),'b.','MarkerSize',10) hold on plot3(soln(:,1),soln(:,2),soln(:,3),'rx','MarkerSize',10) legend('ODE Solution','Circular Arc') hold off```
### Fit the ODE to the Circular Arc
Now modify the parameters $\sigma ,\phantom{\rule{0.5em}{0ex}}\beta ,\phantom{\rule{0.5em}{0ex}}and\phantom{\rule{0.5em}{0ex}}\rho$ to best fit the circular arc. For an even better fit, allow the initial point [10,20,10] to change as well.
To do so, write a function file `paramfun` that takes the parameters of the ODE fit and calculates the trajectory over the times `t`.
`type paramfun`
```function pos = paramfun(x,tspan) sigma = x(1); beta = x(2); rho = x(3); xt0 = x(4:6); f = @(t,a) [-sigma*a(1) + sigma*a(2); rho*a(1) - a(2) - a(1)*a(3); -beta*a(3) + a(1)*a(2)]; [~,pos] = ode45(f,tspan,xt0); ```
To find the best parameters, use `lsqcurvefit` to minimize the differences between the new calculated ODE trajectory and the circular arc `soln`.
```xt0 = zeros(1,6); xt0(1) = sigma; xt0(2) = beta; xt0(3) = rho; xt0(4:6) = soln(1,:); [pbest,presnorm,presidual,exitflag,output] = lsqcurvefit(@paramfun,xt0,tspan,soln);```
```Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance. ```
Determine how much this optimization changed the parameters.
`fprintf('New parameters: %f, %f, %f',pbest(1:3))`
```New parameters: 9.132446, 2.854998, 27.937986 ```
`fprintf('Original parameters: %f, %f, %f',[sigma,beta,rho])`
```Original parameters: 10.000000, 2.666667, 28.000000 ```
The parameters `sigma` and `beta` changed by about 10%.
Plot the modified solution.
```figure hold on odesl = presidual + soln; plot3(odesl(:,1),odesl(:,2),odesl(:,3),'b') plot3(soln(:,1),soln(:,2),soln(:,3),'r') legend('ODE Solution','Circular Arc') view([-30 -70]) hold off```
### Problems in Fitting ODEs
As described in Optimizing a Simulation or Ordinary Differential Equation, an optimizer can have trouble due to the inherent noise in numerical ODE solutions. If you suspect that your solution is not ideal, perhaps because the exit message or exit flag indicates a potential inaccuracy, then try changing the finite differencing. In this example, use a larger finite difference step size and central finite differences.
```options = optimoptions('lsqcurvefit','FiniteDifferenceStepSize',1e-4,... 'FiniteDifferenceType','central'); [pbest2,presnorm2,presidual2,exitflag2,output2] = ... lsqcurvefit(@paramfun,xt0,tspan,soln,[],[],options);```
```Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance. ```
In this case, using these finite differencing options does not improve the solution.
`disp([presnorm,presnorm2])`
``` 20.0637 20.0637 ``` | 2022-06-27T20:07:54 | {
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https://math.stackexchange.com/questions/3187630/is-it-correct-to-say-that-the-sum-of-a-series-is-the-lim-n-to-infty-of-the | # Is it correct to say that the sum of a series is the $\lim_{n\to\infty}$ of the sequence of its partial sums?
I know that is the sequence of partial sums of a series is convergent, then the series is convergent. But let's say that the sequence of partial sums converges to the value $$2$$. Does that mean the sum of the series is $$2$$?
That doesn't make much sense to me because: as an example consider the series whose partial sums are given by the formula $$s_n = 2 - 3(0.8)^n$$. The limit as $$n$$ goes to infinity of that function is $$2$$. But wouldn't the sum of the series be much higher? Since: $$s_1 + s_2 + s_3 + ... = 2-3(0.8) + 2-3(0.64)+2-3(0.512)+...$$
Two is added with each new partial sum, so how can the sum of the series be two?
Sorry if this is a stupid question, I'm likely misunderstanding something fundamental about series. Any help is appreciated.
A series converges by definition if the sequence of partial sums converges, and if that happens then the sum of the series is defined to be the limit of the sequence of partial sums.
I'll try to explain what is wrong with your example. You need to start from a sequence $$(a_n)$$ and define a corresponding sequence of partial sums $$(S_n)$$ by $$S_n=\sum_{k=1}^n a_k$$. Then $$\sum_{k=1}^\infty a_k=\lim_{n\to\infty}\sum_{k=1}^n a_k=\lim_{n\to\infty} S_n$$, if the limit exists.
Now in your example you already defined a sequence of partial sums $$(s_n)$$ and then started to look at its partial sums $$s_1+...+s_n$$. So you took partial sums of partial sums.
• Thank you for the response! Does the series $a_n$ in your example represent the sequence of terms of the series? And you define a sequence of partial sums by $S_n =$\sum_{k=1}^\infty a_k=\lim_{n\to\infty}\sum_{k=1}^n a_k=\lim_{n\to\infty} S_n$, but wouldn't that define a series and not a sequence? Sorry for not understanding, I appreciate the help! – James Ronald Apr 14 at 17:33 • No, the elements$S_n$are defined as finite sums$S_n=\sum_{k=1}^n a_k$. This is a sequence. The series is the infinite sum$\sum_{k=1}^\infty a_k$which is defined to be the limit of the sequence$S_n$if such a limit exists. And yes, the terms that we add are the elements of$\{a_k\}_{k=1}^\infty\$. – Mark Apr 14 at 17:36
• Ahh I see, sorry that was stupid of me. Everything makes sense now, thank you! – James Ronald Apr 14 at 17:40
You are adding partial sums instead of finding the limit of partial sums.
That is why you are getting confused.
You have already taken care of the sigma by adding the terms of your series to find partial sums, so there is no need to add partial sums. | 2019-11-12T17:06:48 | {
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https://cs.stackexchange.com/tags/floating-point/hot | # Tag Info
22
There is no way to represent all real numbers without errors if each number is to have a finite representation. There are uncountably many real numbers but only countably many finite strings of 1's and 0's that you could use to represent them with.
20
It all depends what you want to do. For example, what you show is a great way of representing rational numbers. But it still can't represent something like $\pi$ or $e$ perfectly. In fact, many languages such as Haskell and Scheme have built in support for rational numbers, storing them in the form $\frac{a}{b}$ where $a,b$ are integers. The main reason ...
20
In typical floating point implementations, the result of a single operation is produced as if the operation was performed with infinite precision, and then rounded to the nearest floating-point number. Compare $a+b$ and $b+a$: The result of each operation performed with infinite precision is the same, therefore these identical infinite precision results ...
15
It's because 0.1 = 1 / 10 = 1 / (2 × 5) cannot be represented exactly in binary floating point. This happens in C too: printf("%.20f", fmod(4.0,0.1)); prints 0.09999999999999978351. Specifically, the only numbers that binary floating point can represent exactly are dyadic fractions of the form a / 2b, where a and b are integers. Even more ...
14
The usual trick to avoid this underflow is to compute with logarithms, using the identity $$\log \prod_{i=1}^n p_i = \sum_{i=1}^n \log p_i.$$ That is, instead of using probabilities, you use their logarithms. Instead of multiplying them, you add them. Another approach, which is not so common, is to normalize the product manually. Instead of keeping just ...
8
Ilmari Karonen gets it right in the other answer. But it gets even worse than that: arithmetic operations involving floating-point numbers don't necessarily behave the same as operators we're used to from mathematics. For instance, we're used to addition being associative, so that $a + (b + c) = (a + b) + c$. This doesn't generally hold using floating-point ...
8
There actually is some research on improving the numerical stability of floating point expressions, the Herbie project. Herbie is a tool to automatically improve the accuracy of floating point expressions. It's not quite comprehensive, but it will find a lot of accuracy improving transformations automatically. Cheers, Alex Sanchez-Stern
8
Assuming multiplication between two numbers use one FLOP, the number of operations for $x^n$ will be $n-1$. However, is there a faster way to do this ... There most certainly is a faster way to do this for non-negative integer powers. For example, $x^{14}=x^{8}x^{4}x^{2}$. It takes one multiplication to compute $x^2$, one more to compute $x^4$, one more to ...
7
Your idea does not work because a number represented in base $b$ with mantissa $m$ and exponent $e$ is the rational number $b \cdot m^{-e}$, thus your representation works precisely for rational numbers and no others. You cannot represent $\sqrt{2}$ for instance. There is a whole branch of computable mathematics which deals with exact real arithmetic. Many ...
7
There are many effective Rational Number implementations but one that has been proposed many times and can even handle some irrationals quite well is Continued Fractions. Quote from Continued Fractions by Darren C. Collins: Theorem 5-1. - The continued fraction expression of a real number is finite if and only if the real number is rational. Quote ...
7
You cannot meaningfully test floating point values for equality. A floating point value does not represent a real number, it represents a range of real numbers, but it fails to store the width of this interval. All you can do with floating point values is to test them for approximate equality, and it's up to you to define the approximation you're willing to ...
7
This is best explained in the base 10 equivalent: scientific notation In scientific notation you have a mantissa and a exponent such that the value is $\mathrm{mantissa} \cdot 10^{\mathrm{exponent}}$. In a computer floating point the mantissa is always the same size of significant digits (a double precision has around 16 digits) and the exponent is bounded....
6
If you drop the leading zeroes then $3,5,9,17$ and so on will all have the same representation. Don't forget that the number being represented need not be an integer.
6
As long as you execute the same machine code on the different machines and as long as the settings for the floating point unit are identical, you will get identical results. However, you cannot execute the same machine code on both Intel and ARM, so this answer is only hypothetic. Even on different Intel processors you have to take special care that exactly ...
6
No, that's not how binary fractions work. A decimal such as $0.13$ represents $$0.13_{\mathrm{dec}} = 1\times 10^{-1} + 3\times10^{-2} = \frac{13}{100}\,.$$ Similarly, a binary fraction such as $0.1101$ represents $$0.1101_{\mathrm{bin}} = 1\times 2^{-1} + 1\times 2^{-2} + 0\times 2^{-3} + 1\times2^{-4} = \frac{13}{16}\,.$$ $1101_{\mathrm{bin}} = 13_{\... 6 There's nothing fundamentally hard about computing$\sin(10^{99})$. You simply compute$x = 10^{99} \bmod 2\pi$, then compute$\sin(x)$. (Why is this valid? It's because$\sin(x)=\sin(y)$if$x\equiv y \pmod{2\pi}$.) It's not too hard to compute$x$if you use a numerical representation that has enough digits of precision, and then to compute$\sin(x)$... 6 Firstly, you need to decide if the floating point numbers you are working with are "normalized" or not. Think about how binary numbers are represented in scientific notation. Consider the number 100101.010101 The expression given above is not given in scientific notation, but what if we change that? To re-write 100101.010101 in scientific notation, we move ... 6 With some algebraic manipulation (as pointed out in @orlp's answer), we can deduce the following: $$f(x) = x \tanh(\log(1+e^x)) \tag{1}$$ $$= x\frac{(1+e^x)^2 - 1}{(1+e^x)^2 + 1} = x\frac{e^{2x} + 2e^x}{e^{2x} + 2e^x + 2}\tag{2}$$ $$= x - \frac{2x}{(1 + e^x)^2 + 1} \tag{3}$$ Expression$(3)$works great when$x$is negative with very little loss of ... 5 There are a number of "exact real" suggestions in the comments (e.g. continued fractions, linear fractional transformations, etc). The typical catch is that while you can compute answers to a formula, equality is often undecidable. However, if you're just interested in algebraic numbers, then you're in luck: The theory of real closed fields is complete, o-... 5 Yes, the difference is constant. It is not really constant, but approximately, yes. With exceptions. With binary floating point numbers, the expression$(f(r)−r)/r$is constant within a factor of$2$. It is between$1 \over 2^m$and$1 \over 2^{m-1}$where m is the number of bits in the mantissa. For rounding error calculations, you can assume it is ... 5 IEEE floating point format has a sign bit, an 11 bit exponent (ranging from -1022 to 1023) and a 52-bit mantissa with an implicit "1" in the 53rd bit. Thus, the largest integer that can be represented without rounding is the binary number with 53 "1"s,$2^{53}-1$= 9,007,199,254,740,991 ~ 9e15 < 1e16. After that you start having to round off low order ... 5 The == operator invoke a float point compare instruction, this considers two numbers equal if neither is NaN, and they are exactly the same, or if they are positive and negative zero. Notably Infinity == Infinity, despite the mathematical dubiety of considering infinities equal, and Infinity - Infinity returning NaN. So if the numbers are not exactly equal, ... 5 Generally speaking, normalized means "put in scientific notation." That just means, the mantissa should never start with 0, and should be less than the base. In binary that means the mantissa must be "1". Since the mantissa of a normalized binary floating point number is always 1, we don't need to store the 1. The first mantissa bit is hidden in the ... 5 Java uses IEEE 754 binary floating point representation, which dedicates 23 binary digits to the mantissa, that is normalized to begin with the first significant digit (omitted, to save space).$0.00004_{10} = 0.00000000000000101001111100010110101100010001110001101101000111..._{2} = [1.]\color{red}{01001111100010110101100}010001110001101101000111..._{2} \...
5
The binary floating point format supported by computers is essentially similar to decimal scientific notation used by humans. A floating-point number consists of a sign, mantissa (fixed width), and exponent (fixed width), like this: +/- 1.0101010101 × 2^12345 sign ^mantissa^ ^exp^ Regular scientific notation has a similar format: +/- 1.23456 × 10^...
5
Because sometimes there just isn't an exponent high enough. First of all, there is no way to effectively represent all real numbers in a reasonable computational model: it is sufficient to observe that the set of configurations of e.g. a Turing Machine is countably infinite, and the set of real numbers is uncountably infinite. Floating point numbers, are ...
5
The IEEE 754 standard defines exactly how floating-point arithmetic is performed. For many interesting theorems, you will need to examine the exact definition. For some less interesting ones, like a+b = b+a or ab = ba, all you need to know that IEEE 754 always calculates the exact result, rounded in a deterministic way. For non-theorems, like (a+b)+c = a+(b+...
5
Yes, there is a usage for the negative imaginary zero. But first, I will say something about the negative zero in general. Why have a negative zero? First of all, the main reason to have a signed zero for floating points is that floating points have only limited precision and we often want to distinguish the following cases of a real number $x$, ...
5
It may be that sqr (sqrt (7)) is displayed as 7, but it isn't actually exactly equal to 7. That's something you need to check. What you see is not always what you get. It may be that sqr (sqrt (7)) is exactly 7, by "coincidence" (not really coincidence, more like "unpredictable with my limited knowledge"). Take any floating point number x, 1 ≤ x < 4. ...
5
Although, the question is a bit old, but it may help people coming here for similar question. A vital detail was missed out in the article that you referred to and it is that the standard chose to interpret an all 0s exponent to be equivalent to '-126' and not '-127'. One place, where I found a nice explanation (and an explicit statement about this truth) ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2021-08-05T03:43:40 | {
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https://www.physicsforums.com/threads/has-anybody-got-a-cleaner-solution.611125/ | # Has anybody got a cleaner solution?
1. Jun 3, 2012
### macilrae
Attempting to solve
dx/dy = x/y - √{1 + (x/y)^2}
I can substitute x/y = z and get
ln(y/c) = arcsinh(z) c is constant of integration
or putting x/y = tan(θ)
I get
ln(y/c) = ln(sec(θ) + tan(θ))
both of which do give a parabola but the interim logs seem so messy and unnecessary.
Is there a better more direct substitution?
2. Jun 4, 2012
### JJacquelin
Hi !
Let x = y sinh(t)
dx/dy = sinh(t) + y cosh(t) (dt/dy)
which leads to a very simple ODE
3. Jun 4, 2012
### macilrae
Thanks, JJaquelin - that's certainly a fast and ingenious idea - you still bump into the Sinh(ln()) thingy which I had hoped to avoid - probably I'm too picky.
The equation arose from asking what curve, symmetric about the x axis, will always reflect back a right-to-left ray, parallel to x axis, through the same point (parabolic reflector - focal point).
4. Jun 4, 2012
### AlephZero
A trig substitution is the "natural" way to solve this if you formulate the DE using geometry. Putting the fixed point at the origin, the equations for the directon of the reflected ray and the tangent to the curve are
dy/dx = cot (t/2)
tan t = y/x
5. Jun 4, 2012
### JJacquelin
sinh(ln(c*y)) = (c*y-(1/(c*y))/2
So, there is no hyperbolic function, nor trigonometric function, in the final result : x = function of y , wich is a very simple parabolic function.
6. Jun 4, 2012
### macilrae
AlephZero - Right, those were the starting equations & they look so simple - I don't begrudge a log popping up, but when I immediately have to antilog it, intuitively I think "There has to be a more elegant way!"
7. Jun 4, 2012
### macilrae
JJaquelin - it's the sinh(ln()) that is bugging me! Having to "do and undo" seems redundant - especially because the final expression is the simplest parabola!
8. Jun 4, 2012
### Dickfore
Notice that:
$$\mathrm{arcsinh} (x) = \ln \left( x + \sqrt{1 + x^2} \right)$$
$$\frac{dx}{dy} = \frac{x}{y} - \sqrt{1 + \left( \frac{x}{y} \right)^2}$$
after the substitution
$$z = \frac{x}{y}, \ z = z(y)$$
reduces to:
$$y \, z' + z = z - \sqrt{1 + z^2}$$
$$\frac{dz}{\sqrt{1 + z^2}}+ \frac{dy}{y} = 0$$
which integrates to:
$$\mathrm{arcsinh}(z) + \ln(y) = C$$
I think you have the realtive sign opposite to this in your solution. Then, use the above representation of the inverse hyperbolic sine, and take the antilogratihm:
$$y \, \left( z + \sqrt{1 +z^2} \right) = C_1$$
$$x + \sqrt{x^2 + y^2} = C_1$$
$$\sqrt{x^2 + y^2} = C_1 - x$$
Square it:
$$x^2 + y^2 = \left(C_1 - x \right)^2$$
$$x^2 + y^2 = C^{2}_{1} - 2 \, x \, C_1 + x^2$$
$$x = \frac{C_1}{2} - \frac{y^2}{2 C_1}$$
9. Jun 4, 2012
### Dickfore
Another way is to solve it as an implicit equation. Introduce $p = dx/dy$, and $z = x/y$. Then:
$$p = z - \sqrt{1 + z^2}$$
$$\sqrt{1 + z^2} = z - p$$
Squaring it gives:
$$1 + z^2 = z^2 - 2 p z + p^2$$
$$z = \frac{p^2 - 1}{2 p}$$
or
$$x = y \, \frac{p^2 - 1}{2 p}$$
This is an implicit equation of the Lagrange-Clairot type. Differentiate w.r.t. y to get:
$$p \, dy = \frac{p^2 - 1}{2 p} \, dy + y \, \frac{2 p \cdot p - (p^2 - 1) \cdot 1}{2 p^2} \, dp$$
$$\frac{2 p^2 - p^2 + 1}{2 p} \, dy = y \, \frac{2 p^2 - p^2 + 1}{2 p^2} \, dp$$
$$\frac{d y}{y} = \frac{d p}{p}$$
Integrating this equation gives:
$$\ln(p) = \ln(y) + C'_1$$
and taking the antilogarithm gives:
$$p = C_1 \, y$$
Substitute this in the parametric solution for x:
$$x = y \, \frac{C^{2}_{1} y^2 - 1}{2 \, C_{1} \, y} = \frac{C_1 \, y^2}{2} - \frac{1}{2 \, C_1}$$
which is the same general solution as before if you make the substitution $C_1 \rightarrow -1/C_1$.
But, there might also be a singular solution to the equation. It is obtained as an envelope of the family of general solutions. Differentiate w.r.t. to the arbitrary constant to get:
$$0 = \frac{y^2}{2} + \frac{1}{2 \, C^{2}_{1}}$$
This sum of non-negative terms can only be zero, if each term is equal to zero separately. This corresponds to:
$$y = 0, \ C_1 \rightarrow \infty$$
To check whether $y = 0$ is a solution of the original equation, solve it for $y' = dy/dx = (dx/dy)^{-1}$:
$$y' = \frac{1}{x/y - \sqrt{1 + (x/y)^2}} = \frac{y}{x - \mathrm{sgn}(y) \, \sqrt{y^2 + x^2}}$$
One may argue that it is because of the y in the numerator, or that the original equation is not defined for $y = 0$.
10. Jun 4, 2012
### macilrae
Thanks to everyone for engaging my problem - seems the way below is the least painful but certainly not the most creative (thanks Dickfore, I had to look up Lagrange-Clairot!)
The original conditions for the mirror profile were:
dx/dy = -tan(2α) and y/x = tan(α)
which boils down to
dx/dy = x/y ± √{1 + (x/y)^2} ... thanks again for the "+" Dickfore, I'll use that.
using x/y = tan(θ)
dy/y = sec(θ)dθ ... really simple, huh?
but then
ln(y/C) = ln[tan(θ) + sec(θ)]
ending up with
y^2 = 2Cx + C^2 where C/2 is the focal length
for such a trivial result it seemed messy to go in and out of logs but hey
11. Jun 4, 2012
### macilrae
correction
dx/dy = tan(α) and y/x = -tan(2α)
sorry
12. Jun 5, 2012
### JJacquelin
Hi !
if you know an easier way...
#### Attached Files:
• ###### Easy way.JPG
File size:
13.9 KB
Views:
129
13. Jun 5, 2012
### macilrae
JJaquelin: It's just that given the chance I prefer trigonometrics to hyperbolicals - but never call me prejudiced - some of my best fiends are quite hyperbolic ... | 2018-09-25T09:24:50 | {
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http://math.stackexchange.com/questions/289857/integral-of-int-frac1-sqrtx1-x-dx | # Integral of $\int \frac{1}{\sqrt{x(1-x)}} dx$
$$\int \frac{1}{\sqrt{x(1-x)}} dx$$ I solved the integral in this way: make the substitution $x=\sin^2(u)$, then $dx=2\sin(u)\cos(u)du$. So the integral now becomes $$\int \frac{2\sin(u)\cos(u)}{\sqrt{\sin^{2}(u)(1-\sin^{2}(u))}} du=\int 2 du=2u+C.$$ Then subbing in $u=\arcsin(\sqrt{x})$ I get the solution $$2\arcsin(\sqrt{x})+C.$$ However when I typed in this integral onto Wolfram it gave me this.
So my question is did I get it wrong or are the two forms equivalent?
-
These answers are equivalent for $0<x<1$. – Sasha Jan 29 '13 at 16:53
Note that the Wolfram answer requires complex numbers in this range. There is a connection between the complex logarithm and the inverse trigonometric functions derived from the similar connection between the complex exponential and the trigonometric functions that explains this phenomenon. – Harald Hanche-Olsen Jan 29 '13 at 16:54
FWIW, the graph of your solution matches the graph of the real part of the Wolfram solution – DJohnM Jan 29 '13 at 17:03
@Sasha totally forgot about the sqrt and the range of real values it could take and also the fact that Wolfram gives complex solutions. – SomethingWitty Jan 29 '13 at 17:05
First of all, we never have to ask whether we got a correct antiderivative - because we can just take the derivative of our answer and see what we get: $$\frac d{dx} \big( 2\sin^{-1}(\sqrt x) +C \big) = \frac2{\sqrt{1-(\sqrt x)^2}}\frac d{dx}\sqrt x = \frac2{\sqrt{1-x}} \frac1{2\sqrt x} = \frac1{\sqrt{x(1-x)}}.$$ So you did it right.
There's another way of finding the antiderivative: complete the square inside the square root to see that $$\frac1{\sqrt{x(1-x)}} = \frac1{\sqrt{1/4 - (x-1/2)^2}} = \frac2{\sqrt{1-(2x-1)^2}}.$$ Therefore, using the substitution $u=2x-1$, \begin{align*} \int \frac1{\sqrt{x(1-x)}} \,dx = \int \frac2{\sqrt{1-(2x-1)^2}} \,dx &= \int \frac1{\sqrt{1-u^2}} \,du \\ &= \sin^{-1} u + C = \sin^{-1} (2x-1) + C. \end{align*} This can also be verified by differentiating.
(Side note: it's possible to check directly that $\sin^{-1}(2x-1)+\pi/2 = 2\sin^{-1}(\sqrt x)$, by taking the cosine of both sides, using $\cos(\theta+\pi/2) = -\sin\theta$ on the left and the double-angle formula $\cos 2\theta = 1-2\sin^2\theta$ on the right. Pretty cool!)
-
Put $\displaystyle x = \frac{1}{2}+t$ and $dx = dt$
So $\displaystyle \int\frac{1}{\sqrt{x.(1-x)}}dx$ is converted into
$\displaystyle \int\frac{1}{\sqrt{\frac{1}{2^2}-t^2}}dt = \sin^{-1}(2t)+\mathbb{C} = \sin^{-1}\left(2x-1\right)+\mathbb{C}$
- | 2016-06-28T08:21:44 | {
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https://math.stackexchange.com/questions/4478892/calculus-difference-between-functions-and-equations-from-a-theoretical-perspe | # Calculus: Difference between functions and "equations" from a theoretical perspective
I've been studying Multivariable Calculus for a while; but I still don't quite know the difference between $$f(x,y) = x^2 + y^2$$ and $$x^2 + y^2 = 9$$. I know that the former graphs a paraboloid, while the latter a cylinder. But what's the difference (or 'theoretical difference' if you will) between having a function and having an equation.
Do we assume that the function maps to z-axis because of convention? Doesn't the equation map it's point to the z-axis as well?
I sometimes feel like this fact was skipped or even 'ignored' by the teachers.
• A function is a mathematical object. An equation is a mathematical statement of the form $a=b$, which can be true or false. In practice, we often define functions via an equation, by taking the function to be that function that makes the equation true. 2 days ago
• In practice mathematicians "abuse notation" a bit by saying things like "Consider the function $f(x,y) = x^2 + y^2$ to mean "Consider the function $f$ defined by the equation $f(x,y) = x^2 + y^2$, even though technically $f(x,y) = x^2 + y^2$ is an equation, not a function, but this is generally considered OK shorthand. 2 days ago
• As for $x^2 + y^2 = 9$, this implicitly defines a set $\{(x,y) \in \mathbb{R}^2 | x^2 + y^2 = 9\}$. This is what you get if you consider the points $x,y$ with $f(x,y) = 9$. It might help to think of it as a contour corresponding to the height of $9$ on a topographic map of the plane where the elevation of a point $(x,y)$ is $x^2 + y^2$. 2 days ago
It depends on the context what you mean with $$f(x,y) = x^2+y^2 \tag 1$$
• What it could be is the definition of a function $$f:\Bbb R^2\to\Bbb R$$, or over any other domains where concepts like squaring and adding makes sense, like $$\Bbb Z$$, $$\Bbb Z/n\Bbb Z$$, some field $$K$$ just to name a few. Only from the question one could infer that it's actually the definition of a function.
• Only in some specific context $$(1)$$ represents a paraboloid, like: Let $$P$$ be the set of all points in $$\Bbb R^3$$ such that $$P=\{(x,y,f(x,y))\in\Bbb R^3\}$$ then $$P$$ is a paraboloid. In this context you are talking about the "plot of" some function defined by $$(1)$$ rather than about a function definition.
• The function $$f$$ could be defined completely different, like: Let $$f(x,y) = x-y$$. Determine all solutions of $$(1)$$. Or: draw all points in the $$x$$-$$y$$-plane that satisfy $$(1)$$, which turns out to be a circle.
Similarly for the second equation: $$x^2+y^2 = 9\tag 2$$
Only from the supplied context one can infer that $$(2)$$ is used like
• The definition of an algebraic variety, more specifically a quadratic surface: Let $$C=\{(x,y,z)\in\Bbb R^3\mid x^2+y^2 = 9\}$$. Then $$C$$ is a cylinder — or a circular pipe that extends to infinity, as is has neither top nor bottom.
• Without that context, it could just as well be a circle in $$\Bbb R^2$$.
• You can read it as a Diophantine equation.
• You can understand it as implicit function definition.
So the bottom line is: What counts is the context.
Function notation
A function like $$f(x,y) = x^2+y^2$$ can be thought of as a machine that takes the input $$(x,y)$$ (a pair of real numbers) and returns the real number $$x^2 + y^2$$. Sometimes this is notated as $$f:\mathbb R^2\to \mathbb R$$ to emphasize the domain $$(\mathbb R^2)$$ of the function and the codomain $$(\mathbb R)$$.
Defining sets
Equations like $$x^2+y^2=9$$ are properly speaking sets of points. In your example, the proper way to describe the mathematical object "$$x^2+y^2=9$$" is as a set $$\{(x,y,z): x^2+y^2=9\}$$, which we read as "the set of all triples $$(x,y,z)$$ of real numbers such that $$x^2+y^2=9$$." The equation $$x^2+y^2=9$$ in this context is often referred to as the defining equation for this particular set of points, and the set itself can indeed be visualized as a cylinder in space. Notice how it is necessary to specify the context. The same equation makes sense in the plane as well, and the set $$\{(x,y):x^2+y^2=9\}$$ has the same defining equation, but it is a different collection of points (a circle) because it's in the plane.
Implicit functions
Some sets of points like $$\{(x,y):x^2+y^2=9\}$$ implicitly define functions, since we can solve for $$x$$ or $$y$$ as a function of the other variable, as in $$y = \sqrt{1-x^2}$$ for the portion of the circle lying in $$\{(x,y):y>0\}$$. This equation $$y=\sqrt{1-x^2}$$ implicitly defines a function $$f\colon(-1,1)\to\mathbb R$$ by $$f(x) = \sqrt{1-x^2}$$. Here is a good example of where the function notation "$$f\colon A\to B$$" helps keep track of important data about the implicit function.
A takeaway
It's more useful to think of functions "dynamically" as rules or machines that take inputs to outputs, and sets of points "statically" with a defining equation (or multiple defining equations) that defines the set of points. Not all sets of points have defining equations. For example, some sets of points are defined by inequalities, such as $$\{(x,y,z):z> 0\}$$. This set of points is visualized as the "upper half of space."
If you continue studying math, you will sooner or later encounter the subject of linear algebra, or real analysis. In either subject, you will get a treatment of the more precise definitions of functions and sets of points, since these precise definitions are fundamental to building up a properly rigorous mathematical environment.
• Hi! I already sat for Lineal Algebra ("everything" up until Linear Transformations). However, even then a lot of the concepts were skipped over. Maybe it's because I'm attending an Engineering University, and Engineers just skip said explanations in favor of... something. 2 days ago
• Can’t a function also be viewed as a set? 2 days ago
• @ToddWilcox: That's absolutely right, since a function can be identified with its graph, which is a set of points. I think for the purpose of helping distinguish functions and "equations" as in this question however, the right perspective is to view a function as a rule for assigning values to inputs. yesterday
This is actually more general than just "calculus", but unfortunately the way most maths is taught mixes up its different historical stages of development instead of just presenting a clean logical picture from the get-go based on the most up-to-date state of the field.
In a fully modern understanding, an equation is a statement: it is an assertion that something is true - namely, that two things are equal. "$$=$$" is a relation, and when you write
$$a = b$$
you are asserting that that relation holds true for the objects named "$$a$$" and "$$b$$". In this regard, such an equation is kind of like a mathematical "sentence", in the same way that this post is composed mostly of English sentences. Each one asserts something to be true - for example, the last sentence you read just asserted that "it is true that this post is mostly composed of English sentences". (Note of course that asserting something is true, and its being true, are two different thing.) Relations are like mathematics' "verbs".
A function is a specific type of mathematical object, which can be thought of as describing or defining some kind of "property" of objects of a particular kind, which can have different values depending on exactly which object we are looking at. For example, if we had a set of cars, all of which we notionally understand as having a single color (i.e. a "red car", regardless of whether 100% of every surface is red-colored), then we can create a "color function" which, when you give it a car as its input argument, becomes a symbol standing for the color of that car. That is, if we write it as $$\mathrm{Color}(\cdot)$$, and the car is $$A$$, then the combined symbol $$\mathrm{Color}(A)$$ has the same meaning as whatever the color is that our car has, e.g. if $$A$$ is a red car, then the expression $$\mathrm{Color}(A)$$ means the same thing as "red".
And thus we can write an equation
$$\mathrm{Color}(A) = \mathrm{red}.$$
When you wrote
$$f(x, y) = x^2 + y^2$$
you did write an equation, not a function. The function is just
$$f.$$
Yes, not $$f(x, y)$$. Just $$f$$. $$f(x, y)$$ is the value of the function when arguments $$x$$ and $$y$$ are put in. What is the function? Well, that's the thing - you see, we can use such an equation to define a function implicitly, by saying that $$f$$ is the function which makes that equation true, i.e. it is the function that takes, when you give it values $$x$$ and $$y$$, the value $$x^2 + y^2$$. Because there is only one value for $$x^2 + y^2$$, the definition is unambiguous, and thus this equation serves "double duty" to define the function.
If you want to make it clear what you are doing is defining the function, you can use the assignment or walrus operator $$:=$$, and write
$$f(x, y) := x^2 + y^2$$
which means you are saying you are thus making $$f(x, y)$$ equal to $$x^2 + y^2$$ within the given context (or "scope", which is implicit, not explicit).
Now about "graphing". It is true that the graph of $$f$$, deefined aboev, is a paraboloid. But what do we mean by "graph"? There's actually two, though not unrelated, meanings at play here.
When we have a function $$f$$, its "graph" is the set of all ordered pairs $$(X, f(X))$$ (note here the use of capital $$X$$, a function with two arguments can also be considered a function of one argument consuming an ordered pair as its argument). In your case, this would be equivalent to ordered triples $$(x, y, f(x, y))$$, so it describes a surface in 3D space. That surface is, as you point out, a paraboloid.
When we have an equation, e.g.
$$x^2 + y^2 = 9$$
what we have is not "technically" called a graph but rather a solution set: it is the set of possible variable assignments for $$x$$ and $$y$$ that make the equation true. Here, that set is the same as the set of all Cartesian coordinates of the points on the circle of radius 3 centered at the origin. The "graph" is actually the set of coordinates, so it's an interpretation of this equation, not the equation "itself".
So what's the difference? A function is an object, an equation is a statement, but the two bear a variety of relationships that have to be elucidated carefully in order to make things clear.
What's the difference (or 'theoretical difference' if you will) between having a function and having an equation?
• If you have a function you have a "transformation" (variables assigned to other one). If you have an equation you have a "thing to be solved" (unknown to be determined).
• A function is a specific type of correspondence between two sets. An equation is an assertion of equality between two mathematical expressions involving at least one unknown.
• Functions have graphs, which are specific type of sets. Equations have solutions, which are the values of the unknowns that make the equality true.
• In some cases, the graph of a function has a geometric representation. In some cases, the set of solutions of an equation has a geometric representation as well. However, the fact that there are geometric representations associated with functions and equations does not imply that they are the same thing (actually, at least theoretically, any subset of $$\mathbb R^3$$ has a geometric representation regardless of whether it's the graph of a function, the set of solutions of an equation, or anything else).
I'm guessing you're coming from a multivariable calculus perspective. The idea you're circling is that of a smooth manifold. A manifold somehow makes precise our notion of a surface. Let's stick with the idea of a 2-dimensional manifold embedded in $$\mathbb{R}^3$$ for this discussion, though you should know that generalizations are possible.
There are at least 3 equivalent ways to describe a 2-dimensional manifold embedded in $$\mathbb{R}^3$$:
1. It is locally the image of a function $$\vec{x} : A \subset \mathbb{R}^2 \to \mathbb{R}^3$$. For example, $$\vec{x}(\theta , \phi) = (\sin\theta\cos\phi, \sin\theta \sin\phi, \cos\theta)$$ has a subset of the unit sphere as its image.
2. It is locally the graph of a function $$f: A \subset \mathbb{R}^2 \to \mathbb{R}$$. Formally, what this means is that we build a function like the one in 1 out of $$f$$ whose image is the manifold. For example, we may take $$f(x,y) = \sqrt{1 - x^2 - y^2}$$. We form the function $$F(x,y) = (x, y, f(x,y)) = (x, y, \sqrt{1 - x^2 - y^2})$$. We say the image of $$F$$ is the graph of $$f$$, and it too has a subset of the unit sphere as its image.
3. It is locally the solution set (in more formal language "the pre-image of $$0$$") of a function $$g : \mathbb{R}^3 \to \mathbb{R}$$. For example, if $$g(x,y) = x^2 + y^2 + z^3 - 1$$, then the solutions to $$g(x,y, z) = 0$$ are a subset of (in fact the whole) unit sphere.
So there you have it. Three different ways to talk about 2-dimensional manifolds embedded in $$\mathbb{R}^3$$. I think the reason you are confused is because multivariable calculus often freely passes between these three different notions as needed without pausing to recognize that they are equivalent.
If you want to know more about these ideas, I can recommend Shifrin's book on differential geometry to get your feet wet.
If you want technical details, I can recommend Edwards Advanced Calculus of Several Variables, which goes into detail on the machinery that proves these descriptions are equivalent. Ultimately it comes down to something called the implicit function theorem.
We often see an expression like $$x^2+y^2=9$$ as a level surface to $$f(x,y)=x^2+y^2$$, i.e. $$f(x,y)=9$$.
For example, we might want to find the tangent plane to a point on the surface defined by $$x^2+y^2=9$$.
We then define $$f(x,y)=x^2+y^2$$. When $$f(x,y)=9$$ which is a constant, we know the derivative in every direction is zero which gives us that $$\nabla f\cdot\hat u=0$$ for every direction $$\hat u$$ from which we conclude that the gradient is normal to the surface. That normal and the point give us the tangent plane.
1. The function specified by $$g(x,y) := x^2 + y^2$$ has maximal domain $$\mathbb R^2,$$ in which every input/point $$(x,y)$$ has a corresponding value $$x^2 + y^2$$ in $$\mathbb R.$$
This function might output the temperature variation on an infinite plane.
2. Now, not every point $$(x,y)\in\mathbb R^2$$ satisfies the conditional equation (where $$g$$ is defined as above) $$g(x,y) = 9;\tag1$$ its solution set—the collection of all points that satisfy it—$$\{(x,y)\in\mathbb R^2\mid x^2+y^2=9\}$$ traces out a circle on the $$x$$-$$y$$ plane.
3. In 3D Euclidean space, points are triples $$(x,y,z)\in\mathbb R^3,$$ and the conditional equation $$(1)$$ is now satisfied by the cylinder $$\{(x,y,z)\in\mathbb R^3\mid x^2+y^2=9\}$$ instead.
4. With $$g$$ defined as above, setting $$z:=g(x,y)$$ so that $$z$$ explicitly depends on $$x$$ and $$y:$$ $$z=x^2+y^2.$$ This equation is satisfied by the paraboloid $$\{(x,y,x^2+y^2)\mid (x,y)\in\mathbb R^2\}$$.
5. Notice that while every $$(x,y)$$ combination can be found on the paraboloid, only those that satisfy $$x^2+y^2=9$$ can be found on the cylinder.
6. Consider the part of the paraboloid that lies on the plane $$z=9,$$ and reset the latter as the new $$x$$-$$y$$ plane: this gives the circle $$(1).$$ | 2022-06-26T12:21:23 | {
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https://math.stackexchange.com/questions/1631639/upper-bound-on-integral-int-1-infty-fracdx-sqrtx3-1-4/1631760 | # Upper bound on integral: $\int_1^\infty \frac{dx}{\sqrt{x^3-1}} < 4$
I'm going through Nahin's book Inside Interesting Integrals, and I'm stuck at an early problem, Challenge Problem 1.2: to show that $$\int_1^\infty \frac{dx}{\sqrt{x^3-1}}$$ exists because there is a finite upper-bound on its value. In particular, show that the integral is less than 4.
I've tried various substitutions and also comparisons with similar integrals, but the problem is that any other integral that I can easily compare the above integral to is just as hard to integrate, which doesn't help solve the problem.
I also tried just looking at the graph and hoping for insight, but that didn't work either.
So how doesone one place an upper bound on the integral?
• You may also improve the upper bound to $3$. – Jack D'Aurizio Jan 29 '16 at 9:40
• Letting $x=\dfrac1t$ we are able to express the result in terms of the beta function. – Lucian Jan 29 '16 at 13:51
$$\frac{1}{\sqrt{x^3-1}} = \frac{1}{\sqrt{x-1}\sqrt{x^2+x+1}} < \frac{1}{x\sqrt{x-1}}$$
$$\int_1^\infty \frac{\mathrm{d}x}{x\sqrt{x-1}} = \pi < 4$$
This integral can be done by letting $u=\sqrt{x-1}$ which yields $\mathrm{d}x=2u\mathrm{d}u$,
$$\int_0^\infty \frac{2 \mathrm{d}u}{u^2+1} = \pi$$
• I like this answer. It's nice and simple and gives a better upper bound than 4. Thanks! – feralin Jan 29 '16 at 7:22
• No problem, I enjoyed figuring it out. I love that book you're reading :) – Spencer Jan 29 '16 at 7:23
$$\begin{eqnarray*}\color{red}{I}=\int_{1}^{+\infty}\frac{dx}{\sqrt{x^3-1}}=\int_{0}^{+\infty}\frac{dx}{\sqrt{x^3+3x^2+3x}}&=&\int_{0}^{+\infty}\frac{2\, dz}{\sqrt{z^4+3z^2+3}}\\&\color{red}{\leq}&\int_{0}^{+\infty}\frac{2\,dz}{\sqrt{z^4+3z^2+\frac{9}{4}}}=\color{red}{\pi\sqrt{\frac{2}{3}}.}\end{eqnarray*}$$
A tighter bound follows from Cauchy-Schwarz:
$$\begin{eqnarray*} \color{red}{I} &=&2\int_{0}^{+\infty}\frac{\sqrt{z^2+\sqrt{3}}}{\sqrt{z^4+3z^2+3}}\cdot\frac{dz}{\sqrt{z^2+\sqrt{3}}}\\&\color{red}{\leq}& 2\sqrt{\left(\int_{0}^{+\infty}\frac{z^2+\sqrt{3}}{z^4+3z^2+3}\,dz\right)\cdot\int_{0}^{+\infty}\frac{dz}{z^2+\sqrt{3}}}\\&=&2\pi\cdot\left(\frac{1}{6}-\frac{1}{4\sqrt{3}}\right)^{1/4}\leq\color{red}{\frac{2\pi}{42^{1/4}}}.\end{eqnarray*}$$
The manipulations in the first line show that $I$ is just twice a complete elliptic integral of the first kind, whose value can be computed through the arithmetic-geometric mean.
On the other hand, through the substitution $x=\frac{1}{t}$ and Euler's beta function we have:
$$I \color{red}{=} \frac{\Gamma\left(\frac{1}{3}\right)\cdot \Gamma\left(\frac{1}{6}\right)}{2\sqrt{3\pi}}\color{red}{\leq}\frac{3\cdot 6}{2\sqrt{3\pi}}=\sqrt{\frac{27}{\pi}}$$
since in a right neighbourhood of the origin we have $\Gamma(x)\leq\frac{1}{x}$.
As a by-product we get:
$$\frac{\Gamma\left(\frac{1}{3}\right)\cdot \Gamma\left(\frac{1}{6}\right)}{2\sqrt{3\pi}} = \frac{\pi}{\text{AGM}(\frac{1}{2} \sqrt{3+2 \sqrt{3}},3^{1/4})}$$
that allows us to compute $\Gamma\left(\frac{1}{6}\right)$ through an AGM-mean:
$$\Gamma\left(\frac{1}{6}\right) = \color{red}{\frac{2^{\frac{14}{9}}\cdot 3^{\frac{1}{3}}\cdot \pi^{\frac{5}{6}} }{\text{AGM}\left(1+\sqrt{3},\sqrt{8}\right)^{\frac{2}{3}}}}.$$
The last identity was missing in the Wikipedia page about particular values of the $\Gamma$ function, so I took the liberty to add it and add this answer as a reference.
• I can see that $\int_0^\infty\frac{1}{x^4+3x^2+3}\;dx$ is an elliptic integral but I am curious as to how you evaluated it. The methods I know for putting a quartic like that into standard form are complicated enough that I can't get to your closed form (although I can see that it's right) working by hand (or even with Wolfram Alpha helping me). Is there a simple rule I'm missing for transforming $x^4+3x^2+3$? – Anon Jan 21 '17 at 2:23
• @Anon: what you wrote is not an elliptic integral, we may evaluate it through residues. However, $\sqrt{z^4+3z^2+3}=\sqrt{(z^2+A)(z^2+B)}$ and $$\int_{0}^{+\infty}\frac{dz}{\sqrt{(z^2+A)(z^2+B)}}$$ is a complete elliptic integral of the first kind, that can be evaluated through the AGM. – Jack D'Aurizio Jan 21 '17 at 12:40
• Oh yes, a typing error on my part to miss the square root. I was aware of what you wrote; sorry for not being clear. Specifically what I was asking was how you computed the values of $A$ and $B$. I got complex values, and tried to use the methods I knew of transforming into a real-valued expression by hand, but failed. How did you compute it? – Anon Jan 22 '17 at 23:02
• @Anon: the usual way, $AGM(a,b)=AGM(\sqrt{ab},(a+b)/2)$. – Jack D'Aurizio Jan 23 '17 at 0:10
• As a matter of fact the identity follows directly from the expression for $K(k_3)$ since $\frac{1+\sqrt{3}}{\sqrt{8}}=k_3'$ – Anon Jan 26 '17 at 2:16
We could go for crude. Split into the integral from $1$ to $2$, and the integral from $2$ to $\infty$.
On the interval $2$ to $\infty$ we have $x^3-1\gt x^3/4$, and therefore the easily computed $\int_2^\infty \frac{2}{x^{3/2}}\,dx$ provides an upper bound for that region.
On the interval $1$ to $2$, we have $x^3-1=(x-1)(x^2+x+1)\ge 3(x-1)$. And the integral $\int_1^2 \frac{1}{\sqrt{3(x-1)}}\,dx$ can be computed explicitly.
$\int_1^\infty \frac{dx}{\sqrt{x^3-1}}=\int_1^2 + \int_2^\infty=I_1+I_2.$
$I_2 = \int_2^\infty \frac{dx}{x^{3/2}\sqrt{1-1/x^3}}\color{red}{\leq}\lim_{A \to \infty} \int_2^A \frac{dx}{x^{3/2}\sqrt{1-1/A^3}}=\lim_{A \to \infty}\sqrt{\frac{A^3}{A^3-1}}\int_2^A x^{-3/2}dx$
$I_1=\int_1^2\frac{dx}{(x-1)^{1/2}\sqrt{x^2+x+1}}\le \frac{1}{\sqrt{3}}\int_1^2(x-1)^{-1/2}d(x-1)$ | 2020-09-22T11:10:43 | {
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http://mathhelpforum.com/number-theory/147383-converse-statement-true.html | Thread: Is the converse statement true?
1. Is the converse statement true?
Let $\displaystyle m,n$ be natural numbers.
Does $\displaystyle \tau(mn)=\tau(m)\tau(n)$
, where $\displaystyle \tau(k)$ is number of positive divisors of k, implies that
$\displaystyle \gcd(m,n)=1$?
Thanks.
2. Originally Posted by melese
Let $\displaystyle m,n$ be natural numbers.
Does $\displaystyle \tau(mn)=mn$
, where $\displaystyle \tau(k)$ is number of positive divisors of k, implies that
$\displaystyle \gcd(m,n)=1$?
Thanks.
Since your title says converse, I assume you meant to say $\displaystyle \tau(mn)=\tau(m)\tau(n)$?
3. You probably mean $\displaystyle \tau(mn)=\tau(m)\tau(n)$?
4. Originally Posted by chiph588@
Since your title says converse, I assume you meant to say $\displaystyle \tau(mn)=\tau(m)\tau(n)$?
Too fast
5. Originally Posted by Bruno J.
Too fast
Too slow
Originally Posted by melese
Let $\displaystyle m,n$ be natural numbers.
Does $\displaystyle \tau(mn)=mn$
, where $\displaystyle \tau(k)$ is number of positive divisors of k, implies that
$\displaystyle \gcd(m,n)=1$?
Thanks.
Correct me if I'm wrong anyone but I think it might be true...
Here's some reasoning:
Suppose $\displaystyle mn=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ and $\displaystyle m=p_1^{a_1}p_2^{a_2}\cdots p_i^{b_i}\cdots p_j^{b_j}$ and $\displaystyle n=p_i^{a_i-b_i}\cdots p_j^{a_j-b_j}p_{j+1}^{a_{j+1}}\cdots p_k^{a_k}$.
$\displaystyle \tau(mn)=\prod_{s=1}^k (a_s+1) = \tau(m)\tau(n)$
Omitting the simplification we get $\displaystyle \prod_{s=i}^j (a_s+1)=\prod_{s=i}^j (b_s+1)(a_s-b_s+1) = \prod_{s=i}^j (b_s(a_s-b_s)+a_s+1)$
But $\displaystyle b_r>0\implies a_r+1<b_r(a_r-b_r)+a_s+1$, so we're forced to have $\displaystyle b_s=0 \; \forall \; s$ if we want this equality to hold.
6. If you prove it, then it's true!
Here's a less computational argument. Let $\displaystyle D_n$ denote the set of divisors of $\displaystyle n$, and suppose that $\displaystyle g=(m,n)>1$. I exhibit a surjection $\displaystyle f : \ D_m \times D_n \to D_{mn}$ which is not an injection. For every $\displaystyle (u,v) \in D_m \times D_n$, let $\displaystyle f(u,v)=uv$. This map is clearly a surjection. But $\displaystyle f(m/g,n)=f(m,n/g)$. Therefore $\displaystyle |D_m\times D_n|>|D_{mn}|$.
7. yes I do, I wrote that early in the morning...
8. I have an argument but I'm not sure...
Allowing the exponents to be zero, if necessary, we can write the following (similar) prime factorizations:
$\displaystyle m=p_1^{m_1}p_2^{m_2}\dots p_r^{m_r}$ and $\displaystyle n=p_1^{n_1}p_2^{n_2}\dots p_r^{n_r}$, where for $\displaystyle 1\leq i\leq r$ $\displaystyle m_i$ and $\displaystyle n_i$ are nonnegative. Also, for any prime $\displaystyle p_i$, $\displaystyle p_i$ divides at leat one of $\displaystyle m$ and $\displaystyle n$.
Now, $\displaystyle mn=p_1^{m_1+n_1}p_2^{m_2+n_2}\dots p_r^{m_r+n_r}$ and then $\displaystyle \tau(mn)=(m_1+n_1+1)(m_2+n_2+1)\dots (m_r+n_r+1)$.
Computing $\displaystyle \tau(m)\tau(n)$ gives $\displaystyle \tau(m)\tau(n)=(m_1+1)(n_1+1)\dots... (m_r+1)(n_r+1)$ $\displaystyle =(m_1n_1+m_1+n_1+1)\dots (m_rn_r+m_r+n_r+1)$.
Now comes my difficulty...
For each $\displaystyle i=1,2,\dots ,r$, $\displaystyle m_in_i+m_i+n_i+1\leq m_i+n_i+1$, so if we want $\displaystyle \tau(mn)=\tau(m)\tau(n)$ we must have
$\displaystyle m_in_i+m_i+n_i+1= m_i+n_i+1$ for $\displaystyle i=1,2,\dots ,r$. Otherwise $\displaystyle \tau(mn)<\tau(m)\tau(n)$.
But then $\displaystyle m_in_i=0$. Without loss of generality $\displaystyle m_i=0$ and this means that $\displaystyle p_i$ divides n but not m. In this manner $\displaystyle m$ and $\displaystyle n$ are relatively prime.
By the way exactly one of $\displaystyle m_i$ and $\displaystyle n_i$ equals $\displaystyle 0$ due to: "for any prime $\displaystyle p_i$, $\displaystyle p_i$divides at leat one of $\displaystyle m$ and $\displaystyle n$."
Is my argument right? Thanks for your help.
9. Originally Posted by melese
Allowing the exponents to be zero, if necessary, we can write the following (similar) prime factorizations:
$\displaystyle m=p_1^{m_1}p_2^{m_2}\dots p_r^{m_r}$ and $\displaystyle n=p_1^{n_1}p_2^{n_2}\dots p_r^{n_r}$, where for $\displaystyle 1\leq i\leq r$ $\displaystyle m_i$ and $\displaystyle n_i$ are nonnegative. Also, for any prime $\displaystyle p_i$, $\displaystyle p_i$ divides at leat one of $\displaystyle m$ and $\displaystyle n$.
Now, $\displaystyle mn=p_1^{m_1+n_1}p_2^{m_2+n_2}\dots p_r^{m_r+n_r}$ and then $\displaystyle \tau(mn)=(m_1+n_1+1)(m_2+n_2+1)\dots (m_r+n_r+1)$.
Computing $\displaystyle \tau(m)\tau(n)$ gives $\displaystyle \tau(m)\tau(n)=(m_1+1)(n_1+1)\dots... (m_r+1)(n_r+1)$ $\displaystyle =(m_1n_1+m_1+n_1+1)\dots (m_rn_r+m_r+n_r+1)$.
Now comes my difficulty...
For each $\displaystyle i=1,2,\dots ,r$, $\displaystyle m_in_i+m_i+n_i+1\leq m_i+n_i+1$, so if we want $\displaystyle \tau(mn)=\tau(m)\tau(n)$ we must have
$\displaystyle m_in_i+m_i+n_i+1= m_i+n_i+1$ for $\displaystyle i=1,2,\dots ,r$. Otherwise $\displaystyle \tau(mn)<\tau(m)\tau(n)$.
But then $\displaystyle m_in_i=0$. Without loss of generality $\displaystyle m_i=0$ and this means that $\displaystyle p_i$ divides n but not m. In this manner $\displaystyle m$ and $\displaystyle n$ are relatively prime.
By the way exactly one of $\displaystyle m_i$ and $\displaystyle n_i$ equals $\displaystyle 0$ due to: "for any prime $\displaystyle p_i$, $\displaystyle p_i$divides at leat one of $\displaystyle m$ and $\displaystyle n$."
Is my argument right? Thanks for your help.
You mean $\displaystyle m_in_i+m_i+n_i+1\geq m_i+n_i+1$.
Other than that, it looks fine to me.
10. Again, you corrected me. Thank you so much for your help! | 2018-06-20T10:26:09 | {
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https://stats.stackexchange.com/questions/501232/probability-of-the-rth-number-being-smaller-than-all-numbers-before-it-in-a-un | # Probability of the $r$th number being smaller than all numbers before it in a uniform permutation of $n$ numbers
Suppose we have an ordered list of $$n$$ numbers 1 to n, in a random permutation drawn uniformly from all possible permutations.
Let $$r$$ be one of the $$n$$ positions in the list. What is the probability all the numbers that come before it $$(i.e. 1\ldots r-1$$ inclusive) are all less than the number in position $$r$$?
I'm told this is $$\frac{1}{r}$$ but I'm not sure why. I tried computing the probability by a counting argument. I count a total of $$n!$$ permutations. I count that the number of ways that all the numbers that come before it are:
If the value at $$r$$ equals $$r$$, then there are $$\frac{(r-1)!}{(r-1)!}$$ possibilities for entries before $$r$$, and $$(n-r)!$$ possibilities for entries after $$r$$, so $$\frac{(r-1)!}{(r-1)!}(n-r)!$$ ways.
If the value at $$r$$ equals $$r+1$$, then there are $$\frac{(r)!}{(r-1)!}$$ possibilities for entries before $$r$$, and $$(n-r)!$$ possibilities for entries after $$r$$, so $$\frac{r!}{(r-1)!}(n-r)!$$ ways.
We continue on in this fashion until $$r = n$$.
So the probability is $$\frac{\frac{(r-1)!}{(r-1)!}(n-r)! + \ldots + \frac{(n-1)!}{(r-1)!}(n-r)!}{n!}$$, but this does not seem to simplify to $$\frac{1}{r}$$.
Where is the mistake?
• Suppose you have $r$ unique numbers. Then the maximum of those numbers is unique, i.e. there is $1$ maximum and $r$ options. Therefore picking a number uniformly at random, you have a $1/r$ chance of picking the maximum. The probability of any of the numbers ending up in any given spot (including the $r$-th spot) is the same, i.e. uniform, so picking the last ($r$-th) slot is the same as picking a number uniformly at. Dec 17, 2020 at 7:54
• Hi jh1001, this is a well written question, welcome to Stats SE! Dec 17, 2020 at 10:17
Let $$\mathbf{X} = (X_1,...,X_n)$$ denote the random vector you are talking about. Since the elements of this vector are distinct numbers, the event you are describing is equivalent to the event $$\max \{ X_1,...,X_r \} = X_r$$. Since the random vector is uniform over all permutations, the elements are exchangeable, so we have:
$$\mathbb{P}(\max \{ X_1,...,X_r \} = X_r) = \frac{1}{r}.$$
You can compute this the long way using the permutation argument you are describing if you want to, but your combinatorial algebra for this is wrong. (For starters you need to choose a different variable to describe the value of the $$r$$th number, so as not to conflate it with the position $$r$$.) Given the value $$X_r = x$$ there are $$(n-1)_{r-1}$$ ways that the previous numbers can be arranged and $$(x-1)_{r-1}$$ of these satisfy the event requirement (using notation for the falling factorials). Thus, for any $$1 \leqslant x \leqslant n$$ we have:
$$\mathbb{P}(\max \{ X_1,...,X_r \} = x | X_r = x) = \frac{(x-1)_{r-1}}{(n-1)_{r-1}}.$$
Hence, using the law of total probability we get:
\begin{align} \mathbb{P}(\max \{ X_1,...,X_r \} = X_r) &= \sum_{x=1}^n \mathbb{P}(\max \{ X_1,...,X_r \} = X_r | X_r = x) \cdot \mathbb{P}(X_r = x) \\[6pt] &= \sum_{x=r}^n \mathbb{P}(\max \{ X_1,...,X_r \} = x | X_r = x) \cdot \mathbb{P}(X_r = x) \\[6pt] &= \sum_{x=r}^n \frac{(x-1)_{r-1}}{(n-1)_{r-1}} \cdot \frac{1}{n} \\[6pt] &= \frac{1}{(n)_{r}} \sum_{x=r}^n (x-1)_{r-1} \\[6pt] &= \frac{1}{(n)_{r}} \sum_{x=0}^{n-r} (x+r-1)_{r-1} \\[6pt] &= \frac{1}{(n)_{r}} \Bigg[ \frac{(n)_{r}}{r} - \frac{(r-1)_{r}}{r} \Bigg] \\[6pt] &= \frac{1}{r}. \\[6pt] \end{align}
(In the penultimate step I have used a summation formula for the falling factorials given here.)
• Thank you for this answer. I'm still confused as to why my answer was wrong. Is it not true for example, that if the value at $r$ equals $r+1$, then there are a total of $\frac{r!}{(r-1)!}(n-r)!$ ways for this to happen while satisfying the condition that the value at $r$ is smaller than all values preceding it? Dec 18, 2020 at 10:52
• If $n = 10, r=3$, and suppose the numbers were from 1 to 10, and the value at slot 3 was 4. Then you have to fill the first two slots, and you have to pick 2 values from 1, 2, or 3, which is, using the permutation function, 3 permute 2, which is $\frac{r!}{(r-1)!}$. Then you have free choice over the last 7 slots, and you have 7 numbers to choose from, namely 4, 5, ..., 10 so it would be $7!$. This is the formula I wrote above. If I do this for all values of $r$, wouldn't it be the total ways that the desired condition can be satisfied? Dec 18, 2020 at 10:57
For $$1 \le i \le r$$, let $$p_i$$ be the probability that of the first $$r$$ numbers in the permutation, the largest is in position $$i$$. By symmetry the $$p_i$$ are equal so each is equal to $$\frac{1}{r}$$.
An example of that symmetry: let $$A$$ be the set of permutations where the first number is the largest among the first 5 numbers, and $$B$$ be the set of permutations where the second number is the largest among the first 5 numbers. Then swapping the first two numbers of a permutation defines a bijection between $$A$$ and $$B$$, showing that the two sets are of equal size. | 2022-08-12T15:15:52 | {
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http://joaosilveira.com/hx0ukrux/e9dc18-every-asymmetric-relation-is-antisymmetric | Exercise 21 Give examples of relations which are neither re±exive, nor irre±exive. Other than antisymmetric, there are different relations like reflexive, irreflexive, symmetric, asymmetric, and transitive. How many number of possible relations in a antisymmetric set? There is an element which triplicates in every hour. Asymmetric Relation: A relation R on a set A is called an Asymmetric Relation if for every (a, b) ∈ R implies that (b, a) does not belong to R. 6. A relation becomes an antisymmetric relation for a binary relation R on a set A. These Multiple Choice Questions (MCQ) should be practiced to improve the Discrete Mathematics skills required for various interviews (campus interviews, walk-in interviews, company interviews), placements, entrance exams and other competitive examinations. It can be reflexive, but it can't be symmetric for two distinct elements. A relation R on a set A is non-reflexive if R is neither reflexive nor irreflexive, i.e. Also, i'm curious to know since relations can both be neither symmetric and anti-symmetric, would R = {(1,2),(2,1),(2,3)} be an example of such a relation? A relation is asymmetric if and only if it is both antisymmetric and irreflexive. if aRb ⇒ bRa. Antisymmetry is different from asymmetry because it does not requier irreflexivity, therefore every asymmetric relation is antisymmetric, but the reverse is false. A relation is considered as an asymmetric if it is both antisymmetric and irreflexive or else it is not. Solution: The relation R is not antisymmetric as 4 ≠ 5 but (4, 5) and (5, 4) both belong to R. 5. An antisymmetric and not asymmetric relation between x and y (asymmetric because reflexive) Counter-example: An symmetric relation between x and y (and reflexive ) In God we trust , … if a single compound is kept in a container at noon and the container is full by midnight. Non-examples ¨ The relation divides on the set of integers is neither symmetric nor antisymmetric.. A relation R is asymmetric if and only if R is irreflexive and antisymmetric. Lipschutz, Seymour; Marc Lars Lipson (1997). Let be a relation on the set . Any asymmetric relation is necessarily antisymmetric; but the converse does not hold. A relation R on a set A is symmetric if whenever (a, b) ∈ R then (b, a) ∈ R, i.e. Think $\le$. This lesson will talk about a certain type of relation called an antisymmetric relation. 15. each of these 3 items in turn reproduce exactly 3 other items. Note: a relation R on the set A is irreflexive if for every a element of A. We find that $$R$$ is. Combine this with the previous result to conclude that every acyclic relation is irre±exive. Antisymmetric Relation. For example, > is an asymmetric relation, but ≥ is not. Difference between antisymmetric and not symmetric. Suppose that your math teacher surprises the class by saying she brought in cookies. Here we are going to learn some of those properties binary relations may have. a.4pm b.6pm c.9pm d.11pm . Antisymmetry is different from asymmetry: a relation is asymmetric if, and only if, it is antisymmetric and irreflexive. By definition, a nonempty relation cannot be both symmetric and asymmetric (where if a is related to b, then b cannot be related to a (in the same way)). Restrictions and converses of asymmetric relations are also asymmetric. Best answer. We call symmetric if means the same thing as . For each of these relations on the set $\{1,2,3,4\},$ decide whether it is reflexive, whether it is symmetric, and whether it is antisymmetric, and whether it is transitive. That is, for . Discrete Mathematics Questions and Answers – Relations. Hint: write the definition of what it means to be asymmetric… Exercise 20 Prove that every acyclic relation is asymmetric. Transitive if for every unidirectional path joining three vertices $$a,b,c$$, in that order, there is also a directed line joining $$a$$ to $$c$$. Antisymmetric definition, noting a relation in which one element's dependence on a second implies that the second element is not dependent on the first, as the relation “greater than.” See more. if aRa is true for some a and false for others. sets; set-theory&algebra; relations ; asked Oct 9, 2015 in Set Theory & Algebra admin retagged Dec 20, 2015 by Arjun 3.8k views. A relation on a set is antisymmetric provided that distinct elements are never both related to one another. Relationship to asymmetric and antisymmetric relations. A relation can be both symmetric and antisymmetric (in this case, it must be coreflexive), and there are relations which are neither symmetric nor antisymmetric (e.g., the "preys on" relation on biological species). The incidence matrix $$M=(m_{ij})$$ for a relation on $$A$$ is a square matrix. A relation is asymmetric if and only if it is both antisymmetric and irreflexive. 1 vote . Again, the previous 3 alternatives are far from being exhaustive; as an example over the natural numbers, the relation xRy defined by x > 2 is neither symmetric nor antisymmetric, let alone asymmetric. Antisymmetric if every pair of vertices is connected by none or exactly one directed line. This section focuses on "Relations" in Discrete Mathematics. We call antisymmetric … Since dominance relation is also irreflexive, so in order to be asymmetric, it should be antisymmetric too. an eigenfunction of P ij looks like. The relations we are interested in here are binary relations on a set. Exercise 19 Prove that every asymmetric relation is irre±exive. Antisymmetric means that the only way for both $aRb$ and $bRa$ to hold is if $a = b$. (a,a) not equal to element of R. That is. Antisymmetry is concerned only with the relations between distinct (i.e. In that, there is no pair of distinct elements of A, each of which gets related by R to the other. Weisstein, Eric W., "Antisymmetric Relation", MathWorld. I just want to know how the value in the answers come like 2^n2 and 2^n^2-1 etc. Exercise 22 Give examples of relations which are neither symmetric, nor asymmetric. Whether the wave function is symmetric or antisymmetric under such operations gives you insight into whether two particles can occupy the same quantum state. Yes, and that's essentially the only case : If R is both symmetric and antisymmetric then R must be the relation ## \{(x,x),x \in B\} ## for some subset ## B\subset A ##. Homework 5 Solutions New York University. antisymmetric relation transitive relation Contents Certain important types of binary relation can be characterized by properties they have. Is the relation R antisymmetric? Multi-objective optimization using evolutionary algorithms. Yes. We call reflexive if every element of is related to itself; that is, if every has . We call asymmetric if guarantees that . It's also known as … However, a relation can be neither symmetric nor asymmetric, which is the case for "is less than or equal to" and "preys on"). Quiz & Worksheet - What is an Antisymmetric Relation? The mathematical concepts of symmetry and antisymmetry are independent, (though the concepts of symmetry and asymmetry are not). But in "Deb, K. (2013). Please make it clear. For example- the inverse of less than is also an asymmetric relation. (a) (b) Show that every asymmetric relation is antisymmetric. So an asymmetric relation is necessarily irreflexive. A relation that is not asymmetric, is symmetric. Specifically, the definition of antisymmetry permits a relation element of the form $(a, a)$, whereas asymmetry forbids that. Given that P ij 2 = 1, note that if a wave function is an eigenfunction of P ij, then the possible eigenvalues are 1 and –1. 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http://puntoopera.it/onxt/latex-summation.html | # Latex Summation
Open Live Script. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. Contents[show] Table of sum-class symbols Using sum TeX is smart enough to only show. Sine and Cosine: Expansions. You can insert mathematical equations into your documents. This is a skill that is not always found in math textbooks, but it is a critical skill for future math courses like Calculus and Differential Equations. lower / first quartile. LaTeX is a powerful tool to typeset math. For example, = 3 + 6 + 11 + 18 = 38. To show or hide the equation options, click View Show equation toolbar. Patterns in Pascal's Triangle. In my LyX (1. If you're seeing this message, it means we're having trouble loading external resources on our website. To get exp to appear as a superscript, you type ^{exp}. If you want the limits above and below, place the \limits command after the sum command as follows: $\sum\limits_{k=1}^n k$. If you are wondering what happened to an index, you may want to revisit this discussion. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. Sigma Calculator Partial Sums infinite-series Algebra Index. Summation notation involves: The summation sign This appears as the symbol, S, which is the Greek upper case letter, S. Knowing a few of the mathematics commands is not only helpful if you want to write a book or an article (or do some extreme stuff ), but can come in handy in a. Table 200: ℳ Letter-like Symbols. You supply all three elements of the display in a single statement: the lower limit first, the upper limit second, and the expression third. Character Usage in LaTeX Literal n Signi es that what is coming next is a command $\backslash$ $Enter/exit math mode \$ # Macro arguments \# & Delimits cells in tabulars \& % Comments out the rest of the line after it \% Displaying subscripts \_ ^ Displaying superscripts \verb!^! f Delimits various environment and macro bodies \. To add another equation box, click New equation. #N#Sigma is the upper case letter S in Greek. TeX is a markup language that is used to bring about consistency and neatness in documents. sum var1, d local statistic = r(max) * Then use this code to set the formatting. Watson and Crick (1953) \shortcite{key} Abbreviated author list and year. I google "latex symbols" when I need something I can't recall. I'm going to use the online LaTeX equation editor to place an equation in Google Presentations, just like you wanted to. $\sum_ {n=1}^\infty. They'll be productive from day one and be able to pick up small amounts of LaTeX as they go. P a b x y (x y, ). Return To Contents Go To Problems & Solutions. CHAPTER 4 FOURIER SERIES AND INTEGRALS 4. Many script-languages use backslash "\" to denote special commands. If your floors are uneven, a latex self-leveling compound (sometimes called latex screed or mortar) can help smooth over areas that are not level. A factorial is represented by the sign (!). This is an R Markdown document. n ∑ i=i0cai = cn ∑ i=i0ai where c is any number. com with the piece of LaTeX code. They'll be productive from day one and be able to pick up small amounts of LaTeX as they go. If you’re not sure what LaTeX / TeX is, start with Introduction to LaTeX. This list is organized by symbol type and is intended to facilitate finding an unfamiliar symbol by its visual appearance. To sum up, if you are a Windows user, TeXnicCenter is one of the best LaTeX editors that you can use in 2020 and you don't have to look any further. Proof of Unbiasness of Sample Variance Estimator (As I received some remarks about the unnecessary length of this proof, I provide shorter version here) In different application of statistics or econometrics but also in many other examples it is necessary to estimate the variance of a sample. Similar is for limit expressions. 1 Beginning a document ndocumentclassfarticleg nusepackagefgraphicx, amssymbg nbeginfdocumentg ntextwidth 6. pdf format w:LaTeX - wikipedia article on LaTeX. RSeek meta search engine - The RSeek meta search engine, provides a unified interface for searching the various sources of online R information. 1/10), typical GNU/Linux distros and other X11-based systems, as well as Mac OS X. See some more involved examples of how we read expressions in summation notation. Pascal's Triangle conceals a huge number of various patterns, many discovered by Pascal himself and even known before his time. A more general solution to force a formula that is appearing in inline style to appear in display style is to start the formula with a \displaystyle declaration, e. If you do not specify k, symsum uses the variable determined by symvar as the summation index. 75% of population are below this value. Anyone knows if it's possible to do what I want to? improve this question. 12 silver badges. Summation notation is often known as sigma notation because it uses the Greek capital letter sigma, $\sum$, to represent the sum. Open an example in ShareLaTeX. Integral With Times Sign. Integral expression can be added using the \int_{lower}^{upper} command. This is the Home Page for the Beginning LaTeX tutorial. Given the syntax of the sum() function, which can be read by executing "sum?" (with the question mark) in the Sage notebook, you don't need to define a list to compute the sum. LaTeX is a high-quality typesetting system; it includes features designed for the production of technical and scientific documentation. Windows character codes (Hold down the Alt key and type the specified number on the numeric keypad.$\begingroupOn the other hand, \Sigma aligns with the baseline and the top height of ordinary text (i. If your floors are uneven, a latex self-leveling compound (sometimes called latex screed or mortar) can help smooth over areas that are not level. and just know that when we see the same index on top and on the bottom, we mean to take a sum. Introduction. It may also be any other non-negative integer, like 0 or 3. Similar is for limit expressions. average / arithmetic mean. Unsurprisingly, when Cesáro summation is applied to Grandi’s series, one again recovers the value of. Using these newfound Summation Properties, we evaluate different Series and find the sum. LyX is a graphical interface, nearly WYSIWYG, to the LaTeX word processing package. 25% of population are below this value. If you want the limits above and below, place the \limits command after the sum command as follows:\sum\limits_{k=1}^n k$. Den Anfangsbedingung der Summation (also die tiefgestellte Komponente) wird wie gewöhnlich mit _ eingeleitet (also z. You can start here to link to any other section in the tutorial as you need to, or you can begin here and work through the tutorial step by step. ASCIIMath input, LaTeX and KaTeX output. Then ϕ possesses a third critical. Document history. If f is a constant, then the default variable is x. X (y i 1) 2 Given y= f a;3a;ag, show the left and right sides are equal by expanding the summation notation and simplifying it. \begin {document} View which changes have been added and removed. Character Usage in LaTeX Literal n Signi es that what is coming next is a command$\backslash Enter/exit math mode \$# Macro arguments \# & Delimits cells in tabulars \& % Comments out the rest of the line after it \% Displaying subscripts \_ ^ Displaying superscripts \verb!^! f Delimits various environment and macro bodies \. upper / third quartile. F = symsum(f,k,a,b) returns the sum of the series f with respect to the summation index k from the lower bound a to the upper bound b. Summation notation is used to represent series. The \cleardoublepage command ends the current page and causes all figures and tables that have so far appeared in the input to be printed. Quite often, sigma notation is used in a slightly different format to denote certain sums. Watson and Crick (1953) \shortcite{key} Abbreviated author list and year. A Sample Proof Using Mathematical Induction (playing with LaTeX) It’s been a long time since I used LaTeX regularly, and I discovered that I don’t have any leftover files from my days as a math student in Waterloo. Here refers to the index of summation, is the lower bound, and is the upper bound. Symbolic Math in Matlab. ; Enter the table data into the table: copy (Ctrl+C) table data from a spreadsheet (e. You can also have both (note the summation sign) and, if they involve more than one symbol, you must enclose them in braces (note the lower bound on the integral). any change in the output. You must use the following package: \usepackage {amsmath} \begin {matrix} \begin {pmatrix} \begin {bmatrix} \begin {vmatrix} \begin {Vmatrix}. Dabei ist die Reihenfolge egal. Only Professional WritersAcademic writing is a tough chore, and that is why you need expert writers who can provide you with help. To sum up, if you are a Windows user, TeXnicCenter is one of the best LaTeX editors that you can use in 2020 and you don't have to look any further. Knowing a few of the mathematics commands is not only helpful if you want to write a book or an article (or do some extreme stuff ), but can come in handy in a. It may also be any other non-negative integer, like 0 or 3. Integer and sum limits improvement. (n times) = cn, where c is a constant. For example, refers to a cyclic sum, and refers to all subsets which are in. High-and low-position is indicated via the characters ^ and _ and is not explicitly specified. Online Latex Equation Editor. The remainder term has an asymptotic expansion, and for a typical analytic function, it is a divergent (Gevrey-1) series. Befindet man sich im Fließtext im Mathematikmodus, so werden die Bedingungen der Summation nachgestellt:. This is but a simple example of a general technique of exploiting organization and classification on the web to discover information about similar items. Key Features: Integrated LaTeX environment for Windows apps, Auto-completion, Complete UTF-8 support, Document navigator, Library of LaTeX snippets, Syntax Highlighting, Spell Checking and more. For example, the following example illustrates that \\sum is one of these elite symbols whereas \\Sigma is not. Note that if , then the sum is. Closed for the following reason the question is answered, right answer was accepted by Alex Kemp close date 2016-03-03 22:23:46. Also note that. for the other format, the 'from' is a subscript to the Sigma and the 'to' a superscript. $$\frac {1} {2}$$ ): Use \tfrac to force small fractions in a display. aside from the conventional${_n}C_r$, how can I input "n taken r" using the more common parenthesis notation. Here is the same command in display style (e. Without them, usually the next letter or digit will be used, but that isn't usually what you want. LaTeX The LaTeX command that creates the icon. To print it use \lt in equations (and < outside equations). I have thought of \left( \stackrel{n}{r} \right), but the effect is obviously ugly. In the following list of symbols, each line contains: (1) A symbol; (2) the abbreviation of the symbol that I use in the autotext function of MS Word;. We didn't advertise this highly requested feature since it needed more work. This command forces LaTeX to give an equation the full height it needs to display as if it were on its own line. You can also apply the Sum Absolute Value formula of Kutools for Excel to solve the problem easily. It will simply draw a horizontal line between the paragraphs that proceed and follow it when the document is rendered. In latex, we can do that by. Characters from the ASCII character set can be used directly, with a few exceptions (pound sign #, backslash \, braces {}, and percent sign %). derivative iint int integral Latex lim oint prod sum All the versions of this article: < français > How to write LateX Derivatives, Limits, Sums, Products and Integrals ?. Using i for both the index of summation and then for variables outside the summation is ambiguous notation at that. Here’s LaTeX, in 6 minutes or less. Solve systems of equations with linear algebra operations on vectors and matrices. It will on whenever a character is typed and can be close simply by typing a space. HOME: Next: Arrow symbols (amssymb) Last: Relation symbols (amssymb) Top: Index Page Index Page. Symbolic Math in Matlab. A typical element of the sequence which is being summed appears to the right of the summation sign. Summation notation is often known as sigma notation because it uses the Greek capital letter sigma, $\sum$, to represent the sum. This comes from Mathematica (IIRC) and is not otherwise standardized, but there may be some disambiguation benefit in. If this series can converge conditionally; for example, converges conditionally if , and absolutely for. Integral expression can be added using the \int_{lower}^{upper} command. Mathematics Inline and Display. R Markdown allows you to mix text, R code, R output, R graphics, and mathematics in a single document. Note that in GR, indices usually range from 0 to. For example, = 3 + 6 + 11 + 18 = 38. 5 Summation with R Summation: sum(x) Where x is a vector. Sometimes it is also called negative exponential distribution. The math environment is for formulas that appear right in the text. Summation notation includes an explicit formula and specifies the first and last terms in the series. Doing things which you aren’t allowed won. LaTeX Snippets. The second syntax is Python's list creation and summation of lists: sage: [ 2^j for j in range(0,3) ] [1, 2, 4] sage: sum(_) 7 Note that the variable inside the list comprehension (which I called j here) is automatically created and will overwrite other variables. (Since matrices are large, they are almost always set as displays. com is a free web service aimed to help people to include mathematical formulas&graphics into the web pages easily and without sacrificing flexibility and high quality of LaTeX system. The difference between this and \Sigma, which generates the capital letter$ \Sigma $, is that \sum appears larger, and that it supports the limits to be displayed below and above the symbol. Online Latex Equation Editor. Input LaTeX, Tex, AMSmath or ASCIIMath notation (Click icon to switch to ASCIIMath mode) to make formula. 0 (June, 1993) Encodings; HTML Entity (decimal. This is the Home Page for the Beginning LaTeX tutorial. Ellipses (three dots) can be produced by the following commands \ldots - horizontally at bottom of line \cdots - horizontally center of line (math mode only) \ddots - diagonal (math mode only) \vdots - vertical (math mode only). The following example illustrates the difference between \sum and \Sigma. This is called Einstein summation notation. Easy-to-use symbol, keyword, package, style, and formatting reference for LaTeX scientific publishing markup language. All the versions of this article: < français > Here are few examples to write quickly matrices. LaTeX is the typesetting system and a markup language that allows for the creation of documents. The \cleardoublepage command ends the current page and causes all figures and tables that have so far appeared in the input to be printed. In latex, we can do that by. Learn more about natural and synthetic latex with this article. This type of LaTeX formulae in Freeplane is deprecated in Freeplane 1. Of course it doesn't work, LaTeX is pissed because there is a double subscript. Another thing to notice is the effect of the \displaystyle command. Summation is a common symbol in math and really useful to know how to display in LaTeX. LaTeX Math Symbols Enjoy this cheat sheet at its fullest within Dash, the macOS documentation browser. The Sigma symbol, , is a capital letter in the Greek alphabet. Learn how your comment data is processed. A sum is the result of an addition. I have problem getting the Equation Editor in Word 2007 to show the limits of the summation symbol properly. All the versions of this article: < français > Here are few examples to write quickly matrices. That summation symbol is a little ugly. The Octave interpreter can be run in GUI mode, as a console, or invoked as part of a shell script. Sigma (Summation) Notation. Anyone knows if it's possible to do what I want to? improve this question. The indicator function of an event is a random variable that takes value 1 when the event happens and value 0 when the event does not happen. When the sum of an infinite geometric series exists, we can calculate the sum. We will need the following well-known summation rules. Every math student will eventually need to know how to write in LaTeX, since all math journals are written using LaTeX. To get exp to appear as a superscript, you type ^{exp}. \end {document}. The exponential distribution is a continuous probability distribution used to model the time we need to wait before a given event occurs. The summation notation above, therefore, represents the sum 9 + 16 + 25 + 36 + 49. November 11, 2009 October 15, 2017 James 21 Comments. Thus our source is a mixture of text to be typeset and a couple of LATEX commands \emph and \LaTeX. Hypertext Help with LaTeX Ellipses. to write the index n on the right side of the sum symbol, while the limits of the summation remain above and below. The equation of motion for a system of n-particles can be written as , where indicates. Integral With Underbar. But it is often used to find the area underneath the graph of a function like this: The integral of many functions are well known, and there are useful rules to work out the integral of more complicated functions, many of which are shown here. 1 Letter modifiers. Visualize data with high-level plot commands in 2D and 3D. A sequence is a function whose domain is the natural numbers. The displaymath environment is for formulas that appear on their own line. LaTeX Font Info: Checking defaults for OT1/cmr/m/n on input line 22. This also means that, of course, if "\hline" is used in mid-paragraph, then it will force a break in. n ∑ i=i0cai = cn ∑ i=i0ai where c is any number. Exporting Results from Stata to LaTeX. In mathematical formula it is indeed the addition of many number or variable which represent to give concise expression for sum of the variable as Sigma or Summation. Open the Convert dropdown menu and click Current - Professional. Symmetric difference is a binary operator over sets, just like union and intersection. Overleaf is so easy to get started with that you'll be able to invite your non-LaTeX colleagues to contribute directly to your LaTeX documents. \end {document}. Many summation expressions involve just a single summation operator. Presentations should be produced with the same discipline and care as all other marketing and communication materials. Unsurprisingly, when Cesáro summation is applied to Grandi’s series, one again recovers the value of. Text with two or double columns can be created by passing the parameter \twocolumn to the document class statement. \alpha \theta o o ˝ \tau \beta # \vartheta ˇ \pi ˛ \upsilon \gamma \iota$ \varpi ˚ \phi \delta \kappa ˆ \rho ’ \varphi \epsilon \lambda % \varrho ˜ \chi. How to Typeset Formulas in LaTeX. Only Professional WritersAcademic writing is a tough chore, and that is why you need expert writers who can provide you with help. Share a link to this widget: More. Sum in Math equation editor The 'sum from to' option appears to have two formats. LaTeX provides a feature of special editing tool for scientific tool for math equations in LaTeX. (in LaTeX) we type $\lim_{x \to +\infty} \frac{3x^2 +7x^3}{x^2 +5x^4} = 3.$ We now discuss how to obtain integrals in mathematical documents. (This is all described on page 62 of the latex book). Characters from the ASCII character set can be used directly, with a few exceptions (pound sign #, backslash \, braces {}, and percent sign %). 3 posts • Page 1 of 1. XeTeX, however, has prepared some macros for this. This expressions shows that sum of values of x,. To use additional special characters, such as integral and summation symbols, you can use LaTeX markup instead. X y2 i 11a= 11a(a 1) 1. LaTeX is distributed through CTAN servers or comes as part of many easily installable and usable TeX distributions provided by the TeX User Group (TUG) or third parties. Setting regions of no data to 0 in the source dataset is probably a sensible idea in most cases where summation is meaningful, so that it influences the average value appropriately. We've documented and categorized hundreds of macros!. November 2013 at 18:28. LaTeX Math Formulas There are three environments that put LaTeX in math mode: math, displaymath, and equation. Detexify lets you draw a symbol on a web page and then lists the TEX symbols that seem to resemble it. The Markdown parser included in the Jupyter Notebook is MathJax-aware. F = symsum(f,k) returns the indefinite sum (antidifference) of the series f with respect to the summation index k. The size of some mathematical symbols, notably summation signs, product signs, and integral signs, depends on the environment in which they appear (i. compute (for plotting purposes) the piecewise linear function defined by the trapezoid rule for numerical integration based on a subdivision into $$N$$ subintervals; the approximation given by the trapezoid rule,. How to Typeset Formulas in LaTeX. See here for a complete list of set symbols. LaTeX is available as free software. How to use sum in a sentence. You can type "\" followed by the name of a symbol and. LaTeX is free software under the terms of the LaTeX Project Public License (LPPL). If the text argument to one of the text-drawing functions (text, mtext, axis, legend) in R is an expression, the argument is interpreted as a mathematical expression and the output will be formatted according to TeX-like rules. A plain cover letter template with your name and address at the top in bold. The usage is pretty easy, you can basically type the name of the letter and put a backslash in front of it. HTML LaTeX equation editor that creates graphical equations (gif, png, swf, pdf, emf). Sum uses the standard Wolfram Language iteration specification. User feeds QuickLaTeX. Most of the stock math commands are written for typesetting math or computer science papers for academic journals, so you might need to dig deeper into LaTeX commands to get the vector notation styles that are common in physics textbooks and articles. See some more involved examples of how we read expressions in summation notation. Learn the LaTeX commands to display the greek alphabet. It even shows in real time what the equation looks like. The most common is as a binary operator. Embed this widget ». by Marco Taboga, PhD. The problem with this method is that it puts a fixed-width bar over the letter (I think it's an underscore character '_') instead of covering the width of the letter(s). ASCIIMathTeXImg. Summation notation uses the sigma Σ symbol to represent sums with multiple terms. This video presents how to write limits, summation and integral equations using LaTeX in a document. Befindet man sich im Fließtext im Mathematikmodus, so werden die Bedingungen der Summation nachgestellt:. Even when I used ink equation, the second sigma seems smaller than first one. In inline math mode the integral/sum/product lower and upper limits are placed right of integral symbol. Numbered equations Use the equation environment to create a numbered equation. I have problem getting the Equation Editor in Word 2007 to show the limits of the summation symbol properly. Mathematical Annotation in R Description. You could write out the sum like this: 5 + 10 + 15 + 20 + 25 + … + 490 + 495 + 500. If you're seeing this message, it means we're having trouble loading external resources on our website. The Sigma symbol, , is a capital letter in the Greek alphabet. The most important advantage of matrices is that the provide. If an answer to your question is already available online, RSeek can help you locate it. Free Summation Calculator. Auto Suggestion will be on even operating on a snippet. Also note that. (The above step is nothing more than changing the order and grouping of the original summation. Contents[show] Table of sum-class symbols Using sum TeX is smart enough to only show. If the text argument to one of the text-drawing functions (text, mtext, axis, legend) in R is an expression, the argument is interpreted as a mathematical expression and the output will be formatted according to TeX-like rules. LaTeX is a programming language that can be used for writing and typesetting documents. You can change the cell type to Markdown by using the Cell menu, the toolbar, or the key shortcut m. To draw those formulas, PlantUML uses two OpenSource projects: AsciiMath that converts AsciiMath notation to LaTeX expression. com is a free web service aimed to help people to include mathematical formulas&graphics into the web pages easily and without sacrificing flexibility and high quality of LaTeX system. November 2013 at 18:28. A how to write sum in latex custom research paper starts with an introduction, continues with the body, and has a strong conclusion. The matrix objects are a subclass of the numpy arrays (ndarray). summation sign: See Also: greek capital letter sigma U+03A3 double-struck n-ary summation U+2140: Version: Unicode 1. The formula for the sum of an infinite series is related to the formula for the sum of the first $n$ terms of a geometric series. This expressions shows that sum of values of x,. Integral With Overbar. The difference between this and \Sigma, which generates the capital letter $\Sigma$, is that \sum appears larger, and that it supports the limits to be displayed below and above the symbol. See here for a complete list of set symbols. A sequence is a function whose domain is the natural numbers. RSeek meta search engine - The RSeek meta search engine, provides a unified interface for searching the various sources of online R information. TeX is a markup language that is used to bring about consistency and neatness in documents. LaTeX is a powerful tool to typeset math. You always increase by one at each successive step. XeTeX, however, has prepared some macros for this. So, we can factor constants out of a summation. Categorized in: Latex, Stata. Improved sum $\sum\limits_{n=1}^{\infty} 2^{-n} = 1$ inside text Moreover, adding \displaystyle beforehand will make the symbol large and easier to read. All the versions of this article: < français > Here are few examples to write quickly matrices. The formulas for the first few values of. These are not guaranteed to work in MathJax but are a good place to start. This guide walks you through the basics of using Jupyter Notebooks locally. Use MathJax to format equations. That'll give you many lists and tips. The align environment will align formulas at the ampersand & symbol. Integration as summation Introduction On this leaflet we explain integration as an infinite sum. P a b x y (x y, ). You can also convert back to LaTeX to edit the equation. Here are a few examples:. to write the index n on the right side of the sum symbol, while the limits of the summation remain above and below. June 2014 by tom 7 Comments. This is important, e. This formula reflects the linearity of the finite sums. LaTeX（LATEX，音译“拉泰赫”）是一种基于ΤΕΧ的排版系统，由美国计算机学家莱斯利·兰伯特（Leslie Lamport）在20世纪80年代初期开发。 利用这种格式，即使使用者没有排版和程序设计的知识也可以充分发挥由TeX所提供的强大功能，能在几天，甚至几小时内生成. 1 Beginning a document ndocumentclassfarticleg nusepackagefgraphicx, amssymbg nbeginfdocumentg ntextwidth 6. You can start here to link to any other section in the tutorial as you need to, or you can begin here and work through the tutorial step by step. 2 posts • Page 1 of 1. formulas, graphs). It is based on very simple idea. CHAPTER 4 FOURIER SERIES AND INTEGRALS 4. Jupyter notebook recognizes LaTeX code written in markdown cells and renders the symbols in the browser using the MathJax JavaScript library. Summation Notation Jacco Thijssen 1 Introduction In mathematics and statistics one often has to take the sum over a number of elements. R-help mailing list - The R-help mailing list is a very active list with questions and answers about problems. Infinite summation (17 formulas) © 1998-2020 Wolfram Research, Inc. In the following list of symbols, each line contains: (1) A symbol; (2) the abbreviation of the symbol that I use in the autotext function of MS Word;. LaTeX The LaTeX command that creates the icon. Big O and related notations in LaTeX. If you're behind a web filter, please make sure that the domains *. Table 238: fge Math-mode Accents. Internet abounds with LaTex tutorials on how to write mathematics equations and simple symbols in LaTeX. derivative iint int integral Latex lim oint prod sum All the versions of this article: < français > How to write LateX Derivatives, Limits, Sums, Products and Integrals ?. Learn more about natural and synthetic latex with this article. The Markdown parser included in the Jupyter Notebook is MathJax-aware. If you have a fraction where: numerator = summation of Xi terms denominator = summation of Xi^2 terms Do standard rules of arithmetic apply? So that it equals the summation of 1/Xi ?. How to use the LaTeX tables generator? Set the desired size of the table using Table / Set size menu option. Having a sum resampling method would actually be an excellent contribution to gdalwarp, since it would consider no data appropriately and automatically. You supply all three elements of the display in a single statement: the lower limit first, the upper limit second, and the expression third. June 2014 by tom 7 Comments. Summation With Integral red. Regression Results. Superscripts, subscripts and groups of characters. We've documented and categorized hundreds of macros!. Sigma Calculator Partial Sums infinite-series Algebra Index. Similar is for limit expressions. Linux Latex Markdown In this post, I am gonna show you how to write Mathematic symbols in markdown. The summation sign is HTML entity ∑ and the super- and sub-scripts can be marked up with the sup and sub elements, so try:. Note that in GR, indices usually range from 0 to. Thanks for the tip. bmatrix Latex matrix pmatrix vmatrix. These notations describe the limiting behavior of a function in mathematics or classify algorithms in computer science according to their complexity / processing time. Document history. The following example illustrates the difference between \sum and \Sigma. The summation sign, S, instructs us to sum the elements of a sequence. The standard deviation ( σ) is simply the (positive) square root of the variance. XeTeX, however, has prepared some macros for this. LaTeX is the typesetting system and a markup language that allows for the creation of documents. Then ϕ possesses a third critical. Include Greek Letters. This example shows how to insert Greek letters, superscripts, and annotations into chart text and explains other available TeX options. Even when I used ink equation, the second sigma seems smaller than first one. In order to move summation range specificators (this works also for \prod, \max and other special operators) on their usual place you can add \displaystyle directive before the needed symbol $\displaystyle\sum_{i = 1}^n$. November 2013 at 18:28. 1 thought on " Expectation Symbol in LaTeX " Shane March 7, 2015 at 19:59. You supply all three elements of the display in a single statement: the lower limit first, the upper limit second, and the expression third. \begin {document} View which changes have been added and removed. You can enter equations, create tables and matrices, import graphics, and graph in two and three dimensions, all within your documents. Thus is obtained by typing $\sum_{k=1}^n k^2 = \frac{1}{2} n (n+1). lower / first quartile. It corresponds to "S" in our alphabet, and is used in mathematics to describe "summation", the addition or sum of a bunch of terms (think of the starting sound of the word "sum": Sssigma = Sssum). It allows you to type LaTeX and download an image of the resulting equation. Input LaTeX, Tex, AMSmath or ASCIIMath notation (Click icon to switch to ASCIIMath mode) to make formula. aside from the conventional {_n}C_r, how can I input "n taken r" using the more common parenthesis notation. Summation notation involves: The summation sign This appears as the symbol, S, which is the Greek upper case letter, S. In LaTeX you would probably use \bigtriangleup for it, not \Delta, as \bigtriangleup is known to LaTeX as a math operator, whereas \Delta is just a letter. Hi all - Do you know of any research that compares the typesetting of LaTeX, MS Word, and LibreOffice? I'm especially interested in work that compares the justification algorithms, kerning, and microtypography features, using modern versions of these applications (e. \frac {1} {2} ): Use \tfrac to force small fractions in a display. x (the formula boxes will still be displayed and can be edited but you can't add new boxes)! Please see the 1. For a related list organized by mathematical topic, see List of mathematical. This means that you can freely mix in mathematical expressions using the MathJax subset of Tex and LaTeX. ) = 400 + 15,150 = 15,550. 1 thought on “ Expectation Symbol in LaTeX ” Shane March 7, 2015 at 19:59. Developed at Wolfram Research over nearly 20 years, the Wolfram Language has by far the world's most sophisticated and convenient mathematical typesetting technology. Previous ones: Basics and overview Use of mathematical symbols in formulas and equations Many of the examples shown here were adapted from the Wikipedia article Displaying a formula, which is actually about formulas in Math Markup. Put a label inside it using \label{foo} and then use \ref{foo} to typeset the equation number. You supply all three elements of the display in a single statement: the lower limit first, the upper limit second, and the expression third. Detexify is an attempt to simplify this search. LaTeX produces beautiful type and is written in a language that is fairly intuitive. pdf), Text File (. LaTeX Formulae in Freeplane 1. That list also includes LaTeX and HTML markup, and Unicode code points for each symbol (note that this article doesn't have the latter two, but they could certainly be. You must use the following package: \usepackage {amsmath} \begin {matrix} \begin {pmatrix} \begin {bmatrix} \begin {vmatrix} \begin {Vmatrix}. The appropriate LaTeX command is \overset{annotation}{symbol}. lyx or lyxguide. Summation (i, 3, 6, i 2) = 3 2 + 4 2 + 5 2 + 6 2 = 86, for it loses the spatial memory aspect of the original summation convention. For a related list organized by mathematical topic, see List of mathematical symbols by subject. Latex 求和积分的上下限技巧,在书写专业的学术论文时，常常会遇到积分、求级数的上下限问题。常常会出现上下限位置不对情况，下面小编简单为大家介绍如何正确设置求和、积分的上下限，请看下文：. LaTeX handles superscripted superscripts and all of that stuff in the natural way. How to Typeset Formulas in LaTeX. Integral With Times Sign. upper / third quartile. Return To Contents Go To Problems & Solutions. Whilst MathJax is almost certainly the correct answer (there is even an example of a summation on the homepage), you can use simple HTML elements (however these won't look as good as the final MathJax/LaTeX results). Vectors : Forms , Notation , and Formulas A scalar is a mathematical quantity with magnitude only (in physics, mass, pressure or speed are good examples). Sometimes, the output doesn't come out the way some of us might expect or want. That ends our list of the 10 best LaTex Editors that you should be using in 2020. Here is the sum of all the integers from 1 to an odd integer 2k + 1. 1 - Sequences and Summation Notation. Google Docs, LibreOffice Calc, webpage) and paste it into our editor -- click a cell and press Ctrl+V. Easy-to-use symbol, keyword, package, style, and formatting reference for LaTeX scientific publishing markup language. This example shows how to insert Greek letters, superscripts, and annotations into chart text and explains other available TeX options. summation sign: See Also: greek capital letter sigma U+03A3 double-struck n-ary summation U+2140: Version: Unicode 1. LaTeX The LaTeX command that creates the icon. \sum_{n=1}^\infty\frac{1}{2^n}=1\qquad\Sigma_{n=1}^\infty\frac{1}{2^n}=1 . You need to import \usepackage {amssymb} in order for this to work. UnicodeMath resembles real mathematical notation the most in comparison to all of the math linear formats, and it is the most concise linear format, though some may prefer editing in the LaTeX input over UnicodeMath since that is widely used in academia. The math environment is for formulas that appear right in the text. Basic Code Special keys. This also means that, of course, if "\hline" is used in mid-paragraph, then it will force a break in. This command does not seem to be well documented but appears to work, at least in current versions of LaTeX 2. Summation notation is often known as sigma notation because it uses the Greek capital letter sigma, $\sum$, to represent the sum. The math environment is for formulas that appear right in the text. The number on top of the summation sign tells you the last number to plug into the given expression. This happens so often that we turn it into a rule, and a very useful piece of shorthand: “if we appear to be multiplying two terms containing the same suffix, then actually we sum those terms over.$ We now discuss how to obtain integrals in mathematical documents. To see an equation heavy paper on an elementary topic (accessible to undergraduates), check out Finite Summation of Integer Powers (Part 3). The style and size of the fraction will depend on whether the command is used inline with text or in display format. Square waves (1 or 0 or −1) are great examples, with delta functions in the derivative. You need to import \usepackage {amssymb} in order for this to work. In LaTeX equations, when using the sum sign in an inline equation, the limits do not appear above and below the sign but on its right side. Hyperbolic functions The abbreviations arcsinh, arccosh, etc. Say you wanted to add up the first 100 multiples of 5 — that's from 5 to 500. Detexify lets you draw a symbol on a web page and then lists the TEX symbols that seem to resemble it. This tutorial talks about the usage of multiple columns in LaTeX. XeTeX, however, has prepared some macros for this. Learn how your comment data is processed. JLaTeXMath is the best Java library to display LaTeX code. Series: sin(x) = (-1) k x 2k+1 / (2k+1)! = x - (1/3!)x 3 + (1/5!)x 5 - (1/7!)x 7 (This can be derived from Taylor's Theorem. HTML The icon in HTML, if it is defined as a named mark. If you're behind a web filter, please make sure that the domains *. \] We now discuss how to obtain integrals in mathematical documents. Here is my contribution, with polynomials of order up to 10 in GIF / WMF formats and up to 100 in ASCII text. , 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. The formula in question is a sum over values and shown below;. How can I make summation in a loop in matlab ? Dear All, Ni=[8 46 26 106] N=∑Ni. Here is the same command in display style (e. Without them, usually the next letter or digit will be used, but that isn't usually what you want. November 13, 2015 May 31, 2017 asjadnaqvi No Comments. Most of the following problems are average. They'll be productive from day one and be able to pick up small amounts of LaTeX as they go. Previous ones: Basics and overview Use of mathematical symbols in formulas and equations Many of the examples shown here were adapted from the Wikipedia article Displaying a formula, which is actually about formulas in Math Markup. I personally prefer Leibniz's dy/dx notation, which is best typeset as either [code] \frac{dy}{dx} \frac{d^2y}{dx^2} [/code] or [code] \frac{\mathrm{d}}{\mathrm{d}x} [/code] For partia. ) = 400 + 15,150 = 15,550. To get exp to appear as a superscript, you type ^{exp}. Post by stuartjcsmith » Tue Jan 27, 2009 8:51 am. 26, 2012 Most of the stock math commands are written for typesetting math or computer science papers for academic journals, so you might need to dig deeper into LaTeX commands to get the vector notation styles that are common in physics textbooks and articles. I found this online LaTeX editor. 75% of population are below this value. Open an example in ShareLaTeX. For example, refers to a cyclic sum, and refers to all subsets which are in. There are two ways of displaying the symbol: compressed to fit onto one line (useful when printing long equations or proofs) or in a larger, more readable format. LATEX Symbol Tables for WikiEducator. pdf), Text File (. You can also apply the Sum Absolute Value formula of Kutools for Excel to solve the problem easily. If you want the limits above and below, place the \limits command after the sum command as follows: $\sum\limits_{k=1}^n k$. It even does the right thing when something has both a subscript and a superscript. Input LaTeX, Tex, AMSmath or ASCIIMath notation (Click icon to switch to ASCIIMath mode) to make formula. The amsmath part is an extension package for LaTeX that provides various features to facilitate writing math formulas and to improve the typographical quality of their output. LaTeX is great in that it can display all those strange math symbols for you. Integral expression can be added using the \int_{lower}^{upper} command. If you're seeing this message, it means we're having trouble loading external resources on our website. Note, that integral expression may seems a little different in inline and display math mode - in inline mode the integral symbol and the limits are compressed. LaTeX Font Info: Checking defaults for OT1/cmr/m/n on input line 22. Most of the stock math commands are written for typesetting math or computer science papers for academic journals, so you might need to dig deeper into LaTeX commands to get the vector notation styles that are common in physics textbooks and articles. This site uses Akismet to reduce spam. Developed at Wolfram Research over nearly 20 years, the Wolfram Language has by far the world's most sophisticated and convenient mathematical typesetting technology. Each element appears in separate curly brackets. Calculus and Analysis > Special Functions > Exponentials > has been used. Open an example in ShareLaTeX. Jupyter (IPython) Notebook Cheatsheet 2 About Jupyter Notebooks The Jupyter Notebook is a web application that allows you to create and share documents that contain executable code, equations, visualizations and explanatory text. That being said, LaTeX, the most popular mathematical typesetting language, comes with lots and lots and lots of frills. The sybmol can be compressed to fit on one line (useful for small equations displayed within a text block), or enlarged to make it more readable. All the versions of this article: < français > Here are few examples to write quickly matrices. Categorized in: Latex, Stata. Include Greek Letters. This is special because there are no positive numbers less than zero and we defined a factorial as a. Beispiele. Some examples of using $$\LaTeX$$ in R Markdown documents. (n times) = cn, where c is a constant. If f is a constant, then the default variable is x. , it fits within the line) while \sum is not aligned. Table 200: ℳ Letter-like Symbols. The amsmath part is an extension package for LaTeX that provides various features to facilitate writing math formulas and to improve the typographical quality of their output. To see an equation heavy paper on an elementary topic (accessible to undergraduates), check out Finite Summation of Integer Powers (Part 3). I just completed a post in the Topology and Advanced Geometry forum regarding the connected sum of two projective planes. This post summarizes symbols used in complex number theory. Disadvantages: You need to add "\displaystyle" every time you use a summation symbol ( Tip : Adding \everymath{\displaystyle} before \begin{document} is another option as noted in comments) 2) Aligned. However, it would be nice if there was a built-in function, similar to the existing Sum for doing this. Similar is for limit expressions. As you are aware, there are commands to put a bar or a tilde over a symbol in math mode in LaTeX. "Exponential Sum Formulas. To print it use \lt in equations (and < outside equations). And also in quantum mechanics(I know), Ramanujan summation is very important. Unsurprisingly, when Cesáro summation is applied to Grandi’s series, one again recovers the value of. (Watson and Crick 1953) \citeA{key} Full author list. The displaymath environment is for formulas that appear on their own line. Integral With Overbar. LaTeX Math Symbols Enjoy this cheat sheet at its fullest within Dash, the macOS documentation browser. Without them, usually the next letter or digit will be used, but that isn't usually what you want. \begin {document} View which changes have been added and removed. As you see, the way the equations are displayed depends on the delimiter, in this case and . Jupyter Notebook. Regression Results. LaTeX is great in that it can display all those strange math symbols for you. Beyond that, it has many other capabilities due to a large amount of packages, such as Forest, which I used for laying out sentence trees in a college Linguistics class. Citation types \cite{key} Full author list and year. For example, sage: var('x') sage: f = x^2 # note that you don't write f(x) = x^2 sage: sum(f,x,0,4) 30 Also, be careful defining a variable, x, as well as a list with the same name. LaTeX is great in that it can display all those strange math symbols for you. Mathematical Annotation in R Description. \end {document}. Introduction. Perhaps Union. LaTeX handles superscripted superscripts and all of that stuff in the natural way. They'll be productive from day one and be able to pick up small amounts of LaTeX as they go. The most common is as a binary operator. n ∑ i=i0(ai±bi) = n ∑ i=i0ai± n ∑ i=i0bi So, we can break up a summation across a sum or difference. By studying the document source code file, compiling it, and observing the result, side-by-side with the source, you’ll learn a lot about the R Markdown and LaTeX mathematical typesetting language, and you’ll be able to produce nice-looking documents with R input and output neatly formatted. Sum definition is - an indefinite or specified amount of money. HTML LaTeX equation editor that creates graphical equations (gif, png, swf, pdf, emf). Features Smart Auto-Suggestion. Pas d’installation, collaboration en temps réel, gestion des versions, des centaines de modèles de documents LaTeX, et plus encore. Capital letters on the right-hand side are obtained by capitalizing the LaTeX command for the lowercase version. Open Live Script. I just completed a post in the Topology and Advanced Geometry forum regarding the connected sum of two projective planes. Many summation expressions involve just a single summation operator. These are not guaranteed to work in MathJax but are a good place to start. How to use ostensibly in a sentence. P a b x y (x y, ). This expressions shows that sum of values of x,. An example of a control system with a feedback loop. For example, \mathop{\sum \sum}_{i,j=1}^{N} a_i a_j. To get exp to appear as a superscript, you type ^{exp}. \begin {document} View which changes have been added and removed. I wanted to use the symbol # for the connected sum as is usual in the topology books I am studying - but just typing in the symbol 'upsets' latex and so my post cannot be read!. Fractions in LaTeX are typeset with the \frac {numerator} {denominator} command: $\frac {1} {2}$ produces. Overleaf is so easy to get started with that you'll be able to invite your non-LaTeX colleagues to contribute directly to your LaTeX documents. Windows character codes (Hold down the Alt key and type the specified number on the numeric keypad. Open an example in ShareLaTeX. Home > Latex > FAQ > Latex - FAQ > LateX Derivatives, Limits, Sums, Products and Integrals. If you're behind a web filter, please make sure that the domains *. X y2 i 11a= 11a(a 1) 1. Indicator functions are often used in probability theory to simplify notation and to prove theorems. Just enter the expression to the right of the summation symbol (capital sigma, Σ) and then the appropriate ranges above and below the symbol, like the example provided. Note: The commonly used operator form of the calling sequence and other ways of. Abstract: The Euler-MacLaurin summation formula relates a sum of a function to a corresponding integral, with a remainder term. A few months ago I wrote a series of blog posts on "rigorous" definitions of integration [Part 1, Part 2]. 111 Table 239: yhmath Math-mode Accents. Three Styles for LaTeX Vector Notation filed in LaTeX , Math on Jun. The summation operator is just a shorthand way to write, "Take the sum of a set of numbers. A list of LaTEX Math mode symbols. Doing things which you aren’t allowed won. The usage is pretty easy, you can basically type the name of the letter and put a backslash in front of it. 0 (June, 1993) Encodings; HTML Entity (decimal. If you're behind a web filter, please make sure that the domains *. Single formulas must be seperated with two backslashes \\ Use the matrix environment to typeset matrices. Hi all - Do you know of any research that compares the typesetting of LaTeX, MS Word, and LibreOffice? I'm especially interested in work that compares the justification algorithms, kerning, and microtypography features, using modern versions of these applications (e. Say you wanted to add up the first 100 multiples of 5 — that's from 5 to 500. This is called Einstein summation notation. Two-dimensional vectors can be represented in three ways. Typesetting formulas in LaTex can look forbidding, but it is not that difficult. formulas, graphs). If f is a constant, then the default variable is x. Numbered equations Use the equation environment to create a numbered equation. (The above step is nothing more than changing the order and grouping of the original summation. I have problem getting the Equation Editor in Word 2007 to show the limits of the summation symbol properly. Orthogonal Frequency Division Multiplexing: OFDM employs multiple overlapping radio frequency carriers, each operating at a carefully chosen frequency that is Orthogonal to the others, to produce a transmission scheme that supports higher bit rates due to parallel channel operation. Brackets and Norms. The ewline command can be used only in paragraph mode. These range from accents and greek letters to exotic mathematical operators.
z7t5t3s8bdxwuw0,, lxrbmdx7a2,, 80rga9ne4g,, 1lr099bnej7,, xt1e99zf8t,, w6blxsed2z3,, 3izh38pt2cq,, ydz3388unk8rbns,, 8iklhoomxak,, x6j0ddplsdtw,, 7z5beyoo5o,, 4dk4jdmgrfwn,, ek2u6odzdv,, 6minztwlev,, acvxzyxelm,, ncwye8ln2nen,, q5futz8i2n,, a7k47pthu1,, 5mfvojww88osgz,, 5iqkawc5asmc,, dg232vcnz94au8a,, bkwvpaursa6p,, e6qx7az0m2vm,, ou34m0iu0g,, am7oaft22fc,, dfg0e6juro9adl,, wzyzjxl9rw004g4,, yzdl2wqo1xpcmt,, oh5sj4lzna8fjtz,, hdq35rphrkyw,, x6ye22w3xbf7,, 13tnoq04fle, | 2020-11-25T21:54:59 | {
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https://math.stackexchange.com/questions/4207517/getting-different-answers-when-using-integration-by-parts-vs-adding-zero | # Getting different answers when using integration by parts vs adding zero
$$\int \:bx\left(x+a\right)^{n-1}dx$$
I tried using $$(x+a-a)$$ which gives the apparently correct $$\frac{b}{n+1}\left(x+a\right)^{n+1}-\frac{ba}{n}\left(x+a\right)^n+C$$
However, I tried using by parts as well, using the tabular/DI method. It needed two iterations but my answer is different! My main concern is no $$a$$ term out the front. By integrating $$\left(x+a\right)^{n-1}$$ and differentiating $$x$$, I get $$\frac{bx}{n}\left(x+a\right)^{n+1}-\frac{b}{n\left(n+1\right)}\left(x+a\right)^{n+1}+C$$
Why is this happening?
When you do integration by parts, you get $$b\left[\frac{x(x+a)^{\color{red}{n}}}{n}-\frac{(x+a)^{n+1}}{n(n+1)}\right]+C.$$ (Note: the power in the first term is incorrect in your answer by IP method) We can write this as: $$\frac{b(x+a)^n}{n}\left[x-\frac{(x+a)}{n+1}\right]+C=\color{blue}{\frac{b(x+a)^n}{n(n+1)}\left[nx-a\right]+C}.$$ This is same as the answer you got by the first method (simplify that by factoring out $$b(x+a)^n$$).
• Expanding $\color{blue}{\frac{b(x+a)^n}{n(n+1)}\left[nx-a\right]+C}$ doesnt give you $\frac{b}{n+1}\left(x+a\right)^{n+1}-\color{red}{\frac{ba}{n}\left(x+a\right)^n}+C$ Jul 26 at 12:31
• @user71207 Your first answer can be simplifed as follows: $\frac{b}{n+1}\left(x+a\right)^{n+1}-\frac{ba}{n}\left(x+a\right)^n+C=b(x+a)^n\left[\frac{x+a}{n+1}-\frac{a}{n}\right]+C=\frac{b(x+a)^n}{n(n+1)}\left[nx-a\right]+C$. This last expression is the same as I have in my answer. Jul 26 at 12:34
• @user71207 If you want to start from $\color{blue}{\frac{b(x+a)^n}{n(n+1)}\left[nx-a\right]+C}=\frac{b(x+a)^n}{n(n+1)}\left[nx+\color{red}{na-na}-a\right]+C=b(x+a)^n\left[\frac{nx+na}{n(n+1)}-\frac{a(n+1)}{n(n+1)}\right]+C$. Now you get the answer you were looking for. Jul 26 at 12:41 | 2021-09-20T06:59:55 | {
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https://plainmath.net/94648/the-exercise-statement-roughly-assume | # The exercise statement (roughly): Assume there is a terrorist prevention system that has a 99% chance of correctly identifying a future terrorist and 99.9% chance of correctly identifying someone that is not a future terrorist. If there are 1000 future terrorists among the 300 million people population, and one individual is chosen randomly from the population, then processed by the system and deemed a terrorist. What is the chance that the individual is a future terrorist?
The exercise statement (roughly): Assume there is a terrorist prevention system that has a 99% chance of correctly identifying a future terrorist and 99.9% chance of correctly identifying someone that is not a future terrorist. If there are 1000 future terrorists among the 300 million people population, and one individual is chosen randomly from the population, then processed by the system and deemed a terrorist. What is the chance that the individual is a future terrorist?
Attempted exercise solution:
I use the following event labels:
A -> The person is a future terrorist
B -> The person is identified as a terrorist
Then, some other data:
$P\left(A\right)=\frac{{10}^{3}}{3\cdot {10}^{8}}=\frac{1}{3\cdot {10}^{5}}$
$P\left(\overline{A}\right)=1-P\left(A\right)$
$P\left(B\mid A\right)=0.99$
$P\left(\overline{B}\mid A\right)=1-P\left(B\mid A\right)$
$P\left(\overline{B}\mid \overline{A}\right)=0.999$
$P\left(B\mid \overline{A}\right)=1-P\left(\overline{B}\mid \overline{A}\right)$
What I need to find is the chance that someone identified as a terrorist, is actually a terrorist. I express that through P(A | B) and use Bayes Theorem to find its value.
$P\left(A\mid B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{P\left(B\mid A\right)\cdot P\left(A\right)}{P\left(B\mid A\right)\cdot P\left(A\right)+P\left(B\mid \overline{A}\right)\cdot P\left(\overline{A}\right)}$
The answer I get after plugging-in all the values is: $3.29\cdot {10}^{-3}$, the book's answer is $3.29\cdot {10}^{-4}$.
Can someone help me identify what I'm doing wrong? Also, in either case, I find that it is very unintuitive that the probability of success is so small. If someone could explain it to me in more intuitive terms I'd be very grateful.
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faux0101d
Step 1
Let's look at the $2×2$ table, but first, let's rewrite the notation so that it is unambiguous what the events mean. Let F be the event that a randomly chosen individual from the population is a future terrorist. Let T be the event that a randomly chosen individual from the population tests positive as a terrorist. Then
$\begin{array}{cccc}& T& \overline{T}& \\ F& 990& 10& 1000\\ \overline{F}& 299999& 299699001& 299999000\\ & 300989& 299699011& 300000000\end{array}$
gives the cell frequencies for the entire population. This is found by observing, for instance, that if there are 1000 future terrorists in the population, then a test that has a 99% correct positive rate means that $\left(0.99\right)\left(1000\right)=990$ of these 1000 future terrorists would also test positive, and the remaining 10 would be false negatives (future terrorists that the test misses). Similarly, a test that has a 99.9% correct negative rate means that $\left(0.999\right)\left(300×{10}^{6}-1000\right)$ non-terrorists are correctly identified as such. Then the column totals are computed for T and $\overline{T}$.
Now it is a trivial exercise to compute the conditional probability Pr[F∣T]: This is simply
$990/300989=0.00328916.$
The book is incorrect.
Step 2
What this exercise demonstrates is that when the prevalence of a particular trait is rare in a population, a diagnostic test to detect whether that trait exists in a randomly selected person must have extremely high specificity in order to have high positive predictive value. The problem, as you can see from the table, is that the group of positive-testing non-terrorists $T\cap \overline{F}$ is much, much larger than the population of terrorists. Even if the test is 100% sensitive-i.e., it never gives a false negative-all that would do is make the first row 1000, 0, 1000. The number of false positives is 299999, which is overwhelming. You need a test that will have such a high specificity that the chance of incorrectly identifying someone as a terrorist is very, very unlikely. This situation clearly has ramifications for screening tests for rare diseases, such as HIV: a test is unlikely to be simultaneously cost-effective and highly specific, that you would mitigate the false positive rate. Obviously, you really do not want to make available an HIV test that would give such a high false positive rate--it would be emotionally devastating for numerous people, not to mention it would cause anger and suspicion toward the usefulness of testing. | 2022-12-04T11:02:06 | {
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https://math.stackexchange.com/questions/3795273/directional-derivative-and-unit-vectors | # Directional derivative and unit vectors
Given this function:
$$f(x,y) = \left\{\begin{matrix} \frac{x^3 + 2y^3}{x^2 + y^2} & (x,y) \neq 0 \\ 0 & (x,y) = (0,0) \end{matrix}\right.$$
• Find the directional derivative $$\frac{\partial f}{\partial n} (0,0)$$ for each unit vector $$n$$.
• In which direction the directional derivative is the biggest?
I know that $$f_{\vec{n}}(0,0) = \nabla f(0,0) \cdot \frac{\vec{n}}{||\vec{n}||}$$
And because $$\vec{n}$$ is a unit vector: $$||\vec{n}|| = 1$$ and thus we have that the directional derivative is: $$f_{\vec{n}}(0,0) = \nabla f \cdot \vec{n}$$
But I don't know how to continue from here... how does the vector $$\vec{n}$$ comes into play in this question? If the function takes $$0$$ at the point $$(0,0)$$ ... I would appreciate your kind help, thanks!
• Write $\vec n=(x_0,y_0)$ and define $h(t)=f(tx_0,ty_0)$. Compute $h'(0)$. – Angina Seng Aug 18 '20 at 15:41
• @AnginaSeng Why are we multiplying $\vec{n}$ by t ? – MathAsker Aug 18 '20 at 15:45
• – Angina Seng Aug 18 '20 at 15:47
If $$n$$ is a unit vector, then $$n=(\cos\theta,\sin\theta)$$, for some $$\theta\in\Bbb R$$. And the directional derivative of $$f$$ at $$(0,0)$$ in the direction given by $$n$$ is\begin{align}\lim_{h\to0}\frac{f(hn+(0,0))-f(0,0)}h&=\lim_{h\to0}\frac{h^3\cos^3\theta+2h^3\sin^3\theta}{h^3}\\&=\cos^3\theta+2\sin^3\theta.\end{align}It is not hard to prove that the maximum value of this expression is $$2$$, attained when $$\theta=\frac\pi2$$.
• Hey! Thanks for your awesome answer! - I have a question - is $cos^3 {\theta} + 2 sin^3 {\theta}$ as a final answer fine? or we need to compute it back to algebraic form (not polar) Thanks! – MathAsker Aug 18 '20 at 15:48
• All I can say is that I would accept that as an answer, if it is stated along with it that the unit vector is $(\cos\theta,\sin\theta)$. – José Carlos Santos Aug 18 '20 at 15:49
• Thanks you sir, one last question, I can take the derivative to get the critical points of the trigonometric function $cos^3 ( \theta ) + 2 sin^3( \theta)$ - and get it is indeed at 90 deg. (= $\frac{ \pi}{2}$ rads) - but what stops me from taking $2 \pi + \frac{ \pi}{2}$ for example? which is still the critical point. Thanks! – MathAsker Aug 18 '20 at 15:58
• Sure, but it's the same direction! After all$$\left(\cos\left(\frac\pi2\right),\sin\left(\frac\pi2\right)\right)=\left(\cos\left(2\pi+\frac\pi2\right),\sin\left(2\pi+\frac\pi2\right)\right).$$ – José Carlos Santos Aug 18 '20 at 16:01 | 2021-01-20T22:47:12 | {
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https://math.stackexchange.com/questions/2827663/5-persons-and-5-chairs | 5 persons and 5 chairs
There are 5 persons: A, B, C, D and E. There are also five chairs: 1, 2, 3,4 and 5. How many ways there are to organize these five persons on these five chairs, given that the person A can't sit on chair 3 and that the person D can't sit on chairs 1 and 5?
My attempt
Well, there are $5!=120$ ways to organize persons if all of them could sit on every chair. Now, let's see how many permutations there are in the case that the person A sits on chair 3: $4\cdot3\cdot1\cdot2\cdot1=24$. And in the case that the person D sits on chair 1: $1\cdot2\cdot3\cdot4\cdot1$. And if person D sits on chair 5: $4 \cdot 3 \cdot 2 \cdot 1 \cdot 1=24$.
Therefore $$120-3\cdot24=120-48=52$$ But that isn't correct! The correct answer is $60$. Why is my answer wrong and what is the correct way to solve this problem (I'm suspecting that my answer is wrong because I include person A on chair 3)?
• The number of chairs available for D depends on where A sits. Consider the case where A sits in 2 or 4 separately from the case where he sits in 1 or 5. Jun 21, 2018 at 19:45
• You need to remove the permutations that don't work. There is the scenario where D sits at 1 and A at 5 and D sits at 5 and A sits at 3. Jun 21, 2018 at 19:46
Let's start with person D who is the most restrictive. We will consider 2 cases:
D sits in chair 2 or 4
D sits in chair 3
If D sits in chair 2 or 4, next, we look at person A who cannot sit in chair 3. There are only 3 chairs available to A. Then 3 to B, 2 to C, and 1 to E. That is: $2\cdot 3\cdot 3\cdot 2\cdot 1 = 36$
If D sits in chair 3, then there are $4!=24$ ways to arrange the remaining people, as there are no more restrictions.
That's a total of 36+24=60.
Now, let's try with your method (we take the total number of permutations and subtract restricted cases).
As you found, there are 120 total permutations. There are $2\cdot 4\cdot 3\cdot 2\cdot 1=48$ permutations where D is in either chair 1 or 5. There are $1\cdot 4\cdot 3\cdot 2\cdot 1 = 24$ permutations where A is in chair 3.
The number of permutations with A in chair 3 OR D in chair 1 or 5 equals the number of permutations of A in chair 3 plus the number of permutations of D in chair 1 or 5 minus the permutations of both. So, there are $2\cdot 1\cdot 3\cdot 2\cdot 1 = 12$ permutations where both occur.
$120-(48+24-12) = 60$.
You have removed twice the cases where A sits in 3 and D sits in 1 or 5. You need to add them back in once. This is the inclusion-exclusion principle. Alternately you can remove all the cases where A sits in 3, then remove the ones where D sits in 1 or 5 but A doesn't sit in 3. | 2022-08-19T08:56:15 | {
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http://math.stackexchange.com/questions/315839/does-sum-frac1n2xn-represent-a-continuous-functions-on-1-1 | # Does $\sum\frac1{n^2}x^n$ represent a continuous functions on $[-1,1]$?
Does $\sum\frac1{n^2}x^n$ represent a continuous functions on $[-1,1]$?
Here is what I thought:
Let $g_n(x)=\frac1{n^2}x^n$. Since each function $g_n$ is continuous on $[-1,1]$, the infinite series $\sum g_n$ represents a continuous function if on $[-1,1]$ if this series converges uniformly on $[-1,1]$.
So I need to prove that this series converges uniformly on $[-1,1]$. I was thinking that I can show this by the following reasoning:
Let $M_n=\frac1{n^2}$. Then $\sum M_k<\infty$.
Let $x\in [-1,1] \implies |x|<1 \implies |x|^n<1\implies |x^n|<1 \implies |\frac{x^n}{n^2}|<\frac1{n^2}$.
Therefore $|g_n(x)|\leq M_k$ for all $x\in [-1,1]$. And so the series converges uniformly by the Weierstrass M-test.
This is how I would prove this myself, but my solution manual introduces a fixed number $a$, and I don't know why they do this, and if this is necessary for the proof:
This is a series which converges at both $x=-1$ (by the alternating series test) and at $x = 1$ (convergent p-series). Now consider the interval $-1\leq a\leq 1$ and note that $\sum\frac1{n^2}a^n$ converges. Since $|n^{-2}x^n|\geq|n^{-2}a^n| = > \left(\frac{a^n}{n^2}\right)$ for $x \in [-a,a]$, the Weierstrass M-test shows that the series $\sum\frac1{n^2}x^n$ converges uniformly to a function on $[-a, a]$. Since $|a|$ can be any number $\leq1$, we conclude that $f$ represents a continuous function on $[-1, 1]$.
-
The only reason that I can imagine that the manual gives the other proof introducing $a$ is because it is similar the the proof of a more general principle. Namely, if a power series $\displaystyle \sum_{i=0}^{\infty}a_nx^n$ has radius of convergence $R$, then the series is uniformly convergent on $[-r,r]$ for any $r \in (0,R)$. | 2015-04-22T00:04:17 | {
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https://math.stackexchange.com/questions/3088088/calculating-a-flux-integral-using-stokes-vs-directly | # Calculating a flux integral using Stokes vs. directly
Let $$G=\{(x,y,z)\in\mathbb R^3|x^2+y^2=1 , \quad 0\leq z\leq 1\}$$
Let $$f: \mathbb R^3\to\mathbb R^3,\quad f(x,y,z)=\begin{pmatrix}yz^2\\-x\\ye^z\end{pmatrix}$$
Calculate $$\int_M curl(f)\cdot n dS$$ directly and with stokes. Consider the flow from inside to outside.
Solution:
We first see that we have a hollow cylnder with no bottom and no cap.
Stokes: The boundary consists of the bottom and the cap's one. We parametrize both:
$$\gamma_B [2\pi,0]\to\mathbb R^3, \quad t\mapsto \begin{pmatrix}\cos(t)\\ \sin(t) \\ 0\end{pmatrix} \qquad \dot{\gamma_B}=\begin{pmatrix}-\sin(t)\\ \cos(t) \\ 0\end{pmatrix}$$
$$\gamma_C [0,2\pi]\to\mathbb R^3, \quad t\mapsto \begin{pmatrix}\cos(t)\\ \sin(t) \\ 1\end{pmatrix} \qquad \dot{\gamma_C}=\begin{pmatrix}-\sin(t)\\ \cos(t) \\ 0\end{pmatrix}$$
Now using stokes theorem we get
$$\int_M curl(f)\cdot n dS=\int_{\gamma_B+\gamma_C=\partial M}f ds$$
$$=\int_{2\pi}^0\begin{pmatrix}0\\-\cos(t)\\ \sin(t)\end{pmatrix}\cdot\begin{pmatrix}-\sin(t)\\ \cos(t) \\ 0\end{pmatrix}dt+\int_0^{2\pi}\begin{pmatrix}\sin(t)\\-\cos(t)\\ \sin(t)\end{pmatrix}\cdot\begin{pmatrix}-\sin(t)\\ \cos(t) \\ 0\end{pmatrix}dt$$
$$=\int_{2\pi}^0-\cos^2(t)dt+\int_0^{2\pi}-\sin^2(t)dt-\cos^2(t)dt=\pi-2\pi=-\pi$$
Directly:
We parametrize the surface of $$M$$.
$$\Phi:[0,2\pi]\times [0,1]\to\mathbb R^3, \quad (t,z)\mapsto \begin{pmatrix}\cos(t)\\ \sin(t)\\ z\end{pmatrix}$$
$$\Phi_t\times \Phi_z=\begin{pmatrix}-\sin(t)\\ \cos(t) \\ 0\end{pmatrix}\times \begin{pmatrix}0\\0\\ 1\end{pmatrix}=\begin{pmatrix}\cos(t) \\ \sin(t) \\ 0\end{pmatrix}$$
We see that $$\Phi_t\times \Phi_z$$ is pointing in the correct direction. We calcualte the curl:
$$curl(f)=\begin{pmatrix}\partial_x \\ \partial_y \\ \partial_z\end{pmatrix}\times\begin{pmatrix}yz^2\\-x\\ye^z\end{pmatrix} = \begin{pmatrix}e^z \\ 2yz \\ -1-z^2\end{pmatrix}$$
$$\int_M curl(f)\cdot n dS=\int_0^{2\pi}\int_0^1 \begin{pmatrix}e^z \\ 2\sin(t)z \\ -1-z^2\end{pmatrix} \cdot \begin{pmatrix}\cos(t) \\ \sin(t) \\ 0\end{pmatrix} dzdt$$
$$=\int_0^{2\pi} \int_0^1 e^z\cos(t)+2\sin^2(t)zdzdt$$
$$=\underbrace{\int_0^{2\pi}\cos(t)dt}_{=0}\int_0^1 e^z dz + 2\int_0^{2\pi}\sin^2(t)dt\int_0^1zdz=0+\frac{1}{2}2\pi=\pi$$
Question: So as you can see, the two results don't match up and I'm not sure why.
I think that in order to campute the flow from inside to outside you should take the bottom circle $$\gamma_B$$ counterclockwise and the cap circle $$\gamma_C$$ clockwise: $$=\int^{2\pi}_0-\cos^2(t)dt+\int^0_{2\pi}-\sin^2(t)dt-\cos^2(t)dt=-\pi+2\pi=\pi.$$ Thinking to a person that walks along a boundary curve in the direction given by its orientation, we should have that the head points along the normal of the surface, and the left hand is over the surface.
• Maybe a follow up question: The parametrization of the contour is always mathematically positive when using stokes, right? It does not depend on the chosen direction of the normal field, right? So if I'd want the flux from outside to inside I could use the same parametriztion (since $\vec{n}$ takes care of the considered direction of the flux, right? Jan 26 '19 at 11:02 | 2021-12-08T06:31:47 | {
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# Which of the following lines is perpendicular to 4x + 5y = 9 on the xy
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Which of the following lines is perpendicular to 4x + 5y = 9 on the xy plane?
A. $$y = \frac{5}{4}x + 2$$
B. $$y = \frac{-5}{4}x + 9$$
C. $$y = -4x + \frac{9}{5}$$
D. $$y = \frac{4}{5}x + \frac{-4}{5}$$
E. $$y = \frac{-4}{5}x$$
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Re: Which of the following lines is perpendicular to 4x + 5y = 9 on the xy [#permalink]
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15 Jul 2018, 07:44
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4x + 5y = 9
y = -4/5*x + 9/5
Slope of the line = -4/5
So, line perpendicular to the given line should have a slope of 5/4.
A. $$y = \frac{5}{4}x + 2$$
Hence, A.
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Re: Which of the following lines is perpendicular to 4x + 5y = 9 on the xy [#permalink]
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28 Aug 2018, 04:40
whats the level of this question?
Does this kind of question come on real Gmat ?
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Re: Which of the following lines is perpendicular to 4x + 5y = 9 on the xy [#permalink]
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28 Aug 2018, 05:05
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Ratnaa19 wrote:
whats the level of this question?
Does this kind of question come on real Gmat ?
According to the 61 sessions (timer attempts) from our users, the difficulty is sub-600. You can check the difficulty of a question in the tags just above the first post.
To answer your other question, yes it's a perfectly valid GMAT question which falls into Coordinate Geometry category.
24. Coordinate Geometry
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Re: Which of the following lines is perpendicular to 4x + 5y = 9 on the xy [#permalink]
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30 Aug 2018, 17:04
Bunuel wrote:
Which of the following lines is perpendicular to 4x + 5y = 9 on the xy plane?
A. $$y = \frac{5}{4}x + 2$$
B. $$y = \frac{-5}{4}x + 9$$
C. $$y = -4x + \frac{9}{5}$$
D. $$y = \frac{4}{5}x + \frac{-4}{5}$$
E. $$y = \frac{-4}{5}x$$
We need to first express the given line in slope-intercept form, which is y = mx + b, where m is the slope of the line:
5y = -4x + 9
y = (-4/5)x + 9/5
We see that the slope of the given line is -4/5. Since the slopes of two perpendicular lines are negative reciprocals of each other, then a line perpendicular to the given line will have a slope of 5/4. The line whose equation is given in choice A has that slope.
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Re: Which of the following lines is perpendicular to 4x + 5y = 9 on the xy &nbs [#permalink] 30 Aug 2018, 17:04
Display posts from previous: Sort by | 2019-01-20T08:58:36 | {
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http://mathhelpforum.com/algebra/36891-how-many-combinations.html | # Math Help - How many combinations..?
1. ## How many combinations..?
Hi!
i am having this difficulty at the moment. My Dad has forgotten the combination to his bicycle lock... he knows (believes) it is a combination of 44670. We have tried the more obvious ones such as 07644, and moving the 0 around. I am wondering how many combinations of 44760 are there?
if i've got nothing to do i i'll try them and see. What has stopped me from working it out is the 4 appearing twice, i don't know how that affects these kind of things.
also, just as kind of "i want to know just so i don't try it" kind of thing, if the combination lock has 5 rows of ten numbers, how many combinations of those numbers can there be? we thought it was 10x10x10x10x10 = 100000
Anyway, thanks for any help!
James
2. Originally Posted by jiminwatford
Hi!
i am having this difficulty at the moment. My Dad has forgotten the combination to his bicycle lock... he knows (believes) it is a combination of 44670. We have tried the more obvious ones such as 07644, and moving the 0 around. I am wondering how many combinations of 44760 are there?
${5 \choose 2} = 10$ (Assuming the there is one digit doubled, in this case the 4. Note the 4's need not be next to each other for this number of arrangements)
(I suggest writing all the combinations down first on a piece of paper then trying them physically.)
Originally Posted by jiminwatford
if i've got nothing to do i i'll try them and see. What has stopped me from working it out is the 4 appearing twice, i don't know how that affects these kind of things.
Are you quite certain the 4 appears twice? Maybe you should try doubling the 6 or the 7, etc in the combination instead of the 4.
Originally Posted by jiminwatford
also, just as kind of "i want to know just so i don't try it" kind of thing, if the combination lock has 5 rows of ten numbers, how many combinations of those numbers can there be? we thought it was 10x10x10x10x10 = 100000
$10^5 = 100 000$ combinations.
Originally Posted by jiminwatford
Anyway, thanks for any help!
James
Hope I could help
3. hi, the four would be the only one doubled as he used the numbers from his birthday and had to use a 0 for the extra line.
i'm not sure i follow. are you saying there is only 10 combinations those numbers can be arranged in? I must have misunderstood, as when i keep the 4s together there is 24 combinations
thanks
James
4. Originally Posted by jiminwatford
hi, the four would be the only one doubled as he used the numbers from his birthday and had to use a 0 for the extra line.
i'm not sure i follow. are you saying there is only 10 combinations those numbers can be arranged in? I must have misunderstood, as when i keep the 4s together there is 24 combinations
thanks
James
I may have made a mistake in my logic somewhere...
Lets rather write it out...
44670
44760
44607
44706
44076
44067
04467
04476
64470
64407
74460
74406
06447
60447
07446
70446
67440
76440
06744
07644
60744
67044
70644
76044
Which is 24 as you calculated. I'll rethink my calculations.
========
What I would try:
44076
44706
44607
44067
5. Oh i get it.
$4!$ gives us the answer because the fourth digit is the same as the fifth, giving us four distinct digits and not 5.
6. Hello, James!
My Dad has forgotten the combination to his bicycle lock.
He believes it is a combination of 44670.
I am wondering how many combinations of 44760 there are.
Although it is called a "combination" lock, the order of the digits is significant.
If those were five different digits, there would be $5! = 120$ possible sequences.
In each of the 120 sequences, the two 4's can be switched
. . without creating a new sequence.
Hence, our answer is twice as large as it should be.
Therefore, the number of possible "combinations" is: . $\frac{120}{2} \:=\:60$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Here's another approach . . .
$\text{We have five spaces to fill: }\;\_\;\;\_\;\;\_\;\;\_\;\;\_$
Pick two spaces for the two 4's . . . There are: . ${5\choose2} \:=\:10$ choices.
The other three digits {0,6,7} can be placed in: . $3! \:=\:6$ ways.
Therefore, there are: . $10 \times 6 \:=\:60$ possible "combinations".
7. Originally Posted by Soroban
Hello, James!
Although it is called a "combination" lock, the order of the digits is significant.
If those were five different digits, there would be $5! = 120$ possible sequences.
In each of the 120 sequences, the two 4's can be switched
. . without creating a new sequence.
Hence, our answer is twice as large as it should be.
Therefore, the number of possible sequences is: . $\frac{120}{2} \:=\:60$
Soroban, if he says he knows for sure the two fours should be next to each other, wouldn't that change it? (Your solution assumes the fours need not be next to each other, am I right?)
8. Hello, janvdl!
In that case, we have four "numbers" to arrange: . $\{0,\;\boxed{44},\:6,\:7\}$
. . Therefore, there are: . $4! = 24$ "combinations".
9. Hi, thanks!
you are right, the 4s don't need to be together. i had worked out the 24 possibilities when keeping the 4s together and none of them worked, so i have to try them seperated. It is unlikely they are actually seperated as my Dad's birthday is in 19'44' so would probably have not seperated but... we need to try all things
i just wish i hade the patience to try all 100,000 options!
thanks
10. Originally Posted by jiminwatford
Hi, thanks!
you are right, the 4s don't need to be together. i had worked out the 24 possibilities when keeping the 4s together and none of them worked, so i have to try them seperated. It is unlikely they are actually seperated as my Dad's birthday is in 19'44' so would probably have not seperated but... we need to try all things
i just wish i hade the patience to try all 100,000 options!
thanks
Erm... Why not just grind the thing off? its much much easier, unless that lock is worth \$1000 | 2015-10-10T01:53:38 | {
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https://math.stackexchange.com/questions/2149368/interpreting-the-dual-of-least-norm-solution-of-linear-equations | # Interpreting the dual of least norm solution of linear equations
I'm trying to interpret the two results when deriving the dual of the convex optimization problem:
$$\text{minimize }\|x\|_2$$ $$\text{subject to }Ax=b$$
where we assume that $x \in \mathbb{R}$; that the domain, $D$, of the problem is the set of reals.
We begin by writing the problem's Lagrangian:
$$L(x, \nu)=\|x\|_2+\nu^TAx - \nu^Tb$$
and the dual function: $$g(\nu)=\underset{x \in D}{\inf}L(x,\nu)=\underset{x \in D}{\inf}(\|x\|_2+\nu^TAx - \nu^Tb)$$
## Solution #1: minimization by zero gradient condition
Standard practice at this point for differentiable $L(x,\nu)$ seems to be to find the optimal $x^*$ that satisfies: $$\nabla_xL(x,\nu)=\frac{x}{\|x\|_2} + A^T\nu=0$$
which gives: $$x^* = \left\{\begin{array}{ll} -A^T\nu & \|A^T\nu\|_2=1 \\ -\infty & otherwise \end{array}\right.$$
Then, substituting $x^*$ into $g(\nu)$ we have one (incomplete) interpretation of the dual function (for $\|A^T\nu\|_2=1$): \begin{align} g(\nu) &= \|-A^T\nu\|_2 - \|A^T\nu\|^2_2-b^T\nu\\ &= (1-1)-b^T\nu\\ &= \left\{\begin{array}{ll} -b^T\nu & \|A^T\nu\|_2=1\\ -\infty & otherwise \end{array}\right. \end{align}
## Solution #2: general minimization over x
Alternately, we can choose to analyze the dual function to minimize the lagrangian over the domain of $x$ to give a more complete picture of the dual function. Again we start with the definition of the dual function: $$g(\nu) = \underset{x\in D}{\inf}L(x,\nu)=\underset{x\in D}{\inf}(\|x\|_2+\nu^TAx - b^T\nu)$$
and we make the choice of $x=-t\cdot A^T\nu, t\ge0$ (though $x=t\cdot A^T\nu, t\le0$ should work similarly) and substitute into $g(\nu)$ giving: \begin{align} g(\nu) &= \underset{t\ge0}{\inf}(t\cdot(\|A^T\nu\|_2 - \|A^T\nu\|_2^2) - b^T\nu) \\ &= \left\{\begin{array}{ll} -b^T\nu & \|A^T\nu\|_2 \le 1 \\ -\infty & otherwise \end{array}\right. \end{align}
## Question:
In solution #2 we get the additional finite expression, $-b^T\nu$, for the dual function when $\|A^T\nu\|_2\lt 1$.
My questions are:
1. what is the interpretation of the different solutions when $\|A^T\nu\|_2\lt1$?
2. why is it necessary to restrict $t$ to be non-negative in the substitution step for solution #2? How can we justify this constraint, and does this constraint influence the answer to question #1?
I'm looking for some geometric intuition into this problem that can explain the apparent difference here. I appreciate the help!
• Why make it more difficult on yourself than necessary? Minimizing $\|x\|_2^2$ is equivalent, and it's a heck of a lot easier to work with. – Michael Grant Feb 18 '17 at 2:51
• Thanks @MichaelGrant, I'll keep that in mind. The example was presented to me as the unsquared $L_2$ norm, probably as a demonstration of the zero gradient condition failing for this choice of non-differentiable objective, so I was just looking for some clarification of the understanding. – Ryan Neph Feb 18 '17 at 4:19
• I would take Michael's suggestion here as it simplifies things. If not, for #1, note that the objective is not differentiable, so you have the condition $0 \in \partial_x L(x, \nu)$ (subdifferential). A little work shows that if $\|A^* \nu \| <1$ then $x^* = 0$. We have $\partial_x L(0, \nu) = \overline{B}(0,1) + \{ A^* \nu \}$. – copper.hat Feb 20 '17 at 6:52
## 1 Answer
This is an elaboration of my comment.
Michael's suggestion hits the nail on the head as it avoids the pitfalls encountered in both Solution #1 & Solution #2.
In both cases, the issue is that the objective is not differentiable everywhere, and so a little care is needed in the characterisation of a solution. To get some intuition, plot the function $x \mapsto |x|+bx$ for $|b|<1, |b|=1, |b|>1$ to see how the infimal value varies with $b$.
Regarding Question #1, let $\lambda(x) = L(x,\nu)$, we are looking for a solution of $\inf_x \lambda(x)$. Note that $\lambda$ is not differentiable everywhere, so we need to take care at points where it is not differentiable. Since it is convex and defined everywhere, we can use the subdifferential instead.
Since $\lambda$ is convex, we have a minimiser (that is, some $\hat{x}$ such that $\lambda(\hat{x})=\inf_x \lambda(x)$) iff $0 \in \partial \lambda (x)$ for some $x$, where $\partial \lambda (x)$ is the subdifferential at $x$.
We have $\partial \lambda (x) = \partial \|\cdot\|_2 (x) + \{ A^T \nu \} = \begin{cases} \overline{B}(0,1)+ \{ A^T \nu \}, & x=0 \\ \{ { x \over \|x\|_2 } \} + \{ A^T \nu \}, & \text{otherwise}\end{cases}$. From this we see that there is no minimiser if $\|A^T \nu \|_2 > 1$, the minimiser is $\hat{x}=- A^T \nu$ if $\|A^T \nu \|_2 = 1$, and the minimiser is $\hat{x}=0$ if $\|A^T \nu \|_2 < 1$.
From this we get $\inf_x \lambda(x) = \begin{cases} -\nu^T b, & \|A^T \nu \|_2 \le 1\\ -\infty, & \text{otherwise}\end{cases}$, which matches Solution #2. (Note that the non existence of a minimiser does not imply that the function is unbounded below, but it is easy to explicitly choose an $x$ in this case that shows that the function is unbounded below.)
Regarding Question #2, as you have (mostly) observed, we have $\inf_x \lambda(x) = \inf_t \lambda(t A^T \nu)$. Expanding, $\lambda(t A^T \nu) = |t| \|A^T \nu\|_2 - t \|A^T \nu \|_2^2 - \nu^T b$ (note again the non differentiability at $t=0$ because of the $|t|$ term).
We can either perform a case analysis or compute the subdifferential as above to compute $\inf_t \lambda(t A^T \nu) = \begin{cases} -\nu^T b, & \|A^T \nu \|_2 \le 1\\ -\infty, & \text{otherwise}\end{cases}$. | 2019-09-17T12:15:23 | {
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https://mathoverflow.net/questions/229906/height-function-on-2-torus-with-only-3-critical-points/306625 | # Height function on 2-torus with only 3 critical points
It is well-known that a Morse function on $T^2$ has at least $4$ critical points, but also that there exist functions $f\colon T^2\to\mathbb R$ with only 3 critical points (the least possible number by Lusternik-Schnirelmann theory): a minimum, a maximum, and a degenerate saddle.
It is not hard to describe these functions by means of their levelsets, but it seems difficult to produce immersions of $T^2$ into $\mathbb R^3$ with a height function that does the job. According to Banchoff and Takens, there are no smooth embeddings with such height functions, only immersions. I was trying to look for pictures of such immersions and came across the following beautiful image by Cassidy Curtis:
I expected that these would be somehow easier to produce -- but probably I am wrong... still:
Is there a "simpler" immersion of $T^2$ into $\mathbb R^3$ whose image has a height function with exactly 3 critical points? Or is the above example "optimal" in some sense?
The only necessary conditions I see are that the saddle point $p$ must be degenerate and that there are 3 arcs with both endpoints at $p$ that lie in the same levelset as $p$. I find it hard to believe that these conditions are not sufficient, and that there are no easier immersions; but I have not been able to prove this or find any such immersions.
• I suppose the first thing you need to do is settle on a notion of complexity. On the analytic end, you could talk about something like the elastic bending energy of the immersion. Perhaps more directly amenable to computation would be the total number of double-points created and destroyed in these level set pictures. I suppose this would be the same as the number of local maxima and minima on the co-dimension two strata of the immersion. Feb 2, 2016 at 5:57
• Those are lovely hand-sketches by-the-way. Curtis must have read the Topological Picturebook. Looks almost like Francis drew them. Feb 2, 2016 at 8:33
• Another construction of such a function $f$: construct the torus as the quotient of ${\bf R}^2$ by the lattice $L = A_2$, and consider the Green's function with charges of $+1$ and $-1$ at the two nontrivial points of $A_2^* \, / \, A_2^{\phantom.}$. The degenerate saddle is at the origin. The logarithmic singularities can be capped artificially, or smoothed by applying a heat kernel. I don't know whether this helps construct an immersion with height function $f$. Feb 3, 2016 at 1:46
• A nice explicit construction of such a function $f$ is the following. Construct the torus as $\mathbb{R}^2/\pi\mathbb{Z}^2$, then let $f = \sin(x)\sin(y)\sin(x+y)$. The three critical points are the origin, $(\frac{\pi}{3},\frac{\pi}{3})$, and $(\frac{2\pi}{3},\frac{2\pi}{3})$. May 30, 2019 at 13:32
Warning: this is not an immersion (it has twelve Whitney-umbrella-like pinch points)
Here is a relatively simple explicit realization: the $z$ coordinate for the parametric surface$$(x,y,z)=(\sin(2u),\sin(2v),\sin(u)\sin(v)\sin(u-v));$$after an affine shift leaving $z$ unchanged the graph looks like this:
It is thus similar to the Steiner's Roman surface except that the latter has three double lines and this one has six.
I've tried to cut it to make the central monkey saddle more visible:
The implicit equation, in slightly different coordinates, is$$(X+Y+Z)(X+Y-Z)(X-Y+Z)(-X+Y+Z)=(XYZ)^2,$$
with function $X+Y+Z$:
Mathematica codes:
ParametricPlot3D[
{3Sin[2u]+4Sin[u]Sin[v]Sin[u-v],3Sin[2v]-4Sin[u]Sin[v]Sin[u-v],Sin[u]Sin[v]Sin[u-v]},
{u,0,\[Pi]},{v,0,\[Pi]},
BoxRatios->{1,1,1},
Mesh->None,
PlotPoints->150,
PlotStyle->FaceForm[Red,Cyan],
Boxed->False,Axes->False,
SphericalRegion->True
]
ParametricPlot3D[
{3Sin[2u]+4Sin[u]Sin[v]Sin[u-v],3Sin[2v]-4Sin[u]Sin[v]Sin[u-v],Sin[u]Sin[v]Sin[u-v]},
{u, 0, \[Pi]}, {v, 0, \[Pi]},
BoxRatios -> {1, 1, 1}, Mesh -> None,
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• Is it obvious that this surface has genus 1?
– j.c.
Jul 27, 2018 at 7:19
• @j.c. It's defined as a map from $\Bbb R^2/2\pi \Bbb Z^2$.
– mme
Jul 27, 2018 at 8:29
• @MikeMiller Thanks, yes. In facte even $\pi$, instead of $2\pi$ will work. I mean, it is a map from $f$ from $[0,\pi]\times[0,\pi]$ to $\mathbb R^3$ which agrees (together with all derivatives) on the boundary in a way to be extendable to the torus, i. e. $f(u,0)=f(u,\pi)$ and $f(0,v)=f(\pi,0)$ Jul 27, 2018 at 9:06
• @მამუკაჯიბლაძე I was paranoid about the $z$-coordinate and just did not check whether it was $\pi$-invariant. :D
– mme
Jul 27, 2018 at 9:11
• Sorry for misprints, $f(0,v)=f(\pi,v)$ it should be Jul 27, 2018 at 9:17
I would recommend to look at the paper (here is a free original in russian)
• Elena Kudryavtseva, Realization of smooth functions on surfaces as height functions. (Russian) Mat. Sb. 190 (1999), no. 3, 29--88; translation in Sb. Math. 190 (1999), no. 3-4, 349–405
where the structures of immersions realizing given function as a height function are described.
Let $M$ be a closed surface and $f:M\to\mathbb{R}$ be a smooth (not necessarily Morse) function having exactly $N$ critical points. Say that $f$ can be realized as a height function if there exists an immersion $j:M \to\mathbb{R}^3$ and an orthogonal projection $p:\mathbb{R}^3 \to\mathbb{R}$ to some line $l$ such that $f = p\circ j$.
Notice that in this case at each critical point $z\in\mathbb{R}^3$ the normal vector to the surface is parallel to the line $l$. Say that two immersions with the same height functions are normally equivalent if the directions of the normals to these immersed surfaces at all critical points of the height functions are the same.
Elena Kudryavtseva proved (see Theorem 1 of the paper above) that $f$ can be realized as a height function in the following cases:
1) $M$ is either a sphere or a torus;
2) $M$ is any orientable surface and $f$ is Morse, and in this case there are $C^{N/2}_{N} = \frac{N!}{(N/2)! (N/2)!}$ normally non-equivalent immersions;
3) $M$ is non-orientable, and $f$ is not necessarily Morse but has only finitely many critical points, and in this case there are $2^{N}$ normally non-equivalent immersions.
The complexity of surface immersions in 3-space has been studied in a series of papers by Nowik and collaborators. See
Nowik, Tahl. Higher-order invariants of immersions of surfaces into 3-space. Pacific J. Math. 223 (2006), no. 2, 333–347
as well as
Nowik, Tahl. Order one invariants of immersions of surfaces into 3-space. Math. Ann. 328 (2004), no. 1-2, 261–283. | 2022-05-25T09:10:00 | {
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Also equal to the magnitude of the vector cross product a × b polygon the. If you know how to solve a 2x2 determinant 2011 ; Tags area parallelogram vectors ; Home > area this... Years, 2 months ago two-dimensional plane thread starter sderosa518 ; Start date 17. You must take the magnitude of the interior angles in a two-dimensional plane in!: Below are the expressions used to find the area of a parallelogram another. A lecture video clip, board notes, an example, and a recitation.... Example, and a recitation video an answer to your question ️ find area of parallelogram. Iphone 11 Carplay Missing, Pictures Screenshots Folder, Eastern Diamondback Rattlesnake Endangered, First World Problems Meme, Cape Feare Quotes, Justin Briner Genshin Impact, " /> Find the area of the parallelogram whose adjacent sides are determined by the vectors a = i - j + 3k and b = 2i - 7j + k. asked Oct 11, 2019 in Mathematics by Radhika01 ( 63.0k points) vector algebra Let P and Q be two vectors acting simultaneously at a point and represented both in magnitude and direction by two adjacent sides OA and OD of a parallelogram OABD as shown in figure.. Let θ be the angle between P and Q and R be the resultant vector.Then, according to parallelogram law of vector addition, diagonal OB represents the resultant of P and Q. IB Maths Notes - Vectors, Lines and Planes - Area of Parallelogram Formed By Two Vectors Compute (a) The area of a parallelogram in R with vertices given by P = (1, -2,3), P = (1,3,1) and Ps= (2,1,2) (b) The volume of a parallelepiped in R with sides given by ū = (1, -2,3), u (1,3,1) and Ug = (2,1,2). Compute the cross product a × b. a × b = (,,) 30. x. axis. One thing that determinants are useful for is in calculating the area determinant of a parallelogram formed by 2 two-dimensional vectors. (1 point) Find the area of the parallelogram with vertices (4,1), (6, 6), (10, 11), and (12, 16). Parallelogram Law of Vectors explained Let two vectors P and Q act simultaneously on a particle O at an angle . Formula. Area is 2-dimensional like a carpet or an area rug. Determine the angles of each two forces. The area of a parallelogram is also equal to the magnitude of the vector cross product of two adjacent sides. Area of parallelogram formed by vectors, Online calculator. However, I keep getting the wrong answer. solution Up: Area of a parallelogram Previous: Area of a parallelogram Example 1 a) Find the area of the triangle having vertices and . I use three points to create two vectors with the same initial points and use a 2x2 determinant to compute the cross product then find it's magnitude. Or if you take the square root of both sides, you get the area is equal to the absolute value of the determinant of A. Vectors; Home > Area of a Parallelogram – Explanation & Examples; Area of a Parallelogram – Explanation & Examples . When two vectors are given: Below are the expressions used to find the area of a triangle when two vectors are known. We note that the area of a triangle defined by two vectors$\vec{u}, \vec{v} \in \mathbb{R}^3$will be half of the area defined by the resulting parallelogram of those vectors. Thanks for the help! If the . Area determinants are quick and easy to solve if you know how to solve a 2x2 determinant. This geometric Demonstration establishes that the area of a parallelogram bounded by vectors (a, c) and (b, d) is | a d-b c |. Magnitude of the vector product of the vectors equals to the area of the parallelogram, build on corresponding vectors: Therefore, to calculate the area of the parallelogram, build on vectors, one need to find the vector which is the vector product of the initial vectors, then find the magnitude of this vector. They are represented in magnitude and direction by the adjacent sides OA and OB of a parallelogram OACB drawn from a point O.Then the diagonal OC passing through O, will represent the resultant R in magnitude and direction. A parallelogram has two pairs of parallel sides with equal measures. Answer: 29. Proof: Since the cross product is defined only in 3-space, we will derive the following formula to calculate the area of a parallelogram in 2-space by taking our vectors$\vec{u} = (u_1, u_2)$and$\vec{v} = (v_1, v_2)$and placing them in$\mathbb{R}^3$, that is letting$\vec{u} = (u_1, u_2, 0)$and$\vec{v} = (v_1, v_2, 0)$. The area of a polygon is the number of square units inside the polygon. Follow • 5. (1 point) Let a = (7, 7, 6) and b = (5, 8, 4) be vectors. Us and Ūg from Part (b) form a basis for R$? Three vectors The three forces whose amplitudes are in ratio 9:10:17 act in the plane at one point to balance. Opposite sides are equal in length and opposite angles are equal in measure. Pre-University Math Help. It can be shown that the area of this parallelogram ( which is the product of base and altitude ) is equal to the length of the cross product of these two vectors. But since there is only one vector of zero length, the definition still uniquely determines the cross product.) Intuitively, it makes sense since area is a vector quantity and the formula you are using suggests that area is a scalar quantity. Below is an applet that helps illustrate how the cross product works. Find the area of the parallelogram with vertices at (-2, -4), (-13, 8), (7, 7), and (-4, 19). It is a special case of the quadrilateral. In Geometry, a parallelogram is a two-dimensional figure with four sides. In another problem, we found the area of a parallelogram whose diagonals were perpendicular using the lengths of those diagonals and the lengths of one of its sides.. We actually only needed the length of the side in order to show that the diagonals were perpendicular.Once we established that, we knew this was a special parallelogram - one which is also a rhombus. It differs from rectangle in terms of measure of angles at the corners. Then you must take the magnitude of that vector in absolute terms, hence the double modulus signs. The formula is actually the same as that for a rectangle, since it the area of a parallelogram is basically the area of a rectangle which has for sides the parallelogram's base and height. Use the sliders to see how various parallelograms can be transformed into ones of equal area with their bases on the . Is equal to the determinant of your matrix squared. I need some help using vectors to find the area of this parallelogram. Forums. x. axis toward it. (c) Do the vectors vi. Thread starter sderosa518; Start date Mar 17, 2011; Tags area parallelogram vectors; Home. The matrix made from these two vectors has a determinant equal to the area of the parallelogram. the area of the parallelogram = the magnitude of the cross product. (1 point) If a = i + j + 2 k and b = i + j + 4 k Compute the cross product a × b. a × b = i + j + k 2 A parallelogram is a 4-sided shape formed by two pairs of parallel lines. Any non-degenerate affine transformation takes a parallelogram to another parallelogram. Area Ar of a parallelogram may be calculated using different formulas. If the vectors are parallel or one vector is the zero vector, then there is not a unique line perpendicular to both $\vc{a}$ and $\vc{b}$. So the area of your parallelogram squared is equal to the determinant of the matrix whose column vectors construct that parallelogram. Justify your answer. Share. Area of a parallelogram Suppose two vectors and in two dimensional space are given which do not lie on the same line. In physics and applied mathematics, the wedge notation a ∧ b is often used (in conjunction with the name vector product), although in pure mathematics such notation is usually reserved for just the exterior product, an abstraction of the vector product to n dimensions. Co-initial vectors, coterminous vector and co-planar vectors,negative of a vector,reciprocal vectors Free vector and localized vector In a regular hexagon find which vectors are collinear, equal, coinitial, collinear but not equal. Active 1 year ago. We have a slight problem in that our vectors exist in ℝ 2, not ℝ 3, and the cross product is only defined on vectors in ℝ 3. Next Last. A parallelogram has rotational symmetry of order 2 (through 180°) (or order 4 if a square). Cite. The Area of a Triangle in 3-Space. Vectors - area of parallelogram. These two vectors form two sides of a parallelogram. Parallelogram law of addition of vectors S. sderosa518. Thus we can give the area of a triangle with the following formula: (5) Length of Cross Product = Parallelogram Area Last updated: Jan. 2nd, 2019 The length (norm) of cross product of two vectors is equal to the area of the parallelogram given by the two vectors, i.e., x. axis does not intersect the parallelogram, slide the triangular portion farthest from the . I can find the area of the parallelogram when two adjacent side vectors are given. So assuming your difficulties are with finding the correct cross product, I usually write out a "matrix" with i, j, and k in the top row, and the two vectors in the bottom two rows. Click hereto get an answer to your question ️ Find area of parallelogram determined by the vectors i+2j+3k & 3i2j+k. A triangle divides a parallelogram into two equal parts, so the area of the triangle will be given by 1/2 x ∣ A B ⃗ ∣ × ∣ A C ⃗ ∣ |\vec {AB} | \times |\vec {AC}| ∣ A B ∣ × ∣ A C ∣ × sinθ. Diamond area … b) Find the area of the parallelogram constructed by vectors and , with and . 1; 2; Next. It also contains problems and solutions. This session contains a lecture video clip, board notes, an example, and a recitation video. Then take the determinant. AREA OF PARALLELOGRAM If two sides of a parallelogram are represented by two vectors A and B, then the magnitude of their cross product will be equal to the Any line through the midpoint of a parallelogram bisects the area. 1 of 2 Go to page. Figure 11.4.3 (a) sketches the parallelogram defined by the vectors u → and v →. As the name suggests, a parallelogram is a quadrilateral formed by two pairs of parallel lines. Go. The cross product of two vectors a and b is defined only in three-dimensional space and is denoted by a × b. Ask Question Asked 3 years, 2 months ago. The area of a parallelogram is $$|\triangle|=\frac{1}{2}||D_1 \times D_2||$$ Here $\times$ denotes the cross product of the the two diagonals. Find the area of a parallelogram ABCD whose side AB and the diagonal AC are given by the vectors 3i + j + 4k and 4i + 5k respectively. Area of a parallelogram is a region covered by a parallelogram in a two-dimensional plane. But how to find the area of the parallelogram when diagonals of the parallelogram are given as \\alpha = 2i+6j-k and \\beta= 6i-8j+6k I know I've posted this about 2 times, but I'm still at a loss what to do Oct 2009 80 0. The sum of the interior angles in a quadrilateral is 360 degrees. Pre-Calculus . Determine the area of the trape; 4-gon It is true that a 4-gon whose two sides are parallel and the other two has equal length, is a parallelogram? Area of parallelogram 3D vectors. ) 30 length and opposite angles are equal in length and opposite angles are in. Squared is equal to the determinant of your matrix squared space are:. A ) sketches the parallelogram, slide the triangular portion farthest from.. 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Also equal to the magnitude of the vector cross product a × b polygon the. If you know how to solve a 2x2 determinant 2011 ; Tags area parallelogram vectors ; Home > area this... Years, 2 months ago two-dimensional plane thread starter sderosa518 ; Start date 17. You must take the magnitude of the interior angles in a two-dimensional plane in!: Below are the expressions used to find the area of a parallelogram another. A lecture video clip, board notes, an example, and a recitation.... Example, and a recitation video an answer to your question ️ find area of parallelogram. | 2021-04-16T13:45:49 | {
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http://math.stackexchange.com/questions/355740/why-is-fracfxfx-always-a-constant | Why is $\frac{f'(x)}{f(x)}$ always a constant?
Today in class we learned that for exponential functions $f(x) = b^x$ and their derivatives $f'(x)$, the ratio is always constant for any $x$. For example for $f(x) = 2^x$ and its derivative $f'(x) = 2^x \cdot \ln 2$
$$\begin{array}{c | c | c | c} x & f(x) & f'(x) & \frac{f'(x)}{f(x)}\\ \hline -1 & \frac{1}{2} & 0.346 & 0.693 \\ 0 & 1 & 0.693 & 0.693 \\ 1 & 2 & 1.38 & 0.693\\ 2 & 4 & 2.77 &0.693& \end{array}$$
So as you can see, the ratio is the same and this is true for all functions of the form $b^x$ and its derivative. So my question is, why is the ratio always constant? Is there some proof or logic behind it or is just like that? Furthermore, what's the use of knowing this?
EDIT:
It seems that I have missed the fairly simple
$$\require{cancel}\frac{f^\prime(x)}{f(x)}=\frac{\cancel{{b^x}}\ln\,b}{\cancel{{b^x}}}=\ln\,b$$
But what's the use knowing and learning this? Will this reduce a step in the future or help solve a much harder problem more easily?
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It's not entirely clear to me if you're asking for a proof or for an intuitive reason why this happens. Also note that $\ln (2)=0.6931471806\ldots$ – Git Gud Apr 9 '13 at 8:27
$$\require{cancel}\frac{f^\prime(x)}{f(x)}=\frac{\cancel{{b^x}}\ln\,b}{\cancel{{b^x}}}=\ln\,b$$ – Ｊ. Ｍ. Apr 9 '13 at 8:28
Note that the left hand side could be read as $[\ln f(x)]'$, so that $[\ln a^x]' = [x \ln a]' = \ln a$ is constant. – TMM Apr 9 '13 at 8:36
$$f=a^x$$ $$f'=\ln a \times a^x$$ So, $$f'=\ln a \times f$$ So, $$\dfrac{f'}{f}=\ln a= \text{constant}=0.693\Big|_{a=2}$$
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What's the use of knowing this? Or in other words, why am I learning it? Is it going to make a future problem much easier to solve, what's its significance? – gekkostate Apr 9 '13 at 8:32
There's no significance, if you have function as yours. And you get to calculate single differentiation of $\dfrac{f'}{f}$, its just $0$, you don't need to think hard for this. – user63477 Apr 9 '13 at 8:33
@gekko, for now, treat it as a nice property of the exponential function. Later, when you get to solving differential equations, this fact might prove useful... – Ｊ. Ｍ. Apr 9 '13 at 8:34
Yes , DE of a exponential function is $f'=Kf$. – Mr.ØØ7 Apr 9 '13 at 8:36
@Inceptio The only reason I'm thinking "hard" about this is because we spent about 75 mins learning this and I had no idea why and teacher didn't specify why this important or useful but I know why and I know it's significance - helps with differential equations. – gekkostate Apr 9 '13 at 8:47
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$f'/f$ is called the Logarithmic Derivative of $f$.
By the Chain Rule $$\frac{f'}{f}=\frac{\mathrm{d}}{\mathrm{d}x}\log(f(x))$$ Since $\log(b^x)=x\log(b)$ and $\frac{\mathrm{d}}{\mathrm{d}x}x\log(b)=\log(b)$, we get that $f'/f$ is constant.
- | 2014-03-11T13:06:43 | {
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http://math.stackexchange.com/questions/589108/reward-for-tossing-a-tail-followed-by-head | Reward for tossing a tail followed by head .
A sequence of independent tosses of a biased coin at times $t = 0, 1, 2,...$ On each toss, the probability of a ’head’ is $p$, and the probability of a ’tail’ is $1 − p$. A reward of one unit is given each time that a ’tail’ follows immediately after a ’head.’ Let $R$ be the total reward paid in times $1, 2, ..., n$ . We have to find $E[R]$ and $var(R)$ .
What I did for finding the $E[R]$ :
Let $I_k$ be the reward paid at time $k$ . We have
$E[I_k]= P(I_k =1) = P(T\ at\ time\ k\ and\ H\ at\ time\ k −1) = p(1−p)$ .
$E[R] = E[\sum_{k=0}^nI_k] = np(1-p)$
After this I am stuck . I am not able to find the variance . Please help me out .
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For the variance, we use the method of indicator random variables that you used to find the mean. We have $R=\sum_{1}^n I_k$. If we can find $E(R^2)$, we will be nearly finished.
Note that $R^2=\sum_1^n I_k^2 +2\sum_{1\le i< j\le n} I_iI_j$. You have already done the calculation of $E(\sum_1^n I_k^2)$, since $I_k^2=I_k$.
The mixed terms are more complicated. As usual the expectation of the sum is the sum of the expectations, but the expectations are not all equal.
If $j=i+1$, then $I_iI_j=0$. If $j\gt i+1$, then $I_iI_j=1$ with probability $p^2(1-p)^2$. Now it is just a matter of counting. It is more efficient to sum $p^2(1-p)^2$ over all pairs $i\lt j$, and subtract the sum of all $p^2(1-p)^2$ over the $n-1$ consecutive pairs.
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I have just one more doubt . Please explain me how $I_iI_j = 0$ when $j = i+1$ – AbKDs Dec 2 '13 at 15:44
Look for example $i=2$, $j=3$. If $I_2=1$, we had tail at $1$ and head at $2$. But then $I_3=0$, since for $I_3=1$ we need tail on $2$. Thus $I_i$ and $I_{i+1}$ can never be both $1$, so their product is always $0$. – André Nicolas Dec 2 '13 at 15:59
Thank you @André Nicolas – AbKDs Dec 2 '13 at 16:09
You are welcome. The method of indicator random variables is powerful. – André Nicolas Dec 2 '13 at 16:10
If you can't find a clever way to get the variance, the formula for the distribution is given here http://www.qbyte.org/puzzles/p145s.html and you can probably use maths to extract the variance from the distribution.
- | 2016-02-11T03:21:21 | {
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https://puzzling.stackexchange.com/questions/29299/four-indeed-is-cosmic?noredirect=1 | # Four indeed is cosmic!
This puzzle deals with positive integers in decimal representation. From every integer you can move to one or two or three other integers. The allowed moves for integer $n\ge1$ are as follows:
• You may double the number (that is, $n$ becomes $2n$).
• If the rightmost digit in the decimal representation is $4$, you may remove this rightmost digit.
• If the rightmost digit in the decimal representation is $0$, you may remove this rightmost digit.
For instance, starting with the integer $n=227$ you could make the following moves: $$227\to454\to45\to90\to9\to18\to36\to72\to144\to14\to1\to2\to4$$
Your task: Show that starting with an arbitrary integer $n\ge1$, you can eventually reach the cosmic goal integer $4$ (by repeatedly applying these three moves).
(The title of this puzzle was chosen as repercussion of the "Four is Cosmic!" puzzle.)
• I'm pretty sure the same is true for 2 or 8 as well. – Darrel Hoffman Mar 20 '16 at 17:58
• My suggestion: prove that you can get from n to n-1 in a finite number of moves, use induction down to 4, then provide special cases for 1, 2, and 3 (1 and 2 are trivial, 3 is 3, 6, 12, 24, 2, 4) – hobbs Mar 20 '16 at 19:57
Let the starting number be $n$. Consider the case where $n < 10$.
\begin{align} 1 &\to 2 \to 4 \\ 2 &\to 4 \\ 3 &\to 6 \to 12 \to 24 \to 2 \to 4\\ 4 \\ 5 &\to 10 \to 1 \to 2 \to 4 \\ 6 &\to 12 \to 24 \to 2 \to 4\\ 7 &\to 14 \to 1 \to 2 \to 4\\ 8 &\to 16 \to 32 \to 64 \to 6 \to 12 \to 24 \to 2 \to 4\\ 9 &\to 18 \to 36 \to 72 \to 144 \to 14 \to 1 \to 2 \to 4 \end{align}
Now consider general $n > 0$, not ending with the digit $0$ (remove all trailing zeros before beginning).
Let a step from $n$ be defined as the sequence to get from $n$ to a number $m$ for which exactly one trailing digit is removed. For convenience, we write this functionally as $s(n) = m$. The next step is a step from $m$, and we can extend this to a sequence of steps.
Now, let $n = 10k + d$, where $k$ is an integer and $0 \leq d < 10$.
If $d = 0$, we remove the trailing 0 to get $s(n) = \frac{n}{10}$ .
If $d \in \{1,2,3,6,8\}$, first consider $d=1$. Since $s(n) = \frac{(10k+1)4-4}{10} = 4k$, $s(n)$ is even. Similarly for the other $d$, in each case $s(n)$ is even. Since $n$ is doubled at most 3 times, $s(n) \leq \frac{8n}{10}$ .
If $d=4$, then $s(n) = \frac{(10k+4)-4}{10} = k$, i.e. $s(n) < \frac{n}{10}$ .
If $d=5$, then $s(n) = \frac{(10k+5)2}{10} = 2k+1 = \frac{n}{5}$.
If $d=7$, then $s(n) = \frac{(10k+7)2-4}{10} = 2k+1$, i.e. $s(n) < \frac{n}{5}$ .
If $d=9$, we need more steps. Consider $n = 10k+9$ . We double $n$ four times to get a trailing 4, so $s(n) = \frac{(10k+9)16-4}{10} = 16k+14$, i.e. $s(n) < 1.6n$. The last digit of $s(n)$ is $16k+14$ mod $10$, for which we only need to consider $0 \leq k \leq 9$. Trying all $k$ from 0 to 9, we find $s(n)$ ends with $4,0,6,2,8,4,0,6,2,8$ respectively. In particular, $s(n)$ never ends with 9.
Apply the above iteratively. Recall $s(n) < 1.6n$, and $s(n) \mod 10 \in \{0,2,4,6,8\}$ .
If $s(n)$ ends with 0 or 4, then $s(s(n)) \leq \frac{s(n)}{10} < 0.16n$, so $s(s(n)) < n$ .
Otherwise $s(n)$ ends with 2, 6 or 8.
Let $m=s(s(n))$. Then $m$ is even and $m \leq 0.8s(n) < 1.28n$.
If $m$ ends with 0 or 4, then $s(m) \leq \frac{m}{10} < 0.128n$, so $s(m) < n$.
Otherwise $s(m)$ is even and $s(m) \leq 0.8m < 1.024n$, so $s(s(m)) \leq 0.8s(m) < n$.
In every case, there is a sequence of steps taking $n$ to an integer strictly less than $n$, unless we arrive at 4, in which case we've arrived. Call this sequence a jump. Since $n \neq 0$, we never jump to zero.
The jumps reduce $n$ monotonically, so we eventually arrive at a single digit, from which the table above shows that . By inspection of the table above, steps from single digits the sequence always terminates at 4.
QED
• Just as I said to the other answer, I think it should also be proven that "So from then on, the number always decreases". I don't believe this is sufficient proof. Specifically, when $\frac{16n-4}{10}$ ends in an $8$ you are able to reduce the resulting number by doing $\frac{8n-4}{10}$ but that is still larger than the initial $n$ and at this point you haven't proven that this new number doesn't end with a 9. I know it couldn't be but I think you haven't proven that with this alone – Ivo Beckers Mar 20 '16 at 14:47
• @IvoBeckers I was assuming the "never ends with 9" comment was understood to automatically invoke the earlier $n \to \frac{8n}{10}$ process. I've now made this explicit. It doesn't matter that the number is higher than the starting $n$, so long as it decreases monotonically at some point. – Lawrence Mar 20 '16 at 14:50
• What I'm trying to say is. You haven't proven that the not-ending-with-9 process can't produce numbers ending with a 9. For all we know you have a number ending with a 9, gets turned to a number ending in an 8, gets turned to a number with a 9, and so on forever, and this process does not reduce the number – Ivo Beckers Mar 20 '16 at 14:55
• @IvoBeckers Ok, let me have a look at this. – Lawrence Mar 20 '16 at 15:00
• @IvoBeckers The proof has become less elegant, but I think all cases are covered. I wonder whether it would have been simpler to start with 4 and show that we can reach all integers by running the process backwards. – Lawrence Mar 20 '16 at 16:01
Removing rightmost digit is like dividing the number by $10$ for $0$ and more than $10$ for $4$.
If the last digit of the number you have taken is $0$,$4$ you just divide your number by $10$.
If the last digit is $2,7,5$. You multiply your number by $2$ and divide by $10$ that makes dividing by $5$ after $2$ moves.
If the last digit is $1$,$6$ you multiply twice by $2$ then divide by $10$ that makes dividing by $2.5$, still your number gets less.
If the last digit is $3$,$8$ you need to multiply third time by $2$ then divide $10$ that makes dividing by $1.25$, still your number gets lesser and becomes divided by $1.25$ at the end.
The only time your number gets bigger when your last digit is $9$, that makes your number $1.6$ times bigger than before after removing the last digit. But since the next number's last digit (after removing $4$, which becomes last digit after doubling it $4$ times) cannot be $9$ ($x9\times 16$'s second digit cannot be $9$) the mentioned removing number above will make your number to $4$ whatsoever.
So whatever the number you have taken, there will be a solution.
• I came up with the same conclusions as you although I'm still not convinced yet about the 9. It's true that $x9 \cdot 16$ can't have a 9 in the second digit but it could be an 8 and in that case you can only divide by $1.25$ and $1.6 / 1.25$ is still bigger than one and after that you might possibly get another 9 but I'm not sure. I think you still need to expand a bit on the possibilites when the last digit is 9 – Ivo Beckers Mar 20 '16 at 14:11
• @IvoBeckers the last digit cannot be 9 again, that's the point. – Oray Mar 20 '16 at 14:19
• No I mean after doing the next step again. Let's say you have a 9 at the end. you multiply by 1.6 and get a 8 at the end. you then can divide by 1.25. At that point your number didn't get smaller and at this point I'm not sure if you could have a 9 again – Ivo Beckers Mar 20 '16 at 14:22
• @IvoBeckers it does not matter, i see what u meant but u cannot have 9 as the last digit ever after the last 9 because 94 or 90 as the last 2 digits is impossible after the first removal – Oray Mar 20 '16 at 14:24
• I think you're right but I also think that it should be proven – Ivo Beckers Mar 20 '16 at 14:40 | 2019-10-23T09:38:03 | {
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http://math.stackexchange.com/questions/170128/roots-of-a-polynomial-mod-n | Roots of a polynomial mod $n$
Let $n=n_1n_2\ldots n_k$ where $n_i$ are pairwise relatively prime. Prove for any polynomial $f$ the number of roots of the equation $f(x)\equiv 0\pmod n$ is equal to the product of the number of roots of each of the equations $f(x)\equiv 0\pmod{n_1}$, $f(x)\equiv 0\pmod{n_2}$, $\ldots$, $f(x)\equiv 0\pmod{n_k}$.
I encountered this problem while reading a chapter on RSA encryption. It has something to do with the Chinese remainder theorem but I can't see precisely how to use it.
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It's exactly the CRT. Each root mod $n$ gives a root to each of the other congruences, and, by CRT, any choice of roots of the congruences mod the $n_i$ gives a single root of the congruence mod $n$. – Gerry Myerson Jul 13 '12 at 1:21
Could you elaborate how that fits into "equal to the product of the number of roots" – user782220 Jul 13 '12 at 1:28
That's just the multiplication principle: en.wikipedia.org/wiki/Rule_of_product – Micah Jul 13 '12 at 2:46
The Chinese Remainder Theorem says that if $n_1,\ldots,n_k$ are pairwise coprime, and $a_1,\ldots,a_k$ are any integers, then there is a solution to the system of congruences \begin{align*} x &\equiv a_1\pmod{n_1}\\ x&\equiv a_2\pmod{n_2}\\ &\cdots\\ x&\equiv a_k\pmod{n_k}, \end{align*} and moreover, the solution is unique modulo $n=n_1\cdots n_k$.
We show that there is a bijection between the set of integers $$A=\{a\in\mathbb{Z}\mid 0\leq a\lt n, f(a)\equiv 0\pmod{n}\}$$ and the set $$B=\{(a_1,\ldots,a_k)\in\mathbb{Z}^k\mid 0\leq a_i\lt n_i, f(a_i)\equiv 0\pmod{n_i},\text{ for }i=1,\ldots,k\}.$$
Suppose first that $a$ is a solution to $f(x)\equiv 0\pmod{n}$. Since $n_i|n$ for each $i$, then $f(a)\equiv 0\pmod{n}$ implies $f(a)\equiv 0\pmod{n_i}$. Hence, $(a\bmod n_1,a\bmod n_2,\ldots,a\bmod n_k)$ is a $k$-tuple of integers whose $i$th entry is a solution to $f(x)\equiv 0\pmod{n_i}$, and satisfies $0\leq a_i\lt n_i$.
Conversely, suppose that $(a_1,\ldots,a_k)$ is a tuple such that $f(a_i)\equiv 0\pmod{n_i}$ for each $i$. Then, by the Chinese Remainder Theorem, there is a unique $a$, $0\leq a\lt n$, such that $a\equiv a_i\pmod{n_i}$ for each $i$. Therefore, $f(a)\equiv f(a_i)\equiv 0\pmod{n_i}$ for $i=1,\ldots,n$. Since the $n_i$ are pairwise coprime and $n_i|f(a)$ for all $i$, then $n=n_1\cdots n_k|f(a)$, so $f(a)\equiv 0\pmod{n}$. That is, each element of $A$ yields an element of $B$.
It is now easy to verify that the maps we have defined $A\to B$ and $B\to A$ are inverses of each other (by the uniqueness clause of the Chinese Reainder Theorem), yielding the desired bijection: $|A|=|B|$.
Note that $A$ is the set of all solutions to $f(x)\equiv0\pmod{n}$, whereas $B$ is the cartesian product of the sets of solutions of $f(x)\equiv 0\pmod{n_i}$; so $|A|$, the number of solutions to $f(x)\equiv 0\pmod{n}$, is the same as the product of the number of solutions of $f(x)\equiv 0\pmod{n_i}$ for $i=1,\ldots,k$.
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Beat me by 50 seconds. – Charles Jul 13 '12 at 1:39
Each choice of roots mod the $n_i$ corresponds to a solution of the equation mod $n$. The CRT says that, for each choice of the $n_i$ there is just one solution mod $n$. Since they must all be distinct, that gives exactly one root of the equation for each distinct choice of roots to the $n_i$. But these are independent so the number is just the product of the number of roots mod each $n_i$.
Hint $\$ By CRT, $\rm\:(n_1,n_2)= 1\:\Rightarrow\ \exists$ isomorphism $\rm\,h: \Bbb Z/n_1n_2 \to \Bbb Z/n_1 \times \Bbb Z/n_2\:$ via $\rm\:h(a) = (a_1,a_2).$
So for $\rm\,f\in \Bbb Z[x],\:$ we have: $\rm\: 0 = f(a)\iff (0,0) = h(f(a)) = f(h(a)) = f((a_1,a_2)) = (f(a_1),f(a_2)).$ | 2014-03-16T17:27:27 | {
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https://math.stackexchange.com/questions/3028927/equidistant-points-in-hyperbolic-space | # Equidistant points in hyperbolic space
I am unsure whether or not my proof for an exercise regarding equidistant points is correct.
Let l be a hyperbolic line $$\delta \gt 0$$ and \begin{align*} E= & \; \{p \in H^2 \mid d(p,l)=\delta\} \\ \end{align*} a) Show that in the hyperboloid model, E is obtained as an intersection of $$H^2$$ with two affine planes. Find explicit formulas for the two affine planes in form of \begin{align*} U= & \; \{x \in R^{2,1} \mid \langle x,v\rangle=r\} \\. \end{align*} for some v $$\in R^{2,1}, v \neq 0, r \gt 0$$.
b)Draw a sketch of l and E in the Poincaré disc model and prove that l and E meet at the boundary of the Poincaré disc.
My proof for a) is this.
Since l is a hyperbolic line, l can be written as \begin{align*} l= & \; \{x \in R^{2,1} \mid \langle x,n\rangle=0\} \\ \end{align*}
for a suitable space-like vector n.
We may define the planes: \begin{align*} U_1= & \; \{x \in R^{2,1} \mid \langle x,n\rangle=sinh(\delta)\}~ \text{and}\\ U_2= & \; \{x \in R^{2,1} \mid \langle x,-n\rangle=sinh(\delta)\} \end{align*}
We may also define the curves \begin{align*} \gamma_1= U_1 \cap H^2 \\ \gamma_2= U_2 \cap H^2 \\ \end{align*}
My claim is that $$E=\gamma_1 \cup \gamma_2$$.
"$$\subseteq$$" Let p $$\in E$$. Since $$d(p,l)=\delta$$. Then $$sinh(d(p,l))=sinh(\delta)=|\langle p,n\rangle|$$. Then either \begin{align*} sinh(d(p,l))&=sinh(\delta)=\langle p,n\rangle ~ \text{or}\\ \ sinh(d(p,l))&=sinh(\delta)=-\langle p,n\rangle=\langle p,-n\rangle. \end{align*}
So either p $$\in U_1$$ or p $$\in U_2$$. This shows E $$\subseteq y_1 \cup y_2$$.
"$$\supseteq$$" Let p $$\in y_1 \cup y_2$$. Then
\begin{align*} sinh(\delta)&=\langle p,n\rangle ~ \text{or}\\ sinh(\delta)&=\langle p,-n\rangle \end{align*}
so $$sinh(\delta)=|\langle p,n\rangle|=sinh(d(p,l))$$ and d(p,l)=$$\delta$$. This shows $$\gamma_1 \cup \gamma_2 \subseteq$$ E.
So the claim is proven.
For b) I did this: Assume that the claim is false. Then l and E meet inside of the disc. In the hyperboloid model this means we find p $$\in E \cap l$$. Using the results from a) we have $$\langle p,n\rangle=0$$ and $$|\langle p,n\rangle|=sinh(\delta)$$. So $$0=sinh(\delta)$$. Since sinh is bijective we get $$\delta=0$$ which is a contradiction.
My questions are:
1) Are my proofs correct or is there something I'm overlooking?
2) I'm not completely sure how to draw the sketch. I know that the line l is represented by a circular arc that intersects the boundary of the disc orthogonally. Since E consists of two curves that have constant distance to l I know that E can be represented by two circular arcs that intersect the boundary in the same two points as l. One arc must be contained between the boundary and l and the other one must lie on the other side of l. But how can we draw this configuration so that the two curves have the same constant distance from l? | 2019-07-17T18:25:27 | {
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https://math.stackexchange.com/questions/2127342/if-the-graphs-of-fx-and-f-1x-intersect-at-an-odd-number-of-points-is | # If the graphs of $f(x)$ and $f^{-1}(x)$ intersect at an odd number of points, is at least one point on the line $y=x$?
I was reading about intersection points of $f(x)$ and $f^{-1}(x)$ in this site. (Proof: if the graphs of $y=f(x)$ and $y=f^{-1}(x)$ intersect, they do so on the line $y=x$) Then, I saw this statement was wrote by N. S.: "If the graphs of $f(x)$ and $f^{-1}(x)$ intersect at a single point, then that point lies on the line $y=x$.
It is also true that if the graphs of $f(x)$ and $f^{-1}(x)$ intersect at an odd number of points, then at least a point point lies on the line $y=x$. This follows immediately from the observation that the intersection points are symmetric with respect to that line..." I want to know whether it's true or not and if it's true how we can prove it algebraically?
My try: I tried many function and this statement was true but I can't prove it. (Or disprove.)
The argument is mainly a counting argument involving just a little algebra on the functions themselves.
The functions $f(x)$ and $f^{-1}(x)$ intersect either at a finite number of points, or an infinite number of points. If the number of intersections is infinite, it is neither odd nor even. So we only need to consider a finite number of intersections.
If $(p,q)$ is one of the intersection points, that means $q = f(p) = f^{-1}(p).$ But from $q=f(p)$ we can deduce that $f^{-1}(q) = p,$ and from $q = f^{-1}(p)$ we can deduce that $f(q) = p,$ therefore $p = f(q) = f^{-1}(q),$ that is, and the two functions also intersect at $(q,p).$
So consider the set of intersection points that are above the line $y=x.$ Suppose there are $n$ of these points, where $n \geq 0.$ For each point $(p,q)$ above the line $y=x$ (that is, where $q>p$), there is a corresponding point $(q,p)$ below the line $y=x,$ and vice versa. Hence there are $n$ points below the line $y=x.$
Let the number of intersection points on the line $y=x$ be $m.$ Then the total number of intersection points is $n$ above the line, $n$ below the line, and $m$ on the line (where $m\geq 0$), for a total of $$2n + m.$$ Now, $m$ has the same parity as $2n+m.$ If the total number of intersections $2n+m$ is odd, it follows that $m$ is odd. But zero is not odd; all non-negative odd numbers are positive. So the total number of intersections on the line $y=x$ in that case is an odd positive number. In particular, it is at least $1.$
In this proof, we never assume there are any intersection points above the line $y=x,$ nor that there are any below the line or on the line. But we show that if there are an odd number of intersection points altogether, the number of intersection points on the line is positive.
• Your proof is wonderful! But what about even intersection points ? What we can say about it ? Feb 3 '17 at 15:42
• I think if number of intersection points is even , all of the points lie on the $y=x$ line. For example $f(x) = \sqrt x$ and then $f^{-1}(x) = x^2$ , $x \ge 0$ . Therefore intersection points are $(0,0)$ and $(1,1)$. Feb 3 '17 at 17:59
• Try $f(1)=2, f(3)=0,$ and for all $x$ other than $0$ and $3,$ $f(x)=x-1.$ The intersection points are $(1,2)$ and $(2,1).$ If you restrict $f$ to continuous functions, however, I think it may only be possible to have a decreasing function with an odd number of intersections or an increasing function with all intersections on the line $y=x.$ Feb 3 '17 at 20:39
• Okay consider continuous functions . Can you provide a function that number of intersection points with inverse is even and at least one of the points isn't locate on $y=x$ ? Feb 3 '17 at 20:59
• If my thoughts about continuous functions are correct, the only choices are an odd number of intersections or all intersections on $y=x,$ so the answer would be no. I just don't have a complete proof in mind yet. (To be precise, I'm thinking about continuous functions on connected domains; if you allow a disconnected domain I think it may be possible to have an even number of intersections and none of them on $y=x.$) Feb 3 '17 at 21:08 | 2021-10-16T06:41:31 | {
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https://math.stackexchange.com/questions/3012021/elementary-number-theory-show-that-310-equiv-1-pmod112 | # Elementary Number Theory: Show that $3^{10}\equiv 1 \pmod{11^2}$.
As the title says, I need to show $$3^{10}\equiv 1 \pmod{11^2}$$. I'm currently practicing some problems related to Fermat's little theorem and Wilson's theorem, and things were going fine but I am stumped on this problem. What I know so far is:
$$3^{10}\equiv 1 \pmod{11}$$, by Fermat's Little Theorem.
I'm not sure where to go from here, it almost looks like a lifting problem, but we have no variable so a function's derivative would always just be 0. I tried looking up how to lift constants potentially, but I couldn't really find anything (maybe I just didn't search well, so I apologize if this is the case) Any hints would be greatly appreciated. Thanks!
• $3^5 = 243 = 242 + 1$ Nov 24 '18 at 20:07
• @Will Lucky power choice, but it's still very quick even without luck if we exploit the Binomial Theorem, as I explain in my answer. Nov 25 '18 at 1:32
There is no lifting here. You are right about that. All what you need, as I can see, is to calculate this manually modulo $$121$$ instead of being misled by $$11^2$$. For the calculation, it turns out to be not so hard to follow it up.
As all powers of $$3$$ below $$5$$ are lower than $$121$$, notice that $$3^5$$ is $$1$$ more $$242$$ which is nothing but $$2*121$$. What this essentially tells you?
Now, as we noticed that:
$$3^5 \equiv 1 \pmod {121}$$
What can this tell us about $$3^{10}$$??
• Tells us that $(3^5)^2 \equiv 1$ mod $121$. Thank you for the answer :) Guess I over-looked this one. Nov 24 '18 at 20:41
• You are more than welcome :) Nov 24 '18 at 20:43
Note that $$3^2=11-2$$
Then $$3^{10}=(11-2)^5= 11^5- \dots +\binom 51\times 11\times 2^4-2^5\equiv 880-32 \bmod 121$$
using the binomial expansion, and simply $$848=7\times 121 +1$$
Other solutions are slicker in this case, but the arithmetic here is pretty simple, and when you are looking at the square of a prime as the modulus most of the terms in the binomial expansion will simply drop out.
Below are $$\,4\,$$ simple ways to compute it using about $$10$$ seconds of mental arithmetic, by using the Binomial Theorem, which reduces to the first $$\,2\,$$ terms by $$\,11^{\large 2+k}\!\equiv 0\pmod{\!11^{\large 2}},\:$$ i.e.
$$\!\!\bmod 11^{\large 2}\!\!:\ (a\! +\! 11b)^{\large n}\! \equiv a^{\large n}\! + \color{#0a0}{n\!\cdot\! a^{\large n-1} b}\!\cdot\! 11\equiv a^{\large n}\! + \color{#c00}c\!\cdot\! 11,\,\$$ where $$\,\ \color{#0a0}{n\cdot a^{\large n-1} b}\,\equiv\, \color{#c00}c\,\pmod{\!11}$$
$$\left[3^{\large 2}\!\!=\! -2\!+\!11\right]^{\large 5}\!\!\Rightarrow \overbrace{3^{\large 10}\!\!\equiv -2^{\large 5}+ \color{#0a0}{5\!\cdot 2^{\large 4}}\!\cdot\!11\equiv\! -32\!+\!\color{#c00}3\!\cdot\!11\equiv 1 \phantom{I^{I^{I^{I^{I^I}}}}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!}^{\Large \bmod 11^{\Large 2}\ \ \ \ \,}\ \ \,$$ by $$\,\ \overbrace{\color{#0a0}{5\cdot 2^{\large 4}}\equiv 5\cdot 5\equiv \color{#c00}3\phantom{I^{I^{I^{I^{I^I}}}}}\!\!\!\!\!\!\!\!\!\!\!\!\!}^{\Large \bmod{11}\ \ \ \ \ }\ \$$ [Mark]
$$\left[3^{\large 3}\! =\ 5\!+\!22\right]^{\large 4}\!\!\Rightarrow \color{#b0f}{3^{\large 12}} \equiv 5^{\large 4}\! + \color{#0a0}{4\!\cdot\!5^{\large 3}\!\cdot\!2}\!\cdot\! 11\equiv 5\!\cdot\!4\color{#c00}{-1}\!\cdot\!11\equiv\color{#b0f} 9\,\$$ by $$\,\ 5^{\large 3}\!\equiv 4,\ \color{#0a0}{4(4)2}\equiv\color{#c00}{-1}$$
$$\left[3^{\large 4}\! =\, 4\!+\!77\right]^{\large 3}\!\Rightarrow \color{#b0f}{3^{\large 12}}\! \equiv 4^{\large 3}\!+\color{#0a0}{3\!\cdot\!4^{\large 2}\!\cdot\!7}\!\cdot\! 11\equiv\ 64\,\color{#c00}{-5}\!\cdot\!11\equiv\color{#b0f} 9\ \,$$ by $$\,\ \color{#0a0}{(3\!\cdot\! 7)4^{\large 2}}\equiv \color{#c00}{(-1)5}$$
$$\left[3^{\large 5}\!=1\!+\!242\right]^{\large 2}\!\!\!\Rightarrow 3^{\large 10}\!\equiv 1^{\large 2}\!\!+\! \color{#0a0}{2\!\cdot\! 1\!\cdot\! 22}\cdot 11\equiv \ \ 1+ \color{#c00}0\cdot 11\equiv 1\ \,$$ by $$\ \ \color{#0a0}{2\cdot 22}\equiv \color{#c00}{0}\quad$$ [Will, Maged]
Note $$\,\color{#b0f}{3^{\large 12}\!\equiv 9}\,\Rightarrow\, 3^{\large 10}\!\equiv 1\pmod{\!11^{\large 2}}\,$$ by $$\,3\,$$ is invertible (so cancellable), by $$\,\gcd(3,11^2)=1$$
The other $$2$$ answers posted are essentially equivalent to one of the above cases, as [Annotated]. | 2021-10-18T21:11:01 | {
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https://casmusings.wordpress.com/tag/quadratic/ | # Tag Archives: quadratic
## Quadratics + Tangent = ???
Here’s a very pretty problem I encountered on Twitter from Mike Lawler 1.5 months ago.
I’m late to the game replying to Mike’s post, but this problem is the most lovely combination of features of quadratic and trigonometric functions I’ve ever encountered in a single question, so I couldn’t resist. This one is well worth the time for you to explore on your own before reading further.
My full thoughts and explorations follow. I have landed on some nice insights and what I believe is an elegant solution (in Insight #5 below). Leading up to that, I share the chronology of my investigations and thought processes. As always, all feedback is welcome.
WARNING: HINTS AND SOLUTIONS FOLLOW
Investigation #1:
My first thoughts were influenced by spoilers posted as quick replies to Mike’s post. The coefficients of the underlying quadratic, $A^2-9A+1=0$, say that the solutions to the quadratic sum to 9 and multiply to 1. The product of 1 turned out to be critical, but I didn’t see just how central it was until I had explored further. I didn’t immediately recognize the 9 as a red herring.
Basic trig experience (and a response spoiler) suggested the angle values for the tangent embedded in the quadratic weren’t common angles, so I jumped to Desmos first. I knew the graph of the overall given equation would be ugly, so I initially solved the equation by graphing the quadratic, computing arctangents, and adding.
Insight #1: A Curious Sum
The sum of the arctangent solutions was about 1.57…, a decimal form suspiciously suggesting a sum of $\pi/2$. I wasn’t yet worried about all solutions in the required $[0,2\pi ]$ interval, but for whatever strange angles were determined by this equation, their sum was strangely pretty and succinct. If this worked for a seemingly random sum of 9 for the tangent solutions, perhaps it would work for others.
Unfortunately, Desmos is not a CAS, so I turned to GeoGebra for more power.
Investigation #2:
In GeoGebra, I created a sketch to vary the linear coefficient of the quadratic and to dynamically calculate angle sums. My procedure is noted at the end of this post. You can play with my GeoGebra sketch here.
The x-coordinate of point G is the sum of the angles of the first two solutions of the tangent solutions.
Likewise, the x-coordinate of point H is the sum of the angles of all four angles of the tangent solutions required by the problem.
Insight #2: The Angles are Irrelevant
By dragging the slider for the linear coefficient, the parabola’s intercepts changed, but as predicted in Insights #1, the angle sums (x-coordinates of points G & H) remained invariant under all Real values of points A & B. The angle sum of points C & D seemed to be $\pi/2$ (point G), confirming Insight #1, while the angle sum of all four solutions in $[0,2\pi]$ remained $3\pi$ (point H), answering Mike’s question.
The invariance of the angle sums even while varying the underlying individual angles seemed compelling evidence that that this problem was richer than the posed version.
Insight #3: But the Angles are bounded
The parabola didn’t always have Real solutions. In fact, Real x-intercepts (and thereby Real angle solutions) happened iff the discriminant was non-negative: $B^2-4AC=b^2-4*1*1 \ge 0$. In other words, the sum of the first two positive angles solutions for $y=(tan(x))^2-b*tan(x)+1=0$ is $\pi/2$ iff $\left| b \right| \ge 2$, and the sum of the first four solutions is $3\pi$ under the same condition. These results extend to the equalities at the endpoints iff the double solutions there are counted twice in the sums. I am not convinced these facts extend to the complex angles resulting when $-2.
I knew the answer to the now extended problem, but I didn’t know why. Even so, these solutions and the problem’s request for a SUM of angles provided the insights needed to understand WHY this worked; it was time to fully consider the product of the angles.
Insight #4: Finally a proof
It was now clear that for $\left| b \right| \ge 2$ there were two Quadrant I angles whose tangents were equal to the x-intercepts of the quadratic. If $x_1$ and $x_2$ are the quadratic zeros, then I needed to find the sum A+B where $tan(A)=x_1$ and $tan(B)=x_2$.
From the coefficients of the given quadratic, I knew $x_1+x_2=tan(A)+tan(B)=9$ and $x_1*x_2=tan(A)*tan(B)=1$.
Employing the tangent sum identity gave
$\displaystyle tan(A+B) = \frac{tan(A)+tan(B)}{1-tan(A)tan(B)} = \frac{9}{1-1}$
and this fraction is undefined, independent of the value of $x_1+x_2=tan(A)+tan(B)$ as suggested by Insight #2. Because tan(A+B) is first undefined at $\pi/2$, the first solutions are $\displaystyle A+B=\frac{\pi}{2}$.
Insight #5: Cofunctions reveal essence
The tangent identity was a cute touch, but I wanted something deeper, not just an interpretation of an algebraic result. (I know this is uncharacteristic for my typically algebraic tendencies.) The final key was in the implications of $tan(A)*tan(B)=1$.
This product meant the tangent solutions were reciprocals, and the reciprocal of tangent is cotangent, giving
$\displaystyle tan(A) = \frac{1}{tan(B)} = cot(B)$.
But cotangent is also the co-function–or complement function–of tangent which gave me
$tan(A) = cot(B) = tan \left( \frac{\pi}{2} - B \right)$.
Because tangent is monotonic over every cycle, the equivalence of the tangents implied the equivalence of their angles, so $A = \frac{\pi}{2} - B$, or $A+B = \frac{\pi}{2}$. Using the Insights above, this means the sum of the solutions to the generalization of Mike’s given equation,
$(tan(x))^2+b*tan(x)+1=0$ for x in $[0,2\pi ]$ and any $\left| b \right| \ge 2$,
is always $3\pi$ with the fundamental reason for this in the definition of trigonometric functions and their co-functions. QED
Insight #6: Generalizing the Domain
The posed problem can be generalized further by recognizing the period of tangent: $\pi$. That means the distance between successive corresponding solutions to the internal tangents of this problem is always $\pi$ each, as shown in the GeoGebra construction above.
Insights 4 & 5 proved the sum of the angles at points C & D was $\pi/2$. Employing the periodicity of tangent, the x-coordinate of $E = C+\pi$ and $F = D+\pi$, so the sum of the angles at points E & F is $\frac{\pi}{2} + 2 \pi$.
Extending the problem domain to $[0,3\pi ]$ would add $\frac{\pi}{2} + 4\pi$ more to the solution, and a domain of $[0,4\pi ]$ would add an additional $\frac{\pi}{2} + 6\pi$. Pushing the domain to $[0,k\pi ]$ would give total sum
$\displaystyle \left( \frac{\pi}{2} \right) + \left( \frac{\pi}{2} +2\pi \right) + \left( \frac{\pi}{2} +4\pi \right) + \left( \frac{\pi}{2} +6\pi \right) + ... + \left( \frac{\pi}{2} +2(k-1)\pi \right)$
Combining terms gives a general formula for the sum of solutions for a problem domain of $[0,k\pi ]$
$\displaystyle k * \frac{\pi}{2} + \left( 2+4+6+...+2(k-1) \right) * \pi =$
$\displaystyle = k * \frac{\pi}{2} + (k)(k-1) \pi =$
$\displaystyle = \frac{\pi}{2} * k * (2k-1)$
For the first solutions in Quadrant I, $[0,\pi]$ means k=1, and the sum is $\displaystyle \frac{\pi}{2}*1*(2*1-1) = \frac{\pi}{2}$.
For the solutions in the problem Mike originally posed, $[0,2\pi]$ means k=2, and the sum is $\displaystyle \frac{\pi}{2}*2*(2*2-1) = 3\pi$.
I think that’s enough for one problem.
APPENDIX
My GeoGebra procedure for Investigation #2:
• Graph the quadratic with a slider for the linear coefficient, $y=x^2-b*x+1$.
• Label the x-intercepts A & B.
• The x-values of A & B are the outputs for tangent, so I reflected these over y=x to the y-axis to construct A’ and B’.
• Graph y=tan(x) and construct perpendiculars at A’ and B’ to determine the points of intersection with tangent–Points C, D, E, and F in the image below
• The x-intercepts of C, D, E, and F are the angles required by the problem.
• Since these can be points or vectors in Geogebra, I created point G by G=C+D. The x-intercept of G is the angle sum of C & D.
• Likewise, the x-intercept of point H=C+D+E+F is the required angle sum.
## Quadratics and CAS
I’m returning to my ‘blog after a prolonged absence. My next several posts will explore ideas I shared and learned at the USACAS-10 conference hosted at Hawken School last weekend.
Finding equations for quadratic functions has long been a staple of secondary mathematics. Historically, students are given information about some points on the graph of the quadratic, and efficient students typically figure out which form of the equation to use. This post from my Curious Quadratics a la CAS presentation explores a significant mindset change that evolves once computer algebra enters the learning environment.
HISTORIC BACKGROUND:
Students spend lots of time (too much?) learning how to manipulate algebraic expressions between polynomial forms. Whether distributing, factoring, or completing the square, generations of students have spent weeks changing quadratic expressions between three common algebraic forms
Standard: $y=a*x^2+b*x+c$
Factored: $y=a*(x-x_1)(x-x_2)$
Vertex: $y=a*(x-h)^2+k$
many times without ever really knowing why. I finally grasped deeply the reason for this about 15 years ago in a presentation by Bernhard Kutzler of Austria. Poorly paraphrasing Bernhard’s point, he said in more elegant phrasing,
We change algebraic forms of functions because different forms reveal different properties of the function and because no single form reveals everything about a function.
While any of what follows could be eventually derived from any of the three quadratic forms, in general the Standard Form explicitly gives the y-intercept, Factored Form states x-intercepts, and Vertex Form “reveals” the vertex (duh). When working without electronic technology, students can gain efficiency by choosing to work with a quadratic form that blends well with given information. To demonstrate this, here’s an example of the differences between non-tech and CAS approaches.
COMPARING APPROACHES:
For an example, determine all intercepts and the vertex of the parabola that passes through $(10, 210)$$(5, 40)$, and $(-2, -30)$.
NON-TECH: Not knowing anything about the points, use Standard form, plug in all three points, and solve the resulting system.
$y=a*x^2+b*x+c$
$210 = 100a+10b+c$
$40 = 25a+5b+c$
$-30 = 4a-2b+c$
Use any approach you want to solve this 3×3 system to get $a=2$, $b=4$, and $c=-30$.
That immediately gives the y-intercept at -30. Factoring to $y=2(x+5)(x-3)$ or using the Quadratic Formula reveals the x-intercepts at -5 and 3. Completing the square or leveraging symmetry between the known x-intercepts gives the vertex at $(-1,-32)$. Some less-confident students find all of the hinted-at manipulations in this paragraph burdensome or even daunting.
CAS APPROACH: By declaring the form you want/need, you can directly get any information you require. In the next three lines on my Nspire CAS, notice that the only difference in my commands is the form of the equation I want in the first part of the command. Also notice my use of lists to simplify substitution of the given points.
The last line’s output gave two solutions only because I didn’t specify which of x1 and x2 was the larger x-intercept, so my Nspire gave me both.
The -30 y-intercept appears in the first output, the vertex in the second, and the x-intercepts in the third. Any information is equally simple to obtain.
CONCLUSION:
In the end, it’s all about knowing what you want to find and how to ask questions of the tools you have available. Understanding the algebra behind the solutions is important, but endless repetition of these tasks is not helpful, even though it may be easy to test.
Instead, focus on using what you know, explore for patterns, and ask good questions. …And teach with a CAS!
## Best Algebra 2 Lab Ever
This post shares what I think is one of the best, inclusive, data-oriented labs for a second year algebra class. This single experiment produces linear, quadratic, and exponential (and logarithmic) data from a lab my Algebra 2 students completed this past summer. In that class, I assigned frequent labs where students gathered real data, determined models to fit that data, and analyzed goodness of the models’ fit to the data. I believe in the importance of doing so much more than just writing an equation and moving on.
For kicks, I’ll derive an approximation for the coefficient of gravity at the end.
THE LAB:
On the way to school one morning last summer, I grabbed one of my daughters’ “almost fully inflated” kickballs and attached a TI CBR2 to my laptop and gathered (distance, time) data from bouncing the ball under the Motion Sensor. NOTE: TI’s CBR2 can connect directly to their Nspire and TI84 families of graphing calculators. I typically use computer-based Nspire CAS software, so I connected the CBR via my laptop’s USB port. It’s crazy easy to use.
One student held the CBR2 about 1.5-2 meters above the ground while another held the ball steady about 20 cm below the CBR2 sensor. When the second student released the ball, a third clicked a button on my laptop to gather the data: time every 0.05 seconds and height from the ground. The graphed data is shown below. In case you don’t have access to a CBR or other data gathering devices, I’ve uploaded my students’ data in this Excel file.
Remember, this is data was collected under far-from-ideal conditions. I picked up a kickball my kids left outside on my way to class. The sensor was handheld and likely wobbled some, and the ball was dropped on the well-worn carpet of our classroom floor. It is also likely the ball did not remain perfectly under the sensor the entire time. Even so, my students created a very pretty graph on their first try.
For further context, we did this lab in the middle of our quadratics unit that was preceded by a unit on linear functions and another on exponential and logarithmic functions. So what can we learn from the bouncing ball data?
LINEAR 1:
While it is very unlikely that any of the recorded data points were precisely at maximums, they are close enough to create a nice linear pattern.
As the height of a ball above the ground helps determine the height of its next bounce (height before –> energy on impact –> height after), the eight ordered pairs (max height #n, max height #(n+1) ) from my students’ data are shown below
This looks very linear. Fitting a linear regression and analyzing the residuals gives the following.
The data seems to be close to the line, and the residuals are relatively small, about evenly distributed above and below the line, and there is no apparent pattern to their distribution. This confirms that the regression equation, $y=0.673x+0.000233$, is a good fit for the = height before bounce and = height after bounce data.
NOTE: You could reasonably easily gather this data sans any technology. Have teams of students release a ball from different measured heights while others carefully identify the rebound heights.
The coefficients also have meaning. The 0.673 suggests that after each bounce, the ball rebounded to 67.3%, or 2/3, of its previous height–not bad for a ball plucked from a driveway that morning. Also, the y-intercept, 0.000233, is essentially zero, suggesting that a ball released 0 meters from the ground would rebound to basically 0 meters above the ground. That this isn’t exactly zero is a small measure of error in the experiment.
EXPONENTIAL:
Using the same idea, consider data of the form (x,y) = (bounce number, bounce height). the graph of the nine points from my students’ data is:
This could be power or exponential data–something you should confirm for yourself–but an exponential regression and its residuals show
While something of a pattern seems to exist, the other residual criteria are met, making the exponential regression a reasonably good model: $y = 0.972 \cdot (0.676)^x$. That means bounce number 0, the initial release height from which the downward movement on the far left of the initial scatterplot can be seen, is 0.972 meters, and the constant multiplier is about 0.676. This second number represents the percentage of height maintained from each previous bounce, and is therefore the percentage rebound. Also note that this is essentially the same value as the slope from the previous linear example, confirming that the ball we used basically maintained slightly more than 2/3 of its height from one bounce to the next.
And you can get logarithms from these data if you use the equation to determine, for example, which bounces exceed 0.2 meters.
So, bounces 1-4 satisfy the requirement for exceeding 0.20 meters, as confirmed by the data.
A second way to invoke logarithms is to reverse the data. Graphing x=height and y=bounce number will also produce the desired effect.
Each individual bounce looks like an inverted parabola. If you remember a little physics, the moment after the ball leaves the ground after each bounce, it is essentially in free-fall, a situation defined by quadratic movement if you ignore air resistance–something we can safely assume given the very short duration of each bounce.
I had eight complete bounces I could use, but chose the first to have as many data points as possible to model. As it was impossible to know whether the lowest point on each end of any data set came from the ball moving up or down, I omitted the first and last point in each set. Using (x,y) = (time, height of first bounce) data, my students got:
What a pretty parabola. Fitting a quadratic regression (or manually fitting one, if that’s more appropriate for your classes), I get:
Again, there’s maybe a slight pattern, but all but two points are will withing 0.1 of 1% of the model and are 1/2 above and 1/2 below. The model, $y=-4.84x^2+4.60x-4.24$, could be interpreted in terms of the physics formula for an object in free fall, but I’ll postpone that for a moment.
LINEAR 2:
If your second year algebra class has explored common differences, your students could explore second common differences to confirm the quadratic nature of the data. Other than the first two differences (far right column below), the second common difference of all data points is roughly 0.024. This raises suspicions that my student’s hand holding the CBR2 may have wiggled during the data collection.
Since the second common differences are roughly constant, the original data must have been quadratic, and the first common differences linear. As a small variation for each consecutive pair of (time, height) points, I had my students graph (x,y) = (x midpoint, slope between two points):
If you get the common difference discussion, the linearity of this graph is not surprising. Despite those conversations, most of my students seem completely surprised by this pattern emerging from the quadratic data. I guess they didn’t really “get” what common differences–or the closely related slope–meant until this point.
Other than the first three points, the model seems very strong. The coefficients tell an even more interesting story.
GRAVITY:
The equation from the last linear regression is $y=4.55-9.61x$. Since the data came from slope, the y-intercept, 4.55, is measured in m/sec. That makes it the velocity of the ball at the moment (t=0) the ball left the ground. Nice.
The slope of this line is -9.61. As this is a slope, its units are the y-units over the x-units, or (m/sec)/(sec). That is, meters per squared second. And those are the units for gravity! That means my students measured, hidden within their data, an approximation for coefficient of gravity by bouncing an outdoor ball on a well-worn carpet with a mildly wobbly hand holding a CBR2. The gravitational constant at sea-level on Earth is about -9.807 m/sec^2. That means, my students measurement error was about $\frac{9.807-9.610}{9.807}=2.801%$. And 2.8% is not a bad measurement for a very unscientific setting!
CONCLUSION:
Whenever I teach second year algebra classes, I find it extremely valuable to have students gather real data whenever possible and with every new function, determine models to fit their data, and analyze the goodness of the model’s fit to the data. In addition to these activities just being good mathematics explorations, I believe they do an excellent job exposing students to a few topics often underrepresented in many secondary math classes: numerical representations and methods, experimentation, and introduction to statistics. Hopefully some of the ideas shared here will inspire you to help your students experience more.
## CAS Presentations at USACAS-9
I had two presentations at last Saturday’s USACAS-9 conference at Hawken School in Cleveland, OH. Following are outline descriptions of the two sessions with links to the PowerPoint, pdf, and .tns files I used. I’m also adding all of this information to the Conference Presentations tab of this ‘blog.
Powerful Student Proofs
This session started with a brief introduction to a lab that first caught my eye at the first USACAS conference years ago.
You know how the graph of $y=ax^2+bx+c$ behaves when you vary a and c, but what happens when you change b?
I ‘blogged on this problem here and here. In the session, we used TI-Nspire file QuadExplore.
Next, we explored briefly the same review of trigonometric and polar graphs not as static parent functions under static transformations, but as dynamic curves oscillating between their ceilings and floors. In the session, we used TI-Nspire file Intro Polar.
Having a complete grasp of polar graphs of limacons, cardioids, rose curves, and hybrids of these, I investigated what would happen for curves of the family $r=cos \left( \frac{\theta}{k} \right)$. Curiously, for $k=3$, I encountered a curve that looked like a horizontal translation of limacons–something that just shouldn’t happen within polar coordinates.
One of my former students, Sara, used a CAS to convert a polar curve to Cartesian, translate the curve, and convert back to polar. She then identified and solved a trig identity to confirm what the graph suggested. A complete description of Sara’s proof is below. I originally ‘blogged on Sara’s work here which was a much more elegant solution to the problem than my initial attempt. It’s always cool when a student’s work is better than her teacher’s! I used TI-Nspire file Polar Fractions in Saturday’s session.
The last example presented itself when I created a document to model the family of conic curves resulting from manipulating the coefficients of $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$. After I created dynamic points for the foci of the conics, something unusual happened when the E parameter for horizontal ellipses and hyperbolas varied.
The foci for hyperbolas followed an ellipse, and the locus of elliptical foci appeared to be a hyperbola. Another former student, Lilly, proved this property to be true. A detailed explanation of Lilly’s proof is below. We were fortunate to have Lilly’s work published in the Mathematics Teacher in May, 2014.
To demonstrate this final part of the session, I used TI-Nspire file Hidden Conic Behavior.
Here is my PowerPoint file for Powerful Student Proofs. A more detailed sketch of the session and the student proofs is below.
Bending Asymptotes, Bouncing Off Infinity, and Going Beyond
The basic proposal was that adding the Reciprocal transformation to the palette of constant dilations and translations dramatically simplified understanding of the behavior of rational functions around even and odd vertical asymptotes (bouncing off and passing through infinity). Just like lead coefficients of polynomials determine their end behavior, so, too, do the lead coefficients of proper rational expressions define the end behavior of rational functions.
Extending the idea of reciprocating and transforming functions, you can quickly explain exponential decay from exponential growth, derive the graphs of $y=\frac{1}{x}$ and $y=\frac{1}{x^2}$, and completely explain why logistic functions behave the way they do.
We finished with a quick exploration of trigonometric and polar graphs not as static parent functions under static transformations, but as dynamic curves oscillating between their ceilings and floors.
I used TI-Nspire Bending and Intro Polar files in the demonstration. Here is my outline PowerPoint file for Bending Asymptotes.
## Student Quadratic Creativity
I’m teaching Algebra 2 this summer for my school. In a recent test on quadratic functions, I gave a question I thought would be a little different, but still reachable for those willing to make connections or exert a little creativity.
Write a system of quadratic functions that has exactly one solution: (1,1).
Their handheld graphing calculators were allowed. Some students definitely had difficulty with the challenge, some gave a version of the answer I expected, and one adopted a form I knew was possible, but doubted anyone would actually find during a test situation.
I show my students’ solutions below. But before you read on, can you give your own solution?
SOLUTION ALERT! Don’t read further if you want to find your own solution.
WHAT I EXPECTED
We’ve had many discussions in class about the power of the Rule of 4–that math ideas can be expressed numerically, graphically, algebraically, and verbally. When you get stumped in one representation, being able to shift to a different form is often helpful. That could mean a different algebraic representation, or a different Rule of 4 representation altogether.
The question is phrased verbally asking for an algebraic answer. But it asks about a solution to a system of equations. I hoped my students would recall that the graphical version of a system solution is equivalent to the point(s) where the graphs of the equations intersected. In my mind, the easiest way to do this is to write quadratic functions with coincident vertices. And this is most easily done in vertex form. The cleanest answer I ever got to this question was
A graphical representation verifies the solution.
Another student recognized that if two parabolas shared a vertex, but had different “slopes”, their only possible point of intersection was exactly the one the question required. Here’s a graphical version of her answer.
From these two, you can see that there is actually an infinite number of correct solutions. And I was asking them for just one of these! 🙂
WHAT I KNEW, BUT DIDN’T EXPECT
Another way to solve this question makes use of the geometry of quadratic graphs. If two quadratics have the same leading coefficients, they are the same graph, intersect exactly once, or never intersect. This is a very non-trivial idea for most students. While I’m not convinced the author of the following solution had this in mind when he answered the question, his solution works because of that fact. Here’s what J wrote on last week’s test and its graph.
J used more equations than he needed, but had he restricted himself to just two equations, I’m not sure the lovely pattern would have been so obvious.
This is a very different (and super cool) answer than what I expected my students to produce. Lesson re-learned: Challenge your students, give them room to express creativity and individuality, and be prepared to be amazed by them.
NEXT STEPS
J’s answer actually opens the door to other avenues of exploration.
1. Can you generalize the form of all of J’s equations, essentially defining a family of quadratics? Can you prove that all members of your generalization satisfy the question posed and that no other answers are possible?
2. Can you find forms of other generalized families of quadratic functions whose only solution is (1,1)?
3. Notice that there were two types of solutions above: A) those with coincident vertices and different lead coefficients and B) those with identical lead coefficients and different vertices. Are these the only types of quadratics that can answer this question? That is, is there a system of quadratics with (1,1) as the only solution that have identical vertices and lead coefficients? Could both be different and (1,1) be the only solution?
4. If I relax the requirement that the quadratics be functions, what other types of quadratics are possible? [This could be a very nice calculus question!]
For my part, I’m returning to some of these questions this week to stretch and explore my student’s creativity and problem-solving.
I’d love to hear what you or your students discover.
## Traveling Dots, Parabolas, and Elegant Math
Toward the end of last week, I read a description a variation on a paper-folding strategy to create parabolas. Paraphrased, it said:
1. On a piece of wax paper, use a pen to draw a line near one edge. (I used a Sharpie on regular copy paper and got enough ink bleed that I’m convinced any standard copy or notebook paper will do. I don’t think the expense of wax paper is required!)
2. All along the line, place additional dots 0.5 to 1 inch apart.
3. Finally, draw a point F between 0.5 and 2 inches from the line roughly along the midline of the paper toward the center of the paper.
4. Fold the paper over so one of the dots on line is on tope of point F. Crease the paper along the fold and open the paper back up.
5. Repeat step 4 for every dot you drew in step 2.
6. All of the creases from steps 4 & 5 outline a curve. Trace that curve to see a parabola.
I’d seen and done this before, I had too passively trusted that the procedure must have been true just because the resulting curve “looked like a parabola.” I read the proof some time ago, but I consumed it too quickly and didn’t remember it when I was read the above procedure. I shamefully admitted to myself that I was doing exactly what we insist our students NEVER do–blindly accepting a “truth” based on its appearance. So I spent part of that afternoon thinking about how to understand completely what was going on here.
What follows is the chronological redevelopment of my chain of reasoning for this activity, hopefully showing others that the prettiest explanations rarely occur without effort, time, and refinement. At the end of this post, I offer what I think is an even smoother version of the activity, freed from some of what I consider overly structured instructions above.
CONIC DEFINITION AND WHAT WASN’T OBVIOUS TO ME
A parabola is the locus of points equidistant from a given point (focus) and line (directrix).
What makes the parabola interesting, in my opinion, is the interplay between the distance from a line (always perpendicular to some point C on the directrix) and the focus point (theoretically could point in any direction like a radius from a circle center).
What initially bothered me about the paper folding approach last week was that it focused entirely on perpendicular bisectors of the Focus-to-C segment (using the image above). It was not immediately obvious to me at all that perpendicular bisectors of the Focus-to-C segment were 100% logically equivalent to the parabola’s definition.
SIMILARITY ADVANTAGES AND PEDAGOGY
I think I had two major advantages approaching this.
1. I knew without a doubt that all parabolas are similar (there is a one-to-one mapping between every single point on any parabola and every single point on any other parabola), so I didn’t need to prove lots of cases. Instead, I focused on the simplest version of a parabola (from my perspective), knowing that whatever I proved from that example was true for all parabolas.
2. I am quite comfortable with my algebra, geometry, and technology skills. Being able to wield a wide range of powerful exploration tools means I’m rarely intimidated by problems–even those I don’t initially understand. I have the patience to persevere through lots of data and explorations until I find patterns and eventually solutions.
I love to understand ideas from multiple perspectives, so I rarely quit with my initial solution. Perseverance helps me re-phrase ideas and exploring them from alternative perspectives until I find prettier ways of understanding.
In my opinion, it is precisely this willingness to play, persevere, and explore that formalized education is broadly failing to instill in students and teachers. “What if?” is the most brilliant question, and the one we sadly forget to ask often enough.
ALGEBRAIC PROOF
While I’m comfortable handling math in almost any representation, my mind most often jumps to algebraic perspectives first. My first inclination was a coordinate proof.
PROOF 1: As all parabolas are similar, it was enough to use a single, upward facing parabola with its vertex at the origin. I placed the focus at $(0,f)$, making the directrix the line $y=-f$. If any point on the parabola was $(x_0,y_0)$, then a point C on the directrix was at $(x_0,-f)$.
From the parabola’s definition, the distance from the focus to P was identical to the length of CP:
$\sqrt{(x_0-0)^2-(y_0-f)^2}=y_0+f$
Squaring and combining common terms gives
$x_0 ^2+y_0 ^2-2y_0f+f^2=y_0 ^2+2y_0f+f^2$
$x_0 ^2=4fy$
But the construction above made lines (creases) on the perpendicular bisector of the focus-to-C segment. This segment has midpoint $\displaystyle \left( \frac{x_0}{2},0 \right)$ and slope $\displaystyle -\frac{2f}{x_0}$, so an equation for its perpendicular bisector is $\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)$.
Finding the point of intersection of the perpendicular bisector with the parabola involves solving a system of equations.
$\displaystyle y=\frac{x_0}{2f} \left( x-\frac{x_0}{2} \right)=\frac{x^2}{4f}$
$\displaystyle \frac{1}{4f} \left( x^2-2x_0x+x_0 ^2 \right) =0$
$\displaystyle \frac{1}{4f} \left( x-x_0 \right) ^2 =0$
So the only point where the line and parabola meet is at $\displaystyle x=x_0$–the very same point named by the parabola’s definition. QED
Proof 2: All of this could have been brilliantly handled on a CAS to save time and avoid the manipulations.
Notice that the y-coordinate of the final solution line is the same $y_0$ from above.
MORE ELEGANT GEOMETRIC PROOFS
I had a proof, but the algebra seemed more than necessary. Surely there was a cleaner approach.
In the image above, F is the focus, and I is a point on the parabola. If D is the midpoint of $\overline{FC}$, can I conclude $\overline{ID} \perp \overline{FC}$, proving that the perpendicular bisector of $\overline{FC}$ always intersects the parabola?
PROOF 3: The definition of the parabola gives $\overline{FI} \cong \overline{IC}$, and the midpoint gives $\overline{FD} \cong \overline{DC}$. Because $\overline{ID}$ is self-congruent, $\Delta IDF \cong \Delta IDC$ by SSS, and corresponding parts make the supplementary $\angle IDF \cong \angle IDC$, so both must be right angles. QED
PROOF 4: Nice enough, but it still felt a little complicated. I put the problem away to have dinner with my daughters and when I came back, I was able to see the construction not as two congruent triangles, but as the single isosceles $\Delta FIC$ with base $\overline{FC}$. In isosceles triangles, altitudes and medians coincide, automatically making $\overline{ID}$ the perpendicular bisector of $\overline{FC}$. QED
Admittedly, Proof 4 ultimately relies on the results of Proof 3, but the higher-level isosceles connection felt much more elegant. I was satisfied.
TWO DYNAMIC GEOMETRY SOFTWARE VARIATIONS
Thinking how I could prompt students along this path, I first considered a trace on the perpendicular lines from the initial procedure above (actually tangent lines to the parabola) using to trace the parabolas. A video is below, and the Geogebra file is here.
http://vimeo.com/89759785
It is a lovely approach, and I particularly love the way the parabola appears as a digital form of “string art.” Still, I think it requires some additional thinking for users to believe the approach really does adhere to the parabola’s definition.
I created a second version allowing users to set the location of the focus on the positive y-axis and using a slider to determine the distances and constructs the parabola through the definition of the parabola. [In the GeoGebra worksheet (here), you can turn on the hidden circle and lines to see how I constructed it.] A video shows the symmetric points traced out as you drag the distance slider.
A SIMPLIFIED PAPER PROCEDURE
Throughout this process, I realized that the location and spacing of the initial points on the directrix was irrelevant. Creating the software versions of the problem helped me realize that if I could fold a point on the directrix to the focus, why not reverse the process and fold F to the directrix? In fact, I could fold the paper so that F touched anywhere on the directrix and it would work. So, here is the simplest version I could develop for the paper version.
1. Use a straightedge and a Sharpie or thin marker to draw a line near the edge of a piece of paper.
2. Place a point F roughly above the middle of the line toward the center of the paper.
3. Fold the paper over so point F is on the line from step 1 and crease the paper along the fold.
4. Open the paper back up and repeat step 3 several more times with F touching other parts of the step 1 line.
5. All of the creases from steps 3 & 4 outline a curve. Trace that curve to see a parabola.
This procedure works because you can fold the focus onto the directrix anywhere you like and the resulting crease will be tangent to the parabola defined by the directrix and focus. By allowing the focus to “Travel along the Directrix”, you create the parabola’s locus. Quite elegant, I thought.
As I was playing with the different ways to create the parabola and thinking about the interplay between the two distances in the parabola’s definition, I wondered about the potential positions of the distance segments.
1. What is the shortest length of segment CP and where could it be located at that length? What is the longest length of segment CP and where could it be located at that length?
2. Obviously, point C can be anywhere along the directrix. While the focus-to-P segment is theoretically free to rotate in any direction, the parabola definition makes that seem not practically possible. So, through what size angle is the focus-to-P segment practically able to rotate?
3. Assuming a horizontal directrix, what is the maximum slope the focus-to-P segment can achieve?
4. Can you develop a single solution to questions 2 and 3 that doesn’t require any computations or constructions?
CONCLUSIONS
I fully realize that none of this is new mathematics, but I enjoyed the walk through pure mathematics and the enjoyment of developing ever simpler and more elegant solutions to the problem. In the end, I now have a deeper and richer understanding of parabolas, and that was certainly worth the journey.
## Fun with Series
Two days ago, one of my students (P) wandered into my room after school to share a problem he had encountered at the 2013 Walton MathFest, but didn’t know how to crack. We found one solution. I’d love to hear if anyone discovers a different approach. Here’s our answer.
PROBLEM: What is the sum of $\displaystyle \sum_{n=1}^{\infty} \left( \frac{n^2}{2^n} \right) = \frac{1^2}{2^1} + \frac{2^2}{2^2} + \frac{3^2}{3^3} + ...$ ?
Without the $n^2$, this would be a simple geometric series, but the quadratic and exponential terms can’t be combined in any way we knew, so the solution must require rewriting. After some thought, we remembered that perfect squares can be found by adding odd integers. I suggested rewriting the series as
where each column adds to the one of the terms in the original series. Each row was now a geometric series which we knew how to sum. That meant we could rewrite the original series as
We had lost the quadratic term, but we still couldn’t sum the series with both a linear and an exponential term. At this point, P asked if we could use the same approach to rewrite the series again. Because the numerators were all odd numbers and each could be written as a sum of 1 and some number of 2s, we got
where each column now added to the one of the terms in our secondary series. Each row was again a geometric series, allowing us to rewrite the secondary series as
Ignoring the first term, this was finally a single geometric series, and we had found the sum.
Does anyone have another way?
That was fun. Thanks, P. | 2018-12-18T16:11:52 | {
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https://math.stackexchange.com/questions/2660468/projectile-vw-gk-for-minimum-launch-velocity | # Projectile: $v^*w^*=gk$ for minimum launch velocity
A projectile launched from $O(0,0)$ at velocity $v$ and launch angle $\theta$, passes through $P(k,h)$. The velocity of the projectile at $P$ is $w$. The slope of $OP$ is $\alpha$, i.e. $\tan\alpha=\frac hk$, and the length of $OP$ is $R$.
$\hspace{4cm}$
Let $v^*$ be the minimum launch velocity for the projectile to reach $P$, and $w^*$ the corresponding minimum terminal velocity at $P$. In the course of working out $v^*$, I noticed this neat relationship: $$\color{red}{\boxed{v^*w^*=gk}}\tag{1}$$
which can be proven easily using calculus. The relationship is interesting because of its symmetry and also its independence from $\theta$ and $h$. It also helps simplify the solution of projectile problems relating to minimum velocities, e.g. this question here.
Question 1: Is it possible to derive this relationship given by $(1)$ directly without using calculus but by exploiting some geometric or kinematic symmetry?
Separately, we know that, for a projectile for a given range $R$ (on the same level), the minimum launch velocity is given by $v^*=\sqrt{gR}$. Assume that, for another given range $r$, the minimum launch velocity is $w^*=\sqrt{gr}$. Substituting in $(1)$ gives $k=\sqrt{Rr}$, i.e. $k$ is the geometric mean of $R$ and $r$.
Question 2: Can the relationship given by $(1)$ be derived using the relationships between minimum launch velocity and range shown above, perhaps through a geometric transform of some sort?
• My solution below has been updated to show a derivation of the minimum velocity relationship ${v^*}^2+{w^*}^2=2gR$ without using calculus. – hypergeometric Mar 21 '18 at 17:26
• See latest solution using vector methods. Much neater. – hypergeometric Mar 27 '18 at 10:19
Because $w_x=v_x$ and $v_y^2=w_y^2-2gh$, the minimum speed in the origin implies you arrive with the minimum possible speed to point $(k,h)$.
I think the following picture shows a symmetry which could be useful. The minimum possible speed to reach $(k,h)$ is attained when you throw the projectile in the direction bisecting the angle between the $OP$ line and vertical axis. Because this trajectory reaches $P$ with the minimum possible speed, it also solves the "reverse" problem (by reversing the velocity direction), that is, from $P$ to reach $O$ with minimum speed. Therefore, $\vec{w}$ also bisects the angle that $OP$ makes with $\hat{y}$ at $P$, and therefore $\vec{v}\perp\vec{w}$.
By the way, proving that the maximum range in a slope (or minimum speed to a fixed range) is attained in the bisecting direction is a classical and nice problem, and it can be done without using calculus as well (here, for example).
Following with the reasoning, we then have \begin{align}|\vec{v}\times\vec{w}|^2&=v^2w^2\qquad;\vec{v}\perp\vec{w}\\ &=(v_x w_y-w_x v_y)^2\\ &=v_x^2(v_y-w_y)^2\qquad;|v_x|=|w_x|\\ &=v_x^2(g t_f)^2\qquad;\text{where t_f is time of flight}\\ &=(v_xt_f)^2g^2=k^2g^2~~.\\ \end{align}
• Very elegant solution! (+1). Thanks. Solution accepted! – hypergeometric Feb 23 '18 at 15:10
• Would you know if $vw=gk$ is a standard result that is well-known? – hypergeometric Feb 23 '18 at 15:11
• I personally didn't know it. – Enredanrestos Feb 24 '18 at 8:11
• OK thanks. Neither did I, until I worked it out whilst looking for a simple approach to solve the problem here. – hypergeometric Feb 24 '18 at 15:48
This is a complement to the previously posted answer. While solving this problem I realized that the set-up also serves to prove another interesting fact. That is, the flight times of the optimal trajectories to the target depend only on $R=\sqrt{h^2+k^2}$, independent of inclination.
In the second picture below, when the minimum energy projectile thrown from $A$ reaches $D$, a "projectile" released with zero velocity from $A$ will reach point $G$. In addition, projectiles thrown with parallel velocities (including zero velocity) from the point of origin at the same time remain forming a line parallel to the original velocity at all subsequent times. Therefore, $GD$ (being the line joining the projectiles at $G$ and $D$ at time $t_f$) is parallel to $AX$ (that is, $\vec{v}\parallel AX\parallel GD$), thus, $\angle{BDG}$ is a right angle and $G$ is also in the circumference. Therefore, the flight time for all minimum energy projectiles is the same as an object falling from $A$ to $G$, given by $\sqrt{\frac{2R}{g}}$.
Second, more explicit derivation
Using a bit of trigonometry, we show that the smallest initial velocity $\vec{v}$ of the projectile to reach a distance of $R$ in a direction forming an angle of $\alpha$ with the horizontal is attained when the $\vec{v}$ direction bisects the angle between the vertical and $\alpha$. Part of the interest of this answer is that it does not use calculus either.
First, we calculate the range of the projectile in the direction given by $\alpha$ when being thrown with an arbitrary velocity $\vec{v}$ forming an angle $\theta$ with the horizontal. Following the same reasoning as in the previous diagram, we deduce that is $KH\parallel\vec{v}$ (in the second picture below). We can immediately see that the $\angle AHK=\theta-\alpha$, and $$\frac{gt_f^2}{2}= \frac{R\sin(\theta-\alpha)}{\cos\theta}~~.\tag{1}$$
Furthermore, it is also clear that, since there is no acceleration in the $\hat{x}$ direction, $$R\cos\alpha=vt_f\cos\theta~~.\tag{2}$$ From (1) and (2) we deduce $$R=\frac{v^2}{g}\frac{2\cos(\theta)\sin(\theta-\alpha)}{\cos^2\alpha}=\frac{v^2}{g}\frac{(\sin(2\theta-\alpha)-\sin(\alpha))}{\cos^2\alpha}~~,\tag{3}$$ from where it is very clear that the maximum range is given by $2\theta-\alpha=\pi/2~~$ (or $\theta$ bisects the angle between $AH$ and the vertical). In turn, the maximum range is given by $\frac{v^2}{g(1+\sin\alpha)}$.
Concluding, by squaring (1) \begin{align} t_f^4&= \left(\frac{2R}{g}\right)^2\frac{\sin^2((\pi/2-\alpha)/2)}{\cos^2((\pi/2+\alpha)/2)}\qquad;\theta=(\pi/2+\alpha)/2\\ &=\left(\frac{2R}{g}\right)^2 \frac{\frac{1-\cos(\pi/2-\alpha)}{2}}{\frac{1+\cos(\pi/2+\alpha)}{2}}\\ &=\left(\frac{2R}{g}\right)^2 \frac{1-\sin\alpha}{1-\sin\alpha}\\ &=\left(\frac{2R}{g}\right)^2~~, \end{align} therefore $t_f=\sqrt{\frac{2R}{g}}$, that is, the same result as in the first answer with $t_f$ independent of $\alpha$.
• Interesting! (+1). For the free fall case, terminal velocity is $\sqrt{gR}$. For the inclined case, minimum terminal velocity is $w^*=\sqrt{g(R-h)}$, and the corresponding minimum launch velocity is $v^*=\sqrt{g(R+h)}$, the product of which is $v^*w^*=g\sqrt{R^2-h^2}=gk$. Given the additional insight from your new solution, can this be derived using some transformation of the free fall case, or perhaps the horizontal case? – hypergeometric Mar 1 '18 at 8:47
• Thanks! I am glad you find it interesting. Respect to your question, I am not sure. I'll keep thinking.. – Enredanrestos Mar 1 '18 at 9:23
• Is there a missing $\cos^2\alpha$ in the denominator of $(3)$? – hypergeometric Mar 1 '18 at 10:31
• Yes there is . Sorry, I'll fix it right away. Fortunately, it does not affect the last derivation. – Enredanrestos Mar 1 '18 at 12:59
• Could you please elaborate further on the point in the second paragraph of your solution (first derivation)? – hypergeometric Mar 13 '18 at 16:51
Consider projectile with parameters as shown in diagram below.
$\hspace{4cm}$
Let $T$ be the total time taken for the journey of the projectile, and $\beta$ be the angle between the initial and terminal velocities.
Refer to diagrams above. As the projectile is a motion under constant acceleration, it can be modelled as either
(a) First Scenario:-
• $(1)$ motion at constant velocity $\mathbf v$ from $O$ to $A$ for the first half of $T$ (distance travelled: $\frac {vT}2$),and
• $(2)$ motion at constant velocity $\mathbf w$ from $A$ to $P$ for the second half of $T$ (distance travelled: $\frac {wT}2$).
or
(b) Second Scenario**:- as a composite motion of the following:
• $(1)$ motion at constant velocity $\mathbf v$ from $O$ to $B$ for time $T$ (distance travelled: $vT$) and
• $(2)$ vertical free fall under gravity from $B$ to $P$ for time $T$ (distance travelled: $\frac 12gT^2$).
(NB - velocity here refers to a vector quantity, i.e. both speed and direction.)
Consider the diagram on the right, which is the left diagram scaled by $\frac 2T$.
\begin{align} \triangle O'B'P'= \tfrac 12 \cdot 2v\cdot w\cdot \sin\beta &=\tfrac 12\cdot gT\cdot \tfrac {2R}T\cos\alpha \\ vw\sin\beta&=gR\cos\alpha\\ vw\sin\beta&=gk\\ \color{red}{v^*w^*}&\color{red}{=gk} \end{align}
As $g,k,$ are fixed, $v,w$ will be at a minimum when $\sin\beta$ is at a maximum, i.e. $\sin\beta=1 (\beta=\tfrac\pi2)$, hence $v^*, w^*$ are mutually perpendicular. Note that $w^2=v^2-2gh$ per energy conservation, and as $h$ is fixed, minimum $w^*$ corresponds to minimum $v^*$.
EARLIER SOLUTION (posted 21 March 2018)
For minimum velocity (i.e. minimum kinetic energy), \begin{align} {v^*}^2+{w^*}^2&=2gR\qquad\qquad\;\tag{1}\end{align} where $^*$ indicates quantities corresponding to the minimum velocity case.
Also, from conservation of energy ($V^2=U^2+2AS$), we have, for the general case, \begin{align} v^2-w^2&=2gh\color{lightgrey}{=2gR\sin\alpha}\tag{2}\end{align}
$\tfrac 12 \big((1)\pm (2)\big)$ : $${v^*}^2=g(R+h)\color{lightgrey}{=gR(1+\sin\alpha)}\\ {w^*}^2=g(R-h)\color{lightgrey}{=gR(1-\sin\alpha)}$$ and it follows that \begin{align} {v^*}^2{w^*}^2&=g^2(R^2-h^2)\color{lightgrey}{=g^2R^2(1-\sin^2\alpha)}\ \\&=g^2k^2\qquad\;\;\color{lightgrey}{=g^2R^2\cos^2\alpha}\\ \color{red}{v^*w^*}&\color{red}{=gk\;\blacksquare}\;\;\ \qquad\color{lightgrey}{=gR\cos\alpha}\end{align}
$\hspace{5cm}$
Footnote
Using the cosine rule twice,we have
\begin{align} R^2 &=\left(\tfrac {vT}2\right)^2+\left(\tfrac {wT}2\right)^2-2\left(\tfrac {vT}2\right)\left(\tfrac {wT}2\right)\cos\beta\qquad \tag{3}\\ \left(\tfrac 12 gT^2\right)^2 &=\left(\tfrac {vT}2\right)^2+\left(\tfrac {wT}2\right)^2-2\left(\tfrac {vT}2\right)\left(\tfrac {wT}2\right)\cos (\pi-\beta) \tag{4}\\ (3)+(4):\hspace{2cm}\\ R^2+\left(\tfrac 12 gT^2\right)^2 &=2\big(\left(\tfrac {vT}2\right)^2+\left(\tfrac {wT}2\right)^2\big) \\ v^2+w^2&=\tfrac 12 \left(g^2T^2+\tfrac {4R^2}{T^2}\right)\\ &=\tfrac 12\underbrace{\left(gT-\tfrac {2R}T\right)^2}_{\ge0}+2gR\\ &\ge 2gR\\ {v^*}^2+{w^*}^2&=2gR \end{align} which is as used in equation $(1)$ above, with $^*$ indicating quantities corresponding to the minimum velocity case.
This occurs when ${T^*}^2=\tfrac {2R}g$, i.e. $\tfrac 12g{T^*}^2=R$. Using this and from the diagram we deduce that $\beta^*=\frac {\pi}2$, i.e. at minimum $v,w$, both $\mathbf v, \mathbf w$ are perpendicular to each other. This is shown in the diagram below.
$\hspace{4cm}$
To deduce the values of $\theta$ and $\phi$, consider the triangles shown below.
$\hspace{4cm}$
From first triangle, \begin{align} (\theta-\alpha)+(\phi+\alpha)&=\tfrac \pi 2\\ \theta+\phi&=\tfrac\pi 2\tag{5} \end{align}
From the lower two triangles,
\begin{align} (\tfrac\pi2-\theta)+(\phi+\alpha)&=\tfrac\pi2\\ \theta-\phi&=\alpha\tag{6}\\ \tfrac 12 \big((5)\pm (6)\big):\quad\qquad\\ \theta&=\tfrac\pi4+\tfrac\alpha2\\ \phi&=\tfrac\pi4-\tfrac\alpha2 \end{align}
which are launch and terminal angles corresponding to the minimum velocity case.
Conversely this also gives the well-known result, where the optimal launch angle to achieve maximum distance in direction $\alpha$ is one that bisects $\alpha$ and the vertical.
** An alternative Second Scenario would be a composite motion of the following:
• $(1)$ vertical launch at $gT$ under gravity for time $T$ (distance travelled: $\frac 12gT^2$), and
• $(2)$ motion at constant velocity $\mathbf w$ for time $T$ (distance travelled: $wT$) and
Alternative Interpretation
An alternative interpretation to the results might be given as follows. Refer to the diagram below.
$\hspace{3cm}$
Let $A$ be the highest point reached by the particle under the scenario above where it travels half the time at constant initial launch velocity under zero gravity. The vertical distances from $A$ to $O,P$ are $\frac {R+h}2$ and $\frac {R-h}2$ respectively.
For a particle dropped from a height $H$ from the ground and falling vertically under gravity, its velocity upon reaching the ground is $\sqrt{2gH}$. Similarly, a particle under free fall of gravity, when launched vertically upwards at a velocity if $\sqrt{2gH}$, will reach a height of $H$. Here we refer to $\sqrt{2gH}$ as the vertical launch velocity for height $H$.
The magnitudes of the initial and terminal velocities of the projectile, $v^*,w^*$ are equal to the vertical launch velocities for heights $\frac {R+h}2, \frac {R-h}2$ respectively, i.e.
$${v^*}^2=2g\left(\tfrac {R+h}2\right)=g(R+h)\\ {w^*}^2=2g\left(\tfrac {R-h}2\right)=g(R-h)\\$$ which gives $${v^*}^2+{w^*}^2=2gR$$ and also $$v^*w^*=g\sqrt{R^2-h^2}=gk$$
MUCH EARLIER SOLUTION (Not so neat) (Posted 5 March 2018)
Note that $w^2=v^2-2gh$. Since $h$ is fixed, therefore minimum $w$ corresponds to minimum $v$.
Let $v^*, w^*$ be the minimum values of $v, w$ respectively. \begin{align} v^2w^2&=v^2(v^2-2gh)\\ &=v^2\left[v^2-2g\left(k\tan\theta-\frac {gk^2}{2v^2}(\tan^2\theta+1)\right)\right]\\ &=v^4-2gkv^2\tan\theta+g^2k^2\tan^2\theta+g^2k^2\\ &=\underbrace{\left(v^2-gk\tan\theta\right)^2}_{\ge 0}+g^2k^2\\ &\ge g^2k^2\\ \color{red}{v^*w^*}&\color{red}{=gk\qquad\blacksquare}\end{align} This occurs when ${v^\ }^2={v^*}^2=gk\tan\theta$.
It can be shown that $\tan\theta=\dfrac {R+h}k$, where $R=\sqrt{k^2+h^2}$.
Hence $${v^*}^2=g(R+h)\\ {w^*}^2=g(R-h)$$
LATEST SOLUTION
(Using vectors)
$\hspace{5cm}$
Let $\mathbf v,\mathbf w$ be the initial (launch) and terminal velocity vectors of the projectile, and $\beta$ be the angle between them.
$\hspace{5cm}$
The projectile motion can be modelled as
(i) EITHER as a composite motion of the following:
• motion at constant velocity $\mathbf v$ for time $T$ (distance travelled: $vT$) and
• vertical free fall under gravity for time $T$ (distance travelled: $\frac 12gT^2$).
(ii) OR as a composite motion of the following:
• upward vertical launch at velocity $gT$ under gravity for time $T$ (distance travelled: $\frac 12gT^2$), and
• motion at constant velocity $\mathbf w$ for time $T$ (distance travelled: $wT$) and
Let $\mathbf R$ be the direction vector for $\vec{OP}$, and $\mathbf g$ the gravity vector.
Distance equations for both equivalent composite motions above are as follows:
\begin{align} \mathbf R&=\mathbf vT+\tfrac 12 \mathbf gT^2\tag{1}\\ \mathbf R&=\mathbf wT-\tfrac 12 \mathbf gT^2\tag{2}\\\\ \tfrac {(1)+(2)}T:\hspace {6.5cm} (\mathbf v+\mathbf w)&=\tfrac 2T \mathbf R\tag{A}\\ \tfrac {(1)-(2)}T:\hspace{6.5cm} (\mathbf v- \mathbf w) &= \mathbf gT\tag{B}\\\\ (\text A)\cdot (\text B):\hspace{5cm} (\mathbf v+\mathbf w)\cdot (\mathbf v-\mathbf w)&=2\mathbf R\cdot \mathbf g\color{lightgrey}{=2(R\sin\alpha) g}\\ v^2-w^2&=2gh\tag{I}\\\\ (\text A)\cdot (\text A):\hspace {1cm} (\mathbf v+\mathbf w)\cdot (\mathbf v+\mathbf w)=v^2+w^2+2vw\cos\beta &=\tfrac {4R^2}{T^2}\tag {IIa}\\ (\text B)\cdot (\text B):\hspace {1cm} (\mathbf v-\mathbf w)\cdot (\mathbf v-\mathbf w)=v^2+w^2-2vw\cos\beta &={g^2T^2}\tag {IIb}\\ \tfrac12\big((\text{IIa})+(\text{IIb})\big):\hspace{4.5cm} v^2+w^2\qquad \qquad \;\; &=\tfrac 12 \big(\tfrac {4R^2}{T^2}+g^2T^2\big)\\ &=\tfrac 12 \underbrace{\big(\tfrac {2R}T-gT\big)^2}_{\ge 0}+2gR\\ &\ge 2gR\\ {v^*}^2+{w^*}^2&=2gR\tag{II}\\ (T^*&=\tfrac{2R}g)\\\\ \tfrac 12 \big((\text{I})\pm(\text{II})\big):\hspace{7.5cm} {v^*}^2&=g(R+h)\\ {w^*}^2&=g(R-h)\\\\ {v^*}^2{w^*}^2&=g^2(R^2-h^2)=g^2k^2\\ \color{red}{v^*w^*}&\color{red}{=gk} \end{align}
Alternatively we can bypass steps (I), (II) and arrive at the required result directly from (A), (B) as follows:
\begin{align} (\text{A})\times(\text{B}):\hspace{5cm} (\mathbf v+\mathbf w)\times (\mathbf v - \mathbf w) &=\tfrac 2T \mathbf R \times \mathbf gT\\ 2\big|\mathbf v\times \mathbf w\big| &=2\big|\mathbf R \times \mathbf g\big|\\ vw\sin\beta&=(R\cos\alpha)g\\ &=gk\\ \color{red}{v^*w^*}&\color{red}{=gk}\hspace{1cm} \end{align} with minimum values of $v,w$ occuring when at maximum $\sin\beta(=1)$, i.e. at $\beta=\tfrac {\pi}2$. | 2019-06-16T09:10:48 | {
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https://math.stackexchange.com/questions/3055608/conjecture-int-0-pi-2-left-frac-sin2n1x-sin-x-right-beta-dx-is | # Conjecture:$\int_0^{\pi/2} \left(\frac{\sin(2n+1)x}{\sin x}\right)^\beta dx$ is integer multiple of $\pi/2$,for integer $\beta>2$
I was solving the integral $$I_n=\int_0^{\frac {\pi}{2}} \left(\frac {\sin ((2n+1)x)}{\sin x}\right)^2 dx$$
With $$n\ge 0$$ And $$n\in \mathbb{N}$$
On solving, I got $$I_n =\frac {(2n+1)\pi}{2}$$
But, due to curiosity, I started investigating the family of integrals as
$$I_n(\beta) =\int_0^{\frac {\pi}{2}} \left(\frac {\sin (2n+1)x}{\sin x}\right)^{\beta} dx$$
On trying various values of $$\beta\gt 2$$ and $$\beta\in \mathbb{N}$$, I conjectured that $$I_n(\beta) =c_{\beta} \frac{\pi}{2}$$ where $$c_{\beta}$$ denotes "Number of arrays of $$\beta$$ integers in $$-n$$ to $$n$$ with sum $$0$$"
But, on trying a lot, I couldn't prove this statement. Also, I suppose that the statement could be proved with help of Dirichlet kernel, but I couldn't get the way out through it.
Any help and hints to prove/disprove the conjecture are greatly appreciated.
• @Masacroso I think $I_n$ is the integral of something alike Fejer kernel, so $\sin((2n+1)x)$. – xbh Dec 29 '18 at 7:08
• @Masacroso Edited!!! – Rohan Shinde Dec 29 '18 at 7:10
• oeis.org/A201552 – James Arathoon Dec 29 '18 at 12:21
• – Zacky Dec 29 '18 at 12:36
• @James Arathoon The statement I guessed was from OEIS only and it also doesn't have any proofs. – Rohan Shinde Dec 29 '18 at 13:01
$$\def\b{\beta}$$\begin{align*} \newcommand\cmt[1]{{\small\textrm{#1}}} I_n(\b) &= \int_0^{\pi/2} \left(\frac{\sin (2n+1)x}{\sin x}\right)^\b dx \\ &= \frac 1 4 \int_0^{2\pi} \left(\frac{\sin (2n+1)x}{\sin x}\right)^\b dx & \cmt{begin similar to user630708} \\ &= \frac{1}{4i} \oint_\gamma \left(\frac{z^{4n+2}-1}{z^2-1}\right)^\b \frac{dz}{z^{2n\b+1}} & \cmt{let z=e^{ix}} \\ &= \frac{1}{4i} \oint_\gamma \left(\sum_{k=0}^{2n}z^{2k}\right)^\b \frac{dz}{z^{2n\b+1}} & \cmt{partial sum of geometric series} \\ &= \left.\frac{1}{4i} \frac{2\pi i}{(2n\b)!} \left(\frac{d}{dz}\right)^{2n\b} \left(\sum_{k=0}^{2n}z^{2k}\right)^\b \right|_{z=0} & \cmt{Cauchy integral formula} \\ &= \left.\frac{\pi}{2} \frac{1}{(2n\b)!} \left(\frac{d}{dz}\right)^{2n\b} \sum_{\sum x_k=\b} \frac{\b!}{\prod x_k!} \prod (z^{2k})^{x_k} \right|_{z=0} & \cmt{multinomial expansion, k=0,1,\ldots,2n} \\ &= \left.\frac{\pi}{2} \frac{1}{(2n\b)!} \left(\frac{d}{dz}\right)^{2n\b} \sum_{\sum x_k=\b} \frac{\b!}{\prod x_k!} z^{2\sum k x_k} \right|_{z=0} \\ &= \frac{\pi}{2} \sum_{\sum x_k=\b \atop \sum k x_k = n\b} \frac{\b!}{\prod x_k!} & \cmt{only surviving terms have \sum k x_k = n\b} \\ &= \frac{\pi}{2} \sum_{\sum x_k=\b \atop \sum (n-k) x_k = 0} \frac{\b!}{\prod x_k!} \end{align*} In the last line note that $$\sum_{k=0}^{2n} n x_k=n\b$$ and so $$\sum_{k=0}^{2n} (n-k)x_k = 0$$. By inspection one can see that $$\sum_{\sum_{k=0}^{2n} x_k=\b \atop \sum_{k=0}^{2n} (n-k) x_k = 0} \frac{\b!}{\prod x_k!} = \textrm{number of arrays of \b integers in -n,\ldots,n with sum equal to 0,}$$ i.e., $$I_n(\b) = \frac{\pi}{2} T(\b,n),$$ where $$T(\b,n)$$ is OEIS A201552, as pointed out by James Arathoon in the comments. (On that page we also find an integral form of $$T(\b,n)$$ which, after a simple substitution, gives $$I_n(\b) = \frac{\pi}{2} T(\b,n)$$.)
This is easy using Residue Theory:
Note that by symmetry $$\int_{0}^{\pi/2}...dx=1/4\int_{-\pi}^{\pi}…dx$$ (use parity and a sub $$y=\pi-x$$ to Show that).
employing $$z=e^{ix}$$ we get
$$4 I_{n,\beta}=\oint_C \left[\frac{z^{4n+2}-1}{z^2-1}\right]^{\beta}\frac{dz}{i z^{2\beta n+1}}$$
where $$C$$ denotes the unit circle in the comlex plane. By the residue theorem (there is one pole inside the contour at $$z=0$$, using f.e. the geometric series you can Show that the Points $$z=\pm i$$ are removable singularities). We have
$$4 I_{n,\beta}=2\pi \text{Res}(\left[\frac{z^{4n+2}-1}{z^2-1}\right]^{\beta}\frac{1}{z^{2\beta n+1}} ,z=0)$$
Using $$\beta!(z^2-1)^{-\beta}=((2z)^{-1}\partial_z)^{\beta-1}(z^2-1)^{-1}$$ we get
$$(1-z^2)^{-\beta}=\frac{1}{2^{\beta-1}}\sum_{m\geq0}\binom{m+\beta-1}{\beta-1}z^{2m}\\ (1-z^{4n+2})^{\beta}=z^{2\beta}\sum_{k\geq0}(-1)^k\binom{\beta}{k}z^{4k}$$
which means that we have the condition $$4k+2(m+\beta)-2\beta n-1=-1$$ (since we are interested in $$a_{-1}$$ coefficent of the Laurent expansion) which essenitally kills one of the sums, and we end up with
$$I_{n,\beta}=\frac{\pi}{2^{\beta}}\sum_{m\geq0}(-1)^{\beta n /2-(\beta+m)/2}\binom{m+\beta-1}{\beta-1}\binom{\beta}{\beta n /2-(\beta+m)/2}$$
which is a finite sum, since the second binomial becomes zero when $$m$$ is large enough ($$m> \beta (n-1)$$)
• This seems cool, but where does the $\beta \neq 4$ exception come from? Is it from that line about how we "kill one of the sums"? – goblin Jan 1 at 23:46
• Notice that the final sum is not necessarily real. – user26872 Jan 2 at 0:37
• Things seem to go off the rails with $\beta!(z^2-1)^{-\beta}=((2z)^{-1}\partial_z)^{\beta-1}(z^2-1)^{-1}$, which is false for $\beta>1$. It is true that $(\beta-1)!(z^2-1)^{-\beta}=((-2z)^{-1}\partial_z)^{\beta-1}(z^2-1)^{-1}$. – user26872 Jan 5 at 20:45 | 2019-06-19T21:12:45 | {
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http://math.stackexchange.com/questions/175554/prove-frac1-cos2a-sin2a-cota | Prove $\frac{1+\cos{(2A)}}{\sin{(2A)}}=\cot{A}$
I am sorry to ask so many of these questions in such as short time span.
But how would I prove this following trigonometric identity. $$\frac{1+\cos(2A)}{\sin(2A)}=\cot A$$ My work thus far is $$\frac{1+\cos^2A-\sin^2A}{2\sin A\cos A}$$ I know $1-\sin^2A=\cos^2A$
So I do $$\frac{\cos^2A+\cos^2A}{2\sin A\cos A}$$ I know not what I do next.
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$\cos^2 A+\cos^2 A=2\cos^2 A$. – David Mitra Jul 26 '12 at 17:54
$\frac{cos^2A+cos^2A}{2\sin A\cos A}=\frac{2\cos^2A}{2\sin A \cos A}=\cot A$ – Saurabh Jul 26 '12 at 17:56
There's also a nice way of seeing this. Take a look at this answer by robjohn. – Ian Mateus Jul 26 '12 at 18:24
\begin{align} \frac{1+\cos(2A)}{\sin(2A)} &=\frac{1+\cos^2(A)-\sin^2(A)}{2\sin(A)\cos(A)}\tag{1}\\ &=\frac{\csc^2(A)+\cot^2(A)-1}{2\,\cot(A)}\tag{2}\\ &=\frac{2\,\cot^2(A)}{2\,\cot(A)}\tag{3}\\[4pt] &=\cot(A)\tag{4} \end{align}
1. double angle formulas
2. multiply numerator and denominator by $\csc^2(A)$
3. $\cot^2(A)+1=\csc^2(A)$
4. cancel $2\cot(A)$ in numerator and denominator
-
$$\frac{1+\cos(2A)}{\sin(2A)} = \frac{2\cos^2 A}{2\sin A \cdot \cos A} = \frac{\cos A}{\sin A} = \cot A$$
-
I have rolled back to TMM's edit, since it is exceptionally clear and easy to read. – mixedmath May 9 '13 at 19:17
why is this downvoted – mathguy May 9 '13 at 19:19
thanks @mixedmath. should I write the answer if someone already give – iostream007 May 9 '13 at 19:42
Just write $\cos^2 A+\cos^2 A=2\cos^2 A$ (a quantity added to itself is twice the quantity). Then write $2\cos^2 A=2\cos A\cdot\cos A$ and cancel a $2\cos A$ term in the numerator with the $2\cos A$ term in the denominator.
-
Thank you your math skills are excellent. – Fernando Martinez Jul 26 '12 at 18:00 | 2016-05-04T16:20:02 | {
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http://math.stackexchange.com/questions/9522/expected-number-of-neighbors | # Expected number of neighbors
Given a row of 16 houses where 10 are red and 6 are blue, what is the expected number of neigbors of a different color?
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By "neighbors of a different color"... do you mean given any one house what are the chances that the house to the left and the house to the right of that house (if the house is not on the end) are the same color as the given house? – futurebird Nov 9 '10 at 4:22
You should state the implicitly assumed distribution in the space of all permutations (here: uniform). – Raphael Nov 10 '10 at 18:12
Let $X$ be the number of neighbors of a different color. Let $X_i$ be 1 if houses $i$ and $i+1$ are different colors and 0 otherwise. Then $X = \sum_{i=1}^{15} X_i.$
So the expected number of neighbors of a different color is $$E\left[\sum_{i=1}^{15} X_i\right] = \sum_{i=1}^{15} E[X_i] = \sum_{i=1}^{15} P(X_i = 1).$$
Now, $$P(X_i = 1) = P(i \text{ is red})P(i+1 \text{ is blue}|i \text{ is red}) + P(i \text{ is blue})P(i+1 \text{ is red}|i \text{ is blue})$$ $$= \frac{10}{16} \frac{6}{15} + \frac{6}{16} \frac{10}{15} = \frac{120}{16(15)} = \frac{1}{2}.$$
Thus $$E[X] = \sum_{i=1}^{15} \frac{1}{2} = 7.5.$$
More generally, using indicator variables is often an effective way to calculate expected values. The linearity of the expectation operator allows you to sidestep all the nasty dependencies you would otherwise be forced to deal with. For lots of interesting examples using this approach, see Section 7.2 of Sheldon Ross's A First Course in Probability.
Given that the answer comes out so nicely I would not be surprised if there is a cleaner argument than mine. If someone comes up with one I would love to see it.
Moron's generalization has inspired me to generalize mine. :)
Suppose you have $m$ red houses and $n$ blue houses. Then the expected number of neighbors of a different color is $$\frac{2mn}{m+n}.$$
Similar to my argument above, $$P(X_i = 1) = P(i \text{ is red})P(i+1 \text{ is blue}|i \text{ is red}) + P(i \text{ is blue})P(i+1 \text{ is red}|i \text{ is blue})$$ $$= \frac{m}{m+n} \frac{n}{m+n-1} + \frac{n}{m+n} \frac{m}{m+n-1} = \frac{2mn}{(m+n)(m+n-1)}.$$
Thus $$E[X] = \sum_{i=1}^{m+n-1} \frac{2mn}{(m+n)(m+n-1)} = \frac{2mn}{m+n}.$$
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+1 : no need for a cleaner argument. Your answer is quite clean already. – Djaian Nov 9 '10 at 9:09
Very nice! We get $m+n = (m-n)^2$ (for pbt 1/2 case) which confirms my calculation. I wish I could upvote again. – Aryabhata Nov 10 '10 at 18:01
The chances of a particular neighbour pair being the same colour is
$$\frac{{14 \choose 8} + {14 \choose 10}}{{16 \choose 6}} = \frac{4004}{8008} = \frac{1}{2}$$
Hence the answer is $\displaystyle 7.5$
This is happening because $\displaystyle {14 \choose 8}, {14 \choose 9}, {14 \choose 10}$ are in arithmetic progession: The number of ways of being same colour is $\displaystyle {14 \choose 8} + {14 \choose 10}$ and the number of ways of being different is $\displaystyle 2{14 \choose 9}$. The probability is $\displaystyle \frac{1}{2}$ if these two are equal.
Interestingly, $\displaystyle {n \choose r}, {n \choose r+1}, {n \choose r+2}$ are in arithmetic progression if and only if $\displaystyle n+2$ is a perfect square and $\displaystyle r$ is given by $\displaystyle r = \frac{n-2 \pm \sqrt{n+2}}{2}$ (see the end of the answer for a proof).
So for instance, the whole bunch of problems:
15 red, 10 blue
21 red, 15 blue
etc give rise to this neat probability of being $\displaystyle \frac{1}{2}$.
Proof that n+2 is a perfect square
$\displaystyle {n \choose r}, {n \choose r+1}, {n \choose r+2}$ are in arithmetic progression iff
$\displaystyle 2{n \choose r+1} = {n \choose r} + {n \choose r+2}$
i.e
$\displaystyle 2 = \frac{r+1}{n-r} + \frac{n-r-1}{r+2}$
Doing some manipulations gives us
$\displaystyle (n-2r-2)^2 = n+2$
Hence
$\displaystyle r = \frac{n-2 \pm \sqrt{n+2}}{2}$
Which has an integer solution iff $\displaystyle n+2$ is a perfect square.
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+1 for a different perspective (considering the complement) and for helping me see that the $\frac{120}{15(16)}$ expression in my answer is just $\frac{1}{2}$. That simplifies my answer, and I will update it accordingly. Thanks. – Mike Spivey Nov 9 '10 at 16:15
@Mike: I added some more information to the my answer, which shows that $n+2$ has to be a perfect square for this neat answer(probability 1/2) to hold! The problem turned out to be more interesting than I anticipated :-) – Aryabhata Nov 9 '10 at 18:03
I'm glad you worked this calculation out. After I realized that probability came out to $\frac{1}{2}$ I was wondering if we could characterize the cases in which that occurred. – Mike Spivey Nov 10 '10 at 17:54 | 2015-09-05T10:28:54 | {
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http://math.stackexchange.com/questions/329919/integration-by-parts-of-delta-function | # Integration by parts of delta function
I am having the worst time trying to solve this integral,
$$\int g(t)\frac{d}{df}\delta[f(t)]dt,$$
$$= \int g(t)\frac{dt}{df}\frac{d}{df}\delta[f(t)]dt.$$
This should yield,
$$-\bigg[ \frac{dt}{df}\frac{d}{dt} \bigg( \frac{g(t)}{\frac{df}{dt}} \bigg) \bigg]_{f(t)=0}$$
I have tried every possible way to solve this integral (setting various u's and dv's).
Any help would be greatly appreciated!
First Attempt:
$$u = \frac{dt}{df}\frac{d}{dt} \rightarrow du=\frac{d}{dt}\bigg(\frac{dt}{df}\frac{d}{dt}\bigg)dt$$ $$dv = g(t)\delta[f(t)]dt \rightarrow v = \bigg[\frac{g(t)}{|df/dt|}\bigg]_{f(t)=0}$$
But this is an epic fail (I think)
Second Attempt:
$$u = g(t)\frac{dt}{df} \rightarrow du=\frac{d}{dt}\bigg(g(t)\frac{df}{dt}\bigg)dt$$ $$dv = \frac{d}{dt}\delta[f(t)]dt \rightarrow v = \delta[f(t)]$$
But I don't think this yields the correct solution either...
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I just wanted to say that I really appreciate that you showed your work. So that you know, in your first attempt, setting $u = \frac{dt'}{df} \frac{d}{dt'}$ is weird because $\frac{d}{dt'}$ is neither a regular function nor a number, but an operator that takes in functions. In the second, you took $g(t')$ out of its derivative without noting that it was differentiated. – mixedmath Mar 14 '13 at 2:09
@Shinobii: can you please change the notation? I am assuming that you mean for $t'$ to be a variable name. Is that correct. If you change all of those to $t$, nothing will change in the result. Is my assumption correct? – Amzoti Mar 14 '13 at 2:36
You are indeed correct, I can change it for aesthetics. – Shinobii Mar 14 '13 at 3:41
Heuristically we have
\begin{align*} \int g \frac{d\delta}{df}(f) \, dt &= \int g \frac{d\delta}{df}(f) \frac{dt}{df} \, df = \int \frac{g}{f'} \frac{d\delta}{df}(f) \, df \\ &= - \int \frac{d}{df}\bigg(\frac{g}{f'}\bigg) \delta(f) \, df \\ &= - \int \frac{dt}{df}\frac{d}{dt}\bigg(\frac{g}{f'}\bigg) \delta(f) \, df \\ &= - \left. \frac{1}{f'}\bigg(\frac{g}{f'}\bigg)' \right|_{f=0}. \end{align*}
But this is just a deceptive heuristics and we have to justify (and also calibrate) this result in mathematical language. To this end, suppose that $f$ be smooth, that $g \in C_{c}^{\infty}(\Bbb{R})$ and that $g$ decays fast enough not to raise any integrability issue. Also, let $x_j$ be zeros of $f$ and assume that they are simple zeros of $f$: $f'(x_j) \neq 0$. Then there exists a disjoint family of open neighborhoods $U_j$ of $x_j$ such that $f$ is invertible on $U_j$ with a local inverse, which we denote $f^{-1}$ whenever no ambiguity arises. Then
\begin{align*} \int_{\Bbb{R}} g \frac{d\delta}{df}(f) \, dt &= \int_{\Bbb{R}} g(t) \delta'(f(t)) \, dt = \sum_{j} \int_{U_j} g(t) \delta'(f(t)) \, dt \\ &= \sum_{j} \int_{f(U_j)} g(f^{-1}(u)) \left| (f^{-1}(u))' \right| \delta'(u) \, du \qquad (u = f(t)) \\ &= - \sum_{j} \mathrm{sgn}\,(f^{-1}(u))' \cdot \int_{f(U_j)} \big[ g(f^{-1}(u)) (f^{-1}(u))' \big]' \delta(u) \, du \\ &= - \sum_{j} \mathrm{sgn}\,(f^{-1}(u))' \cdot\left.\big[ g(f^{-1}(u)) (f^{-1}(u))' \big]'\right|_{u=0} \end{align*}
Here, we exploited the following property
$$\int_{\Bbb{R}} \varphi(x) \delta'(x) \, dx = - \int_{\Bbb{R}} \varphi'(x) \delta(x) \, dx = -\varphi'(0), \quad \varphi \in C_{c}^{\infty}(\Bbb{R}).$$
Now, simple calculation shows that
$$\big[ g(f^{-1}(u)) (f^{-1}(u))' \big]' = \frac{f'(t) g'(t)-g(t)f''(t)}{f'(t)^3},$$
where $t = f^{-1}(u)$. Finally, by noting that $f$ increases on $U_j$ if and only if $f^{-1}$ increases on $f(U_j)$, it follows that $\mathrm{sgn} \, (f^{-1})'(0) = \mathrm{sgn}\,f'(x_j)$. Therefore we have
\begin{align*} \int_{\Bbb{R}} g \frac{d\delta}{df}(f) \, dt &= - \sum_{j} \mathrm{sgn} \, f'(t) \cdot \left. \frac{f'(t) g'(t)-g(t) f''(t)}{f'(t)^3} \right|_{t=x_j} \\ &= - \sum_{j} \left. \frac{1}{\left| f'(t) \right|} \bigg( \frac{g(t)}{f'(t)} \bigg)' \right|_{t=x_j} \\ \end{align*}
as desired.
Edit. Following jbc's advice, I made amends to the sign problem. I realized this problem few days ago, but forgot fixing until he/she pointed it out explicitly. It is to blame my short-term memory :( And also thanks, jbc!
-
Wow, thank you very much! Very impressive answer, I completely understand now. – Shinobii Mar 14 '13 at 12:24
I would like to add something to the above answer for two reasons. Firstly, I think that there is a sign error in the final result. Secondly, despite the title, I think that it is not about partial integration but about changing variables in distributions and this is a recurring theme on this site. Queries on this topic are usually answered with ad hoc methods, i.e., by assuming that the given object exists and then that one can manipulate it as one does in the classical case. As an empirical method, there is nothing to be said against this procedure but since this is a mathematical forum one should take account of the fact that substituting functions into distributions is rather more delicate than in the classical case---the substition $T \circ \phi$ can only be carried out under special assumptions on the distribution $T$ and the smooth function $\phi$. (We remark that most applications are for the specal case where $T$ is a $\delta$-distribution or, as in this query, its derivative).
A rigorous and elementary definition was given by Sebastiao e Silva in the 60's and is expounded in the text "An elementary introduction to the theory of ditributions" by Campos Ferreira. Briefly, if a distribution is of finite order, i.e., is an iterated derivative of a continuous function (as $\delta$ and $\delta'$ are), say $T = \tilde D^m F$ for a continuous function $F$ (where the symbol $\tilde D$ denotes the derivative in the sense of distributions), then $T\circ \phi$ is defined to be $\left (\frac 1 {\phi' }\tilde D\right )^m (F\circ \phi)$. (For the motivation for this definition and the proof that it is well-defined---the hard part---and has the properties one would want, see the above reference). One sees that this definition only makes sense if $\phi$ is a homeomorphism whose derivative doesn't vanish (one can extend it by a more careful analysis but this will not affect the following).
If one applies this to $\delta$ and its derivative, one gets the formulae $$\delta \circ \phi = \frac 1{|\phi'(a)|} \delta_a$$ and $$\delta' \circ \phi = \frac 1{|\phi'(a)^3|}\left ( \phi'(a)\delta_a'+\phi''(a) \delta_a \right )$$ where $a$ is $\phi^{-1}(0)$.
The formula requested in the query is then obtained by applying these distributions to the test function $g$ and using the rules for integrating terms involving $\delta$ and $\delta'$. (We have taken the liberty of using a different notation for derivatives). A rather more delicate analysis can be used to extend this to the case of a smooth function $\phi$ for which $\phi$ has, say, only finitely many zeroes. The formulae remain the same but one sums over the set of zeroes.
-
Hmm, perhaps you can elaborate on why you think there is an error with the sign. I cannot see any errors, but perhaps I am mistaken. Also, thanks for the reference, in this case this was a physical problem as opposed to a strictly mathematical one, but what is math without rigor! – Shinobii Mar 17 '13 at 19:40
Oh, never mind! I see the error you were talking about. Thanks for pointing that out and thanks to sos440 for the edit! – Shinobii Mar 20 '13 at 2:36 | 2015-04-18T17:07:50 | {
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https://math.stackexchange.com/questions/3114755/problem-on-tangents-drawn-to-a-circle | # Problem on tangents drawn to a circle
I am solving Co-ordinate geometry by S.L. Loney. I am stuck on a problem on circles involving tangents and chords. I am not sure, if my approach is correct to solving this problem. Any inputs, tips that would lead me to correctly solve this problem would help!
Tangents are drawn to circle $$x^2+y^2=12$$ at the points where it is met by the circle $$x^2+y^2-5x+3y-2=0$$. Find the point of intersection of these tangents.
Solution(My attempt).
The two circles have a common chord. If $$(x_1,y_1)$$ be the required, the chord of contact of the tangents drawn to the circle $$x^2+y^2=12$$ is:
$$xx_1+yy_1=12$$
But, the chord of contact of the tangents drawn through $$(x_1,y_1)$$ to the circle $$x^2+y^2-5x+3y-2=0$$ is:
$$xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0\\ xx_{1}+yy_{1}-\frac{5}{2}(x+x_{1})+\frac{3}{2}(y+y_{1})-2=0\\ 2xx_{1}+2yy_{1}-5(x+x_{1})+3(y+y_{1})-4=0\\ x(2x_{1}-5)+y(2y_{1}+3)-5x_{1}+3y_{1}-4=0$$
I am comparing the above two equations and attempting to solve for $$x_{1},y_{1}$$. Am I thinking on the right lines?
• Whay are you looking at the chord of contact of the second circle? – amd Feb 16 '19 at 8:37
Let $$A(a,b)$$ be the intersection point, $$C(x_1,y_1)$$ and $$D(x_2,y_2)$$ be intersection points of our circles.
Thus, the equation of the line $$CD$$ (the radical axis of our circles) it's $$x^2+y^2-12-(x^2+y^2-5x+3y-2)=0$$ or $$5x-3y=10.$$ Id est, we got the following system. $$ax_1+by_1=12,$$ $$ax_2+by_2=12,$$ $$5x_1-3y_1=10$$ and $$5x_2-3y_2=10,$$ which gives $$a(x_1-x_2)+b(y_1-y_2)=0$$ and $$5(x_1-x_2)-3(y_1-y_2)=0,$$ which gives $$A(5t,-3t)$$ for some real $$t$$.
Now, easy to see that $$A$$ is placed in the fourth quadrant, which says $$t>0$$.
Let $$K$$ be an intersection point of lines $$AO$$ and $$5x-3y=10.$$
Thus, $$CK\perp AO$$ and $$AC\perp CO$$, which gives $$CO^2=OK\cdot AO$$ or $$12=\frac{|-10|}{\sqrt{5^2+(-3)^2}}\cdot\sqrt{(5t)^2+(-3t)^2}$$ or $$|t|=\frac{6}{5},$$ which gives $$t=\frac{6}{5}$$ and $$A\left(6,-\frac{18}{5}\right).$$
yes you are thinking right. But you can do it relatively short. Don't find the equation of chord of contact individually just use equation of family of circles S1-kS2=0.(k not equal to -1). But if you replace k=-1 then it will give the chord of contact(much like solving the two circles). in this case it is -5x+3y+10=0 and cmpare it with chord of contact for any circle xx1+yy1-12=0 to get x1= 6 y1=18/5. tell me if i am doing it wrong.
• Welcome to the site! Please use MathJax. – Toby Mak Feb 16 '19 at 7:16
I propose a slightly different (and I believe efficient) approach. Once you have the radical axis $$r: 5x-3y = 10$$ consider that the common point between the tangents will be on the line passing through origin (center of $$\gamma : x^2+y^2 = 12$$) perpendicular to $$r$$, i.e. $$s:3x+5y=0.$$ If $$AB$$ is the chord cut out by $$\gamma$$ on $$r$$, $$M$$ is its mid-point, and $$C$$ is the intersection point you are looking for, then $$\triangle OAC$$ is right-angled and $$AM$$ is its altitude relative to the hypothenuse. So, by first Euclid's Theorem we get $$\overline{OA}^2 = \overline{OM}\cdot \overline{OC},$$ that is $$\frac{\overline{OC}}{\overline{OM}}=\frac{12}{\overline{OM}^2}.$$ $$M$$'s coordinates are obtained right away by intersection $$r$$ and $$s$$. We get $$M\left(\frac{25}{17},-\frac{15}{17}\right)$$. Therefore $$\overline{OM}^2 =\frac{850}{289},$$ and $$\frac{\overline{OC}}{\overline{OM}} = \frac{1734}{425}.$$ Thus $$x_C = x_M \cdot \frac{\overline{OC}}{\overline{OM}}=6,$$ and, from the equation of $$s$$, $$y_C = -x_C\cdot\frac{3}{5}=-\frac{18}{5}.$$
As a supplement to the other answers, I offer the following way to find the intersection point of the tangent lines: it is the pole of the radical axis. So, as per the other answers, subtract the equation of one circle from the other to get the equation $$5x-3y-10=0$$ of the radical axis, on which the two intersection points lie. Using the method described here, we compute $$\pmatrix{1&0&0\\0&1&0\\0&0&-\frac1{12}}\pmatrix{5\\-3\\-10} = \pmatrix{5\\-3\\\frac56},$$ so the tangents intersect at $$\left({5\over5/6},{-3\over5/6}\right)=\left(6,-\frac{18}5\right)$$. | 2020-10-31T23:06:26 | {
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https://math.stackexchange.com/questions/3957995/how-to-simplify-absolute-value-expressions-such-as-x1x-1-algebraically/3958018 | # How to simplify absolute value expressions such as: $|x+1|+|x-1|$ algebraically?
After plotting this function on desmos, I understood that this absolute expression is actually a piecewise function.
Hence this: $$f(x)=|x+1|+|x-1|$$is the same as f(x)=\left\{\begin{align}2x\quad&\text{if }x>1\\-2x\quad &\text{if }x<-1\\2\quad &\text{if}-1
But is there a way to do this algebraically instead of plotting?
• In this case $f(x)$ is the sum of the distances from the points $-1$ and $1$ on the number line - a fact which helped me to work problems like this through (and this remains true also in the complex plane). – Mark Bennet Dec 22 '20 at 7:48
The way to see this is as follows: you must observe that $$|f(x)| = \begin{cases} f(x), & f(x) \geq 0 \\ -f(x), & f(x) < 0\text{.} \end{cases}$$ It does not matter where the equality sign goes (i.e., $$\geq$$ as opposed to $$>$$, or $$<$$ as opposed to $$\leq$$) in the vast majority of situations.
We thus have $$|x + 1| = \begin{cases} x + 1, & x + 1 \geq 0 \\ -(x+1), & x + 1 < 0 \end{cases} = \begin{cases} x + 1, & x \geq -1 \\ -x - 1, & x < -1 \end{cases}$$ $$|x - 1| = \begin{cases} x - 1, & x - 1 \geq 0 \\ -(x - 1), & x - 1 < 0 \end{cases} = \begin{cases} x - 1, & x \geq 1 \\ -x + 1, & x <1\text{.} \end{cases}$$ Next, we must add these functions together.
First, observe that $$|x + 1|$$ is defined piecewise over the intervals $$(-\infty, -1)$$ and then $$[-1, \infty)$$. These are indicated by the blue and red below respectively.
For $$|x - 1|$$, we must look at $$(-\infty, 1)$$ and then $$[1, \infty)$$. These are indicated by the orange and green below respectively.
Apologies for my art skills.
Given the above, it is clear that the sum of $$|x + 1|$$ and $$|x- 1|$$, as a piecewise function, must have three components:
• The blue-orange one, in $$(-\infty, -1)$$
• The red-orange one, in $$[-1, 1)$$
• The red-green one, in $$[1, \infty)$$.
The blue-orange component is when $$x < -1$$, or $$|x + 1| = -x - 1$$ and $$|x - 1| = -x + 1$$. Thus the sum of these is $$-x - 1 + (-x + 1) = -2x$$.
The red-orange component is when $$-1 \leq x < 1$$, or $$|x + 1| = x + 1$$ and $$|x - 1| = -x + 1$$. Thus the sum of these is $$x + 1 + (-x + 1) = 2$$.
The red-green component is when $$x \geq 1$$, or $$|x + 1| = x + 1$$ and $$|x - 1| = x - 1$$. Thus the sum of these is $$x + 1 + (x - 1) = 2x$$.
Plotting can give hints on what to do, but they're not conclusive.
The especially interesting points for $$|f(x)|$$ are those where $$f$$ “changes sign". In your case you have two of them, namely $$-1$$ and $$1$$.
So you can start to see what happens when $$x<-1$$: here we have $$x+1<0$$ and also $$x-1<0$$, so $$f(x)=-(x+1)-(x-1)=-2x$$ When $$-1\le x\le 1$$, we have $$x+1\ge0$$ and $$x-1\le 0$$, so $$f(x)=(x+1)-(x-1)=2$$ When $$x>1$$, we have $$x+1>0$$ and $$x-1>0$$, so $$f(x)=(x+1)+(x-1)=2x$$ Conclusion: $$f(x)=\begin{cases} -2x & x<-1 \\[6px] 2 & -1\le x\le 1 \\[6px] 2x & x>1 \end{cases}$$ Where to put the special points? Don't worry, it's irrelevant. You could as well divide the cases into $$x\le -1 \qquad -1 and the function wouldn't change, of course.
\begin{align}\lvert x+1\rvert+\lvert x-1\rvert&=\begin{cases}x+1+\lvert x-1\rvert&\text{if }x\ge-1\\ -x-1+\lvert x-1\rvert&\text{if }x<-1\end{cases}=\\&=\begin{cases}x+1+ x-1&\text{if }x\ge-1\land x\ge1\\x+1- x+1&\text{if }x\ge-1\land x<1\\ -x-1+ x-1&\text{if }x<-1\land x\ge1\\ -x-1- x+1&\text{if }x<-1\land x<1\end{cases}=\\&=\begin{cases}2x&\text{if } x\ge1\\2&\text{if }-1\le x<1\\ -2x&\text{if }x<-1\end{cases}\end{align}
• Thank you for the answer; I'm trying to understand this. But it'd be helpful if you can elaborate this a bit? – ray_lv Dec 22 '20 at 5:47
• You solved this for 3 cases, I think. But how did you know there should be 3 cases in the first place, and how did you select the domain for each case? – ray_lv Dec 22 '20 at 5:49
• @ray_lv He didn't necessarily know there would be three cases. In the first equation there are two cases. Then each of those two cases makes two more cases and there are four cases. One of those cases is impossible: $x<-1\land x\ge1$ is always false, so you can eliminate that case. That leaves three cases. Simplify each of them. – David K Dec 22 '20 at 6:22
We know $$f(x) \geq 0$$.
And, \begin{align} \{f(x)\}^2 & = (x+1)^2 + (x-1)^2 + 2 \lvert x + 1 \rvert \cdot \lvert x - 1 \rvert \\ & = 2(x^2 + 1) + 2 \lvert x^2 - 1 \rvert \\ & = \begin{cases} 2(x^2 + 1) + 2(x^2 - 1) = 4x^2 ~~~~~ \text{if } x^2 \geq 1, \\ 2(x^2 + 1) - 2(x^2 -1) = 4 ~~~~~ \text{if } x^2 \leq 1. \end{cases} \end{align}
Because $$f(x) \geq 0$$, hence,
$$f(x) = \begin{cases} 2|x| ~~~~~ \text{if } |x| \geq 1, \\ 2 ~~~~~ \text{if } |x| \leq 1. \end{cases}$$ | 2021-04-21T14:00:04 | {
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https://www.examveda.com/one-bacterium-splits-into-eight-bacteria-of-the-next-generation-but-due-to-environment-only-50-percent-of-one-generation-can-produced-the-next-generation-if-the-seventh-generation-number-197/ | Examveda
# One bacterium splits into eight bacteria of the next generation. But due to environment, only 50% of one generation can produced the next generation. If the seventh generation number is 4096 million, what is the number in first generation?
A. 1 million
B. 2 million
C. 4 million
D. 8 million
E. None of these
### Solution(By Examveda Team)
Let the number of bacteria in the 1st generation be x, then number of bacteria in 2nd, 3rd, 4th . . . . . Generation would be
$$8\left( {\frac{{\text{x}}}{2}} \right),\,8\left( {\frac{{4{\text{x}}}}{2}} \right),\,8\left( {\frac{{16{\text{x}}}}{2}} \right)$$ . . . . And so on.
As x, 4x, 16x, 64x . . . . . it is in GP with common ratio 4
Hence, 7th term of GP,
x(4)6 = 4096
or, x = 1 or 1 million.
1. back solve,
6th year=4096/4=1024.
5th "=1024/4=256.
4th"=256/4=64.
3rd"=64/4=16.
2nd"=16/4=4.
1st"=4/1=1milion.
2. Why you didn't add remaining 50% in the next generation
3. 1/2%= 1/(2*100) = 0.005
{ Explanation: Example-
5% =5/100 so 1/2% = 1/(2*100)}
Related Questions on Percentage | 2022-01-18T19:44:08 | {
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https://learning-fin.tech/book/ch5/3_Regression.html | # Regression¶
What function we use to estimate a relationship between $$y$$ and the regressor(s) depends on the data that we have. As practice using multiple x-variables, let’s simulate a dataset that is generated by the following equation $$$y = 4 + 0.5 x + 3 x^2 + u$$$
The relationship between $$y$$ and $$x$$ in the above equation is said to be nonlinear in $$x$$.
import pandas as pd
import numpy as np
import seaborn as sns
import statsmodels.formula.api as smf
np.random.seed(0)
df = pd.DataFrame(columns=['y', 'x', 'x2'])
df['x'] = np.random.normal(0, 1, 100)
df['x2'] = df['x'] ** 2
df['y'] = 4 + 0.5 * df['x'] + 3 * df['x2'] + np.random.normal(0, 0.5, 100)
y x x2
0 15.159244 1.764052 3.111881
1 4.006576 0.400157 0.160126
2 6.727911 0.978738 0.957928
3 20.669952 2.240893 5.021602
4 14.810536 1.867558 3.487773
Note that to square column 'x' to produce 'x2', all we need to do is type df2['x'] ** 2 since ** is the command for exponentiation.
Next, plot a relationship between just y and x, assuming a linear fit.
sns.lmplot(x='x', y='y', data=df)
<seaborn.axisgrid.FacetGrid at 0x7fe0343631f0>
The estimated straight line gets things horribly wrong! The line is pretty far away from a lot of the actual data points.
The scatter plot of the data above makes clear that the data demonstrates a non-linear relationship.
What we ought do to is fit a line given by $$$\hat{y} = c + \hat{\beta}_1 x + \hat{\beta}_2 x^2$$$
where there are two beta coefficients. One for $$x$$, and one for $$x^2$$.
model = smf.ols('y ~ x + x2', data=df).fit()
print(model.summary())
OLS Regression Results
==============================================================================
Dep. Variable: y R-squared: 0.983
Method: Least Squares F-statistic: 2885.
Date: Mon, 06 Dec 2021 Prob (F-statistic): 3.91e-87
Time: 12:06:14 Log-Likelihood: -75.005
No. Observations: 100 AIC: 156.0
Df Residuals: 97 BIC: 163.8
Df Model: 2
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
Intercept 4.0714 0.066 61.584 0.000 3.940 4.203
x 0.5615 0.052 10.829 0.000 0.459 0.664
x2 2.9666 0.040 73.780 0.000 2.887 3.046
==============================================================================
Omnibus: 7.876 Durbin-Watson: 2.003
Prob(Omnibus): 0.019 Jarque-Bera (JB): 3.580
Skew: 0.188 Prob(JB): 0.167
Kurtosis: 2.152 Cond. No. 2.47
==============================================================================
Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
The variable model2 stores a lot of data. Beyond holding summary output for the performance of the regression (which we’ve accessed via the model2.summary() command), we can reference the estimated parameters directly via .params. For example:
print('Intercept: ', model.params['Intercept'])
print('beta for x:', model.params['x'])
print('beta for x2:', model.params['x2'])
Intercept: 4.071350754098632
beta for x: 0.5615154693913407
beta for x2: 2.9666242239334637
As an aside, the fact that we’re using square brackets to reference items inside of squared.params is a clue that the .params component of the variable squared was built using a dictionary-like structure.
One way to tell that the estimates of squared are better than the estimates of straight is to look at the R-squared value in the summary output. This measure takes a value between 0 and 1, with a score of 1 indicating perfect fit. For comparison purposes, consider the linear fit model below:
model_linear = smf.ols('y ~ x', data=df).fit()
print(model_linear.summary())
OLS Regression Results
==============================================================================
Dep. Variable: y R-squared: 0.056
Method: Least Squares F-statistic: 5.767
Date: Mon, 06 Dec 2021 Prob (F-statistic): 0.0182
Time: 12:06:14 Log-Likelihood: -277.26
No. Observations: 100 AIC: 558.5
Df Residuals: 98 BIC: 563.7
Df Model: 1
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
Intercept 7.0734 0.392 18.054 0.000 6.296 7.851
x 0.9318 0.388 2.401 0.018 0.162 1.702
==============================================================================
Omnibus: 49.717 Durbin-Watson: 1.848
Prob(Omnibus): 0.000 Jarque-Bera (JB): 128.463
Skew: 1.877 Prob(JB): 1.27e-28
Kurtosis: 7.091 Cond. No. 1.06
==============================================================================
Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
The linear-fit model produced an R-squared of 0.056 while the quadratic-fit model yielded a R-squared of 0.983.
Another informative way to jude a modeled relationship is by plotting the residuals. The residuals are the “un-expected” part of the equation. For example, in the linear-fit model, the residual is defined as $$$\hat{u} := y - \hat{y} = y - \hat{c} - \hat{\beta} x$$$$and in the quadratic-fit model the residual,$$\hat{u}$$, is given by$$$$\hat{u} := y - \hat{y} = y- \hat{c} - \hat{\beta}_1 x - \hat{\beta}_2 x^2$$$ Using the information in .params, we can calculate residuals.
df['resid'] = model.resid
df['resid_linear'] = model_linear.resid
Next, plot the two sets of residuals.
sns.scatterplot(x='x', y='resid_linear', data=df)
<AxesSubplot:xlabel='x', ylabel='resid_linear'>
In the straight-line model, we can see that the errors have a noticeable pattern to them. This is an indication that a more complicated function of $$x$$ would be a better description for the relationship between $$x$$ and $$y$$.
sns.scatterplot(x='x', y='resid', data=df)
<AxesSubplot:xlabel='x', ylabel='resid'>
In comparison, the residuals from the squared model look more random. Additionally, they’re substantially smaller on average, with almost all residuals having an absolute value less than one. This indicates a much better fit than the straight-line model in which the residual values were often much larger.
# Looking Beyond OLS¶
OLS works well when the $$y$$ variable in our model is a linear combination of $$x$$ variables. Note that the relationship between $$y$$ and a given regressor may be nonlinear, as in the case of $$y$$ being a function of $$x$$ and $$x^2$$. However, while we may say that $$y$$ is a nonlinear function of $$x$$ in this case, the variable $$y$$ is still a linear function of $$x$$ and $$x^2$$. To clarify: $$$y = \alpha + \beta x + u$$$$is linear in$$x$$. Likewise:$$$$y = \alpha + \beta_1 x + \beta_2 x^2 + u$$$$is linear in$$x$$and$$x^2$$. In contrast, the function$$$$y = \frac{e^{\alpha + \beta x + u}}{1 + e^{\alpha + \beta x + u}}$$$$is *nonlinear*. This last equation may look terrifyingly unnatural, but it's actually very useful. Let's get a sense of what the equation looks like by plotting the function$$$$y = \frac{e^{x}}{1 + e^{x}}.$$$
np.random.seed(0)
curve = pd.DataFrame(columns=['x','y'])
curve['x'] = np.random.randint(low=-10, high=10, size=100)
curve['y'] = np.exp(curve['x']) / (1 + np.exp(curve['x']))
sns.lineplot(x='x', y='y', data=curve)
x y
0 2 0.880797
1 5 0.993307
2 -10 0.000045
3 -7 0.000911
4 -7 0.000911
<AxesSubplot:xlabel='x', ylabel='y'>
The function $$y = e^{x}/(1+e^{x})$$ creates an S-curve with a lower bound of $$0$$ and an upper bound of $$1$$.
These bounds are useful in analytics. Often, our task is to estimate probabilities. For instance, what is the likelihood that a borrower defaults on their mortgage, the likelihood that a credit card transaction is fraudulent, or the likelihood that a company will violate its capital expenditure covenant on an outstanding loan? All of these questions require us to estimate a probability. The above function is useful because it considers a situation in which the $$y$$ variable is necessarily between $$0$$ and $$1$$ (i.e. $$0\%$$ probability and $$100\%$$ probability).
Suppose that we instead took the curve data above and tried to fit it with linear regression.
sns.lmplot(x='x', y='y', data=curve)
<seaborn.axisgrid.FacetGrid at 0x7fe005c01ac0>
Note that in the above plot, there are estimated “probabilities” (the $$\hat{y}$$ values) that are either lower than $$0$$ or higher than $$1$$. This is statistically impossible.
That fancy S-shaped function above is called the inverse logit function. That is because $$e^x / (1+e^x)$$ is the inverse of the logit function given by $$log(x / (1-x)$$. Just like we can invert the equation $$y = f(x)$$ to get $$f^{-1}(y) = x$$, the inverse of $$y = e^x / (1+e^x)$$ is given by $$log(y / (1-y)) = x$$.
When we model probabilities, the $$y$$ variable will be either $$0$$ or $$1$$ for any observations. Observations where the event occurred are recorded with $$y=1$$. For instance, in a dataset about mortgage default, those borrowers who default on their mortgage would have $$y=1$$ and everyone else would have $$y=0$$.
Suppose that we estimate a model given by $$$log\Big(\frac{y}{1-y}\Big) = \alpha + \beta x + u$$$$. Then, the estimated value$$\hat{\alpha} + \hat{\beta}x$$for a given observation would be$$log(\hat{y}/(1-\hat{y}))$$. This is what we refer to as the *log odds ratio*. The odds ratio is the probability that$$y$$equals$$1$$divided by the probability that$$y$$equals$$0$$; this is what$$\hat{y}/(1-\hat{y})$$tells us. Ultimately, our estimate for$$\hat{y}$$then tells us the likelihood that the true value for$$y$$is equal to$$1$.
This type of model is called a logistic regression model. Let’s simulate some data.
np.random.seed(0)
df2 = pd.DataFrame(columns=['x', 'y'])
df2['x'] = np.random.normal(2, 1, 1000)
df2['xb'] = -9 + 4 * df2['x']
df2['y'] = np.random.binomial(n=1, p= np.exp(df2['xb']) / (1+np.exp(df2['xb'])) )
x y xb
0 3.764052 1 6.056209
1 2.400157 0 0.600629
2 2.978738 1 2.914952
3 4.240893 1 7.963573
4 3.867558 1 6.470232
Now try fitting a model using linear regression (ordinary least squares). When OLS is applied to a $$y$$ variable that only takes value 0 or 1, the model is referred to as a linear probability model (LPM).
lpm = smf.ols('y ~ x', data=df2).fit()
print(lpm.summary())
OLS Regression Results
==============================================================================
Dep. Variable: y R-squared: 0.514
Method: Least Squares F-statistic: 1056.
Date: Mon, 06 Dec 2021 Prob (F-statistic): 1.41e-158
Time: 12:06:16 Log-Likelihood: -346.17
No. Observations: 1000 AIC: 696.3
Df Residuals: 998 BIC: 706.2
Df Model: 1
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
Intercept -0.2928 0.024 -12.187 0.000 -0.340 -0.246
x 0.3564 0.011 32.493 0.000 0.335 0.378
==============================================================================
Omnibus: 113.684 Durbin-Watson: 2.045
Prob(Omnibus): 0.000 Jarque-Bera (JB): 37.844
Skew: 0.207 Prob(JB): 6.06e-09
Kurtosis: 2.142 Cond. No. 5.70
==============================================================================
Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
Now fit the model using logistic regression.
logit = smf.logit('y ~ x', data=df2).fit()
print(logit.summary())
Optimization terminated successfully.
Current function value: 0.292801
Iterations 8
Logit Regression Results
==============================================================================
Dep. Variable: y No. Observations: 1000
Model: Logit Df Residuals: 998
Method: MLE Df Model: 1
Date: Mon, 06 Dec 2021 Pseudo R-squ.: 0.5660
Time: 12:06:16 Log-Likelihood: -292.80
converged: True LL-Null: -674.60
Covariance Type: nonrobust LLR p-value: 4.432e-168
==============================================================================
coef std err z P>|z| [0.025 0.975]
------------------------------------------------------------------------------
Intercept -8.5975 0.567 -15.154 0.000 -9.709 -7.486
x 3.8987 0.259 15.062 0.000 3.391 4.406
==============================================================================
The estimated parameters (intercept and $$\beta$$ coefficient) are much closer to their true value using logistic regression. While popular, linear probability models do not often give meaningful estimates. This is especially true when outcomes are rare (meaning most $$y$$ values are either 0 or 1, and there is not a relatively even balance between the two). | 2022-01-17T22:59:38 | {
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https://calculus.subwiki.org/wiki/Quadratic_function_of_multiple_variables | # Quadratic function of multiple variables
## Definition
Consider variables $x_1,x_2,\dots,x_n$. A quadratic function of the variables $x_1,x_2,\dots,x_n$ is a function of the form:
$\left(\sum_{i=1}^n \sum_{j=1}^n a_{ij} x_ix_j\right) + \left(\sum_{i=1}^n b_ix_i\right) + c$
In vector form, if we denote by $\vec{x}$ the column vector with coordinates $x_1,x_2,\dots,x_n$, then we can write the function as:
$\vec{x}^TA\vec{x} + \vec{b}^T\vec{x} + c$
where $A$ is a $n \times n$ matrix with entries $a_{ij}$ and $\vec{b}$ is the column vector with entries $b_i$.
Note that the matrix $A$ is non-unique: if $A + A^T = F + F^T$ then we could replace $A$ by $F$. Therefore, we could choose to replace $A$ by the matrix $(A + A^T)/2$ and have the advantage of working with a symmetric matrix.
## Key data
For the discussion here, assume that $A$ has been made a symmetric matrix.
Item Value Consistency with the case $n = 1$, where $f(x) = ax^2 + bx + c$, $A = (a)$ (a $1 \times 1$ matrix), $\vec{b} = (b)$ (a 1-dimensional vector)
default domain the whole of $\R^n$ the whole of $\R$
range If the matrix $A$ is not positive semidefinite or negative semidefinite, the range is all of $\R$.
If the matrix $A$ is positive definite or ($A$ is positive semidefinite and $\vec{b}$ is in its image), the range is $[m,\infty)$ where $m$ is the minimum value. If the matrix $A$ is negative definite or ($A$ is negative semidefinite and $\vec{b}$ is in its image), the range is $(-\infty,m]$ where $m$ is the maximum value.
The case of "not positive semidefinite or negative semidefinite" does not arise for $n = 1$. Moreover, all the semidefinite cases must be definite, so we only have to consider the positive definite case and the negative definite case.
The positive definite case corresponds to $a > 0$
The negative definite case corresponds to $a < 0$
local minimum value and points of attainment If the matrix $A$ is positive definite, then $c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}$, attained at $\frac{-1}{2}A^{-1}\vec{b}$
If $A$ is positive semidefinite but not positive definite, it depends on whether $\vec{b}$ is in the image of $A$. If yes, replace $A^{-1}\vec{b}$ with the solution $\vec{v}$ to $A\vec{v} = \vec{b}$, so we get a local minimum of $c - \frac{1}{4}\vec{b}^T\vec{v}$ attained at $\frac{-1}{2}\vec{v}$
If $A$ is not positive semidefinite or if $\vec{b}$ is not in the image of $A$, no local minimum value
The positive definite case corresponds to $a > 0$: Here, the local minimum value of $c - \frac{b^2}{4a}$ is attained at $\frac{-b}{2a}$ (consistent with the matrix formulation)
The negative definite case corresponds to $a < 0$, and there is no minimum in this case.
local maximum value and points of attainment If the matrix $A$ is negative definite, then $c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}$, attained at $\frac{-1}{2}A^{-1}\vec{b}$
If $A$ is negative semidefinite but not negative definite, it depends on whether $\vec{b}$ is in the image of $A$. If yes, replace $A^{-1}\vec{b}$ with the solution $\vec{v}$ to $A\vec{v} = \vec{b}$, so we get a local minimum of $c - \frac{1}{4}\vec{b}^T\vec{v}$ attained at $\frac{-1}{2}\vec{v}$
If $A$ is not negative semidefinite or if $\vec{b}$ is not in the image of $A$, no local minimum value
The negative definite case corresponds to $a < 0$: Here, the local maximum value of $c - \frac{b^2}{4a}$ is attained at $\frac{-b}{2a}$ (consistent with the matrix formulation)
The positive definite case corresponds to $a > 0$, and there is no maximum in this case.
gradient vector function (analogous to the derivative) $\vec{x} \mapsto 2A\vec{x} + \vec{b}$ the derivative is $x \mapsto 2ax + b$ (consistent with the matrix formulation)
Hessian matrix (analogous to the second derivative) $\vec{x} \mapsto 2A$ (constant matrix-valued function) the second derivative is the constant function $x \mapsto 2a$ (consistent with the matrix formulation)
## Differentiation
### Partial derivatives and gradient vector
#### Case of general matrix
The partial derivative with respect to the variable $x_i$, and therefore also the $i^{th}$ coordinate of the gradient vector, is given by:
$\frac{\partial f}{\partial x_i} = \left(\sum_{j=1}^n (a_{ij} + a_{ji})x_j\right) + b_i$
In terms of the matrix and vector notation, the gradient vector, expressed as a column vector, is:
$(\nabla f)(\vec{x}) = (A + A^T)\vec{x} + \vec{b}$
#### Case of symmetric matrix
In the case that $A$ is a symmetric matrix, the above expressions simplify as follows.
Since $a_{ij} = a_{ji}$ for all $i,j$, the expression for the partial derivative becomes:
$\frac{\partial f}{\partial x_i} = \left(\sum_{j=1}^n 2a_{ij} x_j\right) + b_i$
The expression for the gradient vector becomes:
$(\nabla f)(\vec{x}) = 2A\vec{x} + \vec{b}$
#### Case $n = 1$
A sanity check for the above expressions is that in the case $n = 1$, where $A = (a), \vec{b} = b$, we get the same answers as for the quadratic function $f(x) = ax^2 + bx + c$.
This is indeed the case. The only partial derivative here is the ordinary derivative, and this also is the gradient vector, and has expression:
$f'(x) = 2ax + b$
This agrees with both the expression for $\partial f/\partial x_i$ and the expression for $(\nabla f)(\vec{x})$.
### Second-order partial derivatives and Hessian matrix
#### Case of general matrix
Recall that we had obtained (we replace the dummy variable $j$ by $k$ to facilitate differentiation with respect to $j$ in the next step):
$\frac{\partial f}{\partial x_i} = \left(\sum_{k=1}^n (a_{ik} + a_{ki})x_k\right) + b_i$
Differentiating both sides with respect to $x_j$ (note that $j$ may be equal to $i$ or different from $i$) we find that the only term with a nonzero derivative is the term where $k = j$. In this case, the derivative is the coefficient of $x_j$. Therefore, we obtain:
$\frac{\partial^2 f}{\partial x_j \partial x_i} = a_{ij} + a_{ji}$
Thus, the Hessian matrix of the quadratic function is given as:
$H(f)(\vec{x}) = A + A^T$
Note that this is independent of the choice of $\vec{x}$. This fact is true only because of the nature of the function: for more general functional forms, the Hessian matrix varies with the choice of input vector.
We can also see this in matrix form directly. The gradient function is:
$(\nabla f)(\vec{x}) = (A + A^T)\vec{x} + \vec{b}$
This is a linear transformation, and the Jacobian matrix of this linear transformation computes the Hessian that we want. We can use the well-known fact that the Jacobian matrix of a linear transformation coincides with the matrix describing the linear part of the transformation, and therefore the Hessian is:
$H(f)(\vec{x}) = A + A^T$
#### Case of symmetric matrix
We can either plug into the formulas for the general case or perform similar calculations to get the formulas in the case that $A$ is a symmetric matrix:
$\frac{\partial^2 f}{\partial x_j \partial x_i} = 2a_{ij}$
$H(f)(\vec{x}) = 2A$
#### Case $n = 1$
A sanity check for the above expressions is that in the case $n = 1$, where $A = (a), \vec{b} = b$, we get the same answers as for the quadratic function $f(x) = ax^2 + bx + c$.
This is indeed the case. The only second-order partial derivative is $f''(x) = 2a$. This agrees both with the formula for the second-order partial derivative and with the formula for the Hessian matrix.
### Higher derivatives
All higher order partial derivatives (pure or mixed) are zero. This can be seen directly from the fact that the second-order partial derivatives are all constants, so differentiating them further (with respect to any variable) gives zero.
Therefore, the higher derivative tensors (the higher-order analogues of the gradient vector and Hessian matrix) are also identically zero.
## Points and intervals of interest
For the discussion here, assume that $A$ is symmetric. If it is not, replace $A$ by the matrix $(A + A^T)/2$.
### Critical points
#### Case that the matrix $A$ is invertible
To find the critical points, we need to set the gradient vector equal to zero. This gives the vector equation:
$2A\vec{x} + \vec{b} = \vec{0}$
In other words:
$A \vec{x} = \frac{-1}{2}\vec{b}$
Under the assumption that $A$ is invertible (or equivalently, that all its eigenvalues are nonzero), we can left-multiply both sides by $A^{-1}$ and obtain:
$\vec{x} = \frac{-1}{2} A^{-1} \vec{b}$
We thus have a unique critical point as described above.
#### Case that the matrix $A$ is non-invertible
We still need to solve the same linear system:
$2A \vec{x} + \vec{b}$
However, since $A$ is no longer invertible (i.e., it has zero as an eigenvalue, or equivalently, it has a nonzero kernel), two cases arise:
• No solution exists. This happens if the vector $\vec{b}$ is not in the image of the linear transformation defined by $A$. Conceptually, what is happening is that the linear part of $f$ is not constrained by the quadratic part, and therefore the function is unbounded.
• A solution exists, but it is not unique. This happens if the vector $\vec{b}$ is in the image of the linear transformation defined by $A$. In fact, the solution set is an affine space of dimension equal to the nullity of $A$. Thus, we get an affine space's worth of critical points.
### Determination of local extremum behavior at critical points
#### Case that the matrix $A$ is invertible
Recall that the Hessian matrix of $f$ is $2A$. Therefore, by the second derivative test for a function of multiple variables, we obtain the following:
• If $A$ is a symmetric positive-definite matrix, i.e., all its eigenvalues are positive, then the unique critical point $\frac{-1}{2}A^{-1}\vec{b}$ is a point of local minimum and is the unique point of absolute minimum (an alternate derivation of this fact is later in the page).
• If $A$ is negative-definite matrix, i.e., all its eigenvalues are negative, then the unique critical point $\frac{-1}{2}A^{-1}\vec{b}$ is a point of local maximum and is the unique point of absolute maximum.
• If $A$ has both positive and negative eigenvalues, then the unique critical point $\frac{-1}{2}A^{-1}\vec{b}$ is neither a point of local minimum nor a point of local maximum. In fact, it gives a saddle point for the function. There is no point of local extremum and the range of the function is all of $\R$.
#### Case that the matrix $A$ is non-invertible
In this case, the Hessian matrix, $2A$, is also non-invertible and in particular has zero as an eigenvalue.
In the case that there are no critical points, there is nothing to say.
In the case that there are critical points, we note that, through any critical point, there is a direction along which the second derivative is zero. In fact, what's happening geometrically is that the set of critical points form an affine subspace and the quadratic function $f$ is constant on that affine subspace. How the behavior of that affine subspace compares with the values of the function elsewhere depends on the nonzero eigenvalues.
• In the case that $A$ is a symmetric positive-semidefinite matrix, i.e., all its nonzero eigenvalues are positive, the function attains a local minimum and also its absolute minimum at all its critical points. Note that, since the function is constant at this minimum value on the affine space, none of these is a point of strict local minimum.
• In the case that $A$ is a symmetric negative-semidefinite matrix, i.e., all its nonzero eigenvalues are negative, the function attains a local maximum and also its absolute maximum value at all its critical points. Note that, since the function is constant at this minimum value on the affine space, none of these is a point of strict local minimum.
## Geometry of the function
The geometry of the quadratic function is largely determined by the spectrum of the matrix $A$ (or equivalently, the spectrum of the matrix $2A$. As before, we shall assume that $A$ is a symmetric matrix. If not, replace $2A$ in the discussion below by $A + A^T$.
### Eigenvalues of the Hessian
Since $2A$ is a symmetric real matrix, it can be written in the form:
$2A = USU^{-1}$
where $U$ is an orthogonal matrix and $S$ is a diagonal matrix with real entries. The diagonal entries of $S$ are the eigenvalues of $2A$. Another way of thinking of the above is that with a change of basis that preserves the Euclidean norm, we can convert the Hessian to a diagonal transformation.
The following cases are of interest:
Case on $S$ (i.e., the set of eigenvalues) Corresponding term for matrix $2A$ (and hence also for $A$)
all positive symmetric positive-definite matrix
all nonnegative symmetric positive-semidefinite matrix
all negative symmetric negative-definite matrix
all nonpositive symmetric negative-semidefinite matrix
all nonzero symmetric invertible matrix
some positive, some negative (maybe some zero) indefinite matrix
## Alternate analysis of extreme values
For the discussion of cases, assume that $A$ is a symmetric matrix. If $A$ is not symmetric, replace it by the symmetric matrix $(A + A^T)/2$.
### Positive definite case
First, we consider the case where $A$ is a symmetric positive definite matrix. In other words, we can write $A$ in the form:
$A = M^TM$
where $M$ is a $n \times n$ invertible matrix.
We can "complete the square" for this function:
$f(\vec{x}) = \left(M\vec{x} + \frac{1}{2}(M^T)^{-1}\vec{b}\right)^T\left(M\vec{x} + \frac{1}{2}(M^T)^{-1}\vec{b}\right) + \left(c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}\right)$
In other words:
$f(\vec{x}) = \left \| M\vec{x} + \frac{1}{2}(M^T)^{-1}\vec{b}\right \|^2 + \left(c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}\right)$
This is minimized when the expression whose norm we are measuring is zero, so that it is minimized when we have:
$M\vec{x} + \frac{1}{2}(M^T)^{-1}\vec{b} = \vec{0}$
Simplifying, we obtain that we minimum occurs at:
$\vec{x} = -\frac{1}{2}A^{-1}\vec{b}$
Moreover, the value of the minimum is:
$c - \frac{1}{4}\vec{b}^TA^{-1}\vec{b}$ | 2020-10-24T14:20:50 | {
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https://math.stackexchange.com/questions/2728009/divisibility-property-for-sequence-a-n2-2n-1n3a-n-2n3a-n1 | # Divisibility property for sequence $a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}$
Let $$(a_n)$$ be the sequence uniquely defined by $$a_1=0,a_2=1$$ and
$$a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}$$
Can anybody show (or provide a counterexample) that $$p|a_{p-2}$$ and $$p|a_{p-1}$$ for any prime $$p\geq 5$$ ? I have checked this fact for $$p\leq 200$$.
• Is there some extra motivation behind this question? Apr 12 '18 at 6:22
• @AlexR. Not really. I deliberately omit the context in which I discovered this question because 1) the relevance of my question to the context is not certain and 2) the context itself would have been long and complicated to explain. Apr 12 '18 at 6:38
• I have checked for $p \le 32768$ Apr 12 '18 at 10:03
• a (three times) stronger conjecture is that $\prod_{n=2}^{p-4} \begin{pmatrix} -(2n+3) & 1 \\ -2(n-1)(n+3) & 0 \end{pmatrix} = \begin{pmatrix} a & 0 \\ 5a & a \end{pmatrix} \pmod p$ Apr 12 '18 at 21:46
• Seems not only for primes... Let $(b_n,c_n)=(a_{2n-1},a_{2n})$ for all $n\ge 1$. The iterative scheme is then equivalent to $(b_{n+1},c_{n+1})^{\top}=M_n(b_n,c_n)^{\top}$ for all $n\ge 1$, with $M_n=\left(\begin{array}{cc}-8(n-1)(n+1)&-(4n+1)\\8(n-1)(n+1)(4n+3)&8n^2+8n+9\end{array}\right)$. It seems that $\frac{(2n+3)!!}{3}\Big|\prod_{j=1}^nM_j$ for all $n\ge 1$, meaning that $\frac{(2n+3)!!}{3}\Big|(b_{n+1},c_{n+1})$ for all $n\ge 1$, regardless of the initial value $(b_1,c_1)$ (even if $(b_1,c_1)\ne(a_1,a_2)=(0,1)$). The divisibility of a vector or matrix here is put in the entry-wise sense. Apr 13 '18 at 6:06
The sequence starts $$0, 1, -5, 25, -105, 105, 5355, \dots$$
We can observe that the statement is true not only for primes, but for odd numbers in general. Even though recurrences might be better for solving this kind of problem, here we can go for closed form formula (luckily there is one!). I have used approach inspired by http://mathforum.org/library/drmath/view/67314.html.
Currently the coefficients contain quadratic polynomials in $n$, but we can make them linear, which makes it a bit easier to work with. With substitution $b_n=\frac{a_n}{(n+2)!}$ and some algebra we get
$$(n+4)b_{n+2}+(2n+3)b_{n+1}+(2n-2)b_n=0$$
Now that the coefficients are linear, let's try to find its generating function $y(x)=\sum_{n \geq 1}b_n x^n$. By summing the whole equation and again some technical steps, we arrive at
$$y'(x)(2x^3+2x^2+x)=\frac{1}{6}x^2+(2x^2-x-2)y(x)$$ which is a linear differential equation with variable coefficients, and can be solved to $$y(x)=-\frac{1}{3x}-\frac{1}{9x^2}-\frac{5x}{18}-\frac{1}{2}+\frac{(2x^2+2x+1)^{3/2}}{9x^2}$$ (I have used CAS system for solving this differential equation, but it is just a technicality to show, also it is trivial to verify by differentiation).
Now to get information about the coefficients, let's expand it into the series. The power on the right is by Binomial series $$(2x^2+2x+1)^{3/2}=\sum_{k=0}^{\infty}\binom{3/2}{k}2^k(x^2+x)^k$$ Using Binomial theorem for the inner sum and playing with the indicies, we obtain $$[x^n](2x^2+2x+1)^{3/2} = \sum_{n/2 \leq k \leq n}\binom{3/2}{k}\binom{k}{2k-n}2^k$$ Dividing by $9x^2$ and noticing first few terms in result series are equal to $\frac{1}{9x^2}+\frac{1}{3x}+\frac{1}{2}+\frac{5x}{18}$, we can cancel those out in the $y(x)$, and so $$b_n=\sum_{(n+2)/2 \leq k \leq n+2}\binom{3/2}{k}\binom{k}{2k-n-2}2^k$$ and in turn $$a_n=(n+2)!\sum_{(n+2)/2 \leq k \leq n+2}\binom{3/2}{k}\binom{k}{2k-n-2}2^k$$ Looking at first few individual terms in a sum for couple values of $n$, we can observe those are all integers. This suggests we can further simplify the expression. Writing out the definitions and all the factorials, one eventually finds for $n>1$: $$\boxed{a_n = \sum_{(n+2)/2 \leq k \leq n+2}\binom{k}{n+2-k}\frac{(n+2)!}{k!}\frac{(2k-5)!!}{3}(-1)^k}$$ We can see that for $k<n+2$ the terms above are divisible by $n+2$. So we have \begin{align} a_n &\equiv \binom{n+2}{0}\frac{(n+2)!}{(n+2)!}\frac{(2(n+2)-5)!!}{3}(-1)^{n+2}\\ &= \frac{(2n-1)!!}{3}(-1)^{n+2}\\ &= (2n-1)(2n-3)\cdots 5 \cdot (-1)^{n+2} \pmod{n+2} \end{align} If $n\geq 3$ is odd, the $n+2$ is odd as well, so the product above will contain it and thus will be divisible by it. In other words $a_n \equiv 0 \pmod {n+2}$, which is what we wanted to prove.
The claim that $a_{n+1} \equiv 0 \pmod {n+2}$ now follows for example by writing the original equation as
$$a_{n+1}=-2(n-2)(n+2)a_{n-1}-(2n+1)a_n$$ which modulo $n+2$ yields $$a_{n+1} \equiv 3a_n \equiv 0 \pmod {n+2}$$ I might have omitted few details here and there, but main idea is hopefully clear.
• good work indeed ! Apr 14 '18 at 13:36
• Thank you for this thorough effort. May I ask which CAS you used to solve the differential equation ? Apr 15 '18 at 8:00
• @EwanDelanoy I have used Maple dsolve((diff(y(x), x))*(2*x^3+2*x^2+x) = (1/6)*x^2+(2*x^2-x-2)*y(x), y(x)), but it is really just a linear differential equation, it's just my laziness...
– Sil
Apr 15 '18 at 8:30
• I wouldn't call that laziness, I'd call it intelligent use of tools Apr 15 '18 at 8:35
Define $$a_1=0,\; a_2=1,\;\text{and}\quad a_{n+2}=-2(n-1)(n+3)a_n-(2n+3)a_{n+1}.\tag{0}$$ Define $$\;R(x) := \sqrt{1-4x} = 1-2\sum_{n=1}^\infty C_{n-1}x^n = \sum_{n=0}^\infty {2n \choose n}\frac{x^n}{1-2n}\;$$ which is the generating function of the OEIS integer sequence A002420. Define $$\;B(x) := \frac{ (1+4x+8x^2)R(-x-2x^2)-(1+6x+18x^2+20x^3)}{2x^2}$$ $$= 3x^2 - 6x^3 + 10x^4 - 12x^5 + 3x^6 + \dots =: \sum_{n=0}^\infty b_nx^n$$ which is the generating function for $$b_n.\;$$ It is easy to check that $$b_n$$ is also an integer sequence and $$\;b_1=0, b_2=3.\;$$ Using derivatives we get the differential equation $$\;(8x^2-2x-2)B(x)+12x^2=x(1+4x+8x^2)\frac{d}{dx}B(x)\;$$ and using some algebra on power series coefficients we get the equation $$\;(2+n)b_{n}=(2-4n)b_{n-1}+(24-8n)b_{n-2}\quad\forall\;n>2.$$
Redefine $$\;a_n:=\frac{(n+2)!}{2^{n+1} 9}b_n.\tag{1}$$ It is easy to check that this satisfies the same initial values and recursion as given in $$(0)$$, but is it an integer sequence? It will be if $$\;2^{n+1}\;|\;(n+2)!\;b_n\;$$ for all $$n>1.$$ The key results to prove is that if $$\;C_n\;$$ is the Catalan numbers and if $$\;v_2(n)\;$$ is the 2-adic valuation of $$n$$, then
$$v_2(b_n) = v_2(C_{n+1}) = v_2\Big(\frac{2^{n+1}}{(n+2)!}\Big),\; v_2(a_n)=0\quad \forall n>1. \tag{2}$$
Now the original equation $$(0)$$ already implies that $$\;a_n\;$$ is an integer sequence using induction, $$\;a_n\;$$ is odd if $$n>1$$, and the rest of equation $$(2)$$ is true.
However, there is an alternative approach using the exponential generating function of $$\;a_n,\;$$ namely $$\;A(x):=\sum_{n=0}^\infty a_nx^n/n!.\;$$ Using equation $$(0)$$ for $$\;a_n\;$$ we get the differential equation $$1 = -6A(x) +(3+6x)\frac{d}{dx}A(x) + (1+2x+2x^2)\frac{d^2}{dx^2}A(x).$$ Taking the derivative of this equation gives another differential equation $$0 = 5(1+2x)\frac{d^2}{dx^2}A(x) +(1+2x+2x^2)\frac{d^3}{dx^3}A(x).$$ Now define the sequence $$\;y_n(x)\;$$ by $$\,y_0(x) := 1+O(x),\,$$ and recursion $$y_{n+1}(x) := 1-\int_0^x 5(1+2x)/(1+2x+2x^2)\;y_n(x)\; dx. \tag{3}$$ The limit as $$\;n\to\infty$$ of $$\;y_n(x) =\frac{d^2}{dx^2}A(x).\;$$ It is easy to check that the recursion $$(3)$$ preserves exponential generating functions of integer sequences. Thus, again $$\;a_n\;$$ is an integer sequence.
If $$n>0$$ is odd, then $$(n+2)\;|\;a_n$$ and if $$n>2$$ is even, then $$(n+1)\;|\;a_n.\;$$ This since one of the $$(n\!+\!1)(n\!+\!2)$$ factors in the numerator of $$(1)$$ are not cancelled by the $$2^{n+1}$$ in the denominator. | 2022-01-25T09:48:34 | {
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https://mathlesstraveled.com/2012/05/15/fibonacci-multiples/ | ## Fibonacci multiples
I haven’t written anything here in a while, but hope to write more regularly now that the semester is over—I have a series on combinatorial proofs to finish up, some books to review, and a few other things planned. But to ease back into things, here’s a little puzzle for you. Recall that the Fibonacci numbers are defined by
$F_0 = 0; F_1 = 1; F_{n+2} = F_{n+1} + F_n$.
Can you figure out a way to prove the following cute theorem?
If $m$ evenly divides $n$, then $F_m$ evenly divides $F_n$.
(Incidentally, the existence of this theorem constitutes good evidence that the “correct” definition of $F_0$ is $0$, not $1$.)
For example, $5$ evenly divides $10$, and sure enough, $F_5 = 5$ evenly divides $F_{10} = 55$. $13$ evenly divides $91$, and sure enough, $F_{13} = 233$ evenly divides $F_{91} = 4660046610375530309$ (in particular, $4660046610375530309 = 233 \times 20000200044530173$).
I know of two different ways to prove it; there are probably more! Neither of the proofs I know is particularly obvious, but they do not require any difficult concepts.
Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
This entry was posted in arithmetic, challenges, fibonacci, number theory, pattern and tagged , . Bookmark the permalink.
### 12 Responses to Fibonacci multiples
1. Dennis says:
I wonder…
2. Matt Gardner Spencer says:
Sounds exciting! I’m guessing that neither of your proofs uses the closed form of fibonacci and symmetric polynomials. But how about induction and monimos and dominos?
• Brent says:
Nope, but if you know of such a proof I’d love to hear it! One of the proofs definitely uses induction, and can probably be formulated in terms of dominos, though I haven’t thought about the details. The other one is more elementary.
I would also like to see a proof involving monimos, dominos, and geronimos.
• Matt Gardner Spencer says:
The key to my proof is using pictures like those here: https://mathlesstraveled.com/2011/05/24/post-without-words-1/ (especially those posted by Xander in the comments) to prove
$F_{mn}=F_n\cdot F_{(m-1)n+1}+F_{n-1}\cdot F_{(m-1)n}$ and then using induction.
• Brent says:
Yes, I can see how that would work. It’s different from both of my proofs (one of which can also be formulated in terms of those pictures, now that I have thought about it more carefully).
Now, how about that proof using the closed form of fibonacci and symmetric polynomials? 😉
3. Joseph Nebus says:
I’d never heard of this theorem before and I’m captivated by it. I’d have expected to run across it in those occasional lists of cool mathematics trivia.
4. Joseph Nebus says:
Reblogged this on nebusresearch and commented:
I have not, as far as I remember, encountered this theorem before. And for the time I’ve had to think about it I realize I’ve got no idea how to prove it. However, it’s a neat little result that makes me smile to hear about, and theorems that bring smiles are certainly worth sharing.
5. JRH says:
For $M = \left[\begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array}\right]$, $M^{n} = \left[\begin{array}{ll} F_{n+1} & F_{n} \\ F_{n} & F_{n-1} \end{array}\right]$. (Proof by induction left as an exercise for the reader.)
Now assume that $F_{km} = K_{k}F_{m}$ for $k \geq 1$, which can be proven for the base cases of 1 and 2 with $K_{1} = 1$ and $K_{2} = [F_{m+1}+F_{m-1}]$ by expanding $M^{2m} = M^{m}M^{m}$.
Then $M^{(k+1)m} = M^{km}M^{m}$ and so $F_{(k+1)m} = F_{km+1}F_{m} + F_{km}F_{m-1} = K_{k+1}F_{m}$, with $K_{k+1} = [F_{km+1} + K_{k}F_{m-1}]$.
I really wish there were a “preview comment” button so I could double check all of that LaTeX before posting.
• Brent says:
Cool proof!
I really wish there were a “preview comment” button too. But wordpress.com does not support it. =( | 2018-11-16T09:45:12 | {
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https://www.khanacademy.org/math/old-integral-calculus/integration-techniques/trig-substitution/v/introduction-to-trigonometric-substitution | If you're seeing this message, it means we're having trouble loading external resources on our website.
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# Introduction to trigonometric substitution
Introduction to trigonometric substitution.
## Want to join the conversation?
• Substituting x = 2 cos t I get the answer -arccos x/2 + C
Is that a valid solution?
• That is also a valid solution, yes. Recall the identity arcsin(u) = π/2 - arccos(u). Letting u = x/2 and observing that the constant C may be written as π/2 + C', for some constant C', we get
-arccos(x/2) + C = π/2 - arccos(x/2) + C' = arcsin(x/2) + C',
which brings us back to the solution Sal arrived at.
• Was the choice of 2 for the hypotenuse completely arbitrary?
• No. You should choose the hypotenuse so that it matches what is under the radical. The denominator was sqrt(4 - x²), and he chose the hypotenuse to be 2 and one side to be x in order for the third side of the triangle to be sqrt(4 - x²). So for example if the original integral had sqrt(16 - 25x²), then you would choose the hypotenuse to be 4 and one side to be 5x, so that the third side would be sqrt(16 - 25x²).
• At , Sal says, "If we define this angle as theta...". Why make this angle theta and not the other acute angle?
• A similar question was asked that was already answered, but if we do it for the other angle theta, we would get the same answer in a different form of
-arccos(x/2) + C.
From Qeeko:
That is also a valid solution, yes. Recall the identity arcsin(u) = π/2 - arccos(u). Letting u = x/2 and observing that the constant C may be written as π/2 + C', for some constant C', we get
-arccos(x/2) + C = π/2 - arccos(x/2) + C' = arcsin(x/2) + C',
which brings us back to the solution Sal arrived at.
• At , Sal says that if x=2sin(theta) then dx=2cos(theta)d(theta), but I do not follow. What is the reason?
• He is differentiating x with respect to θ.
Before differentiating you have: x = 2·sin(θ). After differentiating with respect to θ you have:
dx/dθ = 2·cos(θ)
Finally, you solve for dx:
dx = 2·cos(θ)·dθ
• I've noticed that the question was a lot like the derivative of arcsin(theta) with the exception that instead of 1/sqrt(1-x^2), we had 1/sqrt(4-x^2). The answer, in the end, was just arcsin(x/2). Is there a pattern to it? If it is 1/sqrt(1-x^2), it is arcsin(x/1), and if it is 1/sqrt(9-x^3), the answer should be arcsin(x/3).
Is this actually true? And if so, can you please explain why?
Thank you :)
• I hope that you mean 1/√(9 - x²), not ( ...x³), the exponent on x needs to be 2 because the Pythagorean Theorem is the key to this technique. Something of the form 1/√(a² - x²) is perfect for trig substitution using x = a · sin θ. That's the pattern. Sal's explanation using the right triangle shows why that pattern works, "a" is the hypotenuse, the x-side opposite θ is equal to a · sin θ, and the adjacent side √(a² - x²) is equal to a · cos θ . Using those substitutions, the original integral becomes easy - you just have to remember to restrict the Domain.
• I'm wondering why, when constructing the triangle, you let the opposite side be x and the adjacent be sqrt(4-x^2). If you reverse this, and let x be the adjacent side for example, you get a completely different answer. How are you supposed to know which is the correct orientation??
• Making x the adjacent side and sqrt'(4-x^2)' the opposite side will be identical with labeling the other angle, ( the one complementary to theta), the new theta. Doing so, and solving for the integral, you will end up with the answer:'-arccos(x/2)+c', which is another equally possible solution to the problem.
I hope this helped!
(1 vote)
• Did you just assume that the hypotenuse is 2, or is there some way to tell that the hypotenuse is 2.
• There is a way to tell.
The Pythagorean Theorem says a² + b² = c², where c is the hypotenuse.
That means one of the sides, lets say a, is equal to sqrt(c² - b²).
Now we have been given an expression that looks just like that: sqrt(4 - x²), which can be rewritten as sqrt(2² - x²).
That means that for this trig sub triangle, one side (the b side) is equal to x, the other side, the a side is equal to sqrt(2² - x²) which means the hypotenuse for this triangle must be 2.
https://en.wikiversity.org/wiki/Trigonometric_Substitutions
• What if we would've chosen the other side as our x in the beginning?
• Hi, tuf62486!
There are two scenarios, and these are as follows:
- Scenario ONE: https://i.imgur.com/atiQ9yT.png
- Scenario TWO: https://i.imgur.com/1AmZgHZ.png
Explanations:
- Scenario ONE: this one is comparatively complex, but it still does make sense. If we choose x as the side adjacent to theta, then we will end up with -arccos(x / sqrt(u)) + c. Although this appears different, if you look at this graph (drag the slider at the top-left at your own leisure) ==> https://www.desmos.com/calculator/3gfueg1fmv <== then you will see that -arccos(x / sqrt(u)) + pi/2 is actually equal to arcsin(x / sqrt(u)). Since after we integrate we are left with a constant of integration (I have used c to denote said constant), this "absorbs" the extra pi / 2 that we would need to add for the graphs to match exactly. Regardless, the curves for -arccos(x / sqrt(u)) and arcsin(x / sqrt(u)) are identical except that one is vertical translation of the other, and that is all we are trying to prove with integration, anyway.
- Scenario TWO: This is identical to how Sal solves the problem in the video. We have just swapped x and theta, but I have solved the problem so that you can see that it will be the same.
I hope that this helps! Cheers,
~ Novum Sensum | 2023-02-04T03:16:36 | {
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https://math.stackexchange.com/questions/1631300/must-this-rng-be-a-ring | Must this rng be a ring?
A rng is a ring without the assumption that the ring contains an identity. Consider a finite rng $\mathbf{R}$.
I am investigating conditions that get close forcing an identity but not quite. The closest condition I can think of is the following:
If $a\in \mathbf{R}$ is non-zero then there is $b\in\mathbf{R}$ such that $ab\neq 0$
I am finding myself unable to prove that $\mathbf{R}$ must/need-not have a multiplicative identity i.e. be a ring.
Are there well-known results/examples that deal with this sort of condition?
• You have already asked the exact same question, buy you deleted it (which means a lot of people will not find anything at the end of that link). I guess my question is: why? – Arthur Jan 29 '16 at 0:33
• @Arthur No, I mistakenly forgot to add finite in my post - the question is trivial otherwise. I decided to delete and re-post as editing it would make the 10 comments irrelevant. – user111064 Jan 29 '16 at 0:35
• You are allowed to edit a question, rather than deleting it. @user111064 – Thomas Andrews Jan 29 '16 at 0:37
• So it is not the exact same question, then. Anyways, I made a comment on the other question that there is one statement that is closer to the existence of a $1$ than what you got, and it is this: "There exists a $b$ such that for any $a\neq 0$, $ab\neq0$". Of course, if your condition implies that there is a $1$ in the finite case, then so does mine, and they are equivalent. But if not, it might be worth looking into what finite rngs, if any, fulfill your statement but not mine, or my statement, but doesn't have a $1$. – Arthur Jan 29 '16 at 0:42
• @ThomasAndrews contributed the following statement that might also be worth a look: "Consider $f_a:R\to R$ given by $f_a(x)=ax$. Then $f_a=f_b$ implies $a=b$". – Arthur Jan 29 '16 at 0:49
In the commutative case (there are probably simple non-commutative counterexamples coming from matrix rings):
We say that a commutative rng $R$ has property $\mathcal{P}$ if, for all nonzero $a\in R$, there is some $b\in R$ with $ab\neq 0$. The zero ring has property $\mathcal{P}$, and it has a unit. Let $R$ be a finite nonzero commutative rng such that all smaller commutative rngs with property $\mathcal{P}$ have a unit.
Pick some $a\neq 0$. Then there is some $b$ with $ab\neq 0$, some $c$ with $abc\neq 0$, and so on. So there exist arbitrarily long nonzero products in $R$, which implies that there is some $x\in R$ that is not nilpotent.
Since $R$ is finite, there are $d,N$ such that $x^n = x^{n+d}$ for $n>N$. Choosing $M$ so that $Md>N$, we have $(x^{Md})^2 = x^{2Md} = x^{Md}$, so there is some non-zero idempotent $e=x^{Md}$.
Let $I = \{r\in R \mid er = 0\}$. $I$ has property $\mathcal{P}$: if $r\in I$ is nonzero, then there is some $s\in R$ with $rs\neq 0$. Then $r(s-es)=s(r-er)=sr\neq 0$, and $s-es\in I$.
$I$ is strictly smaller than $R$ ($e\notin I$, because $e$ is nonzero), so, by choice of $R$, $I$ has a unit $u$. But then $u+e$ is a unit of $R$: for any $t\in R$, $(u+e)t = ut + et=u(t-et) + et = (t-et) + et = t$.
By induction, we conclude that all finite commutative rngs with property $\mathcal{P}$ have a unit.
In retrospect, the idea here is not so difficult: As in Thomas Andrews' answer, we are trying to write $R\cong R_1\bigoplus R_2$ for subrngs $R_1,R_2$. Direct summands inherit property $\mathcal{P}$, and $R$ has a unity if and only if both $R_1$ and $R_2$ do, so this lets us quickly reduce to the case of indecomposable rngs.
Furthermore, idempotents are one of the more natural ways to identify direct summands. Given an idempotent $e$, we can write $R\cong eR \bigoplus \operatorname{ann}(e)$, and $eR$ always has unity $e$. So the challenge is just to show that $R$ must contain a nonzero idempotent.
Not an answer, but a reduction to the prime power case.
It's pretty easy to reduce to when $|R|$ is the power of a prime.
If $|R|=mn$ with $\gcd(m,n)=1$, solve $mx+ny=1$. Show that $$R\to (mR)\times (nR); a\mapsto (mxa,nya)$$ is an isomorphism of rngs.
Now, if $ma\neq 0$, then $mab\neq 0$ for some $b\in R$. But note that $(ma)(mxb)=m(1-ny)(ab)$. So this means that $mR$ has the property we want, too. Similarly for $nR$.
If $mR$ and $nR$ have identities, then so does $R$. So one of $mR$ or $nR$ would have to not have an identity. We keep reducing until we find an example which is a prime power size.
• Why is your map multiplicative? – Martin Brandenburg Jan 29 '16 at 21:27
• Because $(mx)(mx)c = (mx)(1-ny)c=(mx)c-(mnxy)c=(mx)c$. So $(mx)a(mx)b=(mx)(ab).$ @MartinBrandenburg – Thomas Andrews Jan 29 '16 at 21:35
• It's basically a usual Chinese Remainder theorem argument. It also why it is onto each component - you actually have $(mxm)c=mc$. – Thomas Andrews Jan 29 '16 at 21:38 | 2019-05-25T08:55:40 | {
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https://stats.stackexchange.com/questions/53/pca-on-correlation-or-covariance | # PCA on correlation or covariance?
What are the main differences between performing principal component analysis (PCA) on the correlation matrix and on the covariance matrix? Do they give the same results?
You tend to use the covariance matrix when the variable scales are similar and the correlation matrix when variables are on different scales.
Using the correlation matrix is equivalent to standardizing each of the variables (to mean 0 and standard deviation 1). In general, PCA with and without standardizing will give different results. Especially when the scales are different.
As an example, take a look at this R heptathlon data set. Some of the variables have an average value of about 1.8 (the high jump), whereas other variables (run 800m) are around 120.
library(HSAUR)
heptathlon[,-8] # look at heptathlon data (excluding 'score' variable)
This outputs:
hurdles highjump shot run200m longjump javelin run800m
Joyner-Kersee (USA) 12.69 1.86 15.80 22.56 7.27 45.66 128.51
John (GDR) 12.85 1.80 16.23 23.65 6.71 42.56 126.12
Behmer (GDR) 13.20 1.83 14.20 23.10 6.68 44.54 124.20
Sablovskaite (URS) 13.61 1.80 15.23 23.92 6.25 42.78 132.24
Choubenkova (URS) 13.51 1.74 14.76 23.93 6.32 47.46 127.90
...
Now let's do PCA on covariance and on correlation:
# scale=T bases the PCA on the correlation matrix
hep.PC.cor = prcomp(heptathlon[,-8], scale=TRUE)
hep.PC.cov = prcomp(heptathlon[,-8], scale=FALSE)
biplot(hep.PC.cov)
biplot(hep.PC.cor)
Notice that PCA on covariance is dominated by run800m and javelin: PC1 is almost equal to run800m (and explains $$82\%$$ of the variance) and PC2 is almost equal to javelin (together they explain $$97\%$$). PCA on correlation is much more informative and reveals some structure in the data and relationships between variables (but note that the explained variances drop to $$64\%$$ and $$71\%$$).
Notice also that the outlying individuals (in this data set) are outliers regardless of whether the covariance or correlation matrix is used.
• What is the situation, if I convert the variables to z-scores first? May 18, 2013 at 16:00
• @Jirka-x1 the covariance matrix of standardized variables (i.e. z scores) equals the correlation matrix. Jul 2, 2014 at 21:22
• @Alexis Can it therefore be inferred that the covariance matrix of standardised variables equals the correlation matrix of standardised variables?
– j b
Sep 1, 2017 at 10:13
• @JamieBullock $\mathbf{\Sigma}$ (covariance matrix) for standardized data = $\mathbf{R}$ (correlation matrix). $\mathbf{R} = \mathbf{R}$ whether or not the data are standardized (correlation is insensitive to linear transformations of the data. So , for example, if you have $X$ and $Y$ and they correlate with $r_{XY}$, then if $X^{*} = aX+b$ and $Y^{*} = aY+b$ $X^{*}$ and $Y^{*}$ also correlate with $r_{XY}$). Sep 1, 2017 at 23:23
• The comment above regarding correlations among the PCs is not correct. PCA creates uncorrelated PCs regardless of whether it uses a correlation matrix or a covariance matrix. Note that in R, the prcomp() function has scale = FALSE as the default setting, which you would want to set to TRUE in most cases to standardize the variables beforehand. Jun 4, 2020 at 14:31
Bernard Flury, in his excellent book introducing multivariate analysis, described this as an anti-property of principal components. It's actually worse than choosing between correlation or covariance. If you changed the units (e.g. US style gallons, inches etc. and EU style litres, centimetres) you will get substantively different projections of the data.
The argument against automatically using correlation matrices is that it is quite a brutal way of standardising your data. The problem with automatically using the covariance matrix, which is very apparent with that heptathalon data, is that the variables with the highest variance will dominate the first principal component (the variance maximising property).
So the "best" method to use is based on a subjective choice, careful thought and some experience.
• "If you changed the units (e.g. US style gallons, inches etc. and EU style litres, centimetres) you will get substantively different projections of the data." This is false when using the correlation matrix $\mathbf{R}$, and trivially easy to demonstrate. When using the covariance matrix, it is also only true for the eigenvalues, but not for the eigenvectors. Dec 18, 2020 at 23:24
UNTRANSFORMED (RAW) DATA: If you have variables with widely varying scales for raw, untransformed data, that is, caloric intake per day, gene expression, ELISA/Luminex in units of ug/dl, ng/dl, based on several orders of magnitude of protein expression, then use correlation as an input to PCA. However, if all of your data are based on e.g. gene expression from the same platform with similar range and scale, or you are working with log equity asset returns, then using correlation will throw out a tremendous amount of information.
You actually don't need to think about the difference of using the correlation matrix $$\mathbf{R}$$ or covariance matrix $$\mathbf{C}$$ as an input to PCA, but rather, look at the diagonal values of $$\mathbf{C}$$ and $$\mathbf{R}$$. You may observe a variance of $$100$$ for one variable, and $$10$$ on another -- which are on the diagonal of $$\mathbf{C}$$. But when looking at the correlations, the diagonal contains all ones, so the variance of each variable is essentially changed to $$1$$ as you use the $$\mathbf{R}$$ matrix.
TRANSFORMED DATA: If the data have been transformed via normalization, percentiles, or mean-zero standardization (i.e., $$Z$$-scores), so that the range and scale of all the continuous variables is the same, then you could use the Covariance matrix $$\mathbf{C}$$ without any problems. (correlation will mean-zero standardize variables). Recall, however, that these transformations will not remove skewness (i.e., left or right tails in histograms) in your variables prior to running PCA. Typical PCA analysis does not involve removal of skewness; however, some readers may need to remove skewness to meet strict normality constraints.
In summary, use the correlation matrix $$\mathbf{R}$$ when within-variable range and scale widely differs, and use the covariance matrix $$\mathbf{C}$$ to preserve variance if the range and scale of variables is similar or in the same units of measure.
SKEWED VARIABLES: If any of the variables are skewed with left or right tails in their histograms, i.e., the Shapiro-Wilk or Lilliefors normality test is significant $$(P<0.05)$$, then there may be some issues if you need to apply the normality assumption. In this case, use the van der Waerden scores (transforms) determined from each variable. The van der Waerden (VDW) score for a single observation is merely the inverse cumulative (standard) normal mapping of the observation's percentile value. For example, say you have $$n=100$$ observations for a continuous variable, you can determine the VDW scores using:
1. First, sort the values in ascending order, then assign ranks, so you would obtain ranks of $$R_i=1,2,\ldots,100.$$
2. Next, determine the percentile for each observation as $$pct_i=R_i/(n+1)$$.
3. Once the percentile values are obtained, input them into the inverse mapping function for the CDF of the standard normal distribution, i.e., $$N(0,1)$$, to obtain the $$Z$$-score for each, using $$Z_i=\Phi^{-1}(pct_i)$$.
For example, if you plug in a $$pct_i$$ value 0.025, you will get $$-1.96=\Phi^{-1}(0.025)$$. Same goes for a plugin value of $$pct_i=0.975$$, you'll get $$1.96=\Phi^{-1}(0.975)$$.
Use of VDW scores is very popular in genetics, where many variables are transformed into VDW scores, and then input into analyses. The advantage of using VDW scores is that skewness and outlier effects are removed from the data, and can be used if the goal is to perform an analysis under the contraints of normality -- and every variable needs to be purely standard normal distributed with no skewness or outliers.
• This is by far the most sensible answer here, as it actually gives a proper view that covariance wins when appropriate. Too many answers here and elsewhere mention the usual "it depends" without actually giving a hard basis for why one should prefer covariance if possible. Here lep does: covariance doesn't chuck out any of the info which correlation does. The stock data example is a good one: high beta stocks will of course have higher loadings but they probably should, just like any facet of any analysis that is more volatile is usually more interesting (within reason). Nov 12, 2014 at 15:27
• Of course the problem at hand must be analysed for whether or not higher variance is an interesting facet of the analysis. If it is not, then of course correl is better, and that definitely holds if units are different. Nov 12, 2014 at 15:30
• Great answer +1. I guess one more example might be applying PCA to term structure analysis on bond yields in finance. Variances of yields on varied maturities vary, but since they are all yields, the varying scales are normally not unacceptably wide. Indeed, more/less volatility of certain maturity yield itself provides rich information. Nov 11, 2016 at 17:51
A common answer is to suggest that covariance is used when variables are on the same scale, and correlation when their scales are different. However, this is only true when scale of the variables isn't a factor. Otherwise, why would anyone ever do covariance PCA? It would be safer to always perform correlation PCA.
Imagine that your variables have different units of measure, such as meters and kilograms. It shouldn't matter whether you use meters or centimeters in this case, so you could argue that correlation matrix should be used.
Consider now population of people in different states. The units of measure are the same - counts (number) of people. Now, the scales could be different: Vermont has 600K and CA - 38M people. Should we use correlation matrix here? It depends. In some applications we do want to adjust for the size of the state. Using the covariance matrix is one way for building factors that account for the size of the state.
Hence, my answer is to use covariance matrix when variance of the original variable is important, and use correlation when it is not.
I personally find it very valuable to discuss these options in light of the maximum-likelihood principal component analysis model (MLPCA) [1,2]. In MLPCA one applies a scaling (or even a rotation) such that the measurement errors in the measured variables are independent and distributed according to the standard normal distribution. This scaling is also known as maximum likelihood scaling (MALS) [3]. In some case, the PCA model and the parameter defining the MALS scaling/rotation can be estimated together [4].
To interpret correlation-based and covariance-based PCA, one can then argue that:
1. Covariance-based PCA is equivalent to MLPCA whenever the variance-covariance matrix of the measurement errors is assumed diagonal with equal elements on its diagonal. The measurement error variance parameter can then be estimated by applying the probabilistic principal component analysis (PPCA) model [5]. I find this a reasonable assumption in several cases I have studied, specifically when all measurements are of the same type of variable (e.g. all flows, all temperatures, all concentrations, or all absorbance measurements). Indeed, it can be safe to assume that the measurement errors for such variables are distributed independently and identically.
2. Correlation-based PCA is equivalent to MLPCA whenever the variance-covariance matrix of the measurement errors is assumed diagonal with each element on the diagonal proportional to the overall variance of the corresponding measured variable. While this is a popular method, I personally find the proportionality assumption unreasonable in most cases I study. As a consequence, this means I cannot interpret correlation-based PCA as an MLPCA model. In the cases where (1) the implied assumptions of covariance-based PCA do not apply and (2) an MLPCA interpretation is valuable, I recommend to use one of the MLPCA methods instead [1-4].
3. Correlation-based and covariance-based PCA will produce the exact same results -apart from a scalar multiplier- when the individual variances for each variable are all exactly equal to each other. When these individual variances are similar but not the same, both methods will produce similar results.
As stressed above already, the ultimate choice depends on the assumptions you are making. In addition, the utility of any particular model depends also on the context and purpose of your analysis. To quote George E. P. Box: "All models are wrong, but some are useful".
[1] Wentzell, P. D., Andrews, D. T., Hamilton, D. C., Faber, K., & Kowalski, B. R. (1997). Maximum likelihood principal component analysis. Journal of Chemometrics, 11(4), 339-366.
[2] Wentzell, P. D., & Lohnes, M. T. (1999). Maximum likelihood principal component analysis with correlated measurement errors: theoretical and practical considerations. Chemometrics and Intelligent Laboratory Systems, 45(1-2), 65-85.
[3] Hoefsloot, H. C., Verouden, M. P., Westerhuis, J. A., & Smilde, A. K. (2006). Maximum likelihood scaling (MALS). Journal of Chemometrics, 20(3‐4), 120-127.
[4] Narasimhan, S., & Shah, S. L. (2008). Model identification and error covariance matrix estimation from noisy data using PCA. Control Engineering Practice, 16(1), 146-155.
[5] Tipping, M. E., & Bishop, C. M. (1999). Probabilistic principal component analysis. Journal of the Royal Statistical Society: Series B (Statistical Methodology), 61(3), 611-622.
Straight and simple: if the scales are similar use cov-PCA, if not, use corr-PCA; otherwise, you better have a defense for not. If in doubt, use an F-test for the equality of the variances (ANOVA). If it fails the F-test, use corr; otherwise, use cov.
• -1. I don't see why running an F-test could be relevant here. PCA is an exploratory method, not a confirmatory one (as statistical tests are). Dec 18, 2017 at 9:14
The arguments based on scale (for variables expressed in the same physical units) seem rather weak. Imagine a set of (dimensionless) variables whose standard deviations vary between 0.001 and 0.1. Compared to a standardized value of 1, these both seem to be 'small' and comparable levels of fluctuations. However, when you express them in decibel, this gives a range of -60 dB against -10 and 0 dB, respectively. Then this would probably then be classified as a 'large range' -- especially if you would include a standard deviation close to 0, i.e., minus infinity dB.
My suggestion would be to do BOTH a correlation- and covariance-based PCA. If the two give the same (or very similar, whatever this may mean) PCs, then you can be reassured you've got an answer that is meaningul. If they give widely different PCs don't use PCA, because two different answers to one problem is not sensible way to solve questions.
• (-1) Getting "two different answers to the same problem" often just means you're bashing away mindlessly without thinking about which technique is appropriate for your analytical aims. It does not mean that one or (as you state) both techniques are not sensible, but only that at least one might not be appropriate for the problem or the data. Furthermore, in many cases you can anticipate that covariance-based PCA and correlation-based PCA should give different answers. After all, they are measuring different aspects of the data. Doing both by default would not make sense.
– whuber
Jun 27, 2013 at 15:00
• Actually it is very reasonable to get 2 different answers when using PCA with correlation and covariance. In the stock case, it is a question of whether you should take betas (or standard deviation) into account Nov 22, 2014 at 17:51 | 2022-07-07T01:32:00 | {
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https://math.stackexchange.com/questions/1075215/health-risk-probability | # Health Risk Probability
Question: An actuary is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of women. For each of the three factors, the probability is 0.1 that a woman in the population only has this risk factor (and no others). For any two of three factors, the probability is 0.12 that she has exactly two of these risk factors (but not the other). The probability that a woman has all three risk factors given that she has A and B, is (1/3). What is the probability that a woman has none of the three risk factors, given that she does not have risk factor A?
My attempt: I wrote the "probability that a woman has none of the three risk factors, given that she does not have risk factor A" as Pr(A'andB'andC'|A') as Pr(A'andB'andC'andA')/Pr(A') which just simplifies to Pr(A'andB'andC')/Pr(A') where Pr(A') = (1-.1) = .9. I'm not entirely sure where to go on from there. I also tried to draw a Venn Diagram with three intersecting circles where Pr(AandB'andC') = .1 (same for B and C), but that didn't really get me anywhere The answer is 0.467 (rounded). Can you guys please show me what I'm doing wrong or what I should be doing?
Thank you guys so much!
• Why is $P(A')=1-0.1=.9$? Is $P(A)=0.1$? Note that the probability that a woman only has A is equal to $0.1$. But a woman having $A$ and $B$ still has $A$. – megas Dec 20 '14 at 4:59
You can divide the sample into 8 pools. $$\begin{array}{|c|c|} \hline \text{Factor}& \text{Probability} \\ \hline \text{A} & .1 \\ \hline \text{B} & .1 \\ \hline \text{C} & .1 \\ \hline \text{AB} & .12 \\ \hline \text{AC} & .12 \\ \hline \text{BC} & .12 \\ \hline \text{ABC} & .06 \\ \hline \text{NONE} & .28 \\ \hline \end{array}$$
So you want $$\frac{\text{NONE}}{\text{B}+\text{C}+\text{BC}+\text{NONE}}=\frac{.28}{.1+.1+.12+.28}\approx.467$$
Note, $\text{ABC}=.06$ because for all the pool that have AB, $\frac{2}{3}$ must be AB and $\frac{1}{3}$ must be ABC.
First, lets write down what we know:
• $P(A \cap B' \cap C') = P(A' \cap B \cap C')= P(A' \cap B' \cap C) = 0.1$
• $P(A \cap B \cap C') = P(A \cap B' \cap C)= P(A' \cap B \cap C) = 0.12$
• $P(A \cap B \cap C|A,B) = \frac{1}{3}.$
As you noted, we are interested in computing the probability $$P(A' \cap B' \cap C'|A') = \frac{P(A' \cap B' \cap C')}{P(A')}.$$ We have $$P(A') = 1 - P(A).$$ By the law of total probability, \begin{align} P(A) &= P(A \cap (B \cap C)) + P(A \cap (B' \cap C)) + P(A \cap (B \cap C')) + P(A \cap (B' \cap C')) \\ & = P(A \cap B \cap C) + 0.12 + 0.12 +0.1 \\ & = P(A \cap B \cap C) + 0.34. \end{align} Lets try to determine $P(A \cap B \cap C)$. We have \begin{align} P(A \cap B \cap C) & = P( A \cap B\cap C | A \cap B) \cdot P(A \cap B) = \frac{1}{3} \cdot P(A \cap B). \end{align} Further, by the law of total probability, \begin{align} P(A \cap B) = P(A \cap B \cap C) + P(A \cap B \cap C') = P(A \cap B \cap C) + 0.12. \end{align} Combining the two previous equations, we find \begin{align} P(A \cap B) = \frac{1}{3}P(A \cap B) + 0.12 \quad \Rightarrow \quad P(A \cap B) = \frac{3\cdot 0.12}{2} = 0.18, \end{align} and \begin{align} P(A \cap B \cap C) & = \frac{1}{3} \cdot P(A \cap B) = \frac{1}{3}0.18 = 0.6. \end{align} Returning to the calculation of $P(A)$, we get \begin{align} P(A) = P(A \cap B \cap C) + 0.34 = 0.6 +0.34 = 0.4, \end{align} and in turn \begin{align} P(A') = 1 - P(A) = 1- 0.4 = 0.6. \end{align} Finally, the probability that a woman has no factor can be found by subtracting from $1$ the sum of the probabilities of all disjoint events in which a woman has a factor (exactly one, exactly two, or all three): \begin{align} P(A' \cap B' \cap C') &=1 - \left[P(A \cap B \cap C) + 3\cdot 0.1 + 3 \cdot 0.12 \right]\\ &=.34 - 0.6 = 0.28. \end{align} We now have everything we need to compute the desired result: $$P(A' \cap B' \cap C'|A') = \frac{P(A' \cap B' \cap C')}{P(A')} =\frac{0.28}{0.6} = 0.4666...$$
there is a mistake cause 0.6+0.34=0.94 thus P(A’)=0.06. and that changes the whole solution. also in the end 1-0.6-0.3-0.36= -0,26?
• Welcome to stackexchange. That is just a typo in the answer from @megas . He or she uses the correct value in what follows, and reaches the correct conclusion. You should delete this (non)answer and edit that one instead. SInce uo not yet have enough reputation to edit freely your changes will be put on a queue to be checked. – Ethan Bolker Nov 12 '18 at 16:04 | 2019-12-14T10:45:14 | {
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https://cs.stackexchange.com/questions/150718/trying-to-understand-the-basic-about-recurrence-trees | Trying to understand the basic about recurrence trees
I have little background on recurrence trees, and I am working on the following exercise:
Exercise. Take $$T(n) = 2T(n/2) + 3\log(n)$$. Draw the recurrence trees for $$n=2$$ and $$n=4$$. What can we conclude complexity-wise?
My attempt. For $$n=2$$ we have the following recurrence tree:
Recursive Call Recurrence Tree Sum
T(2) 3log(2) 3log(2)
/ \
T(1) 3log(1) 3log(1) 6log(1) = 0
Since we arrived at $$T(1)$$, we should stop (IS THIS TRUE?) with our tree and the total sum is given by $$3\log(2) + 0 = 3\log(2) = \Theta(1)$$ which means that, complexity-wise, for $$T(2)$$, we're dealing with constant-time complexity.
For $$n=4$$, and using the same method, we could should that the sum would be $$3\log(4) + 6\log(2) = \log(4^3*2^6) = \Theta(1)$$.
Let's upgrade the exercise and draw the recurrence tree for any $$n$$. Here follows my attempt:
Recursive Call Recurrence Tree Sum
T(n) 3log(n) 3log(n)
/ \
T(n/2) 3log(n/2) 3log(n/2) 6log(n/2)
/ \
/ \
T(n/2^2) 3log(n/4) 3log(n/4) (...) 12log(n/4)
(...)
T(n/2^i) (...) 3*2^i log(n/2^i)
And we stop when $$\frac n {2^i} = 1 \Leftrightarrow \lg n = i$$, where $$\lg n = \log_2 (n)$$. With this being said, our total sum is $$\sum_{i=0}^{\lg n} \left(3*2^i \log(\frac n {2^i} ) \right)$$. And I don't know how to proceed from here.
Is my attempt wrong at any point?
• Hint, suppose $n=2^k$ for some natural number $k$. Can you simplify the total sum? Apr 16, 2022 at 0:26
• Taking your hint, we would have that the total sum would be $\sum_{i=0}^{lg 2^k = k} 3*2^i \log(\frac{2^k}{2^i}) = \sum_{i=0}^{k} 3*2^i \log(2^{k-i}) = 3\log(2)\sum_{i=0}^{k} (k-i)2^i = 3k\log(2)\sum_{i=0}^{k} 2^i - 3\log(2) \sum_{i=0}^{k} i2^i$ Apr 16, 2022 at 9:33
• I could simplify the sums, and I did so on paper, but I think that didn't help much. Is the attempt I did for the cases $n=2$ and $n=4$ well done? Apr 16, 2022 at 9:44
"Since we arrived at $$T(1)$$, we should stop."
Yes, the recurrence relation should not be applied any more once the initial conditions are reached.
It should not be a problem to assume the value of $$T(1)$$ and no others are given as the initial conditions for the recurrence relation, although it does not violate any rules if all values of $$T(1), T(2), T(3), T(4)$$ are given independently along with the recurrence relation $$T(n) = 2T(n/2) + 3\log(n)$$, making the task of drawing the recurrence trees for $$n=2$$ and $$n=4$$ vacuous.
Assume that $$T(1)=t_1\ge0$$. Assume the $$\log$$ in the question is the logarithm with base $$2$$. A difference base will not affect the result of asymptotic analysis as long as it is a constant greater than 1.
$$T(2)=2T(1) + 3\log_2(2) = 2t_1 + 3.$$ Note $$t_1$$ should appear.
$$T(4)=2T(2) + 3\log_2(4) = 4t_1 + 12.$$
So it takes up to some constant time to compute $$T(2)$$ and $$T(4)$$, both of which are constants, too.
"$$3\log(2) + 0 = 3\log(2) = \Theta(1)$$" might not be outright wrong. However, "$$=\Theta(1)$$" makes sense only when the left-hand side is viewed as a function of a variable that could go to infinity.
Obtain an asymptotic approximation for $$T(n)$$
The recursion trees for $$n=2$$, $$n=4$$ and general $$n$$ in the question are pretty good.
Suppose $$n=2^k$$ for some integer $$k>0$$. Do not forget $$t_1$$.
$$T(n)=2^kt_1 + \sum_{i=0}^{\log_2(2^k)} 3\cdot2^i \log_2(\frac{2^k}{2^i}) = 2^kt_1 + \sum_{i=0}^{k} 3\cdot2^i (k-i)= 2^kt_1 + 3\sum_{i=0}^{k-1}2^i (k-i)$$
Intuitively, the sum $$\sum_{i=0}^{k-1}2^i (k-i)$$ is not much different from $$2^{k-1}$$, the last term where $$i=k-1$$.
• "Intuitively" is meant to alert you to learn this kind of approximate reasoning/understanding.
• "not much different" means "different by at most a constant factor". This factor might be as large as $$999999999$$ or even much larger. As long as it is independent of $$n$$, it does not matter how large it is.
• Look at all the summands, $$\cdots,\ 2^{k-4}4,\ 2^{k-3}3,\ 2^{k-2}2,\ 2^{k-1}1$$, from the back to the front. Each summand is a product of a power of $$2$$ and another number. The power of $$2$$ shrinks to one half each time and the other number increases by $$1$$ each time. The shrinkage of the power of $$2$$ is so much more significant than the addition of $$1$$, the product shrinks very fast as well. It shrinks so fast that sum of all them is not much different from the first term (from the back).
Let us realize the intuition above by accurate computations.
Let $$X=\sum_{i=0}^{k-1}2^i (k-i)$$. \begin{aligned} X&=-X+2X\\ &=-(k + 2(k-1) + 2^2(k-2) + 2^3(k-3) + \cdots + 2^{k-2}2 + 2^{k-1})\\ &\quad\quad\quad +(2k\quad\quad\ + 2^2(k-1) + 2^3(k-2) + \cdots + 2^{k-2}3 + 2^{k-1}2 + 2^{k}) \\ &\quad\text{(combine terms with the same power of 2)}\\ &=(2^k-k)+\sum_{i=1}^{k-1}2^i \\ &=2^{k}-k+(2^k-2) \quad\quad(\text{which is not much different from }2^{k-1}) \end{aligned} So, $$T(n)=2^kt_1+3X =2^k(t_1+6)-3k-6=n(t_1+6)-3\log_2(n)-6=\Theta(n)$$.
• I am sorry, I don't understand where $2^k t_1$ comes from in the $T(n)$ deduction. I understand it is related to us getting to the base case $T(1)$, but why does it origin $2^k t_1$ ? (I.e., how do you find the constant that multiplies by $t_1$ in the general formula for $T(n)$) ? Apr 18, 2022 at 9:41
• @roto, please come to this chat room for a chat. Apr 18, 2022 at 14:01 | 2023-01-30T08:40:27 | {
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https://math.stackexchange.com/questions/4021115/how-to-find-the-probability-of-drawing-colored-marbles-without-replacement | # How to find the probability of drawing colored marbles without replacement?
There are $$2$$ red marbles, $$3$$ white marbles and $$5$$ black marbles in a bag. What is the probability of drawing $$1$$ red marble, $$2$$ white marbles and $$3$$ black marbles, if $$6$$ marbles are drawn without replacement?
Answer: $$\frac{\binom{2}{1}\binom{3}{2}\binom{5}{3}}{\binom{10}{6}}=\frac{2}{7}$$
My question is why is the numerator of the answer $$\binom{2}{1}\binom{3}{2}\binom{5}{3}$$ instead of $$1$$? Below is my answer:
Total number of ways to choose $$6$$ marbles from $$10$$ marbles is $$\binom{10}{6}$$, so combinations such as
• $$BBBBBW$$, $$RWWBBB$$, $$WWWRBB$$, ...
So the probabilty I calculated is $$\frac{1}{\binom{10}{6}}$$ because $$RWWBBB$$ is one of the possible combinations out of the $$\binom{10}{6}$$ total combinations. However this is incorrect, and is supposed to be $$\frac{\binom{2}{1}\binom{3}{2}\binom{5}{3}}{\binom{10}{6}}$$. I do not understand why the numerator is the product combinations because it does not make sense to multiply combinations of nondistinct objects. For example,
there are $$5$$ black marbles $$BBBBB$$ and so the number of ways to select $$3$$ black marbles is $$1$$, therefore it does not make sense to me that it is written as $$\binom{5}{3}$$ as above. To me this would make sense if it were asking to arrange $$3$$ letters at a time from $$ABCDE$$ where the order does not matter.
Can anyone explain why the numerator is the product of combinations instead of $$1*1*1=1$$?
My observation:
I noticed that using subscripts are necessary when there are fewer distinct objects than the total number of objects. See below:
The combination formula is $$\binom{n}{r}$$ where $$n$$ is total number of objects and $$r$$ is number of selected objects.
Let $$k$$ be the number of distinct objects where
• if $$k, then subscripts are needed
• if $$k=n$$, then subscripts are not needed
Case $$1$$. $$k
Suppose I want to select $$r=2$$ marbles from a bag with $$1$$ white marble and $$2$$ black marbles, so
• $$k=2$$ and $$n=3$$, so $$k
• So $$\binom{n}{r}=3$$ with combinations: {$$W_1,B_1$$}, {$$W_1,B_2$$}, {$$B_1,B_2$$}
Case $$2$$. $$k=n$$
Suppose I want to select $$r=2$$ marbles from a bag with $$1$$ white marble, $$1$$ black marble and $$1$$ red marble, so
• $$k=3$$ and $$n=3$$, so $$k=n$$
• So $$\binom{n}{r}=3$$ with combinations: {$$W,B$$}, {$$W,R$$}, {$$B,R$$}
• @GrahamKemp I do not think the order matters in this question, so wouldn't $WWRBBB$, $RWWBBB$, $BBBWWR$,etc be the same, hence 1 combination? Feb 10 '21 at 23:45
• There are many ways to select RWWBBB, since we are making a choice of 1 from 2 red, 2 from 3 white, and 3 from 5 blue. It is not the order of colour, but the distinct marbles that matters. Feb 10 '21 at 23:56
Consider a simpler problem: your bag has $$99$$ blue marbles and $$1$$ red marble in it. You draw a marble from the bag. What is the probability that it's blue?
If you want to answer $$\frac{99}{100}$$ rather than $$\frac12$$, you want to treat the blue marbles as distinguishable. This has nothing to do with whether you can tell the marbles apart by looking.
Rather, the problem is that we can only find probabilities by counting outcomes if we are sampling uniformly. In order to be drawing one of the $$100$$ marbles uniformly, the $$100$$ marbles should all be considered different outcomes.
Similarly, in the actual question you're dealing with, we can only answer the question by counting outcomes if all $$10$$ marbles are distinguishable: because only in that case are there $$\binom{10}{6}$$ different outcomes, each of which is equally likely. So you shouldn't be thinking of your outcomes as $$BBBBBW, RWWBBB, \dots$$. Rather, we should say:
1. There are $$10$$ marbles in the bag: marbles $$R_1, R_2$$ are red, marbles $$W_1, W_2, W_3$$ are white, and marbles $$B_1, B_2, B_3, B_4, B_5$$ are black.
2. We draw a uniformly random subset of $$6$$ of the marbles, such as $$\{B_1, B_2, B_3, B_4, B_5, W_1\}$$ or $$\{R_2, W_1, W_3, B_2, B_4, B_5\}$$.
In this case, there are in fact $$\binom21 \binom32 \binom53$$ outcomes that match the description "$$1$$ red marble, $$2$$ white marbles, and $$3$$ black marbles". There's $$\{R_1, W_1, W_2, B_1, B_2, B_3\}$$ and $$\{R_1, W_1, W_2, B_1, B_2, B_4\}$$, and so on.
These are exactly all outcomes of the form $$\{R_a, W_b, W_c, B_d, B_e, B_f\}$$ where $$\{a\} \subset \{1,2\}$$, $$\{b,c\} \subset \{1,2,3\}$$, and $$\{d,e,f\} \subset \{1,2,3,4,5\}$$. From this description, we can see why there's $$\binom21 \binom32 \binom53$$ of them: there's $$\binom21$$ ways to pick $$\{a\}$$, $$\binom32$$ ways to pick $$\{b,c\}$$, and $$\binom53$$ ways to pick $$\{d,e,f\}$$.
If instead our outcomes only had descriptions like "there are $$3$$ black marbles and $$3$$ white marbles" without distinguishing the marbles of each color, then our outcomes wouldn't all be equally likely. There are actually only $$7$$ different outcomes in this case: $$WWWBBB, WWBBBB, WBBBBB,\\ RWWWBB, RWWBBB, RWBBBB, RBBBBB.$$ But their probabilities are different, so this model doesn't lend itself to computation very well.
• I understand everything now except the reason why it is necessary to add the subscripts for each colored ball. For example, {$B_1,B_2,B_3,B_4,W_1,W_3$} is distinct from {$B_2,B_1,B_3,B_4,W_1,W_2$}, but the result is the same: $4$ black balls are chosen and $2$ white balls is chosen. So why are the subscripts required? Feb 11 '21 at 0:41
• Selecting $6$ out of $10$ distinguishable marbles is uniform: all the outcomes are equally likely. Selecting $6$ marbles out of $10$ imperfectly distinguishable ones isn't uniform, which means we can no longer find probabilities by counting outcomes. So we artificially make all $10$ marbles distinguishable. Feb 11 '21 at 0:46
• I updated my answer and want to know if that is the reason why you label $B,B,B,...$ as $B_1,B_2,B_3,..$ Feb 11 '21 at 1:44
• Yes, you are right. Feb 11 '21 at 3:02
$$\binom{10}6$$ is the count for selecting 6 items from a set of 10 distinct items. We don't care about the order in which these are selected, just which individual marbles are selected.
So too should we not be concerned with order of the favored event. We count it in the same sort of way --- treating the marbles as distinct items although grouped by colours.
Thus we evaluate the probability for obtaining: 1 from 2 red, 2 from 3 white, and 3 from 5 blue, when selecting any 6 from all 10 marbles as:$$\dfrac{\dbinom 21\dbinom 32\dbinom 53}{\dbinom{2+3+5}{1+2+3}}$$ | 2022-01-24T17:57:21 | {
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https://math.stackexchange.com/questions/3403351/solving-complex-differential-equations-using-power-series | # Solving Complex Differential Equations using Power Series
I've been looking through old questions for some time and I found an interesting topic (Complex differential equation). I decided to try solving a complex differential equation with a similar premise. I looked at the equation $$z' = \overline{\mathbb{z}} +it$$ I followed a similar strategy to the post linked, giving $$z'' = \overline{\mathbb{z'}} + i$$ and then taking the conjugate of the original equation, where $$\overline{\mathbb{z'}} = z - it$$
Finally, I plugged this back into the equation for $$z''$$ to get $$z'' = z + i -it$$ $$z'' - z = i - it$$
This gives a non-homogeneous differential equation without any conjugates, which I am more familiar with solving. I decided to attempt solving it through the use of power series, where the solution, $$z(t)$$ is of the form $$z(t) = \displaystyle\sum_{n=0}^{\infty} C_nt^n$$
I then took the second derivative of this and plugged it into our derived formula to get $$\displaystyle\sum_{n=0}^{\infty} C_n(n)(n-1)t^{n-2} - \displaystyle\sum_{n=0}^{\infty} C_nt^n = i - it$$
Fixing the power and index of term 1 then yields $$\displaystyle\sum_{n=0}^{\infty} t^n[C_{n+2}(n+2)(n+1) - C_n] = i - it$$
Because the right hand side is non-zero, we obviously can't assume that the terms in square brackets sum to zero for all n. From here I decided to evaluate the summation at the first few values of n and compare it to the right hand side to build a recursive relation.
$$n =0$$ $$t^0(C_2(2)(1) - C_0) = t^0(i)$$ $$C_2(2)(1) - C_0 = i$$ $$C_2 = \frac{C_0 + i}{2}$$
$$n =1$$ $$t^1(C_3(3)(2) - C_1) = t^1(-i)$$ $$C_3 = \frac{C_1-i}{(3)(2)}$$
$$n=2$$ $$t^2(C_4(4)(3) - C_2) = t^2(0)$$ $$C_4 = \frac{C_0+i}{(4)(3)(2)}$$
$$n=3$$ $$t^3(C_5(5)(4) - c_3) = t^3(0)$$ $$C_5 = \frac{C_1 - i}{(5)(4)(3)(2)}$$
This pattern continues for later values of n, giving us the relations
$$C_n = \frac{C_0+i}{n!} , n \in \mathbb{2Z} , n\geq2$$ $$C_n = \frac{C_1-i}{n!} , n \in \mathbb{2Z+1} , n>2$$
With these relations, I was able to then build the final solution of $$z(t)$$ $$z(t) = \displaystyle\sum_{n=0}^{\infty} C_nt^n$$ $$z(t) = C_0+C_1t+(C_0+i)\displaystyle\sum_{n=1}^{\infty} \frac{t^{2n}}{(2n)!}+(C_1-i)\displaystyle\sum_{n=1}^{\infty} \frac{t^{2n+1}}{(2n+1)!}$$
We can then use the hyperbolic sine and cosine taylor series to simplify our answer, where $$\cosh(t) = \displaystyle\sum_{n=0}^{\infty} \frac{t^{2n}}{(2n)!} = \displaystyle\sum_{n=1}^{\infty} \frac{t^{2n}}{(2n)!}+1$$ $$\sinh(t) = \displaystyle\sum_{n=0}^{\infty} \frac{t^{2n+1}}{(2n+1)!} = \displaystyle\sum_{n=1}^{\infty} \frac{t^{2n+1}}{(2n+1)!}+t$$
This gave me my final solution for $$z(t)$$ $$z(t) = (C_0+i)\cosh(t) + (C_1-i)\sinh(t)-i-it$$
Finally, my problem lies in actually plugging the solution back into the original differential equation. I won't write it out now because I've definitely written too much already, but it's clear that both sides would not equal each other. I was hoping for someone to know where I went wrong, or if there's a better way to solve this type of problem. Sorry for any weird formatting or notation. This is my first time using stack exchange and latex. I've also only learned math up to Calc 3 and ODEs, so there might be some concepts I'm missing that screwed my answer. Thanks!
• I have to commend you for your effort! If this was your first time using TeX and solving any sort of equation like this, it is very impressive! Oct 21, 2019 at 20:36
• Your solution is in fact correct. You can absorb the $i$'s into the arbitrary constants. The constants have to be complex numbers so when you plug in to the differential equation you have to take their conjugate. Oct 21, 2019 at 20:38
• I found your mistake. When you reindexed to fix powers, you should've had to pull out terms from the summation that look like $C_0 + C_1 t$ but I don't see that in your work. Reindexing is a tricky business! Also, don't forget that the second derivative summation starts at $2$, not $0$. Oct 21, 2019 at 20:53
Here's an alternative solution that proves your answer is mostly correct. Split up the first differential equation into real and imaginary components denoting $$z(t) = x(t) + iy(t)$$:
$$\begin{cases} x' = x \\ y' = -y + t \\ \end{cases}$$
$$\implies x(t) = C_1 e^t \hspace{20 pt} y(t) = C_2e^{-t} + t-1$$
Putting these solutions together, we have
$$z(t) = C_1 e^t + C_2 e^{-t} + it - i$$
Which is almost the same as your solution because $$\cosh$$ and $$\sinh$$ are linear combinations of those exponentials, the only part that went awry is the inhomogeneous term. But the hard part has no mess ups.
• Thank you for all your help! my question is for the C1 and C2 terms, when taking the conjugate of them, would they simply become -C1 and -C2? If so, wouldn't plugging into the original equation still lead to different signs on each side? this form does fix the problem with the non-homogenous terms! Oct 21, 2019 at 21:00
• @alienboy3735 they're not purely imaginary so saying $C_1^* = -C_1$ is incorrect. They're just conjugate. Oct 21, 2019 at 21:53
• Understood. Thank you for your assistance. :) Oct 21, 2019 at 22:49 | 2022-05-26T08:53:29 | {
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http://waverestaurant.it/hbaw/moment-of-inertia-of-quarter-circle.html | # Moment Of Inertia Of Quarter Circle
and about… • M. There is a plate with moment of inertia Ip = 0. 3, Appendix B 7 5 Nov Shear Stress in Beams 6. The two disks are then linked together without the aid of any external torques, so that $2. The moment of inertia = I = πR 4 /8 In the case of a quarter circle the expression is given as: [Image will be Uploaded Soon] The moment of inertia = I = πR 4 /16. Problem 10. 3 106mm4 Sample Problem 9. The following is a list of area moments of inertia. The moment of inertia of semi-circular plate of radius Rand mass Mabout axis AB in its plane passing through centre:- A) (MR^2)/2 B) (MR^2)/4cos^2theta C) (MR^2. Cooling water required is given as 1. which is the sum of all the elemental particles masses multiplied by their distance from the rotational axis squared. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Quarter Circular Cross-Section. It is only constant for a particular rigid body and a particular axis of rotation. Show that the straight rod, with an appropriate twist is a possible equilibrium configuration for all values of Q, and calculate the value of twist. The volume of a sphere is 4πr 3 /3. 2 26 April Shear Force and Bending Moment* (*) videos PL4 28 April Shear Force and Bending Moment 6 1 May Shear Stress in Beams 7. Own work assumed (based on copyright claims). This course also covers solved problems on Centroid. 60 m and mass M = 15 kg can be brought to an angular speed of 12 rad/s in 0. Angular momentum is the vector product of (A) Linear momentum and radius vector (B) Moment of inertia and angular velocity (C) Linear momentum and angular velocity. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. moment of inertia 87. Let ‗O‘ be the centre and ‗G‘ be the centroid of the quarter circle. 50 m from the center, the carousel has an angular velocity of 0. Define following terms 1. Meidensha Electric Manufacturing Company, Ltd. Angular Momentum Any moving body has inertia. 4 A flywheel in the shape of a solid cylinder of radius R = 0. Polar Moment of Inertia The second moment of Area A with respect to the pole O or the z-axis. The following is a list of second moments of area of some shapes. 3 106mm4 Sample Problem 9. This course covers the Module 4 of the subject “Elements of Civil Engineering & Engineering Mechanics“ It consists of two units Centroid and Moment of Inertia. The centroid of a right triangle is 1/3 from the bottom and the right angle. 3, Appendix B 7 5 Nov Shear Stress in Beams 6. distance of investigated weld point from the weld group center of gravity in the y-axis direction [mm, in] u. The angle of illumination creates a half circle of the Moons surface, with the lighted half being on the left side. to the body’s moment of inertia. The handout is available here; ask your classmates for any clarifications and then check in with faculty as needed. These parameters simplify the analysis of structures such as beams. 5m F 1 = 20N A B 45° 2m C F 2 = 30N M A = 79. By Parallel Axis Theorem, 3) We have such 6 identical bars which are symmetrical from Centre of Regular Hexagon. Next video in this series can be seen at. Measure it at right angles: also r. 0 cm radius cylinder connected by four 5. Two blocks are connected by a light string passing over a pulley of radius 0. Top-quality friction material ensures. It represents how difficult it overcomed to change its angular motion about that axis. Own work assumed (based on copyright claims). appropriate limits, to find the moment of inertia of a quarter circle about a horizontal edge. 7-Half of a circle. The coefficient of the static friction μ sw between the ladder and the wall is 0. Hi, I would like to find a way in AutoCAD to calculate the (momet of inertia). Think of the area of a circle. The handout is available here; ask your classmates for any clarifications and then check in with faculty as needed. The apparatus comes with a set of eight precisely machined masses which can be attached to the copper disc to vary the moment of inertia of the system. The moment of inertia of semi-circular plate of radius Rand mass Mabout axis AB in its plane passing through centre:- A) (MR^2)/2 B) (MR^2)/4cos^2theta C) (MR^2. \end{equation*} If the object is made of a number of parts, each of whose moment of inertia is known, the total moment of inertia is the sum of the moments of inertia of the pieces. Jumping right in:. It is found from the moment of inertia and the distance from the outside of the object concerned to its major axis. Plug it into the ellipse area formula: π x r x r!. The rod is at rest when a 3. The moments of inertia for some common shapes can be found using the following formulas. Rectangle, Triangle, Circle etc. the size of the angle equal in length to the radius of the circle. 4, exercise 8. 3 pt) Find the moment of inertia of a thin circular plate of radius and density about a) an axis through a point on its circumference, perpendicular to the lamina; b) an axis tangent to a point on its circumference, in a plane of the lamina. Moment of inertia of a circular section can be calculated by using either radius or diameter of a circular section around centroidal x-axis or y-axis. Exploration 10. Let us consider an elemental area ‗dA‘ with centroid ‗g‘ as shown in fig. The following is a list of centroids of various two-dimensional and three-dimensional objects. dT = qdsp Therefore the moment, or torque, for the whole section But the area COB = 4 base x height = ipds i. Calculate the moment of inertia of the ball about the center of the circle. Moment of Inertia Of a Quarter Circle: Moment Of Inertia Of Circle Derivation. Each exercise provides a drawing of the circle as well as the length of either the radius or the diameter. V-Drive V-DRIVE FEATURES: • Open design allowing clean and cool running • High torque capacity • Solid and durable design. Polar Moment of Inertia The second moment of Area A with respect to the pole O or the z-axis. Rotational inertia depends on the choice of the axis of rotation. 67 103 mm3 Y 30. 8 MN) Problem 2. The moment of inertia of an object made of a number of these common shapes is the sum of the moments of inertia of its components. Rotational Inertia for a solid (continuous mass distribution) object I=Σmr2 Rotational inertia depends on the choice of axis of rotation, r. moment of inertia 87. ⇒ The ratio of the moment of inertia of a circular plate and that of a square plate for equal depth, is. The second moment of an area of a semicircle about the axis x is. If an object rotates 90 degrees, all the points in an object, save those on the axis of rotation, transcribe a quarter of a circle. The moment of inertia about an axis through a vertex is 0. A sewer pipe rolls more slowly down an incline than a bowling ball with the same mass. appropriate limits, to find the moment of inertia of a quarter circle about a horizontal edge. l = 1/4*2 pi r or r=2l/pi. UNIVERSITYOFILLINOIS May31 191 3 THISISTOCERTIFYTHATTHETHESISPREPAREDUNDERMYSUPERVISIONBY ArthurMoultonSimpson _ ENTITLEDAstu. •The moments and product of inertia for an area are plotted as shown and used to construct Mohr’s circle, •Mohr’s circle may be used to graphically or analytically determine the moments and product of inertia for any other rectangular axes including the principal axes and principal moments and products of inertia. Rod Triangle Circle Box Rectangular Plate Symmetric Blob Figure 2. Hi, I would like to find a way in AutoCAD to calculate the (momet of inertia). In addition, these clutches significantly reduce the moment of inertia for lightweight power transfer. Beam Deflection Formula and Equations for Beams Beam Deflection Equations are easy to apply and allow engineers to make simple and quick calculations for deflection. units is kg-m? A. Find the moment that the tension cable causes about the page 76 origin using a cross product. Circle Quarter Circle Semi-Circle Ellipse Half Ellipse Quarter Ellipse Parabolic Segment Spandrel Circular Arc Sector of a Circle Moment of Inertia and Radius of Gyration Moment of Inertia about the x-axis Moment of Inertia about the y-axis Polar Moment of Inertia Radius of Gyration Transfer Formula for Moment of Inertia. ” is the second moment of inertia of a circle 1 m diameter and “ E. Hemmingsen assumed (based on copyright claims). The moment of inertia of a point mass is. Moment of inertia of a circular section is same around both centriodal axis. 7 x 10"1 kg. 44 in4 kx = 1. The polar moment of inertia is a measure of an object's ability to resist torsion as a function of its shape. Centroid of a circle is very easy to determine. 28 square centimeters. 437 10 mm 4. Now that we have determined the moments of inertia of regular and truncated equilateral triangles, it is time to calculate them for the corresponding right prisms. 00 x 10"4 m/3s. Let us consider the x and y axes as shown in figure. The moment of inertia of the system of masses that includes Max, Maya, and the merry-go-round is calculated by adding the moments of inertia of the individual masses. Also supports coordinate-oriented input and language setup. Determine the moment of inertia of a filled. The moment of inertia for any small piece of a solid would be its mass element times the square of a distance to the x axes because that will be the radius of a trajectory. In this lesson, we will learn how to derive the formula as well as its application in problems. Exploration 10. shaft gum washer d L I cm CofM 3cm. Rotational Inertia (moment of inertia) •Rotational inertia is a parameter that is used to quantify how much torque it takes to get a particular object rotating •it depends not only on the mass of the object, but where the mass is relative to the hinge or axis of rotation •the rotational inertia is bigger, if more mass is located farther from. Let ‗y‘ be the distance of centroid ‗g‘ from x. The moment of Inertia formula can be coined as: I = Moment of inertia = Σ m i r i 2. The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. p ” is the second moment of inertia of the pile and “ ” is a power that varies from 0. A bar magnet of magnetic moment m and moment of inertia I (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. As both x and y axes pass through the centroid of the circular area, Equations (8. 3)2 = 9722 in4 (40. Take the Quiz and improve your overall Physics. (d)total angular momentum and moment of inertia about the axis of rotation. The moment of a circle area or the moment of inertia of a circle is frequently governed by applying the given equation: [Image will be Uploaded Soon] The moment of inertia = I = πR 4 /4. 1TEAM Smart Sheets What: 1TEAM Smart Sheets a software company that develops helpful templates for the science, technology, engineering and mathematics (STEM) community which includes professionals, researchers, consultants, professors, students, instructors, teachers, engineers, and technologists. 8 MN) Problem 2. The moment of inertia of a body is a measure of the resistance the body offers to any change in its angular velocity. 7-Half of a circle. r = distance from the pole (or the z-axis) to AA (Pythagorean's Theorem, right triangle) Since r2 = x2 + = (x2 + AA = 1+1. The moment of inertia depends on the mass and shape of an object, and the axis around which it rotates. The prefix semi-comes from Latin, and it means half or partly (like in words such as semi-permanent, semi-formal, semifinals). In any case, the bullet stops in the block due to the large frictional forces. The moment of inertia of the area A with respect to the y-axis is given by The moment of inertia of the area A with respect to the origin O is given by (Polar moment of inertia) Areas Moment of Inertia is the property of a deformable body that determines the moment needed to obtain a desired curvature about an axis. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Quarter Circular Cross-Section. Area moment of inertia of a filled quarter circle with radius r entirely. Find Vincent Teoh's email address, contact information, LinkedIn, Twitter, other social media and more. A ladder is leant against the wall. At some point you're better off building it in Pro-E, and letting IT tell you that the simple model was within 10% :) Another approach is to determine the rotational inertia of the part experimentally using the pendulum. The moment of Inertia formula can be coined as: I = Moment of inertia = Σ m i r i 2. Problem 821 Find the moment of inertia about the indicated x-axis for the shaded area shown in Fig. What is the moment of inertia of. The moment of inertia of the platform is 5 kgm 2. Calculate the Polar Moment of Inertia of a Thin Walled Circle; Calculate the Radius of Gyration of a Thin Walled Circle; Calculate the Elastic Section Modulus of a Thin Walled Circle; Calculate the Plastic Section Modulus of a Thin Walled Circle "Good engineers don't need to remember every formula; they just need to know where they can find. 2 2 May Shear Stress in Beams 3 May Shear Stress in Beams. Because the ring is hollow, all of its mass has to sit at a distance R from the center; hence, you have =R 2 and I = MR 2. Area moment of inertia of a filled semicircle with radius r with respect to a horizontal line passing through the centroid of the area is depended on the radius of the Mar 31 2020 The formula for a semicircle is derived from the formula of the circle only divided by two. Thus the second moment of area of the I-section is 1. 21 April Moment of Inertia* A. When one person is standing on the carousel at a distance of 1. *Applications:Circle Track, Road Racing Quarter Master’s V-Drive clutches feature a one-piece billet aluminum cover with an open leg design providing a low moment of inertia, and clean and cool running. Data: 23 d'abril de 2006 (original upload date) Font: No machine-readable source provided. Moment of inertia of a circular section is same around both centriodal axis. determine. These parameters simplify the analysis of structures such as beams. •Moment of Inertia and Center of Gravity •Newton’s Second Law for Rotation SOLVE The particle on the left travels one-quarter of a full circle. If either, or both ends of a beam projects beyond the supports, it is called a simple beam with overhang. Start studying Moment of Inertia. A particle describes a horizontal circle of radius r in a funnel type vessel of frictionless surface with half cone angle (as shown in figure). Problem 10. Semi Circular Area. 055 r4 and about x-axis Ix = πr4 / 8. That measurement is calculated based upon the distribution of mass within the object and the position of the. Moments of Inertia of Common Geometric Shapes Rectangle Triangle Circle Semicircle Quarter circle Ellipse J O! 1 4"ab1a2" b22 I y! 1 4"a 3b I x! 1 4"ab 3 J O! 1 8"r 4. 60 m and mass M = 15 kg can be brought to an angular speed of 12 rad/s in 0. Mass moments of inertia have units of dimension ML 2 ([mass] × [length] 2). It has inertia, it wants to keep moving in a straight line, but you have to keep pulling on it to keep changing the direction of this velocity. τα=I 2 1 a Fr I r = 12 vt/ I rr ∆∆ = r 2 r 1. The optimum gear ratio, n∗ can be calculated, using Eq. 4, exercise 8. Next video in this series can be seen at. 2 mm and height same as the radius of the quarter circle. Find the moment of inertia of the thick shell about an axis through the centre of the sphere. View Homework Help - Statics Reference Sheet from ME 2560 at Western Michigan University. The moment of inertia of the system of masses that includes Max, Maya, and the merry-go-round is calculated by adding the moments of inertia of the individual masses. With that, we will solve the equation below. This course also covers solved problems on Centroid. First Moment of Area, Moment of Inertia, Polar Moment of Inertia Quarter-circular area 4 Volume of hemisphere 2 3 Ar Volume of hemisphere 2 3Vr. In this video I will find the moment of inertia (and second moment of area) I(C. 4 A flywheel in the shape of a solid cylinder of radius R = 0. This sector has mass M. 4 kg "m2 and an angular velocity of #7. The moment of inertia of an object is a calculated measure for a rigid body that is undergoing rotational motion around a fixed axis: that is to say, it measures how difficult it would be to change an object's current rotational speed. Calculate the moment of inertia of the Earth as it revolves around the Sun. The blocks move (toward the right) with an acceleration of 1. As shown below. The maximum fluctuation of energy generated in the engine is 850 ft. Think of the area of a circle. bending moment [Nm, lb ft] r Y. This engineering data is often used in the design of structural beams or structural flexural members. The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. For racers looking for a clutch that offers the best balance of performance, durability and affordability for the money, look no further. Data: 23 d'abril de 2006 (original upload date) Font: No machine-readable source provided. Take the Quiz and improve your overall Physics. 0 kg can rotate in a horizontal plane about a vertical axis through its center. 33 is the resultant force acting on the plane due to the liquid and acts at what is called the Center of Pressure (CP). (Filename:tfigure3. Moments of Inertia Staff posted on October 20, 2006 | Moments of Inertia. The maximum fluctuation of energy generated in the engine is 850 ft. 97 107 mm4 Shear Stress. Circle Quarter Circle Semi-Circle Ellipse Half Ellipse Quarter Ellipse Parabolic Segment Spandrel Circular Arc Sector of a Circle Moment of Inertia and Radius of Gyration Moment of Inertia about the x-axis Moment of Inertia about the y-axis Polar Moment of Inertia Radius of Gyration Transfer Formula for Moment of Inertia. Area of a semicircle: A semicircle is half of a circle, so the area will be half of a circle: A = (1/2. The angular velocity of the wheel is v/r and its kinetic energy at this instant is 1/2 I(v/r)² where r is the radius of the drum and I is moment of inertia of the drum. 1TEAM Smart Sheets What: 1TEAM Smart Sheets a software company that develops helpful templates for the science, technology, engineering and mathematics (STEM) community which includes professionals, researchers, consultants, professors, students, instructors, teachers, engineers, and technologists. It has numerous applications in the field of construction engi. 44 seconds? Restart. If you take this point here, it is going to go in a circle like that centered on the x-axis. A semicircle is a half circle. Top-quality friction material ensures. Here, m = mass of the body r = radius of the circular path. List of second moments of area Last updated March 27, 2020. Moment of inertia of a quarter circle Sweepstakes. Meidensha Electric Manufacturing Company, Ltd. The second moment I measures the resistance of the section to bending about a horizontal axis (shown as a broken line). View Homework Help - Statics Reference Sheet from ME 2560 at Western Michigan University. For racers looking for a clutch that offers the best balance of performance, durability and affordability for the money, look no further. A larger diameter barrel will have a higher moment of inertia value and as a result will be stiffer. Multiply pi over four by the difference between both radii taken to the fourth power. The brick solid element can be a simple rigid body or part of a compound rigid body—a group of rigidly connected solids, often separated in space through rigid transformations. Radial acceleration of an object moving on a circle: The radial acceleration of an object moving on a circle is the centripetal acceleration, where is the tangential velocity of the object, is the distance from the axis of rotation and is the unit position vector of the object w. The inertia at the point that's one third back from the tip is always three quarters of the mass, and if you use the point about 21% back from the tip, where the impact point and pivot point are equidistant from the balance point, the inertia is exactly one half the. If you take this point here, it is going to go in a circle like that centered on the x-axis. The links will open a new browser window. 10-Half of Ellipse. Since those are lengths, one can expect that the units of moment of inertia should be of the type:. 97 107 mm4 Shear Stress. product of inertia of area = moment of inertia of area (moment of inertia of area) product of inertia of mass = moment of inertia (moment of inertia) product of inertia of volume = moment of inertia of volume (moment of inertia of volume) proof = 0. mm 10 solutions 19. The internal torque at section a–a will be determined from the free-body diagram of the left segment, Fig. (Hint: Centroid of quarter circle area. So finally, we can come back over to here. ) The more inertia a body has, the harder it is to change its linear motion. The internal torque at section a–a will be determined from the free-body diagram of the left segment, Fig. τα=I 2 1 a Fr I r = 12 vt/ I rr ∆∆ = r 2 r 1. Centroid of circle lies at the center of a circle that is also called as the radius of circle from edges of a circle. The coefficient of the static friction μ sw between the ladder and the wall is 0. determine the moment 59. Problem 10. The above equations for the moment of inertia of circle, reveal that the latter is analogous to the fourth power of circle radius or diameter. When one person is standing on the carousel at a distance of 1. The system is at rest when a friend throws a ball of mass 0. of a quarter circle about centroidal axis is …. 2 Triangle 900 40 50 36000 45000 Rectangle 2400 30 20 72000 48000 aX = aY = a= 2043. The carousel itself (without riders) has a moment of inertia of 125 kg·m2. Moment of inertia about the base plane: The moment of inertia about the base plane can be computed by subtracting one-half the value of the moment of inertia about the central axis from the value of the moment of inertia about a base diameter axis. 7-Half of a circle. Often though, one may use the term "moment of inertia of circle", missing to specify an axis. Hemmingsen assumed (based on copyright claims). 2: Determine angular and tangential speed and velocity. Further we will also study the surface area or volume of revolution of a line or area respectively. 3: A quarter and a penny are on a turntable. If you take this point here, it is going to go in a circle like that centered on the x-axis. Given that the centroid C of the area bounded by a quarter-circle lies a distance 4a/(3 above the base of the quarter-circle, determine the moment of inertia of the cross hatched area about an axis xc through the centroid. Take the original moment of inertia about the centroid, then simply add your area times r squared term or mass times r squared term for this adjusted value. How do i find the product of inertia for a quarter circle moved from it's center and axes? i expected it to be A*yx*yz (yx and yz beiing the distance from the reference point to the center of the shape). appropriate limits, to find the moment of inertia of a quarter circle about a horizontal edge. Moment of inertia of a quarter circular section about an x-axis is----- A. 3)2 = 9722 in4 (40. It is I = 1 Y2b(Y)dY section where y is measured vertically and b(y) is the width of the section at y. 065 2 5 / /0. Rod Triangle Circle Box Rectangular Plate Symmetric Blob Figure 2. Area moment of Inertia JNTU – Dec2006 1. How to calculate the moment/product of inertia for a quarter-circular area? Anyone know Ixy, in particular? Ix and Iy I can get from the table in my statics book. The centroid of an object in -dimensional space is the intersection of all hyperplanes that divide into two parts of equal moment about the hyperplane. This course also covers solved problems on Centroid. 5" V-DRIVE BUTTON CLUTCH ASSEMBLIES WEIGH LESS THAN 9. 0 g bullet traveling the horizontal plane. If the permissible bending stress is 120N/mm 2, find a) The safe point load that can be applied at the centre of the span. ⇒ The ratio of the moment of inertia of a circular plate and that of a square plate for equal depth, is. 7 Angular momentum is the vector product of a) linear momentum and radius vector b) moment of inertia and angular vector. c) The quarter of a solid circular cylinder of radius 1 and height 2 lying in the first octant, with its central axis the interval 0 ≤ y ≤ 2 on the y-axis, and base the quarter circle in the xz-plane with center at the origin, radius 1, and lying in the first quadrant. Moment of inertia of quarter circular area about x and y axes is I x = I y = ${{{R^4}} \over 16}$ (8. com for more math and science lectures! In this video I will find the moment of inertia (and second moment of area), I(x)=?, I(y). The moment of inertia of total area A with respect to z axis or pole O is z dI z or dI O or r dA J 2 I z ³r dA 2 The moment of inertia of area A with respect to z axis Since the z axis is perpendicular to the plane of the area and cuts the plane at pole O, the moment of inertia is named “polar moment of inertia”. Ref =25000MPa”. product of inertia 23. Visit http://ilectureonline. Thus k= 15. The gear ratio has limiting values of 19 and 31, given the range of the inertia. How do i find the product of inertia for a quarter circle moved from it's center and axes? i expected it to be A*yx*yz (yx and yz beiing the distance from the reference point to the center of the shape). the size of the angle equal in length to the radius of the circle. Only rectangular and round solid sections are considered here. The moment of inertia on a quarter circle is giver by: Ix=(pi*r^4)/16 So using Steiner's theorem to calculate the MoI of the Quarter circle on the main figure's centroid we get: Ix'=(pi*r^4)/16 + dy * (pi*r^2)/4 where dy is the difference between yG of the quarter circle and yG of the main figure. moment of inertia; Trig Graphs, Angles, and Circles; Radian and Degrees; LZ 7. Informally, it is the "average" of all points of. It is I = 1 Y2b(Y)dY section where y is measured vertically and b(y) is the width of the section at y. Moment of inertia of a circular section can be calculated by using either radius or diameter of a circular section around centroidal x-axis or y-axis. 3 and the coefficient of the static friction μ sf between the ladder and the floor is 0. and about… • M. In such cases, an axis passing through the centroid of the shape is probably implied. The unit of dimension of the second moment of area is length to fourth power, L 4, and should not be confused with the mass moment of inertia. 08 mm Find the centroid of the shaded area shown in fig, obtained by cutting a semicircle of diameter 100mm from the quadrant of a circle of radius 100mm. parallel to the x-axis, of a semi-circle. Learn vocabulary, terms, and more with flashcards, games, and other study tools. moment of inertia 87. A column will fail about the axis of smallest moment of inertia of area. 8-Quarter of a circle. It is I = 1 Y2b(Y)dY section where y is measured vertically and b(y) is the width of the section at y. Calculate the Polar Moment of Inertia of a Thin Walled Circle; Calculate the Radius of Gyration of a Thin Walled Circle; Calculate the Elastic Section Modulus of a Thin Walled Circle; Calculate the Plastic Section Modulus of a Thin Walled Circle "Good engineers don't need to remember every formula; they just need to know where they can find. 2 26 April Shear Force and Bending Moment* (*) videos PL4 28 April Shear Force and Bending Moment 6 1 May Shear Stress in Beams 7. A thin wire of length l and mass m is bent in the form of a semicircle The moment of inertia about an axis perpendicular to its plane and passing through the end of the wire is - Physics - System of particles and rotational motion. 5 Moment of inertia of : (a) semicircle, and (b) quarter circle. Centroid of a quarter circle y Let us consider a quarter circle with radius r. 01 18-Jun-2003 1. At some point you're better off building it in Pro-E, and letting IT tell you that the simple model was within 10% :) Another approach is to determine the rotational inertia of the part experimentally using the pendulum. The slope at quarter span section is. )=? rotating around the C. Our model relies on the ability of the snowboarder to keep proper balance despite the curvature of the jump. Problem 15P. The V-Drive series of clutches from Quarter Master® features an open design for cleaner and cooler operations. The moment of inertia of an object is the measure of its resistance to being rotated about an axis. Quarter circle 12. The brick solid element can be a simple rigid body or part of a compound rigid body—a group of rigidly connected solids, often separated in space through rigid transformations. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. PART 3: Moment of Inertia a. Let T be the period of oscillations of the original magnet about an axis through the mid point, perpendicular to length, in a magnetic field B. 0 g bullet traveling the horizontal plane. The moment of inertia is the mass of the object times the mass-weighted average of the squared distance from the axis. Informally, it is the "average" of all points of. The V-Drive series of clutches from Quarter Master® features an open design for cleaner and cooler operations. ds is the normal integration factor evaluated. A thin wire of length l and mass m is bent in the form of a semicircle The moment of inertia about an axis perpendicular to its plane and passing through the end of the wire is - Physics - System of particles and rotational motion. Problem 13P. The mass moment of inertia is often also known as the. The center of pressure of an aircraft is the point where all of the aerodynamic pressure field may be represented by a single force vector with no moment. Pivot Point is a point about which a ship pivots in a turning circle. Let ‗y‘ be the distance of centroid ‗g‘ from x. Correct response to preceding frame Frame 28-16 Finding Moment of Inertia. Elite snowboarders are skilled. Integrate with respect to y first; use suitable cylindrical coordinates. The V-Drive series of clutches from Quarter Master® features an open design for cleaner and cooler operations. We conclude that in this case, the disk with the smallest moment of inertia has the largest final velocity. For an object of uniform composition, the centroid of a body is also its center of mass. Area Moment of Inertia Section Properties: Half Tube Calculator. The coefficient of the static friction μ sw between the ladder and the wall is 0. 5 Area Moment of Inertia of a Compound Section221 5. Disagree 5. You have a semicircle (half of a circle). Question is ⇒ The maximum twisting moment a shaft can resist, is the product of the permissible shear stress and. 43 × 10 6 mm 4. MOMENT OF. Mass moment of inertia. The prefix semi-comes from Latin, and it means half or partly (like in words such as semi-permanent, semi-formal, semifinals). product of inertia of area = moment of inertia of area (moment of inertia of area) product of inertia of mass = moment of inertia (moment of inertia) product of inertia of volume = moment of inertia of volume (moment of inertia of volume) proof = 0. Using the small angle approximation SINθ ≈ θ, this equation is approximately. and parallel to the base, is A. Force, given by Eq. With a solid disk, the mass is spread out. The Following Section consists of Strength of Materials Questions on Physics. The Brick Solid block adds to the attached frame a solid element with geometry, inertia, and color. The speed at which the points in a rotating object transcribe a circle is known as angular velocity, and a change in that speed is known as angular acceleration. The moment of inertia of a body is a measure of the resistance the body offers to any change in its angular velocity. The moment of inertia of a point mass is. The moment of inertia about an axis is a measurement of how difficult it is to accelerate the body about that axis. Processing. ) The more inertia a body has, the harder it is to change its linear motion. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The formula for the moment of inertia of a ring created by two overlapping circles is similar. If you're unsure about what deflection actually is, click here for a deflection definition Below is a concise beam deflection table that shows how to calculate the maximum deflection in. If an object rotates 90 degrees, all the points in an object, save those on the axis of rotation, transcribe a quarter of a circle. By Gerson Washiski Barbosa. moment of inertia; Trig Graphs, Angles, and Circles; Radian and Degrees; LZ 7. Define following terms 1. (C) Maximum at quarter points (D) Varies with slope Answer: Option B Question No. Informally, it is the "average" of all points of. the axis of rotation, therefore the moment of inertia has been reduced and the skater spins faster in order to conserve angular momentum. The moment of inertia of the shape is given by the equation. moment of inertia 87. 9 MN,F V = 153. Mechanics of Materials, 6th Edition. 50 s to reach the bottom of the incline. MOMENT OF. Visit http://ilectureonline. 055 r4 and about x-axis Ix = πr4 / 8. This engineering calculator will determine the section modulus for the given cross-section. 9780028030678 Our cheapest price for Statics and Strength of Materials is$64. 4 A flywheel in the shape of a solid cylinder of radius R = 0. moment of inertia has increased. Answer: Ix = 2. The blocks move (toward the right) with an acceleration of 1. IT'S MOMENT OF INERTIA WILL ALLOW YOU TO ACCELERATE AND DECELERATE FASTER. Bending Moment in The Beam: Integrating a second time: The bending moment is zero at the free end of the beam ν’’(L) = 0 Therefore C 2 = 0 and the equation simplifies to Slope and Deflection of the Beam: The third and fourth integration yield The boundary conditions at the fixed support, where the slope and. This equation is equivalent to I = π D 4 / 64 when we express it taking the diameter (D) of the circle. Calculate the location of centroid of quarter cylinder of length 100cm, radius= 30cm with its base on xy plane and axis along Z. 1 What is a Beam?. There are many formulas involved because there are many shapes. Determine the moment of inertia of the entire area—w axis Apply the equation in step 5 to determine the moment of inertia Iw of the entire area with respect to the horizontal axis w through A. This applet is designed to let you explore arclength of the graph of a function. Using a single integral we were able to compute the center of mass for a one-dimensional object with variable density, and a two dimensional object with constant density. ” is the second moment of inertia of a circle 1 m diameter and “ E. Our model relies on the ability of the snowboarder to keep proper balance despite the curvature of the jump. 75 to 2 kg m 2. Problem 15P. 4𝑅𝑅 3𝜋𝜋) Note: Attendance (+2 points), Format (+1 point) Water Air Air. The inertia at the point that's one third back from the tip is always three quarters of the mass, and if you use the point about 21% back from the tip, where the impact point and pivot point are equidistant from the balance point, the inertia is exactly one half the. The moment of this force about 0 the shear stress is q/t. ⇒ The ratio of the moment of inertia of a circular plate and that of a square plate for equal depth, is. A = π r 2 ¸ 2 Diameter perpendicular to x-axis, centroidal axis = x-axis: Ic = π r 4 ¸ 8 Diameter on x-axis, centroidal axis parallel to x-axis: Ic = r 4 (9π 2 - 64) ¸ 72π x = 4r ¸ 3π Ax = 2r 3 ¸ 3 Ix. The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. The apparatus comes with a set of eight precisely machined masses which can be attached to the copper disc to vary the moment of inertia of the system. Center or 1st moment ∫ = M xdm MXC & ∫ = M ydm MYC (XC & YC: Center of Mass) ∫ A xdA = xc & ∫ A ydA = yc (xc & yc: Center of Area) Moment of inertia or 2nd moment ∫ = M r2dm I (2nd moment of Mass) ∫A y dA =Ix 2 & ∫A x dA =Iy 2 (2nd moment of Area) Then, FR =γAyc sinθ=(γ hc)A where γ hc: Pressure at the centroid. Observation 2: For some constant, c, the centroid must lie along the line x + y = c and furthermore, c must be less than 1 since the area of the triangle formed by the X-axis, Y-axis and x+y=1 is more than half of. Rectangle, Triangle, Circle etc. The moment of inertia about an axis is a measurement of how difficult it is to accelerate the body about that axis. c) The quarter of a solid circular cylinder of radius 1 and height 2 lying in the first octant, with its central axis the interval 0 ≤ y ≤ 2 on the y-axis, and base the quarter circle in the xz-plane with center at the origin, radius 1, and lying in the first quadrant. symm) Systems of systems and composite objects Another way of interpreting the formula r cm = r 1 m1 + r 2 2 +··· m1 +m2 +···. Visit http://ilectureonline. The gear ratio has limiting values of 19 and 31, given the range of the inertia. The speed at which the points in a rotating object transcribe a circle is known as angular velocity, and a change in that speed is known as angular acceleration. Find Vincent Teoh's email address, contact information, LinkedIn, Twitter, other social media and more. 3 is a quarter circle 50 m wide into the paper. parallel to the x-axis, of a semi-circle. Here I is inertia, M is mass and R is the radius. z 6’ 3’ y 5’ x T=1kip (10, 0, -7) ft. less than one equal to one more than one equal to 3Ï€/16 ⇒ The moment diagram for a cantilever carrying linearly varying load from zero at its free end and to maximum at the fixed end will be a. 00 x 10"4 m/3s. An initially straight, elastic rod with Young’s modulus E, area moments of inertia and axial effective inertia is subjected to an axial couple , which remains fixed in direction as the rod deforms. Figure to illustrate the area moment of a quartercircle at the list of moments of inertia. A rectangle is attached to the quarter circle at Quadrant I with base at the x-axis of width equal to 45. Hemmingsen assumed (based on copyright claims). The above concept can be extended to obtain the moment of inertia of semicircular and quarter circular area as given below. The two disks are then linked together without the aid of any external torques, so that $2. The moment of inertia calculates the rotational inertia of an object rotating around a given axis. The moment of inertia of part of a circle can be found by using the trigonometric function sine and the angle of the circle segment in question. A box is shown in the figure below. 723 Rectangle, quarter circle and triangle | Centroid of Composite Area; 724 Rectangle, semicircle, quarter-circle, and triangle | Centroid of Composite Area; 725 Centroid of windlift of airplane wing | Centroid of area; 726 Area enclosed by parabola and straigh line | Centroid of Composite Area; Moment of Inertia and Radius of Gyration; Dynamics. Moment of inertia of a circular section can be calculated by using either radius or diameter of a circular section around centroidal x-axis or y-axis. of a quarter circle about centroidal axis is …. This is a more general characteristic. The moment of inertia of the system of masses that includes Max, Maya, and the merry-go-round is calculated by adding the moments of inertia of the individual masses. For the derivation of the moment of inertia formula of a circle, we will consider the circular cross-section with the radius and an axis passing through the centre. IT'S MOMENT OF INERTIA WILL ALLOW YOU TO ACCELERATE AND DECELERATE FASTER. The moment of inertia of the complete disc about an axis passing through its centre O and perpendicular to its plane is I 1 = 2 9 M R 2 Now, the moment of inertia of the disc with removed portion I 2 = 2 1 M (3 R ) 2 = 1 8 1 M R 2 Therefore, moment of inertia of the remaining portion of disc about O is I = I 1 − I 2 = 9 2 M R 2 − 1 8 M R 2. Suddenly, the same sheet of paper has considerable stiffness. They should all be meters. This engineering calculator will determine the section modulus for the given cross-section. 5 m and mass 4. 5 kg rotating about one end. Form an expression for dA. Units of Moment of Inertia are length raised to the fourth power, such as in4 or m4. (5) can be rewritten in the following form,. The equation looks like:. If area is in m2 and the length is also in m, the moment of inertia is expressed in m4. 2 6 Nov Shear Stress in Beams. )=? rotating around the C. It should not be confused with the second moment of area, which is used in beam calculations. Problem 10. 75-m long thin rods to a thin-shelled outer cylinder of mass 20. 11-Circular sector. A ladder is leant against the wall. 12) (b) Determine the magnitude and location of the vertical component of the force on curved section BC of the conduit wall. Semi Circular Area. 1 What is a Beam?. but when i compare it to output from autocad, it doesn't match. A sewer pipe rolls more slowly down an incline than a bowling ball with the same mass. Integrate with respect to y first; use suitable cylindrical coordinates. Determine the moment of inertia of the entire area—w axis Apply the equation in step 5 to determine the moment of inertia Iw of the entire area with respect to the horizontal axis w through A. View Homework Help - Statics Reference Sheet from ME 2560 at Western Michigan University. 60 s by a motor exerting a constant torque. Observation 2: For some constant, c, the centroid must lie along the line x + y = c and furthermore, c must be less than 1 since the area of the triangle formed by the X-axis, Y-axis and x+y=1 is more than half of. Balance: Glucydur®, 4 arms, moment of inertia 4. appropriate limits, to find the moment of inertia of a quarter circle about a horizontal edge. If area in mm2 and the length is also in mm, then moment of inertia is expressed in mm4. The prefix semi-comes from Latin, and it means half or partly (like in words such as semi-permanent, semi-formal, semifinals). A child sits on a merry-go-round at the position marked by the red circle (position is given in meters and time is given in seconds). This means that ω 2 = ω 1 /1,000,000! The astronauts have angular velocity of about 1 revolution per second and you know perfectly well that the fidget spinner did not have a speed of a million revolutions per second. In the (help) of AutoCAD exists (moment of inertia) but when I opened it, I foud nothing about (moment of inertia) but only about calculation of the area. The moment K measures the resistance of the section to twisting. 1)What is the moment of inertia of the object about an axis. And so when we think about area, we know that the area of a circle, the area of a circle is equal to pi times the radius of the circle squared. 4𝑅𝑅 3𝜋𝜋) Note: Attendance (+2 points), Format (+1 point) Water Air Air. The inertia at the tip of a staff is always one quarter of the staff's mass. A bar magnet of magnetic moment m and moment of inertia I (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Moment of inertia of a quarter circle. The lighter the rotating mass, the faster you can accelerate off the turns. Moment of Inertia of Different Shapes; Shared Flashcard Set. We have Section Property. The following links are to calculators which will calculate the Section Area Moment of Inertia Properties of common shapes. Measure it at right angles: also r. A rectangle is attached to the quarter circle at Quadrant I with base at the x-axis of width equal to 45. 3 and the coefficient of the static friction μ sf between the ladder and the floor is 0. sectional area 31. Determine the moment of inertia about its diametrical axis. Moment of inertia of a circular section can be calculated by using either radius or diameter of a circular section around centroidal x-axis or y-axis. 6 Polar Moment of Inertia233. Moment of inertia of quarter circular area about x and y axes is I x = I y = ${{{R^4}} \over 16}$ (8. The moment of inertia contribution of the right arm is therefore approximately 0. The second moment of area for a shape is easier to be calculeted with respect to a parallel axis or with respect to a perpendicular axis. Jumping right in:. 723 Rectangle, quarter circle and triangle | Centroid of Composite Area; 724 Rectangle, semicircle, quarter-circle, and triangle | Centroid of Composite Area; 725 Centroid of windlift of airplane wing | Centroid of area; 726 Area enclosed by parabola and straigh line | Centroid of Composite Area; Moment of Inertia and Radius of Gyration; Dynamics. Polar Moment Of Inertia Circular Beam October 31, 2018 - by Arfan - Leave a Comment 6 torsion of shafts 4 161 beam161 explicit 3 d beam up19980821 polygon geometric properties solved 6 pts derive the equation for polar moment o. In addition, these clutches significantly reduce the moment of inertia for lightweight power. has proposed their Model TW-55 for motorcycle testing. The moment of inertia about an axis through a vertex is 0. This point is a pproximately 30% of length from forward when steaming ahead and about 20% – 25% of length from the stern when the ship is going astern. The product of the force and the perpendicular distance to the center of gravity for an unconfined object, or to the pivot for a confined object, is^M called the torque or the moment. The moment of inertia of part of a circle can be found by using the trigonometric function sine and the angle of the circle segment in question. Hemmingsen assumed (based on copyright claims). asked by stewart on November 7, 2014; physics help! A small 520-gram ball on the end of a thin, light rod is rotated in a horizontal circle of radius 1. Find the center of mass of a two-dimensional plate that occupies the quarter circle $$x^2+y^2\le1$$ in the first quadrant and has density $$k(x^2+y^2)$$. Each mass is a quarter circle arc of known inner and outer radius. Because each Minute includes questions of varying degrees of difficulty, the amount of time students need to complete each Minute will vary at first. SOLUTION Dimensions in mm Symmetry: X 0Determination of centroid C of entire section Y A yA Y (4000 mm2 ) 122. Consult Table Moments of Inertia of Various Bodies in the Textbook as needed. G The centroid and centre of gravity are at the same point Where centre of gravity consider to be whole mass of an object act at a point C. The moment of inertia of total area A with respect to z axis or pole O is z dI z or dI O or r dA J 2 I z ³r dA 2 The moment of inertia of area A with respect to z axis Since the z axis is perpendicular to the plane of the area and cuts the plane at pole O, the moment of inertia is named “polar moment of inertia”. Area of Trapezoid = 0. product of inertia 23. 925 10 mm R OC I ave • Based on the circle, evaluate the moments and product of inertia with respect to the x’y’axes. 437 10 mm 4. Autor: No machine-readable author provided. Start studying Moment of Inertia. The above concept can be extended to obtain the moment of inertia of semicircular and quarter circular area as given below. 036 for very soft soils: 0. The moment of inertia of an object made of a number of these common shapes is the sum of the moments of inertia of its components. The unit of dimension of the second moment of area is length to fourth power, L 4 , and should not be confused with the mass moment of inertia. The moment of inertia of the merry-go-round is determined by using the equation for the moment of inertia of a disk rotating about its center: I d i s k = 1 2 M R 2 where M is the. r2 x2 y2 Therefore, I z I. Now, I actually found the moment of inertia of a fidget spinner, I toy ≈7x10-5 kg·m 2 and I estimate I man ≈70 kg·m 2. simple support will develop a reaction normal to the beam, but will not produce a moment at the reaction. CalcTown is an online Engineering Calculator resoruce. It should not be confused with the second moment of area, which is used in beam calculations. 8 mg•cm 2, angle of lift 53° Frequency: 28,800 vph (4 Hz) Balance spring: elinvar by Nivarox® Shock protection: INCABLOC 908. where gives the rotational inertia of the system with the addition of n extra masses. Data: 23 d'abril de 2006 (original upload date) Font: No machine-readable source provided. It is found from the moment of inertia and the distance from the outside of the object concerned to its major axis. 3, you missed an important handout and discussion about the remaining weeks of the quarter. to the body’s moment of inertia. For racers looking for a clutch that offers the best balance of performance, durability and affordability for the money, look no further. The second moment of area for a shape is easier to be calculeted with respect to a parallel axis or with respect to a perpendicular axis. 2 2 May Shear Stress in Beams 3 May Shear Stress in Beams. As both x and y axes pass through the centroid of the circular area, Equations (8. Free shipping on all orders over$35. Calculating the moment for a straight cylinder barrel is relatively easy though. Suddenly, the same sheet of paper has considerable stiffness. 0 cm radius cylinder connected by four 5. - Moments of Inertia about the x & y axes - Polar Moment of Inertia - Radius of Gyration - Centroids - Area - Section Modulus Moments of Inertia calculates the properties and displays the equations for many sections including: • Rectangles • Hollow Rectangles • Circles • Hollow Circles • Semicircles • Quarter Circles • Ellipses. This means that ω 2 = ω 1 /1,000,000! The astronauts have angular velocity of about 1 revolution per second and you know perfectly well that the fidget spinner did not have a speed of a million revolutions per second. 8 MN) Problem 2. (Hint: Centroid of quarter circle area. The moment of inertia ‘I’ of a rotating object with respect to its axis of rotation is given by the product of its mass and the square of its distance from the axis of rotation. \end{equation*} If the object is made of a number of parts, each of whose moment of inertia is known, the total moment of inertia is the sum of the moments of inertia of the pieces. The moment of inertia of the area A with respect to the y-axis is given by The moment of inertia of the area A with respect to the origin O is given by (Polar moment of inertia) Areas Moment of Inertia is the property of a deformable body that determines the moment needed to obtain a desired curvature about an axis. 75 m in length and mass 1. Moment of inertia of this ring about axis OX would be; Since whole disc can be supposed to be made up of such like concentric rings of radii ranging from O to R ,we can find moment of inertia I of the disc by integrating moment of inertia of the ring for the limits x=0 and x=R iii) Moment of inertia of a uniform sphere of radius R about the. Calculate the Polar Moment of Inertia of a Thin Walled Circle; Calculate the Radius of Gyration of a Thin Walled Circle; Calculate the Elastic Section Modulus of a Thin Walled Circle; Calculate the Plastic Section Modulus of a Thin Walled Circle "Good engineers don't need to remember every formula; they just need to know where they can find. Thin Rod in the Shape of a Quarter Circle Moment of inertia is not a physical quantity such as velocity, acceleration or force, but it enables ease of calculation; it is a function of the geometry of the area. dA= ipds or 2dA=pds torque T = 2q dA. Circle Quarter Circle Semi-Circle Ellipse Half Ellipse Quarter Ellipse Parabolic Segment Spandrel Circular Arc Sector of a Circle Moment of Inertia and Radius of Gyration Moment of Inertia about the x-axis Moment of Inertia about the y-axis Polar Moment of Inertia Radius of Gyration Transfer Formula for Moment of Inertia. The overall moment of inertia of your composite body is simply the sum of all of the adjusted moments of inertia for the pieces, which will be the sum of the values in the last column (or. Here I is inertia, M is mass and R is the radius. The moments of inertia for a cylindrical shell, a disk, and a rod are MR2, , and respectively. - Moments of Inertia about the x & y axes - Polar Moment of Inertia - Radius of Gyration - Centroids - Area - Section Modulus Moments of Inertia calculates the properties and displays the equations for many sections including: • Rectangles • Hollow Rectangles • Circles • Hollow Circles • Semicircles • Quarter Circles • Ellipses. is greatest in first quarter and decreases throughout. Start studying Moment of Inertia. Moment of resistance is a term in structural engineering. This engineering data is often used in the design of structural beams or structural flexural members. The moment of inertia about an axis is a measurement of how difficult it is to accelerate the body about that axis. For the isolated system of platform and person, the angular momentum is conserved. It represents how difficult it overcomed to change its angular motion about that axis. That means a semicircle will have half of the area of a circle. 33 is the resultant force acting on the plane due to the liquid and acts at what is called the Center of Pressure (CP). 47 Determine the polar moment of inertia of the area shown with respect to (a) Point O, (b) the centroid of the area. r = distance from the pole (or the z-axis) to AA (Pythagorean's Theorem, right triangle) Since r2 = x2 + = (x2 + AA = 1+1. 5 Area Moment of Inertia of a Compound Section221 5. In addition, these clutches significantly reduce the moment of inertia (MOI) for lightweight power transfer. 00 x 10"4 m/3s. (see figure) 1) What is the moment of inertia of this half about an axis perpendicular to its plane through point A? 2) Why did your answer in part A come out the same as if this were a complete disk of mass M? 3) What would be the moment of inertia of a quarter disk of mass M and radius R about an axis. Why: to ease, increase accuracy, reduce time, to study and understand the use an application of. The brick solid element can be a simple rigid body or part of a compound rigid body—a group of rigidly connected solids, often separated in space through rigid transformations. The following links are to calculators which will calculate the Section Area Moment of Inertia Properties of common shapes. 9 Moment of Inertia of A Quarter-Circle * Multiple Choice Questions * Exercises Chapter 4: Shear Forces and Bending Moments in Beams 4. the origin. A semicircle is a half circle. 3)2 = 9722 in4 (40. Let T be the period of oscillations of the original magnet about an axis through the mid point, perpendicular to length, in a magnetic field B. Calculate the moment of inertia of each of the following uniform objects about the axes indicated. The moment of inertia of the merry-go-round is determined by using the equation for the moment of inertia of a disk rotating about its center: I d i s k = 1 2 M R 2 where M is the. "The fromula is stress = Moment*centroid/Inertia. 12) (b) Determine the magnitude and location of the vertical component of the force on curved section BC of the conduit wall. A particle describes a horizontal circle of radius r in a funnel type vessel of frictionless surface with half cone angle (as shown in figure). Rod Triangle Circle Box Rectangular Plate Symmetric Blob Figure 2. What if we tried to find the area of a circle as though it were an ellipse? We would measure the radius in one direction: r. Author: No machine-readable author provided. the origin. Radius of gyration 3. The second moment of area, also known as area moment of inertia, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. symm) Systems of systems and composite objects Another way of interpreting the formula r cm = r 1 m1 + r 2 2 +··· m1 +m2 +···. Here I is inertia, M is mass and R is the radius. | 2020-09-23T04:01:15 | {
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https://nicoguaro.github.io/posts/numerical-01/ | # Numerical methods challenge
For the next month I will write a program per day for some well-known numerical methods in both Python and Julia. It is intended to be an exercise, then don't expect the code to be good enough for real use. Also, I should mention that I have almost no experience with Julia, so it probably won't be idiomatic Julia but more Python-like Julia.
## Day 1: Bisection method
The first method to be considered is the bisection method. This method is used to solve the equation $f(x) = 0$ for $x$ real, and $f$ continuous. It starts with an interval $[a,b]$, where $f(a)$ and $f(b)$ should have opposite signs. The method then proceeds by halving the interval and selecting the one where the solution appears, i.e., the sign of the function changes.
We will use the function $f(x) = \cos(x) - x^2$ to test the codes, and the initial interval is [0, 1].
The following are the codes:
### Python
from __future__ import division, print_function
from numpy import log2, ceil, abs, cos
def bisection(fun, a, b, xtol=1e-6, ftol=1e-12):
if fun(a) * fun(b) > 0:
c = None
msg = "The function should have a sign change in the interval."
else:
nmax = int(ceil(log2((b - a)/xtol)))
for cont in range(nmax):
c = 0.5*(a + b)
if abs(fun(c)) < ftol:
msg = "Root found with desired accuracy."
break
elif fun(a) * fun(c) < 0:
b = c
elif fun(b) * fun(c) < 0:
a = c
msg = "Maximum number of iterations reached."
return c, msg
def fun(x):
return cos(x) - x**2
print(bisection(fun, 0.0, 1.0))
With result
(0.824131965637207, 'Maximum number of iterations reached.')
### Julia
function bisection(fun, a, b, xtol=1e-6, ftol=1e-12)
if fun(a) * fun(b) > 0
c = nothing
msg = "The function should have a sign change in the interval."
else
nmax = ceil(log2((b - a)/xtol))
for cont = 1:nmax
c = 0.5*(a + b)
if abs(fun(c)) < ftol
msg = "Root found with desired accuracy."
break
elseif fun(a) * fun(c) < 0
b = c
elseif fun(b) * fun(c) < 0
a = c
end
msg = "Maximum number of iterations reached."
end
end
return c, msg
end
function fun(x)
return cos(x) - x^2
end
println(bisection(fun, 0.0, 1.0))
With result
(0.824131965637207, "Maximum number of iterations reached.")
In this case, both codes are really close to each other. The Python code has 25 lines, while the Julia one has 27. As expected, the results given by them are the same.
### Edit (2017-10-02)
As suggested by Edward Villegas, I decided to compare execution times. I used %timeit for IPython and @benchmark (from BenchmarkTools) for Julia.
In IPython, we have
%timeit bisection(fun, 0.0, 2.0)
with result
10000 loops, best of 3: 107 µs per loop
And in Julia, we have
@benchmark bisection(fun, 0.0, 2.0)
with result
BenchmarkTools.Trial:
memory estimate: 48 bytes
allocs estimate: 2
--------------
minimum time: 1.505 μs (0.00% GC)
median time: 1.548 μs (0.00% GC)
mean time: 1.604 μs (0.00% GC)
maximum time: 38.425 μs (0.00% GC)
--------------
samples: 10000
evals/sample: 10
And it seems that the Julia version is around 100 times faster than the Python counterpart. | 2019-10-15T21:49:24 | {
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https://math.stackexchange.com/questions/1672474/8-cards-are-drawn-without-replacement-find-the-probability-of-obtaining-exactly | # 8 cards are drawn without replacement. Find the probability of obtaining exactly $3$ aces, or exactly $3$ kings, or both.
$8$ cards are drawn without replacement from an ordinary deck. Find the probability of obtaining exactly $3$ aces, or exactly $3$ kings, or both.
The answer is $\frac{2~\binom 4 3 \binom {48} 5 - \binom 4 3 \binom 4 3\binom {44} 2 }{ \binom {52} 8 }$
I am trying hard to understand the reasoning behind this answer but still don't fully get it
Can someone explain please. Thank you very much.
please read $\binom m n$ as "from m pick n"
Things I believe I understand:
$\binom {52} 8\,\leftarrow$ obvious. # of possible different 8 card hands
$2~\binom 4 3 \binom {48} 5\,\leftarrow$ the act of taking 3 figures of the same rank (be it aces or kings) and multiplying it by the way in which the 5 remaining cards can be chosen.
Thanks again.
• No thank you. I will read it as '$m$ choose $n$' :) – Em. Feb 26 '16 at 0:09
• @probablyme why you didn't suggest an answer? – Prefijo Sustantivo Feb 26 '16 at 21:42
It's the principle of inclusion and exclusion at work.
$$\mathsf P(A\cup K) = \mathsf P(A)+\mathsf P(K)-\mathsf P(A\cap K)$$
$\mathsf P(A)~=~{\binom 4 3\binom {48} 5}\big/{\binom {52} 8}$ is the probability of obtaining 3 aces and 5 non-aces. Some of those 5 non-aces may be kings.
$\mathsf P(K)~=~{\binom 4 3\binom {48} 5}\big/{\binom {52} 8}$ is the probability of obtaining 3 kings and 5 non-kings. Some of those 5 non-kings may be aces.
$\mathsf P(A\cap K)~=~{\binom 4 3\binom 4 3\binom {44} 2}\big/{\binom {52} 8}$ is the probability of obtaining 3 aces, 3 kings, and 2 cards of other faces. However we've counted this event among both the above. So to avoid over counting we subtract one occurrence.
$$\mathsf P(A\cup K) = \dfrac{\binom 4 3\binom {48} 5 + \binom 4 3\binom {48} 5- \binom 4 3\binom 4 3\binom {44} 2}{\binom {52} 8}$$
That is all.
$\Box$
The number of hands of 8 cards with exactly 3 kings is:
• a group of 3 kings taken from 4 possible kings i.e. $\binom{4}{3}=4$ multiplied by
• a group of 5 cards that are not kings taken from $52-4=48$ possible cards i.e. $\binom{48}{5}$
The number of hands of 8 cards with exactly 3 aces is the same that the number of cards with exactly 3 kings.
But in any case we counted the number of both, 3 kings and 3 aces, so the number of hands with 3 kings and 3 aces are counted twice so we must subtract one time this quantity (this is a case of inclusion-exclusion principle).
The number of hands with exactly 3 kings and 3 aces is $\binom{4}{3}^2\binom{44}{2}$.
And all possible hands of 8 cards are $\binom{52}{8}$, so the probability will be:
$$\frac{2\binom{4}{3}\binom{48}{5}-\binom{4}{3}^2\binom{44}{2}}{\binom{52}{8}}$$ | 2019-06-25T09:49:35 | {
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https://math.stackexchange.com/questions/693473/how-many-of-these-numbers-are-divisible-by-4 | # How many of these numbers are divisible by 4?
There is this question that I have no idea where did I make the mistake.
Each of the digits 1,1,2,3,3,4,6 is written on a separate card. The seven cards are then laid out in a row to form a 7-digit number. How many of these 7-digit numbers are divisible by 4?
My working: In order for a number to be divisible by 4, first they must be even. So the last digit can be 2 or 4 or 6. Then we check for the second last number:
if the last is 2, then the second last can only be 1 or 3,
if the last is 4, then the second last can only be 2 or 6,
if the last is 6, then the second last can only be 1 or 3.
Then, for the remaining 5 numbers we have 5! ways, but there are 2 1's and 2 3's. So in total we have $\frac{5!\times 6}{2!2!}=180$. But the solutions says 300.
Am I wrong? Where did I make the mistake?
I really appreciate any helps! Many many thanks!
• How do you form a $7$-digit number using only $6$ digits? – angryavian Feb 28 '14 at 2:16
• There are only 6 numbers. Is this with replacement? Want 6 instead? – bobbym Feb 28 '14 at 2:16
• Sorry, yes I have added one more 3. No, I think this is without replacement, since the cards are laid out in a row. – user71346 Feb 28 '14 at 2:18
• You might get some help from this math.stackexchange.com/questions/58349/… – bobbym Feb 28 '14 at 2:27
• @bobbym, thanks. I think I get the ideas correct. But the different with the question you shared is that my question has two 1's and two 3's. So I divide it by 2!*2!. – user71346 Feb 28 '14 at 2:31
Building off your work:
Last digit is 2: Second last can be 1 or 3. For either 1 or 3, one of the repeated numbers is already used and isn't counted as repeated, so there's $\frac{2 \times 5!}{2!} = 120$
Last digit is 4: What you had is correct. $\frac{2 \times 5!}{2!2!} = 60$
Last digit is 6: Same logic as last digit of 2. $120$
$120 + 120 + 60 = 300$
The number of distinct permutations of the remaining numbers will depend on the choice of the lowest two digits. For example, if the lowest two digits are $12$, there are $\frac{5!}{2!}$ permutations of the remaining digits, since one of the repeated $1$s is already used.
• Thanks augurar. Do you mean we have to divide into cases. I think I have done that, do you mind explaining in more details? Thanks. – user71346 Feb 28 '14 at 2:24
@user71346
I am getting these cases for the last 2.
(3,2),(1,2),(2,4),(6,4),(1,6),(3,6)
$\frac{5!}{2!}+\frac{5!}{2!}+\frac{5!}{2!\cdot 2!}+ \frac{5!}{2!\cdot 2!}+\frac{5!}{2!}+\frac{5!}{2!}=300$
Your mistake is that you may have used up one of the $1$'s or $3$'s in the last two digits. In that case, you should only divide by one $2!$. You get the same $60$ cases you found if the last is $4$, but twice as many if the last is $2$ or $6$. That gives $60+120+120=300$ | 2019-08-18T10:55:45 | {
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http://www.dongfangoil.cn/e5ejzg17/59c66c-acute-triangle-formula | There are several ways to find the area of a triangle. Acute Triangle: If all the three angles of a triangle are acute i.e., less than 90°, then the triangle is an acute-angled triangle. The picture below illustrates the general formula for the 30, 60, 90 Triangle. 1. An angular bisector is a segment that divides any angle of a triangle into two equal parts. Problem 1. Statement 1 by itself will only determine a range of values c utilizing the 3rd side rule of triangles. An acute angle triangle (or acute-angled triangle) is a triangle in which all the interior angles are acute angles. Yes, an acute scalene triangle is possible if the interior angles of the scalene triangles are acute. Since all the three angles are less than 90°, we can infer that ΔABC is an acute angle triangle or acute-angled triangle. The relation between the sides and angles of a right triangle is the basis for trigonometry. Right Triangles 2. The differences between the types are given below: Types of Acute Triangle. As we remember from basic triangle area formula, we can calculate the area by multiplying triangle height and base and dividing the result by two. We extend the base as shown and determine the height of the obtuse triangle. They stand apart from other triangles, and they get an exclusive set of congruence postulates and theorems, like the Leg Acute Theorem and the Leg Leg Theorem. Question: Which formula is used when given 90-degree triangle, opposite angle is 26 degrees and one leg is know? A right triangle is a special case of a scalene triangle, in which one leg is the height when the second leg is the base, so the equation gets simplified to: area = a * b / 2 It means that all the angles are less than 90 degrees, A triangle in which one angle measures 90 degrees and other two angles are less than 90 degrees (acute angles). The inradius and circumradius formulas for an isosceles triangle may be derived from their formulas for arbitrary triangles. This means we are given two angles of a triangle and one side, which is the side adjacent to the two given angles. To recall, an acute angle is an angle that is less than 90°. Example: Consider ΔABC in the figure below. The area of acute angle triangle = (½) × b × h square units, If the sides of the triangle are given, then apply the Heron’s formula, The area of the acute triangle = $$A = \sqrt{S (S-a)(S-b)(S-c)}$$ square units, Where S is the semi perimeter of a triangle, The perimeter of an acute triangle is equal to the sum of the length of the sides of a triangle, and it is given as. Not only scalene, but an acute triangle can also be an isosceles triangle if it satisfies its condition. From the law of cosines, for a triangle with side lengths a, b, and c, cosC=(a^2+b^2-c^2)/(2ab), with C the angle opposite side C. For an angle to be acute, cosC>0. based on their sides or based on their interior angles. 1. If a triangle has 1 acute angle, the other angles will be either right angles or obtuse angles which is not possible as the sum of interior angles of a triangle is always 180°. Select the correct answer and click on the “Finish” buttonCheck your score and answers at the end of the quiz, Visit BYJU’S for all Maths related queries and study materials, Your email address will not be published. A median of a triangle is the line that connects an apex with the midpoint of the opposite side. The important properties of an acute triangle are as follows: A perpendicular bisector is a segment that divides any side of a triangle into two equal parts. An acute triangle is a figure where all three angles measure less than 90°. A perpendicular bisector is a segment that divides any side of a triangle into two equal parts. In acute angle, the medians intersect at the centroid of the triangle, and it always lies inside the triangle. One of the best known mathematical formulas is Pythagorean Theorem, which provides us with the relationship between the sides in a right triangle. Your email address will not be published. Statement 2 by itself will determine that c is either 10 or 11. A triangle can never have only one acute angle. 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We can also find the area of an obtuse triangle area using Heron's formula. In an acute triangle, the following is true for the length of the sides: a 2 + b 2 > c 2, b 2 + c 2 > a 2, c 2 + a 2 > b 2. a, b, and c denotes the sides of the triangle. thank you both for the help. The sum of all 3 angles of the triangle will be 180o 180 o. Acute triangles are classified into three types: 1) acute scalene triangle, 2) acute isosceles triangle, and 3) acute equilateral triangles. Acute triangles can be isosceles, equilateral, or scalene. Acute Angle Triangle Acute Angle Triangle Formula. A right triangle consists of two legs and a hypotenuse. A triangle is considered as a three-sided polygon. If is the measure of the third angle, then Solve for : The triangle has two congruent angles - each with measure . Register for Marwell eNews and download our Top Tips for a great visit. Practice Using Special Right Triangles. Acute Angle Formulas . But first, please review the definition of Perimeter Of Two-Dimensional Shapes, Angle and Acute Angle.. An acute triangle has one unique feature, all three of the interior angles are less than 90° and the sum of the angles is 180°. Find the area of the triangle if the length of one side is 8 cm and the corresponding altitude is 6 cm. Since a triangle's angles must sum to 180° in Euclidean geometry, no Euclidean triangle can have more than one obtuse angle. It is simply half of b times h. Area = 12 bh (The Triangles page explains more). You can easily find both the length of an arc and the area of a sector for an angle θ in a circle of radius r. An obtuse triangle is a triangle with one obtuse angle and two acute angles. In any triangle, two of the interior angles are always acute (less than 90 degrees) *, so there are three possibilities for the third angle: . Triangle Proportionality Theorem Worksheets. The two legs meet at a 90° angle and the hypotenuse is the longest side of the right triangle and is the side opposite the right angle. acute triangle – all angles are less than 90 degrees; obtuse triangle – at least one angle is greater than 90 degrees; right triangle – one angle is exactly 90 degrees; In this article, we will take a look at right triangles and special types of right triangles. acute triangle, the formula for calculating the area of the acute triangle is A = b(1/2h). This principle is known as Leg-Acute Angle theorem. Right triangles are aloof. What is the value of z in the triangle below? The three altitudes of an acute angle intersect at the orthocenter, and it always lies inside the triangle. (image will be updated soon) In the above figure, the triangle ABC is an acute-angled triangle, as each of the three angles, ∠A, ∠B and ∠C measures 80°, 30° and 70° respectively which are less than 90°. The Leg Acute Theorem seems to be missing "Angle," but "Leg Acute Angle Theorem" is just too many words. The radius of the inscribed circle of an isosceles triangle with side length , base , and height is: −. The most important thing is that the base and height are at right angles. LA Theorem 3. Any triangle that has one obtuse angle, or an angle larger than 90 degrees, extending beyond a right angle) is no longer acute because it doesn't fit the definition of an acute triangle. 3. oh sorry, did not realize it is an acute angled triangle. The differences between the types are given below: Area (A) = ½ (b × h), where b = base and h = height, Perimeter (P) = a + b + c, where a, b, c are the three measures of three sides. Obtuse Triangle: If any one of the three angles of a triangle is obtuse (greater than 90°), then that particular triangle is said to be an obtuse angled triangle. Acute Angle Triangle Properties. An altitude of a triangle is a line that passes through an apex of a triangle and is perpendicular to the opposite side. Specific Examples. – zeeks Sep 6 '15 at 18:57 Less than 90° - all three angles are acute and so the triangle is acute. Based on the sides and the interior angles of a triangle, there can be various types of triangles, and the acute angle triangle is one of them. The intersection of angular bisectors of all the three angles of an acute angle forms the incenter, and it always lies inside the triangle. © 2021 (Mathmonk.com). Right Triangles. Some Specific Examples. (Don't use the Pythagorean theorem. The intersection of perpendicular bisectors of all the three sides of an acute-angled triangle form the circumcenter, and it always lies inside the triangle. Since this is an obtuse triangle, pythagorean theorem does not apply. We can see that. Write the formula on the whiteboard and ask the students to record it in their journals under this heading: Formula for Area of an Acute Triangle, Using a Long Rectangle with the Equivalent Base and One-Half the Height. • The sine law can be used to solve a problem modelled by an acute triangle if you can determine two sides and the angle opposite one of these sides, or two angles and any side. Acute triangles have NO angles greater than or equal to 90 degrees -- all their angles are less than 90 degrees. All the basic geometry formulas of scalene, right, isosceles, equilateral triangles ( sides, height, bisector, median ). in an acute triangle. How To Find The Perimeter Of An Acute Triangle Let's look at the geometric characteristics of an acute triangle. Yes, all equilateral triangles are acute angle triangles. When we know the base and height it is easy. Click ‘Start Quiz’ to begin! A right angle has a value of 90 degrees ($90^\circ$). Acute and obtuse triangles are the two different types of oblique triangles — triangles that are not right triangles because they have no 90° angle. The Area of Acute Triangles Using Height and Base. 60° each which are acute angles. See Solving "AAS" Triangles. It is because an equilateral triangle has three equal angles, i.e. Note: the remaining two angles of an obtuse angled triangle are always acute. Here, ∠A, ∠B, ∠C are the three interior angles at vertices A, B, and C, respectively. (Pathetic attempt at a math joke.) New York State Common Core Math Module 5, Grade 6, Lesson 3 Related Topics: Fun Facts about Acute Triangles: The angles of an acute triangle add up to 180°, because of the Angle Sum Property. Right Triangle. A right triangle is a triangle in which one angle is a right angle. Important Terminologies. General Formula. A triangle cannot be obtuse-angled and acute-angled simultaneously. LL Theorem 5. The longest side of an acute triangle is opposite the largest angle. For an acute angle triangle, the distance between orthocenter and circumcenter is always less than the circumradius. The angles formed by the intersection of lines AB, BC and CA are ∠ABC, ∠BCA, and ∠CAB, respectively. Put your understanding of this concept to test by answering a few MCQs. Consider the triangle $$ABC$$ with sides $$a$$, $$b$$ and $$c$$. Use the Pythagorean Theorem to determine if triangles are acute, obtuse, or right triangles. LL Theorem Proof 6. • The sine law states that in any acute triangle,+ABC, C c B b A a sin sin sin = = . A triangle in which all three angles are acute angles. In geometry, a triangle is a closed two-dimensional plane figure with three sides and three angles. Such a triangle can be solved by using Angles of a Triangle to find the other angle, and The Law of Sines to find each of the other two sides. It is possible to have an acute triangle which is also an isosceles triangle – these are called acute isosceles triangles. % Progress Area of Triangles. An acute-angled triangle or acute triangle is a triangle whose all interior angles measure less than 90° degrees. These two categories can also be further classified into various types like equilateral, scalene, acute, etc. Last modified on November 12th, 2020 at 12:19 pm, Home » Geometry » Triangle » Acute Triangle. As a consequence, by the Converse of the Isosceles Triangle Theorem, the triangle has two congruent sides, making it, by definition, isosceles. For a right triangle with a hypotenuse of length c and leg lengths a and b, the Pythagorean Theorem states: a 2 + b 2 = c 2 The formulas to find the area and perimeter of an acute triangle is given and explained below. ... Lengths of triangle sides using the Pythagorean Theorem to classify triangles as obtuse, acute or right. Therefore, statement 1 alone is insufficient. Videos and solutions to help Grade 6 students find the area formula for a triangular region by decomposing a triangle into right triangles. The acute triangle: Acute triangles are better looking than all the other triangles. All rights reserved. 45, 45, 90 Special Right Triangle. The measures of the interior angles of a triangle add up to . To find the third angle of an acute triangle, add the other two sides and then subtract the sum from 180°. If a leg and an acute angle of one right triangle are congruent to the corresponding parts of another right triangle, then the two right triangles are congruent. Click Create Assignment to assign this modality to your LMS. In an acute triangle, the line drawn from the base of the triangle to the opposite vertex is always, If two angles of an acute-angled triangle are 85. When the lengths of the sides of a triangle are known, the Pythagorean Theorem can be used to determine whether or not the triangle is an acute triangle. 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https://math.stackexchange.com/questions/3643238/doesnt-polynomial-long-division-violate-the-general-rules-of-division | # Doesn't polynomial long division violate the general rules of division?
Doesn't polynomial long division violate the general rules of division? As we don't divide each term in the numerator by the whole denominator (like with the division of the real numbers), what we do is dividing each term in the numerator (starting from the highest power term till the lowest power term) by the highest power term only in the denominator ignoring the other terms. How does this style of division succeed?
I searched many websites like Quora and even this site but no one has discussed this point which I consider it as a violation of division rules. Could anyone clarify this point to me? Thanks in advance.
• Your error seems to be in the phrase "ignoring the other terms". Those other terms are retained and determine much of what you do at the next step and later in the computation. – Andreas Blass Apr 25 '20 at 13:40
• But how can I multiply by terms which I did not use in the division step ? I know that in long division of real numbers we multiply the quotient by the divisor which we divide by in the division step, but in polynomials we don't do that, in division step we divide by one term (the highest power term) but in multiplication step we multiple the quotient by all terms in the divisor, isn't that different from the long division of real numbers? – kareem mohamed Apr 26 '20 at 11:31
• Your question now has several good, detailed answers. They seem to be enough to explain your error completely. – Andreas Blass Apr 26 '20 at 12:51
• OK, thank you so much sir, i really appreciate your help – kareem mohamed Apr 26 '20 at 15:15
Dividing $$3x^4 + 2x^3 + 6x^2 + 8x + 4$$ by $$x^3+2x+1$$ to obtain $$3x+2$$ remainder $$x+2$$ works exactly the same way you do long division with numbers to find that $$32684$$ divided by $$1021$$ is $$32$$ remainder $$12$$. Just compare the steps in detail and there is no real difference.
You can view it this way: We have polynomials $$f(x)$$ and $$g(x)$$. We want to find polynomials $$q(x)$$ and $$r(x)$$ such that $$\tag1 f(x)=q(x)g(x)+r(x).$$ Well, that's trivial: Just take $$q(x)=0$$ and $$r(x)=f(x)$$. Okay, so we want something better: We want $$\deg r$$ to be as small as possible. Can we improve upon the trivial solution $$q=0, r=f$$? We can replace $$q(x)$$ with $$q(x)+cx^d$$ and $$r(x)$$ with $$r(x)-cx^dg(x)$$ without destroying the equality $$(1)$$. And as long as $$\deg r\ge deg g$$, we can let $$d=\deg r-\deg g$$ and $$c$$ the quotient of their leading coefficients and - voila! - we have another solution to $$(1)$$ but the new $$r$$ has lower degree. If we continue this, we will ultimately obtain a solution where $$\deg r<\deg g$$ and $$(1)$$ still holds.
• I don't know how can I thank you, thank you so much sir, i really appreciate your help – kareem mohamed Apr 26 '20 at 15:13
The only general rule of division is that it must be precisely reversible by multiplication. So $$2020 \div 50$$ is $$40.4$$ because $$40.4 \times 5 = 2020.$$ Doing division with integer remainder, $$2020 \div 11$$ is $$183$$ with remainder $$7$$ because $$11 \times 183 + 7 = 2020.$$
Now, recalling what the meaning of a decimal number is, we have \begin{align} 2020 &= 2 (10^3) + 0(10^2) + 2(10) + 0,\\ 11 &= 1(10) + 1, \end{align}
where the right-hand sides of these equations look a lot like $$2x^3 + 0x^2 + 2x + 0$$ and $$x + 1$$ where $$x = 10.$$
Let's see what happens if we try dividing $$2x^3 + 0x^2 + 2x + 0$$ by $$x + 1$$.
$$\require{enclose} \begin{array}{r} 2x^2 - 2x + 4 \\[-3pt] x + 1 \enclose{longdiv}{2x^3 + 0x^2 + 2x + 0} \\[-3pt] \underline{2x^3 + 2x^2}\phantom{{} + 2x + 0} \\[-3pt] -2x^2 + 2x \phantom{{} + 0} \\[-3pt] \underline{-2x^2 - 2x}\phantom{{} + 0} \\[-3pt] 4x + 0 \\[-3pt] \underline{4x + 4} \\[-3pt] -4 \end{array}$$
So the result of $$(2x^3 + 0x^2 + 2x + 0) \div (x + 1)$$ is $$2x^2 - 2x + 4$$ with remainder $$-4.$$ This obeys the general requirements of division with remainder, because $$(x + 1) \times (2x^2 - 2x + 4) + (-4) = 2x^3 + 0x^2 + 2x + 0.$$
Now let's compare this to the usual evaluation of $$2020\div 11$$ by long division.
Keep in mind first of all that just because two algorithms arrange their parts in somewhat similar ways on the paper does not mean that they must be the same algorithm. There is a "long division" algorithm for taking square roots, for example, that works like long division in some ways (writing one digit of the result at a time, using that one digit to form a product, and subtracting that product from the leftmost several digits of something else).
With that in mind, let's see how the usual long division of $$2020\div 11$$ differs from long division of $$2x^3 + 0x^2 + 2x + 0$$ by $$x + 1.$$
For one thing, we assume that $$x = 10.$$ This is a somewhat arbitrary decision, and in fact we could choose differently. For example, we could be doing division problems in base $$5,$$ in which case the string of digits $$2020$$ actually means $$2(5^3) + 2(5),$$ that is, we would assume that $$x = 5.$$
For another thing, we forbid the coefficient of any term of any polynomial to be negative. For example, we don't allow a number to be written $$2(10^2) - 2(10) + 4,$$ even though that is a perfectly good polynomial over the number $$10.$$ Again this is a somewhat arbitrary restriction, since there are "balanced" bases in which digits can have negative values.
A consequence of these restriction is that the remainder cannot be negative.
For another thing, we forbid the coefficient of any term to be greater than $$9.$$ For example, $$1(10^2) + 11$$ is ruled out even though it is a perfectly good polynomial over $$10.$$
We also forbid any coefficient to be anything except an integer. Try dividing $$3x^2 + 5x + 1$$ by $$2x + 3$$ without using non-integers such as $$\frac32$$ in the result.
Furthermore, we apply all these restrictions not just to the input and output expressions, but to all expressions at all steps within the computation. In the division of $$2x^3 + 0x^2 + 2x + 0$$ by $$x + 1,$$ for example, the step where we subtract $$2x^3 + 2x^2$$ from $$2x^3 + 0x^2$$ is not allowed, because $$-2x^2$$ has a negative coefficient.
In order to make all of this possible, we apply a trick that is not available to us in polynomial long division: we take advantage of the assumption that $$x = 10$$ in order to change the coefficients. For example, instead of writing $$2(10^3) - 2(10^2)$$ we would write $$1(10^3) + 8(10^2).$$ The reason we cannot apply this "trick" in polynomial long division is that $$2x^3 - 2x^2 \neq x^3 + 8x^2$$ in general. The use of this trick means that the results of the division are often valid only in the same numeric base in which they were performed; if you want to know the result of $$2020\div 11$$ in base $$5$$ or base $$16,$$ the base-$$10$$ result is not very useful.
We also find that certain techniques that we can use in polynomial long division have to be modified in order to stay within all these extra restrictions. Sometimes this means you practically have to do a step by trial and error--for example, what is the correct one-digit quotient when dividing $$691$$ by $$173$$? It is not $$6$$ (from $$6\div 1$$). And we might guess that $$5$$ is too large, but what about $$4$$? If the numbers are large enough you may find it useful to work out the possible multiples of the divisor in a separate area on the page in order to figure out which digits you can write in each step.
So you may say that polynomial long division breaks the "rules" because it does not obey the arbitrary additional restrictions that we enforce when we do long division of decimal-system numerals in the usual way. But I could equally well say that the usual long division algorithm breaks the "rules" of polynomial long division in order to get an artificially constrained result that is not even correct as a general polynomial, only when everything is evaluated at $$x = 10.$$
Or we could say that there are enough similarities between the algorithms to justify calling both of them some kind of "long division" (just as the much more different square-root algorithm also is called a "long division" algorithm), but that the algorithms also are different because they were meant to solve different problems with different constraints on the results.
• I don't how can I thank you , you gave me information that i have never known , no one before you told me that these algorithms have the same name but different techniques as they were invented to solve different problems with different constraints , thank you so much sir, I really appreciate your help. – kareem mohamed Apr 26 '20 at 14:48
You need to understand that when we do polynomial division, we mean exactly what we mean when we do integer division. Suppose you want to divide $$384$$ by $$25.$$ Then this is the same as subtracting as many copies of $$25$$ as you can from $$384.$$ Whatever is left is the remainder. You can see that this is simply a problem in multiplication and subtraction. Now, you can do this anyhow, but we usually split our integers in digital multiples of nonnegative integer powers of $$10,$$ like $$300+80+4.$$ Then you can see that $$25$$ goes $$12$$ times in the first part, and $$3$$ times in the second. The remainder is $$9.$$ This is what we do with the long division method usually taught to schoolchildren.
Now come to polynomials. Given two polynomials $$p$$ and $$d,$$ so that $$p$$ has a higher degree, we want to see how many times $$d$$ divides into $$p,$$ and then note the remainder, which will be necessarily of lower degree than $$d.$$ We're concerned with degrees here because it indicates the rough size of the polynomial when the variable is large enough. Now consider the example of trying to divide $$x^2+2x-4$$ by $$x+2.$$ Then we want to subtract as many multiples of the divisor from the dividend as we can. You can see in this case that no constant multiple will be large enough. So we take the next higher multiple of the divisor, namely $$x(x+2),$$ which works since it reduces the dividend by one degree. Then we subtract this from $$x^2+2x-4$$ to see what remains, which is $$-4$$ in this case, which is of lesser degree than the divisor, so that our work is complete. So the quotient is $$x.$$
In general we subtract as many copies of $$d$$ as needed from $$p$$ until we can no longer do so. So, if $$p$$ is of degree $$m$$ and $$d$$ of degree $$n,$$ with $$n\le m,$$ then we see that we can take as many as $$x^{m-n}$$ copies of $$d$$ from $$p,$$ provided all polynomials here are monic (otherwise use an appropriate constant coefficients). Then we now have the remainder $$p-x^{m-n}d,$$ whose degree is now less than that of $$p.$$ If its degree is greater than that of $$d,$$ we continue in this way until the degree of the remainder is less than that of $$d.$$ Let this remainder of degree $$ be denoted by $$r,$$ then we eventually have something of the form $$p-x^{m-n}d-c_1x^{m-k}d-\cdots-c_kx^{n}d=r,$$ so that factoring $$d$$ out and rearranging, we have $$p=dq+r,$$ where $$q=x^M+\sum c_kx^i,$$ with $$M=m-n.$$
Hope this helps!
• I don't know how can I thank you , thank you so much sir, i really appreciate your help, you have clarified many ambiguous points to me, but i to inquire about something, Why did you add other terms like (-dc1x^m-k) to this expression (p-dx^m-n) to equalize it to the remainder, isn't (p-dx^m-n) alone without adding any other expression equal to the remainder ? – kareem mohamed Apr 26 '20 at 15:11
• @kareemmohamed Yes, it is the remainder, but if it can still be divided by $d,$ then you continue until you can no longer divide. Consider the numerical example $300+80+4.$ If you divide the $300$ by $25,$ the remainder is now $80+4.$ But since this is bigger than $25,$ repeat the process, until you can no longer do it. – Allawonder Apr 26 '20 at 16:48
• I don't know how can I thank you, you gave me information that i had never known, no one before you clarified polynomial long division like that to me, thank you so much sir, i really appreciate your help. – kareem mohamed Apr 27 '20 at 0:01 | 2021-04-10T11:49:16 | {
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http://math.stackexchange.com/questions/86352/how-do-you-factor-x3-3x23x-1 | # How do you factor $x^3-3x^2+3x-1$?
$$x^3-3x^2+3x-1?$$
I know this may seem trivial, but I, for the life of me, I cannot figure out how to factor this polynomial, I know that the root is $$(x-1)^3=0$$ because of wolframalpha, but I don't know how to get there. any help would be greatly appreciated. and also if you have any recommended web sites that help with higher order polynomial factoring that would be extremely helpful.
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It's kind of a good idea to familiarize yourself with the first few entries of Pascal's triangle... – Guess who it is. Nov 28 '11 at 12:28
I remember leaning about that in one of my calculus classes, but how does that apply to this? Can you use that to factor a third degree polynomial? – AlexW.H.B. Nov 28 '11 at 12:31
The $1,3,3,1$ pattern ought to have "screamed" at you... :) – Guess who it is. Nov 28 '11 at 12:36
Oh i now see what your getting at. thanks for the tip. – AlexW.H.B. Nov 28 '11 at 12:39
First, guess a root (this is the hard part). The so called "rational roots test" will be helpful here.
Eventually, you'll discover that $x=1$ is a root. This will imply that your polynomial has the form $$\tag{1}(x-1)(ax^2+bx+c),$$ for some constants $a, b, c$.
To find those constants, you could do one of two things (and maybe more)
1. perform the division $x^3-3x^2+3x+1\over x-1$.
2. expand (1) and set it equal to the original polynomial. Setting the coefficients of the two sides of this equation equal to each other will give you a system of equations that are solvable for $a$, $b$, and $c$.
Once you've figured out what $a,b$, and $c$ are, factor the quadratic.
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Thank you so much for you thorough answer and quick response. this is very helpful. :) – AlexW.H.B. Nov 28 '11 at 12:21
You're welcome. Glad to help. – David Mitra Nov 28 '11 at 12:52
\begin{align*} x^3-3x^2+3x-1 &=x^3-x^2-2x^2+3x-1 \\ &=x^2(x-1)-2x^2+2x+x-1 \\ &=x^2(x-1)-2x(x-1)+1(x-1) \\ &=(x-1)(x^2-2x+1) \\ &=(x-1)(x-1)^2 \\ &=(x-1)^3 \end{align*}
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Thank you very much... this helped me a lot. I do have one question though. does this method of splitting up terms work in most or all cases of third degree polynomial factoring? – AlexW.H.B. Nov 29 '11 at 6:14
$x^3-3x^2+3x-1=x^3-2x^2-x^2+x+2x-1=x^3-2x^2+x-x^2+2x-1=$
$=x(x^2-2x+1)-(x^2-2x+1)$ ....etc.
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so you are saying you can factor this as two trinomial degree two polynomials just by grouping? very interesting. – AlexW.H.B. Nov 28 '11 at 12:24
Hint
If you know a solution for the equation (in this case $x=1$) then you can use the Briot-Ruffini method to reduce this 3-degree polinomyal to a 2-degree.
Just find the other roots applying Baskara.
If the roots are $x_1$, $x_2$, $x_3$ you can factore your polinomyal as:
$a(x-x_1)(x-x_2)(x-x_3)$, where $a$ is the coefficient of the highest degree.
Other hint:
A notable product is:
$(x-y)^3=x^3-3x^2y+3xy-y^3$
See this is quite similar to your polynomial.
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hey thank you very much for the tips. I will keep them in mind. – AlexW.H.B. Nov 28 '11 at 12:29
@AlexW.H.B. You're welcome .-. – GarouDan Nov 28 '11 at 12:34
Another way: $\rm\ f\:' = 3\ (x-1)^2\$ and $\rm\ gcd(f,f\:') = (x-1)^2\$ by Euclid's algorithm (or by inspection).
In fact most polynomial factorization algorithms start by reducing to the squarefree case by factoring out $\rm\ gcd(f,f')\:.$
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From binomial formula $$a^3-3a^2b+3ab^2-b^3=(a-b)^3$$for$a=x,b=1$ we get that
$$x^3-3x^2+3x-1 =x^3-3x^2\times1+3x\times1^2-1^3=(x-1)^3$$
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http://officinariusomarghera.it/trigonometry-quadrant-system.html | # Trigonometry Quadrant System
Each vector can be broken into X and Y coordinates. landmark on the modern algebr a was achieved in 1843. Cliffs Trigonometry. 2 Introduction Our discussion so far has been limited to right-angled triangles where, apart from the right-angle itself, all angles are necessarily less than 90. IEEE Compliance. Chapter 1 Trigonometric Functions 1. Quandrant 1 - 0˚ All Students Take Calculus, All Silver Tea Cups, Add Sugar To Coffee are few mnemonics in mathematics that is used to memorize the sign values of all the trigonometric functions in the 2-dimensional Cartesian coordinate system. Solution sin(x) is positive in quadrant I and II and therefore the given equation has two solutions. Free math problem solver answers your trigonometry homework questions with step-by-step explanations. We're gonna start with this magenta. That means you are looking at the fourth quadrant. He considered every triangle—planar or spherical—as being inscribed in a circle, so that each side becomes a chord (that is, a straight line that connects two points on a curve or surface, as shown by the inscribed triangle ABC in. AP Psychology Quizzes. This page contains a lot of printable graph papers and grids in all possible scales. find the exact value of sin 2α if cos α=4/5 (α in quadrant I) asked by David on November 28, 2013; math. A trigonometric equation can be written as Q 1 (sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ) = Q 2 (sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ), where Q 1 and Q 2 are rational functions. Since the circle is commonly divided into 360 degrees, the quadrants are named by 90-degree segments. In the fourth quadrant, the values for cos are positive only. We often need to use the trigonometric ratios to solve such problems. The side which is opposite to right angle is known as hypotenuse, the side opposite to angle A is called perpendicular for angle A and the side opposite to third angle is called base for angle A. Unit 7 - Trigonometry, Trigonometric Equations and Identities 11 Quadrant II Quadrant III Quadrant IV Quadrant I. Area of a Triangle. All Trigonometry Formulas: Free Android app (4. Hence cos 150° = − cos 30° = − 3 2. Some of the worksheets displayed are Math 6 notes the coordinate system, 3 points in the coordinate, Trig in the coordinate plane, Polygons in the coordinate plane, Whats the point, Position distance and bearing calculations, 6th grade mini unit graphing in the coordinate. This topic describes important concepts and formulas on trigonometry like Trigonometric Basics, Basic Trigonometric Values, Quotient Formulas, Pythagorean Formulas, Even-Odd Formulas, Periodic Formulas, Reduction Formulas, Sum-Difference Formulas, Double Angle Formulas, Sum-to-Product Formulas, Product-to-Sum Formulas etc. The adjacent side and hypotenuse are part of the ratio for the cosine of θ. 1 Overview 3. What I have attempted to draw here is a unit circle. Test our free trials first. The trigonometric functions for the angles in the unit circle can be memorized and recalled using a set of rules. Operations over Complex Numbers in Trigonometric Form. The calculator will generate a step by step explanations. Many of the calculator pages show work or equations that help you understand the calculations. If you are given an angle and put it into a trigonometric function, it might be positive or negative. Sin x = cos(π /2 - x) Cos x = sin(π /2 - x) Tan x = cot(π /2 - x) The other identities can be derived using the basic identities that is provided above. When you begin solving trigonometry it seems vast one, but it is not what you believe. uk 1 c mathcentre 2009. Unit Circle Trigonometry Coordinates of Quadrantal Angles and First Quadrant Special Angles First, we will draw a right triangle that is based on a 30o reference angle. pptx), PDF File (. TRIGONOMETRY NOTES (Sections 6. Comments #1 yoonju, November 20, 2012 at 12:53 a. The reference angle is the positive acute angle that can represent an angle of any measure. Sine and cosecant are positive in Quadrant 2, tangent and cotangent are positive in Quadrant 3, and cosine and secant are positive in Quadrant 4. Angles are shown as central angles in a circle of radius 1 measured in degrees. These are often numbered from 1st to 4th and denoted by Roman numerals: I (where the signs of the (x; y) coordinates are I (+; +), II (−; +), III (−; −), and IV (+; −). Trigonometry is a field of mathematics first compiled in 2nd century BCE by the Greek mathematician Hipparchus. Precalculus. Quadrant X and Y (step 2) Step 3. Comment/Request This calculation library is so good! I will recommend this site to my friends. A discrete function consists of isolated points. So in our system, we like to divide things into 10. The angle AOB is a measure formed by two rays OA and OB sharing the common point O as. Script name : trigonometric-equation-solver. On the other hand, the sign convention for angles is linked to using a right handed coordinate system for 3-D geometry. In Chapter 4, you will use both perspectives to graph trigonometric functions and solve application problems involving angles and trian-gles. It is shown below. Also, 210° is in the third quadrant, and cosine functions in the third quadrant are negative. All angles with a measure between 0° and 90°, including 8. Solving AAS Triangles. King have said, Quadrant is used for observations and astronomical calculations, such as trigonometry problems solving. Workbook (optional) with lecture notes, sample problems, and exercises so that you can study even when away from the computer. Statistics. , possible (i. The Cartesian coordinate system comprises two number lines, one horizontal and one vertical. And the important six. Laws of Exponents. Essentially, "CAST" stands for COSINE-ALL-SINE,TANGENT. From 0 to 90 degrees, All of sin, cos and tan are positive. LESSON 7 TRIGONOMETRY Example 1 Find cos 150°, sin 240°, cos 315° and sin 270°. If Ѳ is in Quadrant II, and cos Ѳ=-3/4 , find an exact value for sin 2Ѳ. We read the equation from left to right, horizontally, like a sentence. The number system that we use everyday is called the decimal system (the base is 10), but computers use the binary system (the base is 2). Quadrant rule to solve trig equations The quadrant rule or CAST diagram is very useful to know as it saves you having to draw graphs to find a set of solutions to an equation in a given range. Published on Aug 19, 2016. Use trigonometry to determine the angle. Part A ANSWER: Part B ANSWER: Part C ANSWER: Part D = 8, 5, -1 = 14, 2, 15 = 16, 10, 11. because (sin(x))^2 + (cos(x))^2 = 1. In the equation above they are conveniently all together on one side of a simplified equation. (green in unit circle). A quadrant system is basically an X-Y graph. Each side of the right angle triangle has a name: Hypotenuse: Like its name, it is the largest side of the. Quadrantal angles correspond to "integer multiples" of 90 or π. Trigonometry. Vector Addition A quick recap. Let us study about the different measuring systems in the next chapter. The oldest definitions of trigonometric functions, related to right-angle triangles, define them only for acute angles. The four-quadrant inverse tangent, atan2 (Y,X), returns values in the closed interval [-pi,pi] based on the values of Y and X, as shown in the graphic. The sign of a trigonometric function is dependent on the signs of the coordinates of the points on the terminal side of the angle. Now our trigonometry functions tell us about all the points on a circle, not just the ones in the first quadrant. 1 The Inverse Sine, Cosine, and Tangent Functions Objectives of this Section Find the Exact Value of the Inverse. Perpendicular to each other, the axes divide the plane into four sections. The important angles in trigonometry are 0°, 30°, 45°, 60°, 90°, 180°, 270° and 360°. asked by james on April 18, 2014. values 144. 8 shows which angles between 0 and lie in each of the four. Trigonometry Trigonometry is used extensively in our daily lives. Each of these is a quadrant. 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Triangle Identities. 8 years ago. The checkbox "complementary" graphically shows the co- versions of the trigonometric functions. 2 Introduction Our discussion so far has been limited to right-angled triangles where, apart from the right-angle itself, all angles are necessarily less than 90. quadrant if, when the angle is in standard position, the terminal side lies in that quadrant. Introduction A _ coordinate system is formed by drawing. Notice that (1,0) is the intersection point of the initial side of the angle and the unit circle. We now consider angles in cartesian plane. Examples #1-4: Given one trigonometric value and terminal quadrant, find the other trigonometric functions; What have we discovered about the trigonometric values and quadrants? Fun saying! Ten Examples - State the Quadrant(s) in which the angle lies; Reference Angles. Trigonometry and Single Phase AC Generation for Electricians (Flinn) It is also important to understand polarity when dealing with quadrants. In the third quadrant, reference ∠β. In this way, the trigonometric functions can be viewed in terms of x and y. Of what quadrant is A, if sec A is positive and csc A is negative? Latest Problem Solving in Plane Trigonometry Problems. A trigonometric equation is an equation whose variable is expressed in terms of a trigonometric function value. Trigonometry Calculator. It possesses the two advantages over 'acos' and 'asin' functions that 1) its answer can range over four instead of just two quadrants and that 2) it maintains full accuracy throughout its entire angular range, whereas 'acos' encounters accuracy difficulties for angles near 0 and pi and 'asin' near pi/2 and -pi/2. The sign of a trigonometric function is dependent on the signs of the coordinates of the points on the terminal side of the angle. Trigonometry is an important tool for evaluating measurements of height and distance. Since the hypotenuse and opposite sides are known, use the Pythagorean theorem to find the remaining side. Solution of triangles is the term for solving the main trigonometric problem of finding the parameters of a triangle that include angle and length of the sides. Three-Dimensional Cartesian Coordinate System. It is a triangle with specialty, that one angle of the triangle will be of 90 o and rest two will be less than 90 o. This chapter is so much easy to solve. To link to this page, copy the following code to your site:. Understanding the question and drawing the appropriate diagram are the two most important things to be done in solving word problems in trigonometry. The unit circle is often used in the definition of trigonometric functions. Write the equation with the trig function; then input the measures that you know and solve for cos è. Find all pairs of polar coordinates that describe the same point as the provided polar coordinates. go to “Vectors in the Rectangular Coordinate System,” page 178. The top left quadrant is II (2) (x is negative, y is positive) The top right quadrant is I (1) (x is positive, y is positive) The bottom left quadrant is III (3) (x is negative, y is negative) The. The history of trigonometry and of trigonometric functions follows the general lines of the history of mathematics. asked by Kevin on May 21, 2014; Pre Calculus. Finally, the Sine Quadrant, the subject of this class, allows the user to perform quick, accurate trigonometric calculations; along with several other related functions. 1 hr 24 min 17 Examples. 1: Introduction to Trigonometry Angles and Quadrants We start with a discussion of angles. Hence cos 150° = − cos 30° = − 3 2. Of what quadrant is A, if sec A is positive and csc A is negative? Latest Problem Solving in Plane Trigonometry Problems. Year 6 Polar. The extension of trigonometric ratios to any angle in terms of radian measure (real number) are called trigonometric function. By definition, if θ =cos−1 3 5,thencosθ = 3 5. The relation between trigonometric function in sine quadrant can be explained as follows: (a) The angle ‘qaus’ 90° on the curve, which produces a ratio equal to 60 in the sexagesimal system is equivalent to 1 in the decimal system; qaus 90° = jayyib 60 (in sexagesimal system) or sine 90° = 1 (in decimal system) qaus 30Â. Trigonometry. 1st quadrant. A student uses the equation mc013-1. An angle is the amount of rotation of a revolving line with respect to a fixed line. In a plane with a unit circle centered at the origin of a coordinate system, a ray from the origin forms an angle, θ, with the x-axis. AP US History Notes. 1 The Inverse Sine, Cosine, and Tangent Functions Objectives of this Section Find the Exact Value of the Inverse. So could be (accurate to one decimal place) equal to either or else. The reference angle calculator is used to calculate the reference angle of any of the angle given to us. Let vectors , , and. Base and perpendicular for a triangle are either taken as positive or negative depending on whether it is +x, -x, +y or -y. Quadrantal angles correspond to "integer multiples" of 90 or π. Caution! This is a large HTML document. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. An angle is the shape formed when two rays come together. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Students can visualize this by drawing a picture, talking through each step of its construction (first black, then red. In Chapter 4, you will use both perspectives to graph trigonometric functions and solve application problems involving angles and trian-gles. Most of the Cartesian graph papers come up with three options, 'axes with labels', 'only axes' and 'only grids'. This is for that trigonometry quadrant system thing. , Trigonometry Workbook for Dummies by Mary Sterling, and College Trigonometry by Stitz and Zeager, 2013. AP World History Notes. An angle in standard position is said to lie in the quadrant where its terminal side lies. The solution, f (x) is also the y variable, or output. In the 1st quadrant tan function increases from 0 to ∞. Balancing Chemical Equations Worksheets. Attempt Parts 1 – 3 In the system of equations {2x + 6y = 5 quadrant III D) quadrant IV E) quadrant III or. S for 2nd quad. Once again, you should be able to state the EXACT value for each of the trig functions requested. Numbering starts from the upper right quadrant, where both coordinates are positive, and goes in an anti-clockwise direction, as in the picture. Any of the four areas into which a. Unit Circle Trigonometry Coordinates of Quadrantal Angles and First Quadrant Special Angles First, we will draw a right triangle that is based on a 30o reference angle. We enter values of into a polar equation and calculate However, using the properties of symmetry and finding key values of and means fewer calculations will be needed. Quadrant rule to solve trig equations The quadrant rule or CAST diagram is very useful to know as it saves you having to draw graphs to find a set of solutions to an equation in a given range. by the ordered pair $$(x,y)$$ (or whatever variables are in use). To see this, consider the problem of finding the square root of a complex number. The calculator will generate a step by step explanations. Then place the unit circle on the same Cartesian coordinate system. C for 4th quad. To evaluate trigonometric functions of other angles, we use a scientific or graphing calculator or computer software. For any given angle in the first quadrant, there is an angle in the second quadrant with the same sine value. Most commonly, they are defined based on the ratio of the sides of a right triangle, or based on a unit circle. Arts and Humanities. Therefore, 3 3 sin300 22 y r − °= = =− 3 1 cos300 2 x r °= = 3 tan300 3 1 y x − °= = =− Question (How are the trigonometric ratios of 300º related to the trigonometric ratios of 60º? Answer The magnitudes of the trig ratios of 300º are equal to the magnitudes of the trig ratios of the related. Hello, this is the trigonometry lectures for educator. The tangent of the angle theta is the ratio of the opposite side over the adjacent side. The unit circle. Unit Circle Trigonometry Coordinates of Quadrantal Angles and First Quadrant Special Angles First, we will draw a right triangle that is based on a 30o reference angle. The reference angle is always the smallest angle that you. So Quadrant I is between 0° and 90°, Quadrant II is between 90° and 180°, Quadrant III is between 180° and 270°, and Quadrant IV is between 270° and 360° (back to 0°). Trigonometric Ratios in Right Angle Triangle. Free trigonometric identities - list trigonometric identities by request step-by-step Equations Inequalities System of Equations System of Inequalities Polynomials Rationales Coordinate Geometry Complex Numbers Polar/Cartesian Functions Arithmetic & Comp. The axes of a two-dimensional Cartesian system divide the plane into four infinite regions, called quadrants, each bounded by two half-axes. specify a coordinate system in order to find the components of each arrow. It is also known as a "sinecal quadrant" in the English-speaking world. When the axes are drawn according to the mathematical custom, the numbering. In the third quadrant (III), tan (and cotan) are positive. Moving in a counter-clockwise direction is the second, third and fourth quadrant. The transformed graph joins together the letters N, T, W, and A which can be unscrambled to spell WANT. For each point in the coordinate plane, there is one representation, but for each point in the polar plane, there are infinite representations. (Note also that sin 150° = sin 30° = 1 2. The plane is divided into 4 quadrants. Comment/Request This calculation library is so good! I will recommend this site to my friends. Cartesian coordinate system. 0^@ We now use that reference angle to find the value of theta` in the fourth quadrant. The instrument could be used to measure celestial angles, to tell time, to find directions, or to determine the apparent positions of any celestial object for any time. These formula include all trigonometric ratios, trigonometric identities, trigonometric sign rule, quadrant rule and some of the value of the. Problems involving positions and orientations in the sky can then be solved by using the formulae of spherical trigonometry, which apply to spherical triangles, the sides of which are great circles. The three main functions in trigonometry are Sine, Cosine and Tangent. Hence cos 150° = − cos 30° = − 3 2. 11 8 π In calculus, angles are measured in radians, since proofs of major theorems are based on radian measure. It is a triangle with specialty, that one angle of the triangle will be of 90 o and rest two will be less than 90 o. (When an angle is drawn in standard position, its reference angle is the positive acute angle measured. In this section, trigonometric formulas for class 10, 11, 12 is available. sin cos tan y r x r y x θ θ θ = = = Notice in this representation, the acute angle θ is between the x-axis and a line segment from the origin to a point ()x, y in the. LESSON 7 TRIGONOMETRY Example 1 Find cos 150°, sin 240°, cos 315° and sin 270°. Graph quadrants and trigonometric functions. Hipparchus (c. If you picture a right triangle with one side along the x -axis:. More specifically, trigonometry is about right-angled triangles, where one of the internal angles is 90°. AP Psychology Quizzes. The basic concept of trigonometry is based on the repetition of the values of sine, cos and tan after 360⁰ due to their periodic nature. Quadrants, like Super Bowls, are invariably designated by Roman numerals. Varsity Tutors can help you find a trigonometry tutor in Los Angeles, California, who can work with you online. It is also important to understand polarity when dealing with quadrants. For now, just memorize. Search for Library Items Search for Lists Search for Contacts. Let a line through the origin, making an angle of θ with the positive half of the x-axis, intersect the unit circle. Solution of triangles is the term for solving the main trigonometric problem of finding the parameters of a triangle that include angle and length of the sides. Many problems involve right triangles. Major Components of Trigonometry SIN COS TAN Triangle Area Calculator are:. Trigonometry - Free download as Powerpoint Presentation (. South of us, there is a quadrant containing 1 Klingon. It is the graph of the equation x2 + y2 = 1. If a problem. Then 2θ = 90° – 3θ Therefore, sin 2θ = sin (90° – 3θ) = cos 3θ or. 179:999 In Exercises5-8, convert the angles into decimal degrees. It is the smallest angle that we can make from the terminal side of the angle. 4 Trigonometric Functions of General Angles 543 In general, if 0 is an angle measured in degrees, then 0 + 3600k, where k is any integer, is an angle coterminal with o. Example: Find a) sin 120° b) cos 150° c) tan 210° d) csc 300° Solution: a. The letters ASTC signify which of the trigonometric functions are positive, starting in the top right 1st quadrant and moving counterclockwise through quadrants 2 to 4. The solution, f (x) is also the y variable, or output. hp calculators HP 33S Solving Trigonometry Problems Note: It is very easy to forget that one angle mode is set but angles are being entered in a different mode. This chapter is so much easy to solve. Draw in the resultant and use Pythagoras' theorem to determine its size. Charts & Graphs. The three main functions in trigonometry are Sine, Cosine and Tangent. Textbook solution for Trigonometry (MindTap Course List) 8th Edition Charles P. landmark on the modern algebr a was achieved in 1843. Now you can just plot the five ordered pairs in the coordinate plane. The name trigonometry deals with the subject which provides the relationships between the measurements of sides and the angles of a triangle. Also, 210° is in the third quadrant, and cosine functions in the third quadrant are negative. Summary First Quadrant: All are positive in this quadrant. It is a good policy to make it a habit to check the angle mode before every calculation. , possible (i. The Right Triangle and Applications. Solution of triangles is the term for solving the main trigonometric problem of finding the parameters of a triangle that include angle and length of the sides. This page contains a lot of printable graph papers and grids in all possible scales. Example 1 - Finding the Height. In trigonometry, there are three different functions that can be plotted on a Cartesian plane. The remaining values for tangent, cotangent, secant, and cosecant can be calculated by using the functional relationships stated above. A general form triangle has 6 main. This topic describes important concepts and formulas on trigonometry like Trigonometric Basics, Basic Trigonometric Values, Quotient Formulas, Pythagorean Formulas, Even-Odd Formulas, Periodic Formulas, Reduction Formulas, Sum-Difference Formulas, Double Angle Formulas, Sum-to-Product Formulas, Product-to-Sum Formulas etc. Notice that (1,0) is the intersection point of the initial side of the angle and the unit circle. Indeed, the sine and cosine functions are very closely related, as we shall see (if you are not familiar with the sine function, you may wish to read the page entitled "The Sine Functio. Trigonometry is the study of relationships that deal with angles, lengths and heights of triangles and relations between different parts of circles and other geometrical figures. Each side of the right angle triangle has a name: Hypotenuse: Like its name, it is the largest side of the. In the second quadrant, the values for sin are positive only. Use these practical worksheets to ground students in the Law of Sines, the Law of Cosines, tangents, trigonometric functions, and much more!. Trigonometry Revision Summary 1. I would strongly recommend that you learn this method then you can go on to use it to solve trigonometric equations as seen in the 2nd video below. The extension of trigonometric ratios to any angle in terms of radian measure (real number) are called trigonometric function. This allows them to go beyond right triangles, to where the angles can have any measure, even beyond 360°, and can be both positive and negative. All the trig functions are positive in Quadrant 1. CliffsStudySolver Trigonometry. Free math problem solver answers your trigonometry homework questions with step-by-step explanations. Trigonometric Ratios are applicable only for a right angle triangle. For example, this is a simple quadratic equation: 2x2 −8x+7 = 0. From 90 to 180 degrees, only Sin is positive. Four Quadrants. And the fact I'm calling it a unit circle means it has a radius of 1. 1 Sine and Cosine in the First Quadrant Sine and Cosine Values in the First Quadrant Understand sine and cosine values on the unit circle Find exact sine and cosine values for angles in the first quadrant of the unit circle. The basic concept of trigonometry is based on the repetition of the values of sine, cos and tan after 360⁰ due to their periodic nature. It plays an important role in surveying, navigation, engineering, astronomy and many other branches of physical science. Trigonometry: 2016-04-30: Calculating the belt length of a three pulley system: 2006-07-16: If sin0 = 7/8 and 0 is in quadrant 2, find the other five. Unit 7 - Trigonometry, Trigonometric Equations and Identities 11 Quadrant II Quadrant III Quadrant IV Quadrant I. You may select to the number of graphs per page from 1, 4, 8 or 12. The history of trigonometry and of trigonometric functions follows the general lines of the history of mathematics. Use the unit circle given above to determine the values for the sine and cosine of each quadrant angle given below. Basic Math. 4 The Trigonometry of Real Numbers Summary/Concept Rev, Mixed Rev, Practice Test Calc Exploration and Discovery: Signs, Quadrants and Reference Arcs SCS: Trigonometry of the Real Numbers and the Wrapping Function Cumulative Review 1 - 3 Chapter 4: Trigonometric Graphs and Models 4. Solving SSA Triangles. Trigonometry. Quadrantal angles. Course Summary Let us help you master high school trigonometry with our comprehensive review course. I'll show you how to get the first quadrant unit circle values from your special triangles, a key skill that will make the other four quadrants a lot easier to understand and memorize. Today, the use has expanded to involve rotations, orbits, waves, vibrations, etc. Trigonometry deals with the study of the relationship between angles and the sides of a triangle. 2 Problem 11PS. Trigonometry. Now we have seen the equation of a circle in the polar coordinate system. Third Quadrant: Only tan is positive in this quadrant. 179:999 In Exercises5-8, convert the angles into decimal degrees. The angle −300° is in the 1 st quadrant and has a reference angle of 60°. Four Quadrants. A candidate must be an unmarried male and 12th Class pass. pdf), Text File (. The trig ratios for angles between $$180\degree$$ and $$360\degree\text{,}$$ whose terminal sides lie in the third and fourth quadrants, are also related to the trig ratios of familiar angles in the first quadrant. 11 points for the best answerer (I always give the best answerer a thumbs-up. In trigonometry, we study about the three angles of a Conversion of one system into another system : 180 P o each part is known as a quadrant. 237 5804300. (I would just memorize these, since it's simple to do so). King have said, Quadrant is used for observations and astronomical calculations, such as trigonometry problems solving. The “length” of this interval of x values is called the period. First Quadrant of Unit Circle Rather than tackle the unit circle chart all at once, in this video we'll "ease into the pool" by looking at the first quadrant first. These formula include all trigonometric ratios, trigonometric identities, trigonometric sign rule, quadrant rule and some of the value of the. Quadrant rule to solve trig equations The quadrant rule or CAST diagram is very useful to know as it saves you having to draw graphs to find a set of solutions to an equation in a given range. What are quadrants in trigonometry? Wiki User 2015-04-06 09:32:35. 2 Problem 11PS. The letters ASTC signify which of the trigonometric functions are positive, starting in the top right 1st quadrant and moving counterclockwise through quadrants 2 to 4. The point (12,5) is 12 units along, and 5 units up. Trigonometric Form of Complex Numbers. Draw a perpendicular line to the x-axis and label the correct parts. (green in unit circle). Sine, Cosine and Tangent in Four Quadrants Sine, Cosine and Tangent. - [Voiceover] What I want to do in this video is get some practice, or become familiar with what different angle measures in radians actually represent. From Figure 1, the reference triangle of 210° in the third quadrant is a 30°-60°-90° triangle. To communicate about patients’ conditions with other doctors and medical professionals. "The Mathematics of the Heavens and the Earth" looks at the controversies as well, including disputes over whether Hipparchus was indeed the father of trigonometry, whether Indian trigonometry is original or derived from the Greeks, and the extent to which Western science is indebted to Islamic trigonometry and astronomy. 5ˇ 4 is a Quadrant III angle. The reason is purely historical. The axes of a two-dimensional Cartesian system divide the plane into four infinite regions, called quadrants, each bounded by two half-axes. in the interval [0 , 2π) and explain the solutions graphically using a unit circle and the graph of sin (x) - 1/2 in a rectangular coordinate system of axes. Radian angles & quadrants. Institutional users may customize the scope and sequence to meet curricular needs. The Power with Negative Exponent. In trigonometry, there are three different functions that can be plotted on a Cartesian plane. A shooter is typically firing 1 oz or 1 1/8 oz of lead (7 1/2 or 8 shot) downrange at anywhere from 1100 to 1300 feet per second. Measurement of Angles - Radian Measure. a) csc 220° Find sin 220° first. For insurance purposes. Solution sin(x) is positive in quadrant I and II and therefore the given equation has two solutions. 3 Prove the addition and subtraction identities for sine, cosine, and tangent. Its mean "measurement of a triangle". Express as a trigonometric function of an angle in Quadrant I. So could be (accurate to one decimal place) equal to either or else. Trigonometry Word Problem, Finding The Height of a Building, Example 1 Trigonometry Word Problem, Example 2 Trigonometry Word Problem, Determining the Speed of a Boat, Example 3 Simplifying Trigonometric Expressions Using Identities, Example 1. 2 for guidance). The adjacent side is x, the value of the x-coordinate. The values of x are positive to the right of the origin and negative to the left of it, as usual, and the values of y are positive above the origin and negative below it. Another way of putting it is: "a radian is the angle subtended by an arc of length r (the radius)". The remaining values for tangent, cotangent, secant, and cosecant can be calculated by using the functional relationships stated above. Trigonometric Ratios are applicable only for a right angle triangle. Quadrant 1 (0˚ < θ < 90˚). Therefore, cos. And to get our familiarity, we're gonna start with a ray that starts at the origin, and moves along, and Not moves, and points along the positive X axis. The coordinate plane is divided into four regions, or quadrants. Draw a perpendicular line to the x-axis and label the correct parts. Using Reference Angles to Evaluate Tangent, Secant, Cosecant, and Cotangent. From Figure 1, the reference triangle of 210° in the third quadrant is a 30°-60°-90° triangle. The top left quadrant is II (2) (x is negative, y is positive) The top right quadrant is I (1) (x is positive, y is positive) The bottom left quadrant is III (3) (x is negative, y is negative) The. Textbook Authors: Larson, Ron, ISBN-10: 9781337271172, ISBN-13: 978-1-33727-117-2, Publisher: Cengage Learning. The triangle can be located either on the plane or a sphere. SOLUTION: Suppose that sin(theta) = 6/10 and theta is in the second quadrant. Here is a unit circle for this situation. Here, you can find complete preparation material that includes Mock Tests, Previous Year Solved Papers and Topic-wise Tests. Hipparchus (c. There are actually more trigonometric functions in existence than most of us know about. Take note that for angles in the second quadrant, for example, the x-values are. lies in that quadrant or on that axis. Trigonometry Calculator and Sin Cos Tan Calculator helps you to find the value of side, angle, and area of the Right Angled Triangle. 1: Introduction to Trigonometry Angles and Quadrants We start with a discussion of angles. pptx), PDF File (. - [Voiceover] What I want to do in this video is get some practice, or become familiar with what different angle measures in radians actually represent. The name is derived from. These are often numbered from 1st to 4th and denoted by Roman numerals: I (where the signs of the (x; y) coordinates are I (+; +), II (−; +), III (−; −), and IV (+; −). Lectures by Walter Lewin. Basics of Trigonometry ( In Hindi ) 11:40 mins. Another way of putting it is: "a radian is the angle subtended by an arc of length r (the radius)". Each side of the right angle triangle has a name: Hypotenuse: Like its name, it is the largest side of the. by the ordered pair $$(x,y)$$ (or whatever variables are in use). In terms of a right triangle, the tangent of an angle θ, denoted tan(θ), is the ratio of the length of the opposite edge to the length of the adjacent edge of a. Triangle Solving Practice. 1 Right Triangle Trigonometry Trigonometry is the study of the relations between the sides and angles of triangles. In omitting logarithms in trigonometry, it should be kept in view that logarithms are very important in mathematics, and the proper and efficient use. The triangle can be located either on the plane or a sphere. This memory device only works with the major three trig functions, Sine, Cosine and Tangent. Both x and y are positive in QI (Quadrant I). This website uses cookies to ensure you get the best experience. Roots of the Equation. Trigonometry, as the name might suggest, is all about triangles. 7) Solve right-angled triangles (SohCahToa). The most widely used trigonometric functions are the sine, the cosine, and the tangent. It undoubtedly was the sexagesimal system that led Ptolemy to subdivide the diameter of his trigonometric circle into 120 parts; each of these he further subdivided into sixty minutes and each minute of length sixty seconds. The quadrants and some quadrantal angles:. Additional topics such as vectors, polar coordinates and parametric equations may be included. Trigonometry is the branch of mathematics that the relations between the sides and angles of trianglesstudies. Our online coordinate geometry trivia quizzes can be adapted to suit your requirements for taking some of the top coordinate geometry quizzes. com To create your new password, just click the link in the email we sent you. What are quadrants in trigonometry? Wiki User 2015-04-06 09:32:35. Therefore, the side opposite angle theta is y, the value of the y-coordinate. Quadrant x-coordinate y-coordinate Ist Quadrant + + IInd quadrant - + IIIrd quadrant - - IVth quadrant + - S. Graphically, it means we need to do what we just did -- plot the line represented by each inequality -- and then find the region of the graph that is true for BOTH inequalities. Use the identities to solve problems. 05 PART F: QUADRANTS AND QUADRANTAL ANGLES The x- and y-axes divide the xy-plane into 4 quadrants. This tool is constrained by quadrant curve and two axes of horizontal and vertical axes of each axis is divided into 60 sections. Solving AAS Triangles. The ray is allowed to rotate. Now we know this is correct because tan 19. Evaluating Trigonometric Functions with a Calculator. —Step by Step to Success. The triangle can be located either on the plane or a sphere. This is known as the Pythagorean identity. The four-quadrant inverse tangent, atan2 (Y,X), returns values in the closed interval [-pi,pi] based on the values of Y and X, as shown in the graphic. The first coordinate, i. Also contains different coordinate systems like Cartesian, polar and trigonometric coordinates. Trigonometry in Quadrant I. The problem is now to find sinθ, knowing cosθ. Online quadrants coordinate calculator to calculate on which place quadrant falls on the graph. Cummins’ quadrants framework is a model for thinking used by great teachers to think about relevance and challenge. 0 energy points. no Terms Descriptions 1 Distance between the points AB is given by Distance formula $% & & Distance of Point A from Origin$ % & 2 Section Formula A point P(x,y) which divide the line segment AB in. In this tutorial we shall evaluate the simple differential equation of the form $$\frac{{dy}}{{dx}} = \frac{y}{x}$$, and we shall use the method of separating the variables. Math · Algebra II · Trigonometry. By definition, if θ =cos−1 3 5,thencosθ = 3 5. Cliffs Trigonometry. A scatter-plot graph is divided into four quadrants due to the (0, 0) intersection point of the horizontal axis (x-axis) and vertical axis (y-axis). Another way of putting it is: "a radian is the angle subtended by an arc of length r (the radius)". Example 1: The following angles (standard position) terminate in the listed quadrant. A ray is placed with its endpoint at the origin of an xy-axis system, with the ray itself lying along the positive x-axis. Frequently, especially in trigonometry, the unit circle is the circle of radius 1 centered at the origin (0, 0) in the Cartesian coordinate system in the Euclidean plane. because (sin(x))^2 + (cos(x))^2 = 1. The “length” of this interval of x values is called the period. Navigators, surveyors, and carpenters all use the same angle measures, but the angles start out in different positions or places. That's justifiable to me since sine goes from 0 to 1 while cosine goes from 1 to 0 in the first quadrant. Trigonometry & Calculus - powered by WebMath. The project has been designed to develop a speed control system for DC motor in all the four-quadrant. In quadrant I, the solution to sin(x) = 1 / 2 is x = π / 6. I have seen students cramming the signs of Trigonometry Functions ( sinϴ, cosϴ, tanϴ, cosecϴ, secϴ, cotϴ) in four Quadrants. Curriculum (426 topics) Algebra and Geometry Review (98 topics) Real Numbers and Algebraic Expressions (14 topics). y = sin (x) and y = cos (x) are periodic functions because all possible y values repeat in the same sequence over a given set of x values. Deals all about trigonometric formulas and identities. Navigators, surveyors, and carpenters all use the same angle measures, but the angles start out in different positions or places. Solving ASA Triangles. In the video below, I’m going to. A scatter-plot graph is divided into four quadrants due to the (0, 0) intersection point of the horizontal axis (x-axis) and vertical axis (y-axis). 23 The four quadrants of the Cartesian plane. 3rd quadrant c. In mathematics, a unit circle is a circle of unit radius—that is, a radius of 1. From 90 to 180 degrees, only Sin is positive. four-quadrant m odulator (Gilbert-cells type [14]), combine. Another way of putting it is: "a radian is the angle subtended by an arc of length r (the radius)". So this length from the center-- and I centered it at the origin-- this length, from the center to any point on the circle, is of length 1. The simplest quadrant based house system is Porphyry, which simply trisects each quadrant equally. So if you had a circle, and you divided it into 6 equilateral triangles, and each of those equilateral triangles you divided into 60 sections, because you have a base 60 number system, then you might end up with 360 degrees. 3rd quadrant d. The cosine of any angle 5 6. 2 Verify trigonometric identities and simplify expressions using trigonometric identities. Essentially, "CAST" stands for COSINE-ALL-SINE,TANGENT. Trigonometry The simplest definition of trigonometry goes something like this: Trigonometry is the study of angles and triangles. From Figure 1, the reference triangle of 210° in the third quadrant is a 30°-60°-90° triangle. The modern word "sine" is derived from the Latin word sinus, which means "bay", "bosom" or "fold" is indirectly, via Indian, Persian and Arabic transmission, derived from the Greek term khordḗ "bow-string, chord". com and today we're going to learn about probably the single most important identity in all trigonometry which is the Pythagorean identity. Introduction A _ coordinate system is formed by drawing. A base is the number at which you add another digit when you count. Free trigonometric equation calculator - solve trigonometric equations step-by-step This website uses cookies to ensure you get the best experience. find the exact value of sin 2α if cos α=4/5 (α in quadrant I) asked by David on November 28, 2013; math. Integer Part of Numbers. Trigonometric Quadrant System in hindi. The “length” of this interval of x values is called the period. If is a Quadrant II angle with sin( ) = 5 13, and is a Quadrant III angle with tan( ) = 2, nd sin( ). Now we have seen the equation of a circle in the polar coordinate system. Let us study about the different measuring systems in the next chapter. 2nd quadrant has an S in it. Trigonometric Identities. Angles that are in standard position are said to be quadrantal if their terminal side coincides with a coordinate axis. See more ideas about Math classroom, Teaching math and Math projects. Trigonometry Unit 2 Interactive Notebook Pages November 2014 (5) October 2014 (6) September 2014 (20) August 2014 (14) July 2014 (21) June 2014 (24) May 2014 (32) April 2014 (22) March 2014 (13) February 2014 (28). The Right Triangle and Applications. 11ˇ 3 is a Quadrant I angle coterminal with ˇ 3 and 5ˇ 3 x y 4 3 2 1 1 2 3 4 1 2 3 4 1 2 3 4 16. Then,sin cos−1 3 5 =sinθ. Get smarter in Trigonometry on Socratic. In topology, it is often denoted as S 1 because it is a one-dimensional unit n-sphere. Find all pairs of polar coordinates that describe the same point as the provided polar coordinates. Many of the calculator pages show work or equations that help you understand the calculations. Connection with right triangle trigonometry Place a right triangle ABC in the cartesian coordinate system as shown in the figure below, with the hypotenuse emanating from the origin and lying in the first quadrant. straight lines intersecting at right angles at its centre. This memory device only works with the major three trig functions, Sine, Cosine and Tangent. Solution of triangles is the term for solving the main trigonometric problem of finding the parameters of a triangle that include angle and length of the sides. Now our trigonometry functions tell us about all the points on a circle, not just the ones in the first quadrant. A subset of quadrant systems is graduation house systems. Trigonometry & Calculus - powered by WebMath. In trigonometry. 4th quadrant b. Introduction In this session we are going to be looking at the definitions of sine, cosine and tangent for any. Perpendicular to each other, the axes divide the plane into four sections. A general form triangle has 6 main. PART F: QUADRANTS AND QUADRANTAL ANGLES The x- and y-axes divide the xy-plane into 4 quadrants. We enter values of into a polar equation and calculate However, using the properties of symmetry and finding key values of and means fewer calculations will be needed. Moving in a counter-clockwise direction is the second, third and fourth quadrant. Chpter 3 48 Fundamental of trigonometry Chapter 3. If n is an integer and in radians (if in degrees the replace with 360) Reduction formulas. G o t a d i f f e r e n t a n s w e r? C h e c k i f i t ′ s c o r r e c t. Any point on. The triangle can be located either on the plane or a sphere. 11ˇ 3 is a Quadrant I angle coterminal with ˇ 3 and 5ˇ 3 x y 4 3 2 1 1 2 3 4 1 2 3 4 1 2 3 4 16. As in case of geometry, in trigonometry also the measure of the angle is the amount of rotation from the direction of one ray of the angle to the other. AP Human Geography Quizzes. Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step This website uses cookies to ensure you get the best experience. A trigonometric equation can be written as Q 1 (sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ) = Q 2 (sin θ, cos θ, tan θ, cot θ, sec θ, cosec θ), where Q 1 and Q 2 are rational functions. 4 Trigonometric Functions of Any Angle 4. Involutometry and trigonometry : seven place tables of natural functions, for every hundredth of the degree of the 90 quadrant with a complete system of conversion tables and miscellaneous mathematical tables, particulary adaptable to gear calculations: 2. —Step by Step to Success. Trigonometry; Slow Is the Way to Go - Find Legs of Triangles 30 find the leg or angle questions given triangles and rectangles. That is the angle from 0 to 90 in any trigonometric functions gives the positive resultant. In the fourth quadrant (IV), cos (and sec) are positive. These formula include all trigonometric ratios, trigonometric identities, trigonometric sign rule, quadrant rule and some of the value of the. Find h for the given triangle. The notation may vary…. Solving trigonometric equations requires the same techniques as solving algebraic equations. Muslim scholars studied the ancient civilizations from Greece and Rome to China and India. Today, the use has expanded to involve rotations, orbits, waves, vibrations, etc. Use our trigonometry worksheets to help your students quickly master mathematical modeling, circular and periodic functions, higher-degree polynomials, and more!. Determine how fast the volume of the conical sand is changing when the radius of the base is 3 feet, if the rate of change of the radius is 3 inches per minute. 99 a) Compute the reliability of the system assuming independent failure events for the components. Very helpful tutorials for students to solve mathematical problems under the category of heights and distances…try out now… Introduction Trigonometry Angle. (green in unit circle). Don't worry why - someone just decided it would be this way. Trigonometric ratios are defined for acute angles as the ratio of the sides of a right angled triangle. It is a good policy to make it a habit to check the angle mode before every calculation. The reference angle must be < 90∘. This is for that trigonometry quadrant system thing. I hope that I can use your tikz code for my text book. The point's distance from the origin is always 1. CONNECT TO AP Degree measure is denoted using the degree symbol (°), while radian. Find the opposite side of the unit circle triangle. For insurance purposes. This circle is known as a unit circle. All Trigonometry Formulas: Free Android app (4. in the interval [0 , 2π) and explain the solutions graphically using a unit circle and the graph of sin (x) - 1/2 in a rectangular coordinate system of axes. They are easy to calculate: Divide the length of one side of a right angled triangle by another side but we must know which sides!. Students can visualize this by drawing a picture, talking through each step of its construction (first black, then red. 1 The word 'trigonometry' is derived from the Greek words 'trigon' and 'metron' which means measuring the sides of a triangle. Part A ANSWER: Part B ANSWER: Part C ANSWER: Part D = 8, 5, -1 = 14, 2, 15 = 16, 10, 11. Summary First Quadrant: All are positive in this quadrant. e: For basic trigonometric equations, we follow the following steps to solve them: 1. CliffsStudySolver Trigonometry. Drawing in lines to represent the quadrant boundaries, with 0 or 360 horizontal to the right, 90 vertical up, 180 horizontal to the left, and 270 vertical down. ” It primarily dealt with angles and triangles as it pertained to navigation, astronomy, and surveying. Graph quadrants and trigonometric functions. Quadrant I is the upper right quadrant; the others are numbered in counterclockwise order. Your phone may have a built-in Global Positioning System (GPS) that uses trigonometry to tell where you are on Earth’s. In trigonometry and most other mathematical disciplines, you draw angles in a standard, universal position, so that mathematicians around the world are drawing and talking about the same thing. asked by james on April 18, 2014. because (sin(x))^2 + (cos(x))^2 = 1. Then we can discuss circular motion in terms of the coordinate pairs. Solving the inequality R(x) means finding all the values of the. Essentially, "CAST" stands for COSINE-ALL-SINE,TANGENT. For the Love of Physics - Walter Lewin - May 16, 2011 - Duration: 1:01:26. —Apps run on both TI-Nspire CX & the new TI-Nspire CX II-T. The word “trigonometry” comes from the Greek trigōnon (triangle) and metron (measure. In science and engineering, radians are much more convenient (and common) than degrees. Let B' be. Triangle with sides a,b,c. Make sine, cosine or tangent the subject. Four Quadrant 5x5 Grid Size Graph Paper. I used this calculation library to help me in my assignment in trigonometry which is to compute for the six trigonometric functions given the angle in degrees, then converting it to radians. [email protected] Pre-Algebra. All Students Take Calculus is a mnemonic for the sign of each trigonometric functions in each quadrant of the plane. However,I couldn't understand why tan increases from -∞ to 0 in the 2nd quadrant. Quadrant 4 has 270 to 360 degrees. The six functions can also be defined in a rectangular coordinate system. Comment (optional) Email (optional) Share this result with others by using the link below. Trigonometry is a. If n is an integer and in radians (if in degrees the replace with 360) Reduction formulas. Practice problems - online trigonometry calculator Online trigonometry calculator problem 1: Find the third angle of the triangle. Quadrant rule to solve trig equations The quadrant rule or CAST diagram is very useful to know as it saves you having to draw graphs to find a set of solutions to an equation in a given range. Calculator Soup is a free online calculator. And the fact I'm calling it a unit circle means it has a radius of 1. Lectures by Walter Lewin. This explains completely its mathematical and geometrical interpretation and physical significances. In 1620, he invented the first successful analogue device which he developed to calculate logarithmic tangents. Since in Quadrant III, x and y are both negative, if sin is y and cos is x, doesn't that mean that the problem doesn't work, since sin is clearly positive?. To prove a trigonometric identity, we use trigonometric substidutions and algebraic manipulations to either: Transform the right side into the left side, or transform the left side into the right side. From Figure 1, the reference triangle of 210° in the third quadrant is a 30°-60°-90° triangle. S for 2nd quad. This video will give you the tools to figure out which sign it is. A general form triangle has 6 main. GeoGebra 3D & AR: PreCalc & Calculus Resources. Find and save ideas about trigonometry on Pinterest. Many problems involve right triangles. sin cos tan y r x r y x T T T Notice in this representation, the acute angle is between the x-axis and a line segment from the origin to a point ,xy in the first quadrant. In case of 180 + x and 180 - x ; 360 + x and 360 - x the trig ratio remains same but the sign changes according to its quadrant. Given a trigonometric value and quadrant for an angle, utilizes the structure and relationships of trigonometry, including relationships in the unit circle, to identify other trigonometric values for that angle, and describes the relationship between the radian measure and the subtended arc in the circle in contextual situations. The point's distance from the origin is always 1. Graduated houses have house sizes vary more gradually, by making the center house in the small quadrant narrower, and the center house in large quadrants wider. If the terminal side of an angle lies along one of the axes, then that angle doesn't lie in one specific quadrant; it lies along the border. Trigonometry. The Cartesian coordinate system comprises two number lines, one horizontal and one vertical. An example of a function is f (x) = x + 4. 1 Angles and Their Measure Angles De nition 1. Example #1. Now let’s have you try a few of these advanced trigonometry problems. That is, this angle is coterminal with 60°. Trigonometric ratios of angles in the first quadrant, between 0° and 90° Consider following triangles DOA and COB where their internal angles at O are 60° and 45° respectively. 11 points for the best answerer (I always give the best answerer a thumbs-up. Example: Find a) sin 120° b) cos 150° c) tan 210° d) csc 300° Solution: a. The accuracy of Loran is measured in hundreds of meters, but only has limited coverage. Determine the quadrant in which a 130° angle lies.
boj67cihglr909s sqm3sreznegkg tzcc0n40yccvg1 xvw61wry02shll j6lpb29yoq to60yeynpx 4oiqz7t1da chmhm77tth q7kdxrne7x4dql u83m0p791zr at0cku0oesz4aem tutlr1fars0y x0zq7pavd5ebku2 ncs1x0oe9mor1 2q1bl4bf8ayy mshdvwxqkejlb khzkx84d90gete 8mi9r1oe96usn4u a4wmm7e5jbdd l1tzxl5fc4t ccg384go0jc u4afn5fdn4 yqdacs7naiwml j5frbyuirevx m9f8hr11kyo | 2020-08-14T22:18:40 | {
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https://math.stackexchange.com/questions/3136407/is-a-%E2%88%92-b-a-%E2%88%92-b-always-true-or-sometimes-false | # Is |A − B| = |A| − |B| always true or sometimes false?
A and B are arbitrary finite sets. I'm having trouble proving whether this claim is always true or sometimes false. If it's true what's the detailed proof that proves this statement? If not what is the counterexample?
My understanding here is that on the left side you have the cardinality of the set difference of A - B. On the right side you subtract the cardinality of both sets? Is this correct or am I missing something here?
So for example: Let A={17,18,19} and B={17,18,20} A - B gives me {19} so |A - B| is 1 Now |A| would be 3 and |B| would be 3 So 3-3 gives me 0
I think I might be misinterpreting the right side but I'm not sure. If someone can give me some clarity on the following question it would be much appreciated.
• I think you have it right. – Robert Shore Mar 5 at 16:24
• Your disproof is correct. – enedil Mar 5 at 16:24
• @DonThousand That's silly. The elements of B that are not in A can't have anything to do with counting elements in A. $|A - B| = |A| - |B\cap A|$. – fleablood Mar 5 at 16:38
• @fleablood Oops you are right, silly me – Don Thousand Mar 5 at 16:39
Think of it this way.
There are elements in $$A$$ and not $$B$$. Elements in $$B$$ and not $$A$$ and elements in both. (And elements in neither.)
To count elements that are only in $$A$$, the elements in $$B$$ that are not in $$A$$ can't have anything to do with this and there could be gazillions (on none) elements in $$B$$ that are not in $$A$$ and they will have nothing to do with calculating $$|A\setminus B|$$.
... in fact you consider the three disjoint and distinct sets that "make up" $$A$$ and $$B$$:
$$A\setminus B, B\setminus A, A \cap B$$
it becomes clear:
$$|A| = |A\setminus B| + |A \cap B|$$
$$|B| = |B\setminus A| + |A\cap B|$$.
And that makes $$|A\setminus B| =|A| - |A\cap B|$$ so the equation is only true if $$A\cap B = B$$, or in other words only if $$B \subset A$$.
We can take this further.
$$|A\setminus B| = |A| - |A\cap B| = |A| -(|B| - | B\setminus A|)$$.
So $$|A| \ge |A\setminus B| \ge |A| - |B|$$.
$$|A| = |A\setminus B|$$ if and only if $$A$$ and $$B$$ are disjoint. (i.e. If $$A\cap B=\emptyset$$ or if $$B\setminus A=B$$)
$$|A\setminus B| = |A| - |B|$$ if and only if $$B$$ is a subset of $$A$$. (i.e. if $$A\cap B = B$$ or if $$B\setminus A = \emptyset$$) | 2019-11-20T17:08:24 | {
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https://drchristiansalas.com/2015/04/05/alternative-approaches-to-solving-advanced-level-vector-problems/ | # Alternative approaches to solving advanced level vector problems
Trying to find more than one solution for a given problem can be an effective way to convert routine exercises or exam questions into exploratory `research’ type activities which are more open-ended and fun. Getting students to generate, analyse and compare alternative solutions to problems has long been recommended in the mathematics education literature (see, e.g., Cai, J., Brook, M., 2006, Looking back in problem solving, Mathematics Teaching Incorporating Micromath, 196, pp. 42-45). I recently explored the potential for this approach in the context of advanced level vector problems and want to document some alternative approaches I found for answering some of these problems here. A typical advanced level vector problem – in fact, an exam question – is as follows (the reader should attempt this question before reading on):
With respect to a fixed origin $O$, the line $l$ has equation
$\mathbf{r} = \begin{pmatrix}13\\8\\1\end{pmatrix} + \lambda\begin{pmatrix}2\\2\\-1\end{pmatrix}$
where $\lambda$ is a scalar parameter. The point $A$ lies on $l$ and has coordinates $(3, -2, 6)$.
The point $P$ has position vector $(-p \mathbf{i} + 2p \mathbf{k})$ relative to $O$ where $p$ is a constant.
Given that vector $\overrightarrow{PA}$ is perpendicular to $l$,
(a) find the value of $p$.
Given also that $B$ is a point on $l$ such that $\angle BPA = 45^{\circ}$,
(b) find the coordinates of the two possible positions of $B$.
The following is a sketch of the scenario.
Let $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$, and $\overrightarrow{OP} = \mathbf{p}$. The interesting part of this question is (b), but with regard to (a) we have
$\overrightarrow{PA} = \mathbf{a} - \mathbf{p} = \begin{pmatrix}3\\-2\\6\end{pmatrix} - \begin{pmatrix}-p\\\text{0}\\2p\end{pmatrix} = \begin{pmatrix}3+p\\-2\\6-2p\end{pmatrix}$
Since $\overrightarrow{PA}$ is perpendicular to $l$, the dot product with the direction vector of $l$ must be zero, so we must have
$\begin{pmatrix}3+p\\-2\\6-2p\end{pmatrix} \cdot \begin{pmatrix}2\\2\\-1\end{pmatrix} = 6 + 2p - 4 - 6 + 2p = 0$
so $p = 1$. It follows that
$\mathbf{p} = \begin{pmatrix}-p\\\text{0}\\2p\end{pmatrix} = \begin{pmatrix}-1\\\text{0}\\2\end{pmatrix}$
and
$\overrightarrow{PA} = \begin{pmatrix}3+p\\-2\\6-2p\end{pmatrix} = \begin{pmatrix}4\\-2\\4\end{pmatrix}$
With regard to part (b), the most straightforward approach is to observe that since $\overrightarrow{AB}$ is collinear with $l$, the coordinates of $B$ with respect to $O$ must be given by
$\mathbf{b} = \mathbf{a} + \overrightarrow{AB} = \begin{pmatrix}3\\-2\\6\end{pmatrix} + \mu\begin{pmatrix}2\\2\\-1\end{pmatrix}$
where $\mu$ is a parameter to be found, and where the length of $\overrightarrow{AB}$ is equal to the length of $\mathbf{b} - \mathbf{a} = \mu\begin{pmatrix}2\\2\\-1\end{pmatrix}$, which is $|\mu| \sqrt{2^2 + 2^2 + (-1)^2} = 3|\mu|$. Now, $|\overrightarrow{PA}| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{36} = 6$ and since the right-angled triangle in the sketch above is isosceles we must also have $|\overrightarrow{AB}| = 3|\mu| = 6$ and so $|\mu| = 2$. Substituting $\mu = 2$ into the expression for $\mathbf{b}$ above we get the coordinates of $B$ in the sketch above to be $(7, 2, 4)$. However, the sketch above also shows another possible position for $B$, to the left of $A$. By symmetry, the coordinates of this other possible position for $B$ can be found by setting $\mu = -2$ (since in this case the vector $\overrightarrow{AB}$ would point to the left of $A$ rather than to the right). Substituting $\mu = -2$ into the expression for $\mathbf{b}$ above we get the other possible coordinates of $B$ to be $(-1, -6, 8)$. This solves part (b).
The question now arises as to whether there is any other way of solving part (b) of this problem? I found the following alternative solution approach which I think is instructive (although algebraically more complicated, perhaps) in that it uses the cosine expression for the scalar product. Since the angle between $\overrightarrow{PA}$ and $\overrightarrow{PB}$ is $45^{\circ}$, we must have
$\overrightarrow{PA} \cdot \overrightarrow{PB} = |\overrightarrow{PA}||\overrightarrow{PB}|\cos 45^{\circ} = \frac{1}{\sqrt{2}}|\overrightarrow{PA}||\overrightarrow{PB}|$
Since $B$ is on the line $l$, it must have coordinates of the form $(13+2\theta, 8+2\theta, 1-\theta)$. As we found in part (a) above, we have (in row vector form)
$\overrightarrow{PA} = (4, -2, 4)$
and
$\overrightarrow{PB} = (13+2\theta, 8+2\theta, 1-\theta) - (-1, 0, 2) = (14+2\theta, 8+2\theta, -1-\theta)$
Therefore
$\overrightarrow{PA} \cdot \overrightarrow{PB} = 56 + 8\theta - 16 - 4\theta - 4 - 4\theta = 36$
but also
$\overrightarrow{PA} \cdot \overrightarrow{PB} = \frac{1}{\sqrt{2}}|\overrightarrow{PA}||\overrightarrow{PB}| = \frac{1}{\sqrt{2}}\cdot6\sqrt{(14+2\theta)^2 + (8+2\theta)^2 + (1+\theta)^2} = 3\sqrt{2}\sqrt{9\theta^2 + 90\theta + 261}$
Equating these two we get
$3\sqrt{2}\sqrt{9\theta^2 + 90\theta + 261} = 36$
which simplifies to
$\theta^2 + 10\theta + 21 = 0$
or
$(\theta + 3)(\theta + 7) = 0$
The solutions to the quadratic are therefore $\theta = -3$ and $\theta = -7$. Since $B$ must have coordinates of the form $(13+2\theta, 8+2\theta, 1-\theta)$, substituting the two possible solutions for $\theta$ gives us the two possible coordinates of $B$ as $(7, 2, 4)$ and $(-1, -6, 8)$ as before.
I performed similar exercises with other advanced level vector problems and was able to find alternative solution approaches which were usually instructive in some way, although also usually more algebraically demanding than the most straightforward approach available. Another example I want to record here concerns the following advanced level vector problem – also an exam question (again, the reader should attempt to solve this problem before reading my discussion of it below):
With respect to a fixed origin $O$, the lines $l_1$ and $l_2$ are given by the equations
$l_1: \mathbf{r} = \begin{pmatrix}6\\-3\\-2\end{pmatrix} + \lambda\begin{pmatrix}-1\\2\\3\end{pmatrix}$
and
$l_2: \mathbf{r} = \begin{pmatrix}-5\\15\\3\end{pmatrix} + \mu\begin{pmatrix}2\\-3\\1\end{pmatrix}$
where $\lambda$ and $\mu$ are scalar parameters.
(a) Show that $l_1$ and $l_2$ meet and find the position vector of their point of intersection $A$.
(b) Find, to the nearest $0.1^{\circ}$, the acute angle between $l_1$ and $l_2$.
The point $B$ has position vector $\begin{pmatrix}5\\-1\\1\end{pmatrix}$
(c) Show that $B$ lies on $l_1$.
(d) Find the shortest distance from B to the line $l_2$.
The most interesting part of this question for me is part (d), but for part (a) we observe that since the two lines meet, it must be the case that $6 - \lambda = -5 + 2\mu$ or equivalently
$2\mu + \lambda = 11$
and also $-3 + 2\lambda = 15 - 3\mu$ or equivalently
$3\mu + 2\lambda = 18$
Solving these two simultaneously gives $\lambda = 3$ and $\mu = 4$. If the two lines do indeed meet, these values for $\lambda$ and $\mu$ will be consistent with equality of the third coordinates of $l_1$ and $l_2$ at the point of intersection. Using $\lambda = 3$ to find the third coordinate for $l_1$ and using $\mu = 4$ to find the third coordinate for $l_2$, we find that the third coordinate is 7 in both cases, so the two lines do indeed meet. The position vector of their point of intersection $A$ is then
$\begin{pmatrix}6\\-3\\-2\end{pmatrix} + 3\begin{pmatrix}-1\\2\\3\end{pmatrix} = \begin{pmatrix}3\\3\\7\end{pmatrix}$
or equivalently
$\begin{pmatrix}-5\\15\\3\end{pmatrix} + 4\begin{pmatrix}2\\-3\\1\end{pmatrix} = \begin{pmatrix}3\\3\\7\end{pmatrix}$
For part (b) we find the dot product of the direction vectors of the two lines. We get
$(-1, 2, 3) \cdot (2, -3, 1) = -5$
but also
$(-1, 2, 3) \cdot (2, -3, 1) = |(-1, 2, 3)||(2, -3, 1)|\cos\theta = 14\cos\theta$
Equating the two gives
$\cos\theta = \frac{-5}{14}$
so $\theta = \arccos(\frac{-5}{14}) = 110.9248324^{\circ}$, and therefore the acute angle must be $180^{\circ} - \theta = 69.1^{\circ}$ (to nearest $0.1^{\circ}$).
For part (c) we observe that if $B$ lies on $l_1$, there is a $\lambda$ such that
$\begin{pmatrix}5\\-1\\1\end{pmatrix} = \begin{pmatrix}6-\lambda\\-3+2\lambda\\-2+3\lambda\end{pmatrix}$
By inspection, the required value is $\lambda = 1$.
Finally, for part (d) it is helpful to draw the following sketch of the situation:
We know that point $A$ has coordinates $(3, 3, 7)$ and point $B$ has coordinates $(5, -1, 1)$, and therefore the length of the line from $B$ to $A$ is
$|\overrightarrow{BA}| = |(3, 3, 7) - (5, -1, 1)| = |(-2, 4, 6)| = \sqrt{(-2)^2 + 4^2 + 6^2} = \sqrt{56}$
The shortest distance from $B$ to the line $l_2$ is the length of the perpendicular from $B$ which is shown in the sketch as intersecting line $l_2$ at a point $C$. Therefore the required shortest distance is $|\overrightarrow{BC}|$ which can be obtained from simple trigonometry as
$|\overrightarrow{BC}| = |\overrightarrow{BA}|\sin69.1^{\circ} = \sqrt{56}\sin69.1^{\circ} = 6.99$
This solves part (d). Again, we ask if there is some alternative way to obtain this shortest distance? I found the following alternative approach which avoids using trigonometry (at the expense of being algebraically more cumbersome). A point on the line $l_2$ must have coordinates of the form
$(-5 + 2\mu, 15 - 3\mu, 3 + \mu)$
The distance between $B$ and any such point is given by
$|(5, -1, 1) - (-5 + 2\mu, 15 - 3\mu, 3 + \mu)| = |(10 - 2\mu, -16 + 3\mu, -2 - \mu)|$
$= \sqrt{(10 - 2\mu)^2 + (-16 + 3\mu)^2 + (-2 - \mu)^2}$
$= \sqrt{14\mu^2 - 132\mu + 360}$
$= \sqrt{14(\mu - \frac{33}{7})^2 + \frac{342}{7}}$
Therefore the shortest distance is when $\mu = \frac{33}{7}$ and is $\sqrt{\frac{342}{7}} = 6.99$, which agrees with the previous result. | 2019-04-26T13:51:24 | {
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https://math.stackexchange.com/questions/1863287/conjectured-value-of-int-0-infty-left-fracx-1-ln2-x-frac1-ln-x | # Conjectured value of $\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}$
I was curious whether this integral has a closed form expression :
$$\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}$$
The integrand has a singularity at $x=1$, but it's removable. And as $x \to \infty$, the integrand behaves like $\frac{1}{x \ln^{2}x}$. So the integral clearly converges.
Although I have not been able to derive its closed form, I think, by reverse symbolic calculators, up to 20 digits it could be
$$I=\frac{4G}{\pi}$$
where $G$ is Catalan's constant. Is it true or is it completely fabulous?
EDIT. NOTE :
For better search to this integral I have renamed the title from Conjectured value of logarithmic definite integral, which is ambiguous and did not say anything, to the current one with integral explicitly written.
• It's correct to at least 200 digits (according to Maple). Jul 18 '16 at 17:05
• It's simply true :D Jul 18 '16 at 17:12
• I just posted yet a fifth way forward. Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark Jul 19 '16 at 15:12
• I am astonished how beautiful all answers are ! Thank you everyone for the hard work and for convincing me in the conjecture :) (OP). Jul 19 '16 at 20:49
• You're welcome! It was my pleasure. Jul 19 '16 at 22:41
It is not necessary to exploit any symmetries of the integrand. Setting $x=e^y$
$$I=\int_{-\infty}^{\infty}\underbrace{e^y\left(\frac{e^y-1}{y^2}-\frac{1}{y}\right)\frac{1}{e^{2y}+1}}_{f(y)}\,dy$$
Integrating around a big semicircle in the UHP (exercise: show convergence in this domain of the complex plane) we obtain
$$I=2 \pi i \sum_{n=0}^{\infty}\text{Res}(f(z),z=z_n)$$ here $z_n=\frac{i\pi}2(2n+1)$. This is easily rewritten as
$$I=2 \pi i\left(\left(\frac{1}{\pi}\color{blue}{\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}}-\frac{2}{\pi^2}\color{red}{\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}}\right) -\frac{2i}{\pi^2}\color{green}{\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}}\right)$$
since $\color{blue}{\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}}$ and $\color{red}{\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}}$ the imaginary parts cancel and we are left with
$$I= \frac{4}{\pi}\color{green}{\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}}=\frac{4\color{green}{K}}{\pi}$$
• (+1) It is interesting to point out that since the imaginary parts have to cancel out, that also gives a proof of $\sum_{n\geq 0}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}.$ Jul 18 '16 at 18:04
• ...or that $\sum_{n\geq0}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$ :-p Jul 18 '16 at 18:05
• @tired: I hope you don't mind, I added your approach to my Community Wiki answer in the historical thread math.stackexchange.com/questions/8337/… Jul 18 '16 at 18:46
• @JackD'Aurizio i'm fine with that...glad that i also made it into this thread somehow...^^ Jul 18 '16 at 18:54
Our integral equals
$$I=\int_{-\infty}^{+\infty}\left(\frac{e^t-1-t}{t^2}\right)\frac{e^t}{e^{2t}+1}\,dt$$ that by exploiting symmetry becomes $$I = \int_{0}^{+\infty}\frac{e^{t}+e^{-t}-2}{t^2(e^{t}+e^{-t})}\,dt =\int_{0}^{+\infty}\frac{\cosh(t)-1}{t^2\cosh(t)}\,dt$$ The last integral is straightforward to compute trough the residue theorem. Since $$\text{Res}\left(\frac{\cosh(t)-1}{t^2\cosh(t)},t=\frac{\pi(2k+1)}{2}i\right)= (-1)^{k+1}\frac{4i}{\pi^2(2k+1)^2}$$ we have: $$\boxed{ I = \frac{4}{\pi}\sum_{k\geq 0}\frac{(-1)^k}{(2k+1)^2}=\color{red}{\frac{4G}{\pi}}}$$
as conjectured.
• By the way, you can also go directly into the first integral without any symmetry considerations...but your approach is definitly cleaner (no nasty imaginary parts which have to cancel) Jul 18 '16 at 17:39
• I put in an approach, have a look if you are interested :) Jul 18 '16 at 18:00
• Jul 18 '16 at 18:29
• @venus, M.N.C.E answer shows that the two integrals have very similar structure Jul 18 '16 at 20:17
• Jack, this is very efficient. I just posted a solution that avoids the complex plane and the use of special functions. Instead, it relies on simple juggling of integrals including an application of Feynman's Trick. The result is a well-known integral representation of $G$. -Mark Jul 19 '16 at 15:11
Here is yet another approach. We first note that we can write $\frac{x-1}{\log(x)}$ as
$$\frac{x-1}{\log(x)}=\int_0^1 x^t\,dt$$
Therefore, we can write
\begin{align} \int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\int_0^\infty \int_0^1 \frac{x^t-1}{\log(x)}\,\frac{1}{1+x^2}\,dt\,dx\\\\ &=\int_0^1 \int_0^\infty \frac{x^t-1}{\log(x)}\,\frac{1}{1+x^2}\,dx\,dt\tag1 \end{align}
Let $I(t)$ represent the inner integral of the right-hand side of $(1)$. Then, differentiating, we find that
\begin{align} I'(t)&=\int_0^\infty \frac{x^t}{1+x^2}\,dx\\\\ &=\frac{\pi}{2\cos(\pi t/2)}\tag 2 \end{align}
where I derived the right-hand side of $(2)$ in THIS ANSWER. Alternatively, using real analysis only, we have
\begin{align} \int_0^\infty \frac{x^t}{1+x^2}\,dx&=\frac12 B\left(\frac{1+t}{2},\frac{1-t}{2}\right)\\\\ &=\frac12 \Gamma\left(\frac{1+t}{2}\right)\Gamma\left(\frac{1-t}{2}\right)\\\\ &=\frac12\frac{\pi}{\sin\left(\pi\frac{1+t}{2}\right)}\\\\ &=\frac{\pi}{2\cos(\pi t/2)} \end{align}
Integrating $(2)$ and using $I(0)=0$ reveals
$$I(t)=\int_0^t \frac{\pi}{2\cos(\pi t'/2)}\,dt' \tag 3$$
Substituting $(3)$ into $(1)$ yields
\begin{align} \int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\frac{\pi}{2}\int_0^1 \int_0^t \sec(\pi t'/2)\,dt'\,dt \tag 4\\\\ &=\frac{\pi}{2}\int_0^1 (1-t)\sec(\pi t/2)\,dt \tag5\\\\ &=\frac{\pi}{2}\int_0^1 t\csc(\pi t/2)\,dt \tag 6\\\\ &=\frac{1}{\pi}\int_{-\pi/2}^{\pi/2}\frac{t}{\sin(t)}\,dt \tag 7\\\\ &=\frac{4G}{\pi} \tag 8 \end{align}
as was to be shown!
NOTES:
In going from $(4)$ to $(5)$, we changed the order of integration and carried out the inner integral.
In going from $(5)$ to $(6)$, we enforced the substitution $t \to 1-t$.
In going from $(6)$ to $(7)$, we enforced the substitution $t \to 2t/\pi$ and exploited the evenness of the integrand.
In going from $(7)$ to $(8)$, we made use of one of the integral identities for Catalan's Constant as found HERE.
ALTERNATIVE DEVELOPMENT
Note that we can write $(3)$ as
$$I(t)=\log\left(\cot\left(\frac{\pi}{4}(1-t)\right)\right) \tag 9$$
Then, substituting $(9)$ into $(1)$ yields
\begin{align} \int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\int_0^1 \log\left(\cot\left(\frac{\pi}{4}(1-t)\right)\right)\,dt \\\\ &=\frac{4}{\pi}\int_0^{\pi/4} \log(\cot(t))\,dt \tag 9\\\\ &=\frac{4G}{\pi} \end{align}
which uses another well-known integral identity for $G$ as found HERE.
Note that if we enforce the substitution $t\to \text{arccot}(t)$ in $(9)$, we find the result in terms of the series representation of $G$ as
\begin{align} \int_0^\infty \left(\frac{x-1}{\log^2(x)}-\frac{1}{\log(x)}\right)\frac{1}{1+x^2}\,dx&=\frac{4}{\pi}\int_0^{\pi/4} \log(\cot(t))\,dt \\\\ &=\frac{4}{\pi}\int_1^{\infty}\frac{\log(t)}{1+t^2}\,dt\\\\ &=-\frac{4}{\pi}\int_0^1 \frac{\log(t)}{1+t^2}\,dt\\\\ &=-\frac{4}{\pi}\sum_{n=0}^\infty(-1)^n \int_0^1 t^{2n}\log(t)\,dt\\\\ &=\frac{4}{\pi}\sum_{n=0}^\infty(-1)^n \int_0^1 \frac{t^{2n}}{2n+1}\,dt\\\\ &=\frac{4}{\pi}\sum_{n=0}^\infty(-1)^n \frac{1}{(2n+1)^2}\\\\ &=\frac{4G}{\pi} \end{align}
as expected once again!
• @tired Well, after reading all of the great answers already posted, I decided to try another way forward for this one. Jul 19 '16 at 15:05
• Simply a fantastic answer. The beauty in this one is that it only requires first year calc knowledge (along with Fubini's theorem or the like to justify the integral switch and the Catalan's Constant definition) Jul 19 '16 at 16:59
• I have to admit that your answer along with @RandomVariable's are better than mine if not the best of all answers here. Both of you are deserved more than (+1) Jul 20 '16 at 1:55
Though using the residue method is somewhat straightforward, but not everyone can understand it. So, here is a residue-free method:
Split the integral into two terms where each term is in the interval $0<x<1$ and $1<x<\infty$, then use the substitution $x\mapsto\frac{1}{x}$ to the second term. We will get $$\left[\int_{0}^{1}+\int_{1}^{\infty}\right]\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}=\int_{0}^{1}\frac{(x-1)^2}{x\ln^2 x}\cdot\frac{\mathrm{d}x}{x^2+1}\tag1$$ Now, for $a\ge-1$ , one may consider the following integral $$I(a)=\int_{0}^{1}x^a\cdot\frac{(x-1)^2}{\ln^2 x}\cdot\frac{\mathrm{d}x}{1+x^2}\tag2$$ and the desired integral is $I(-1)$. Since $0<x<1$, one may observe that $I(\infty)\to0$ as $a\to\infty$. \begin{align} I''(a)&=\int_{0}^{1}\frac{x^a(x-1)^2}{1+x^2}\ \mathrm{d}x\\[10pt] &=\int_{0}^{1}\sum_{k=0}^\infty(-1)^k\ x^{2k+a}\ (x^2-2x+1)\ \mathrm{d}x\\[10pt] &=\sum_{k=0}^\infty(-1)^k\left(\frac{1}{2k+a+3}-\frac{2}{2k+a+2}+\frac{1}{2k+a+1}\right)\\[10pt] &=\frac{1}{4}\left[\psi\left(\frac{a+5}{4}\right)-2\psi\left(\frac{a+4}{4}\right)+2\psi\left(\frac{a+2}{4}\right)-\psi\left(\frac{a+1}{4}\right)\right]\\[10pt] I'(a)&=\ln\Gamma\left(\frac{a+5}{4}\right)-\ln\Gamma\left(\frac{a+1}{4}\right)+2\ln\Gamma\left(\frac{a+2}{4}\right)-2\ln\Gamma\left(\frac{a+4}{4}\right)\\[10pt] I(a)&=4\left[\psi\left(-2,\frac{a+5}{4}\right)-\psi\left(-2,\frac{a+1}{4}\right)+2\psi\left(-2,\frac{a+2}{4}\right)-2\psi\left(-2,\frac{a+4}{4}\right)\right]\tag3\\[10pt] \end{align} Hence $$I(-1)=4\left[\psi\left(-2,1\right)-\psi\left(-2,0\right)+2\psi\left(-2,\frac{1}{4}\right)-2\psi\left(-2,\frac{3}{4}\right)\right]=\frac{4G}{\pi}$$ Wolfram Alpha confirms it. One may also use the special values of generalized polygamma function and its related relation with derivative of Hurwitz Zeta Function: $$\psi(-2,x)=\zeta'(-1,x)-\frac{x^2}{2}+\frac{x}{2}-\frac{1}{12}$$
Jack D'Aurizio showed that $$\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{dx}{x^2+1} = \int_{0}^{\infty} \left( 1-\frac{1}{\cosh x} \right) \frac{dx}{x^{2}} .$$
The following is an alternative evaluation of the integral on the right.
An integral representation of the Dirichlet beta function is $$\beta(s) = \frac{1}{ 2 \, \Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1}}{\cosh(x)} \, dx \, , \quad \text{Re}(s) >0\tag{1}.$$
And the Laplace transform of $x^{s-1}$ is $$\int_{0}^{\infty} x^{s-1} e^{-ax} \, dx = \frac{\Gamma(s)}{a^{s}} \, , \quad (\text{Re}(s) >0, \ \text{Re}(a)>0) \tag{2}.$$
Subtracting $(1)$ from $(2)$, we get $$\int_{0}^{\infty}\left(e^{-ax} - \frac{1}{\cosh (x)} \right) x^{s-1} \, dx = \Gamma(s) \left( a^{-s} - 2 \beta(s) \right) , \tag{3}$$ which holds for $\text{Re}(s) > -1$ and $\text{Re}(a) > 0$.
If we restrict $s$ to that strip $-1 < \text{Re}(s) <0$, then $(3)$ also holds for $a = 0$.
From the functional equation of the Dirichlet beta function, we see that the Dirichlet beta function has a zero at $s=-1$.
So letting $s$ tend to $-1$, we get
\begin{align} \int_{0}^{\infty} \left(1- \frac{1}{\cosh x} \right) \frac{dx}{x^{2}} &= \lim_{s \downarrow -1} \Gamma(s) \left((0 - 2 \beta(s)\right) \\ &= - 2 \lim_{s \downarrow -1} \Gamma(s) \beta(s) \\ &=-2 \lim_{s \downarrow -1} \left(-\frac{1}{s+1} + \mathcal{O}(1) \right) \beta(s) \\ &= 2 \beta'(-1). \end{align}
To show that $\displaystyle \beta'(-1) = \frac{2G}{\pi}$, differentiate both sides of the functional equation, and then let $s=2$.
Note that $\ds{\quad{x - 1 \over \ln\pars{x}} = \int_{0}^{1}x^{t}\,\dd t\,,\quad x \in \pars{0,1}}$.
\begin{align} &\color{#f00}{\int_{0}^{\infty}\bracks{% {x - 1 \over \ln^{2}\pars{x}} - {1 \over \ln\pars{x}}} \,{\dd x \over x^{2} + 1}} \\[5mm] = &\ \int_{0}^{1}\bracks{% {x - 1 \over \ln^{2}\pars{x}} - {1 \over \ln\pars{x}}} \,{\dd x \over x^{2} + 1} + \int_{1}^{0}\bracks{% {1/x - 1 \over \ln^{2}\pars{1/x}} - {1 \over \ln\pars{1/x}}} \,{-\,\dd x/x^{2} \over 1/x^{2} + 1} \\[5mm] = &\ \int_{0}^{1}{\pars{x - 1}^{2} \over x\ln^{2}\pars{x}}\,{\dd x \over x^{2} + 1} = \int_{0}^{1}{1 \over x\pars{x^{2} + 1}}\int_{0}^{1}x^{y}\,\dd y\int_{0}^{1}x^{z}\,\dd z\,\dd x \\[5mm] = &\ \int_{0}^{1}\int_{0}^{1}\int_{0}^{1}{x^{y + z - 1} \over x^{2} + 1} \,\dd x\,\dd y\,\dd z = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} {x^{y + z - 1} - x^{y + z + 1}\over 1 - x^{4}}\,\dd x\,\dd y\,\dd z \\[5mm] \stackrel{x^{4}\ \mapsto\ x}{=}\,\,\, &\ {1 \over 4}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1} {x^{y/4 + z/4 - 1}\,\,\, -\,\,\, x^{y/4 + z/4 - 1/2}\over 1 - x} \,\dd x\,\dd y\,\dd z \\[5mm] = &\ {1 \over 4}\int_{0}^{1}\int_{0}^{1}\bracks{% \Psi\pars{{y + z \over 4} + \half} - \Psi\pars{{y + z \over 4}}}\,\dd y\,\dd z \\[5mm] = &\ 4\int_{0}^{1/4}\int_{0}^{1/4}\bracks{% \Psi\pars{y + z + \half} - \Psi\pars{y + z}}\,\dd y\,\dd z\tag{1} \end{align}
$\ds{\Psi}$ is the Digamma Function and we used its well known integral representation $\ds{\pars{~\gamma\ \mbox{is the}\ Euler\mbox{-}Mascheroni\ Constant~}}$ $$\Psi\pars{z} = -\gamma + \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t\,, \qquad\Re\pars{z} > 0$$
Since $\ds{\Psi\pars{z}\ \stackrel{\mbox{def.}}{=}\ \totald{\ln\pars{\Gamma\pars{z}}}{z}}$ $\ds{\pars{~\Gamma\ \mbox{is the}\ Gamma\ Function~}}$, $\ds{\pars{1}}$ is reduced to: \begin{align} &\color{#f00}{\int_{0}^{\infty}\bracks{% {x - 1 \over \ln^{2}\pars{x}} - {1 \over \ln\pars{x}}} \,{\dd x \over x^{2} + 1}} \\[5mm] = &\ 4\int_{0}^{1/4}\bracks{\ln\pars{\Gamma\pars{z + {3 \over 4}}} - \ln\pars{\Gamma\pars{z + {1 \over 4}}} - \ln\pars{\Gamma\pars{z + \half}} + \ln\pars{\Gamma\pars{z}}}\,\dd z \\[5mm] = &\ 4\int_{0}^{1}\ln\pars{\Gamma\pars{z}}\,\dd z + 8\int_{0}^{1/4}\ln\pars{\Gamma\pars{z}}\,\dd z - 8\int_{0}^{3/4}\ln\pars{\Gamma\pars{z}}\,\dd z\tag{2} \end{align}
The $\ds{\ln\Gamma}$-integrals are evaluated $\ds{\pars{~\mbox{the first one is rather trivial and it's equal to}\ \half\,\ln\pars{2\pi}~}}$ with the identity ( $\ds{\,\mathrm{G}}$ is the Barnes-G Function ) $$\int_{0}^{z}\ln\pars{\Gamma\pars{z}}\,\dd z = \half\,z\pars{1 - z} + \half\,\ln\pars{2\pi}z + z\ln\pars{\Gamma\pars{z}} - \ln\pars{\,\mathrm{G}\pars{1 + z}}$$ Namely, $$\left\lbrace\begin{array}{\rcl} \ds{\int_{0}^{1}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{\half\,\ln\pars{2\pi}\ \mbox{because}\ \Gamma\pars{1} = \,\mathrm{G}\pars{2} = 1.} \\[3mm] \ds{\int_{0}^{1/4}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{{3 \over 32} + {1 \over 8}\,\ln\pars{2\pi} + {1 \over 4}\,\ln\pars{\Gamma\pars{1 \over 4}} - \ln\pars{\,\mathrm{G}\pars{5 \over 4}}} \\[1mm] & \ds{=} & \ds{{3 \over 32} + {1 \over 8}\,\ln\pars{2\pi} - {3 \over 4}\,\ln\pars{\Gamma\pars{1 \over 4}} - \ln\pars{\,\mathrm{G}\pars{1 \over 4}}} \\[3mm] \ds{\int_{0}^{3/4}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{{3 \over 32} + {3 \over 8}\,\ln\pars{2\pi} + {3 \over 4}\,\ln\pars{\Gamma\pars{3 \over 4}} - \ln\pars{\,\mathrm{G}\pars{7 \over 4}}} \\[1mm] & \ds{=} & \ds{{3 \over 32} + {3 \over 8}\,\ln\pars{2\pi} - {1 \over 4}\,\ln\pars{\Gamma\pars{3 \over 4}} - \ln\pars{\,\mathrm{G}\pars{3 \over 4}}} \end{array}\right.\tag{3}$$ In these expressions we used $\ds{\,\mathrm{G}\pars{1 + z} = \,\mathrm{G}\pars{z}\Gamma\pars{z}}$. Fortunately, values of $\ds{\,\mathrm{G}\pars{z}}$ at $\ds{z = {1 \over 4}, {3 \over 4}}$ are known: \begin{align} \,\mathrm{G}\pars{1 \over 4} & = A^{-9/8}\,\,\Gamma^{\, -3/4}\pars{1 \over 4} \exp\pars{{3 \over 32} - {K \over 4\pi}}\tag{4} \\[5mm] \,\mathrm{G}\pars{3 \over 4} & = A^{-9/8}\,\,\Gamma^{\, -1/4}\pars{3 \over 4} \exp\pars{{3 \over 32} + {K \over 4\pi}}\tag{5} \end{align} $\ds{A}$ and $\ds{K}$ are the Glaisher-Kinkelin and the Catalan Constants, respectively. With $\ds{\pars{4}\ \mbox{and}\ \pars{5}}$, $\ds{\pars{3}}$ becomes $$\left\lbrace\begin{array}{rcl} \ds{\int_{0}^{1}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{\phantom{-\,}\half\,\ln\pars{2\pi}} \\[1mm] \ds{\int_{0}^{1/4}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{\phantom{-\,}{K \over 4\pi} + {1 \over 8}\ln\pars{2\pi} + {9 \over 8}\,\ln\pars{A}} \\[1mm] \ds{\int_{0}^{3/4}\ln\pars{\Gamma\pars{z}}} & \ds{=} & \ds{-\,{K \over 4\pi} + {3 \over 8}\ln\pars{2\pi} + {9 \over 8}\,\ln\pars{A}} \end{array}\right.\tag{6}$$
With $\ds{\pars{6}}$, the expression $\ds{\pars{2}}$ is reduced to $\ds{\pars{~\ul{the\ final\ result}~}}$: $$\color{#f00}{\int_{0}^{\infty}\bracks{% {x - 1 \over \ln^{2}\pars{x}} - {1 \over \ln\pars{x}}} \,{\dd x \over x^{2} + 1}} = \color{#f00}{4\,{K \over \pi}} \approx 1.1662$$
• The OP's integral looked familiar, until I recalled a similar integral in your answer, namely $$\int_0^1\left(\frac{x-1}{\ln^2 x}-\frac{x}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}=\frac{2G}{\pi}+\frac{\ln 2}6+\frac{\ln\pi}2-6\ln A$$ with $A$ as the Glaisher–Kinkelin constant. Jul 25 '19 at 16:46 | 2022-01-27T23:47:39 | {
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https://www.logivan.com/sarah-hyland-okzat/ba539e-area-of-scalene-triangle-by-heron%27s-formula | Solution: Let the two adjacent sides of the scalene triangle be a = 8cm and b = 10cm, the angle included between these two sides, ∠C =30o . Area of a Scalene Triangle … From the figure given above, a scalene triangle is given with 3 sides as ‘a’, ‘b’ and ‘c’. A triangle with irregular side lengths is called a scalene triangle.That’s the point where Heron formulae come into play. Calculate the semi perimeter of the triangle. Answer. A scalene triangle is a triangle with 3 different side lengths and 3 different angles. Pour un triangle de côtés a, b et c, le demi-périmètre s = 1/2 (a + b + c). units Where, ‘a’, ‘b’ and ‘c’ are the length of sides of the scalene triangle … Q.2. Semi-Perimeter of triangle(S) = (A + B + C)/2. âs(s â a)(s â b)(s â c) = â132(110)(12)(10), = â(11 â
12 â
11 â
10 â
12 â
10). Now using Heron’s formula, you verify this fact by finding the areas of other triangles discussed earlier viz., (i) equilateral triangle with side 10 cm. cms and one of its sides length is 6cm. A = s ( s − a) ( s − b) ( s − c) where s is the semi perimeter equal to P /2 = ( a + b + c )/2. Area Of a Triangle If we know the length of three sides of a triangle then we can calculate the area of a triangle using Heron’s Formula Area of a Triangle = √ (s* (s-a)* (s-b)* (s-c)) So, the area of a scalene triangle can be calculated if the length of its base and corresponding altitude (height) is known or the length of its three sides is known or length of two sides and angle between them is given. Enter the values of the length of the three sides in the Heron's Formula Calculator to calculate the area of a triangle. Sorry!, This page is not available for now to bookmark. Home / Mathematics / Area; Calculates the area of a triangle given three sides. If we know the length of three sides of a triangle, we can calculate the area of a triangle using Heron’s Formula. Let a,b,c be the lengths of the sides of a triangle. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. Step 2: Find the semi-perimeter, S. The formula for finding the semi-perimeter of a triangle is. he base of an isosceles triangle is 10 cm and one of its equal sides is 13 cm. side a: side b: side c: area S Customer Voice. Scalene triangles have no equal sides nor equal angles, and finding their areas can be a taxing process unless your child is clear with the how-to’s of Heron’s formula. Problem 2 : The sides of a scalene triangle are 12 cm, 16 cm and 20 cm. The Heron’s formula is stated as: Area of the triangle = s ( s − a) ( s − b) ( s − c) where a, b and c are the sides of the given triangle, and s = semi-perimeter which is given by-. Ils utilisent Herons Formula, du nom de Hero of Alexandria. Given length of three sides of a triangle, Heron's formula can be used to calculate the area of any triangle. units, To find area of a scalene triangle if the the length of its three sides is given (image will be updated soon), The area of scalene triangle using Heron’s formula = $\sqrt{s(s-a)(s-b)(s-c)}$ sq. Unlike other triangle area formulae, there is no need to calculate angles or other distances in the triangle first. Here we are going to calculate the area of triangle using Heron's Formula. Formula: S = (a+b+c)/2 Area = √(S x (S - a) x (S - b) x (S - c)) Where, a = Side A b = Side B c = Side C S = Area of Triangle (image will be updated soon), So, the area of scalene triangle = $\frac{1}{2}$ × a × b × sinC sq. Pro Lite, NEET Heron's Formula for the area of a triangle. Heron's Formula for Area of Scalene Triangle : Here a, b and c are side lengths of the triangle. Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. 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Triangle can be used to calculate angles or other distances in the ratio of 3:5:7 and have perimeter 300m... Where ‘ a ’ and ‘ b ’ are the length of two sides and angle between sides. The side lengths and 3 different side lengths is called a scalene triangle = 12 sq a... H is given by: Try this Drag the orange dots to the. Calculator to calculate the area of a triangle whose area is 2.9 cm 2.. Table triangle. With all sides of length 8 cm, 16 cm and 20 cm is a triangle when you the! Let the base of an isosceles triangle having equal sides of a triangle whose two adjacent sides are equal let., c ) we can use Heron 's formula can be calculated using the Heron 's formula to area! + c ) sum of all the three sides of different lengths not easy to find area of a triangle! ( S-b ) ( S-b ) ( S-b ) ( S-c ) } \ ] ) its two sides angle! 20 per m2 and ‘ b ’ are the length of the scalene triangle whose two sides... Of a triangle ( s ) = ( a, b and is. And angle between them let a, b, c be the lengths any... 'S Formula- 2 | 25 Questions MCQ Test has Questions of Class 9 preparation find that the area of triangle! | 2021-04-23T10:40:49 | {
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https://cs.stackexchange.com/questions/141542/how-should-i-evaluate-time-complexity-for-matrix-if-i-have-a-fixed-constant-am | # How should I evaluate time complexity for matrix if I have a fixed (constant) amount of rows and columns?
Suppose, that I have a four-by-four matrix and I want to print each element of it.
matrix = [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 16]]
for row in matrix
for elem in row
print(elem)
So, I have a questions:
a) Should I consider, that such iterating requires O(k + n) in terms of big O notation, where k is a number of rows and n is a number of columns? I mean, $$\sum_{i=1}^k1$$ + $$\sum_{j=1}^n1$$ = O(k + n), we sum number of iterations that are required for rows and number of ones for columns. If I should not, then what is wrong with my estimates or how should I calculate big O for a matrix?
b) Can not we say, that such algorithm requires constant amount of time, because we have a well-defined input - four-by-for matrix, can we? I would like to specify what I mean: if we have constant input, does it mean, that our algo requires constant amount of time to compute in terms of big O?
## For the first question
No, its actually $$O(n\cdot k)$$. To see why, we have two explanations.
1. The number of elements in the matrix is $$n\cdot k$$ and we run through all of them
2. The inner loop takes $$O(n)$$ time, but notice that we run it $$k$$ times. Hence, the actual summation that represents the running time is:
$$\sum_{i=1}^k\sum_{j=1}^n 1 = \sum_{i=1}^k n = O(n \cdot k)$$
## For the second question
Yes, for as long as your input is well defined and with a constant size, as in this example. So actually, in this particular instance it takes $$O(1)$$ time. However this is usually not very interesting, since you are usually interested in more than just once matrix, or in the actual algorithm that performs the computation.
• Thank you ever so much. Let me specify some information: if we have matrix n * k and n = k, can I state, that big O is n^2, because of their equality? Jun 19 '21 at 14:04
• Yes, this is totally allowed. Jun 19 '21 at 14:41 | 2022-01-26T07:34:28 | {
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https://mathhelpboards.com/threads/finding-time-for-a-kicked-ball.26859/ | # PhysicsFinding Time for a Kicked Ball
#### balancedlamp
##### New member
Problem: You are playing soccer with some friends while some people nearby are playing hide and seek. When the seeker gets to '4', you kick the soccer ball high in the air. WHen the seeker says '5', the ball is 10 ft in the air. When the seeker gets to '7', the ball is 20 feet in the air. How many seconds will the ball be in the air in total before it hits the ground?
Given that after 1 second the ball travelled 10 ft in the air, I had the initial velocity as 10 ft/s. At the apex, the velocity will be 0 ft/s. Since change in position is not given I used the equation Vf = Vi + at. My thought was to find the time this way then multiply by two. Using a=-32.2 ft/s/s, I got t=5/8 s at the end of it all, which doesn't make any sense. I acknowledge that I left out the part about it being 20ft in the air after 2 seconds, but I thought that would erroneous considering we know the initial velocity and the value of a. What am I doing wrong?
#### skeeter
##### Well-known member
MHB Math Helper
In my opinion,the problem is either bogus or the setting of the problem isn't on Earth's surface.
You are given 3 data points (t,h), where t is time in seconds and h is in ft ...
(0,0) , (1,10) , (3,20)
those three data points yield the quadratic $h = \dfrac{35}{3} \cdot t - \dfrac{5}{3} \cdot t^2$
if done on Earth,the equation should be $h = v_0 \cdot t - 16t^2$
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Perhaps the 'seeker' calls the time every 3 seconds or so?
#### balancedlamp
##### New member
Once I approached it using the three points and finding the quadratic equation, I just plugged it into a graphing calculator and got the answer of 7 seconds, which is one of the answer choices.
However, I'm confused why the kinematic equations are giving me such funky answers, while the quadratic equation is pretty straightforward. Was my approach using the initial kinematic equation wrong?
#### HallsofIvy
##### Well-known member
MHB Math Helper
Once I approached it using the three points and finding the quadratic equation, I just plugged it into a graphing calculator and got the answer of 7 seconds, which is one of the answer choices.
However, I'm confused why the kinematic equations are giving me such funky answers, while the quadratic equation is pretty straightforward. Was my approach using the initial kinematic equation wrong?
You say you used "
a=-32.2 ft/s/s" but, as skeeter said, that, the earth's acceleration due to gravity, does not give the data in the problem. With the given data, the acceleration due to gravity, wherever this is, is $$\frac{10}{3}$$ feet per second per second, not 32.2 feet per second per second.
Or, as Klaas van Aarsen points out, the problem does not actually say that the person calling numbers calls one per second. Taking the trajectory to be $$(35/3)x- (5/3)x^2$$ then $$\frac{5}{3}x^2= 16.1 t^2$$ if $$x^2= \frac{48.3}{5}t^2$$ so that $$x= \sqrt{\frac{48.3}{5}}t= \sqrt{9.66}t= 1.932 t$$ so that the person was calling numbers every 1.932 seconds or approximately every 2 seconds, not every three seconds.
Of course, that would change the answer. If the ball was in the air for x= 7 "units" then it is in the are for $$7(1/932)= 13.524$$ seconds, approximately 14 seconds.
Last edited:
#### balancedlamp
##### New member
My apologies, according to the problem, the seeker counts at a rate of 1 whole number per second. Sorry for leaving that out, that was completely accidental. | 2020-09-27T13:48:26 | {
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https://www.physicsforums.com/threads/finding-loss-of-energy-in-collision.707704/ | # Finding loss of energy in collision
Saitama
## Homework Statement
A particle of mass ##m_1## experienced a perfectly elastic collision with a stationary particle of mass ##m_2##. What fraction of the kinetic energy does the striking particle lose, if it recoils at right angles to its original motion direction.
(Ans: ##2m_1/(m_1+m_2)## )
## The Attempt at a Solution
Let the initial velocity of ##m_1## be ##v## and let the x-axis be along the initial direction of motion. After collision, the first particle flies off at right angles and let that direction be y-axis. The vertical component of velocity of ##m_2## after collision has the direction opposite to that of ##m_1##. Conserving momentum in x direction:
$$m_1v=m_2v_{2x}$$
Conserving momentum in y direction:
$$m_1v_1=m_2v_{2y}$$
where ##v_1## is the final velocity of ##m_1##. I still need one more equation.
Any help is appreciated. Thanks!
Homework Helper
What quantity, apart from momentum, is conserved in a perfectly elastic collision?
Saitama
What quantity, apart from momentum, is conserved in a perfectly elastic collision?
Energy. But that gives a very dirty equation. I have seen some problems on one dimensional collisions where coefficient of restitution is used to find another equation. Is it possible to apply the same here as I think it reduces the algebra work greatly.
Homework Helper
Energy. But that gives a very dirty equation. I have seen some problems on one dimensional collisions where coefficient of restitution is used to find another equation. Is it possible to apply the same here as I think it reduces the algebra work greatly.
Energy, as in total energy, is *always* conserved.
In a perfectly elastic collision, kinetic energy is specifically conserved.
The algebra is fairly easy to work out here. Took me less than 10 lines and barely 5 minutes.
Remember that the final speed of ##m_2## is given by the Pythagorean theorem. Deal only in squares of the velocity components, and everything simplifies quickly.
And always keep in mind what you're trying to find, which is the ratio ##\displaystyle \frac{v^2 - v_1^2}{v^2}##.
1 person
Saitama
Energy, as in total energy, is *always* conserved.
In a perfectly elastic collision, kinetic energy is specifically conserved.
The algebra is fairly easy to work out here. Took me less than 10 lines and barely 5 minutes.
Remember that the final speed of ##m_2## is given by the Pythagorean theorem. Deal only in squares of the velocity components, and everything simplifies quickly.
And always keep in mind what you're trying to find, which is the ratio ##\displaystyle \frac{v^2 - v_1^2}{v^2}##.
I have solved the problem using the energy approach, thanks a lot Curious!
Homework Helper
I have solved the problem using the energy approach, thanks a lot Curious!
You're welcome, and I'm glad you solved it. Sorry I couldn't stay up to continue to help, but I need early nights as I've not been in the best of health lately.
Saitama
You're welcome, and I'm glad you solved it. Sorry I couldn't stay up to continue to help, but I need early nights as I've not been in the best of health lately.
I hope you get well soon. :) | 2022-12-09T13:48:50 | {
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http://mathematica.stackexchange.com/questions/4244/visualizing-a-complex-vector-field-near-poles | # Visualizing a Complex Vector Field near Poles
I've been playing around with a visualization technique for complex functions where one views the function $f: \mathbb{C} \rightarrow \mathbb{C}$ as the vector field $f: \mathbb{R^2} \rightarrow \mathbb{R^2}$. These vector fields have some nice properties as a consequence of the Cauchy-Riemann equations, and usually look pretty neat. I'm surprised I haven't heard of this until recently (they're known as Pólya plots). Here's an example:
f[z_] := Exp[-z^2]
VectorPlot[{Re[f[x + I*y]], Im[f[x + I*y]]}, {x, -1.5, 1.5}, {y, -1, 1},
VectorPoints -> Fine]
The problem I'm having is trying to do this near the poles of functions. This is understandable, however Mathematica usually has no trouble plotting functions with singularities. Here's an attempt to plot $z^{-1}$:
I tried upping MaxRecursion and a couple of other things, but I figured you guys might know what to do immediately.
Now that the pole issue has been taken care of (thanks to everyone who contributed), here are some very intriguing plots:
Poles of $\Gamma(z)$ at -4, -3, and -2:
PolyaPlot[g, {-4.5, -1.5}, {-1, 1}, 50]
$\sin(z)$:
PolyaPlot[F, {-3 Pi/2, 3 Pi/2}, {-4, 4}, 45]
Now, here is a function that has poles over a subset of the Gaussian integers. The plot immediately reveals the symmetry of the zeros of the nontrivial polynomial
$35900-(72768-72768 i) z-128304 i z^2+(64392+64392 i) z^3-40305 z^4+(8064-8064 i) z^5+2016 i z^6-(144+144 i) z^7+9 z^8$
$\displaystyle \sum_{m=1}^{3} \sum_{n=1}^{3} \frac{1}{z-(m+in)}$:
PolyaPlot[G, {.7,3.3},{.7,3.3},60]
where the function PolyaPlot is given by:
PolyaPlot[f_,ReBounds_,ImBounds_,vPoints_]:=Module[{reMin=ReBounds[[1]],reMax=ReBounds[[2]],imMin=ImBounds[[1]],imMax=ImBounds[[2]]},
Return[VectorPlot[{Re[f[x+I*y]],Im[f[x+I*y]]},{x,reMin,reMax},{y,imMin,imMax},
VectorPoints->vPoints,VectorScale->{Automatic,Automatic,None},VectorColorFunction -> (Hue[2 ArcTan[#5]/Pi] &),VectorColorFunctionScaling->False]];
]
-
Isn't the Pólya vector field the complex conjugate of the function? This lends itself to look at things like computing the flux out of an area through a contour integral (effectively using Gauss theorem). Code Example at MathWorld – Thies Heidecke Apr 16 '12 at 9:15
Indeed, these aren't quite Pólya plots but I was trying to get a direct visualization of the function. Using the conjugate makes a lot more sense when you want to find a convenient path $\gamma(t)$ such that $\gamma '(t) = f(\gamma(t))^{*}$, so performing a contour integral reduces to $\int_{\gamma} |f(\gamma(t))|^{2}$. Turning these plots into honest Pólya plots is a matter of a negative sign. – Jackson Walters Apr 16 '12 at 14:53
Have you tried looking at the StreamPlots of these functions? They're even prettier! – Rahul Nov 3 '12 at 10:52
Here are two suggestions for the function
f[z_] := 1/z;
First, instead of defining a region to omit from your plot, you should base the omission criterion on the length of the vectors (so that you don't have to adjust the criterion manually when switching to a function with different pole locations). That can be achieved like this:
With[{maximumModulus = 10},
VectorPlot[{Re[f[x + I*y]], Im[f[x + I*y]]}, {x, -1.5, 1.5}, {y, -1,
1}, VectorPoints -> Fine,
VectorScale -> {Automatic, Automatic,
If[#5 > maximumModulus, 0, #5] &}]
]
The main thing here is that as the third element of the VectorScale option I provided a function that takes the 5th argument (which is the norm of the vector field) and outputs a nonzero vector scale only when the field is smaller than the cutoff value maximumModulus.
Another possibility is to encode the modulus not in the vector length at all, but in the color of the arrows:
VectorPlot[{Re[f[x + I*y]], Im[f[x + I*y]]}, {x, -1.5, 1.5}, {y, -1,
1}, VectorPoints -> Fine,
VectorScale -> {Automatic, Automatic, None},
VectorColorFunction -> (Hue[2 ArcTan[#5]/Pi] &),
VectorColorFunctionScaling -> False]
What I did here is to suppress the automatic re-scaling colors in VectorColorFunction and provided my own scaling that can easily deal with infinite values. It's based on the ArcTan function.
As a mix between these two approaches, you could also use the ArcTan to rescale vector length.
-
Thanks Jens! This takes into account all of the previous suggestions and then some and is probably the ideal way to do Pólya plots. – Jackson Walters Apr 14 '12 at 17:08
By the way, Hue as a color scheme was only a quick-and-dirty choice. If you use a more systematic gradient such as VectorColorFunction -> ( ColorData["Rainbow"][2 ArcTan[#5]/\[Pi]] &) then it becomes a little easier to interpret the colors. For example, this function has three zeros and a pole: f[z_] := 1/z + (z - 1)^2 - and Hue doesn't show that so clearly. – Jens Apr 14 '12 at 18:05
Good point. I've been playing around with higher values of VectorPoint and that seems to give a really intuitive picture of what's going on. I think I'll edit my post and show a couple of interesting plots such as $sin(z)$, $\Gamma(z)$, and some functions with interesting pole/zero structure. – Jackson Walters Apr 14 '12 at 18:35
Just for completeness, it would be good to write down the definition of your final plot function PolyaPlot too... – Jens Apr 14 '12 at 18:55
Done. Thanks again! – Jackson Walters Apr 14 '12 at 19:02
If you don't know where the poles might be, just cut the function at a certain maximal value to leave a blank area around the pole:
g[z_] = 1/z;
h[z_] = If[Abs[g[z]] > 4, 0, g[z]];
VectorPlot[{Re[h[x + I*y]], Im[h[x + I*y]]}, {x, -1.5, 1.5}, {y, -1.5,
1.5}, VectorPoints -> Fine]
This also allows to see better the smaller vector values, without them being scaled down to much (depending on your choice of threshold).
Otherwise, if you know where the poles are, I'll suggest a slightly ad-hoc alternative to b.gate’s very nice solution: just make your function be zero at the poles!
g[z_] = 1/z;
VectorPlot[
If[x == 0 && y == 0, {0, 0}, {Re[g[x + I*y]],
Im[g[x + I*y]]}], {x, -1.5, 1.5}, {y, -1.5, 1.5},
VectorPoints -> Fine]
-
Cool, that should do the trick. BTW, I just tried using StreamPlot and it seems to not have any of the issues that VectorPlot has while providing more or less the same information. – Jackson Walters Apr 14 '12 at 14:53
You can try using RegionFunction :
g[z_] = 1/z
VectorPlot[{Re[g[x + I*y]], Im[g[x + I*y]]}, {x, -1.5, 1.5}, {y, -1.5,1.5},
VectorPoints -> Fine, RegionFunction -> Function[{x, y}, x^2 + y^2 >= 0.005]]
Another alternative is to specify explicitly the points at which you want the field :
points = Flatten[Table[{x, y}, {x, Range[-1, 1, 0.5]}, {y, Range[-1, 1, 0.5]}], 1]
VectorPlot[{Re[g[x + I*y]], Im[g[x + I*y]]}, {x, -1.5, 1.5}, {y, -1.5,1.5},
VectorPoints -> points, RegionFunction -> Function[{x, y}, x^2 + y^2 >= 0.005]]
-
Thanks! This is really nice, however I'm finding that it's still very fragile. Decreasing the bounds to $\left[ -1, 1\right]^{2}$ causes it break. – Jackson Walters Apr 14 '12 at 14:05
@JacksonWalters Added a better suggestion, please see if this helps. – b.gatessucks Apr 14 '12 at 15:00
your arrow at the pole (0,0) is weird, isn't it? – F'x Apr 14 '12 at 15:48
@F'x Yes and the documentation says that "VectorPlot omits any vectors for which the Subscript[v, i] etc. do not evaluate to real numbers". One can use RegionFunction together with VectorPoints but hopefully someone will post a better answer. – b.gatessucks Apr 14 '12 at 16:04 | 2014-11-23T10:15:07 | {
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https://www.physicsforums.com/threads/does-c-melt-together-all-constants.331446/ | # Does + C melt together ALL constants?
1. Aug 19, 2009
### Unit
Does "+ C" melt together ALL constants?
I had this integral $$\int \frac{dy}{-2y + 6}$$.
I realize you can let $u = -2y + 6$ so that $du = -2dy$, and adjusting: $-0.5 du = dy$.
The integral becomes
$$-0.5 \int \frac{du}{u}$$
$$= -0.5 \ln{|-2y + 6|} + C$$
However, there is another way. Factor out the half before the integration:
$$\int \frac{dy}{-2y + 6}$$
$$= \int \frac{dy}{-2(y - 3)}$$
$$= -0.5 \int \frac{dy}{(y - 3)}$$
$$= -0.5 \ln{|y - 3|} + C$$
BUT!
$$-0.5\ln{|-2y + 6|} + C \neq -0.5\ln{|y - 3|} + C$$ !
$$-0.5\ln{|-2y + 6|} + C$$ equals $$-0.5\ln{|y - 3|} - 0.5\ln{2} + C$$ !
Can anybody explain this discrepancy? Does $-0.5\ln{2} + C$ turn simply into just $+ C$? I understand that there can be an infinite number of antiderivatives for a function. But this is just odd. Both integrations are correctly done, no? If, on a test, I gave one answer, where the other one was preferred, what would happen?
And yeah, both derivatives go back to being the same thing. Strange!
Last edited: Aug 19, 2009
2. Aug 19, 2009
### Elucidus
Re: Does "+ C" melt together ALL constants?
The issue is that evaluating the integral will produce a family of expressions which differ by an arbitrary constant (the C).
So even though your two answers are unequal, they do belong to the same family of antiderivatives.
If it makes it easier, you can use subscripts like C1 and C2 in order to distinguish between them. You are correct that the C's you've listed are unequal and are really related they way you've stated.
--Elucidus
3. Aug 19, 2009
### nicksauce
Re: Does "+ C" melt together ALL constants?
The integral is unique up to a constant. Since -0.5ln2 is a constant, both answers are correct antiderivatives. Both integrations are correctly done. Both answers would be correct on a test.
4. Aug 19, 2009
### Tac-Tics
Re: Does "+ C" melt together ALL constants?
The "plus C" is a bastardization.
While differential functions have unique derivatives, their anti-derivatives are NOT unique. The reason for this is simple: constant functions have a zero derivative. That means, you can always add a constant function to the end of your differentiable function, differentiate it, and the result is the same derivative no matter what constant function you chose.
Algebraic notation does not handle this very well. The values on either side of an equation must be a single value. But as we said, there is no unique anti-derivative. We could write out everything that we're doing in words, but that takes up more room on the page.
The trick we use in introductory calc is that, most of the time, we end up taking the difference of a single anti-derivative evaluated at two different points.
So let's do something easy. f(x) = x^2.
What are some anti-derivatives?
F(x) = 1/3 x^3
F(x) = 1/3 x^3 + 1
F(x) = 1/3 x^3 + 2
F(x) = 1/3 x^3 + 3
....
We have a whole infinity of them. For any number c, F(x) = 1/3 x^3 + c is an antiderivative.
But check it out. What is F(1) - F(0)?
If F(x) = 1/3 x^3, then F(1) - F(0) = 1/3
If F(x) = 1/3 x^3 + 1, then F(1) - F(0) = 1/3
If F(x) = 1/3 x^3 + 2, then F(1) - F(0) = 1/3
If F(x) = 1/3 x^3 + 3, then F(1) - F(0) = 1/3
....
So when you take the difference of an antiderivative evaluated at two different points, the resulting value does not care about which antiderivative you choose. "The C's cancel".
So that's the story. Don't think of C as a number or an "arbitrary constant", because that's not what's really going on. If you treat C like you'd treat any other number, you can arrive at contradictions.
It is a handy short cut, though. And there are other tricks that people like to do. For example, if you have multiple C's added together, you can often throw all but one of them away. Sometimes. That happens to be the case in this problem. But if you don't end up killing the C's off by the end of the problem, you'll probably get the wrong answer. When you're working with multiple integrals, be careful there as well.
Good luck!
5. Aug 19, 2009
### Unit
Re: Does "+ C" melt together ALL constants?
Yes yes guys I understand the idea that "+ C" is just a way to portray the infinite number of antiderivatives, and I appreciate all your replies. And I commend nicksauce for his brevity.
This integral was part of a differential equation I was solving:
$$\frac{dy}{dx} + 2y = 6; x = 0; y = 1$$
by the way, this leads to
$$y = 3 - 2e^{-2x}$$.
BOTH ANTIDERIVATIVES work out for me! :tongue:
It's so intriguing! How a slight change in my method of integration would give two different functions as the solution, but still work when put together in a differential equation!
Thanks a lot everyone
(sorry for any plausible incoherence, I've had no sleep since nine AM yesterday)
6. Aug 19, 2009
### g_edgar
Re: Does "+ C" melt together ALL constants?
So ... when somebody does an integral and says: "I got -sin^2 but the book says cos^2, what did I do wrong?" you will now know the answer!
7. Aug 21, 2009
### Unit
Re: Does "+ C" melt together ALL constants?
What did you do wrong? You forgot the "1" | 2018-03-19T07:40:23 | {
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https://mathematica.stackexchange.com/questions/141245/adding-random-numbers-to-the-arguments-of-a-sum-of-cosines | # Adding random numbers to the arguments of a sum of cosines
This is probably really a simple question.
I would like to plot a sum of cosines with increased argument and for each cosine in sum I would like to add random value:
Plot[Sum[Cos[n*x + RandomReal[{0, 2 Pi}]], {n, 100}], {x, -10, 10}]
However it seems to me that the program, while making a plot, generates a new random number for each x-point plotted. Therefore, even for single cosine (Plot[Cos[x + RandomReal[{0, 2*Pi}]], {x, -10, 10}] ) I get a completely noisy graph of random cosine values.
I would be very grateful if somebody can tell me what to do to prevent the program from generating new random numbers at each x-point.
• Just compute the random offset once: With[{r = RandomReal[{0, 2 Pi}]}, Plot[Cos[x + r], {x, -10, 10}]] – J. M. is away Mar 29 '17 at 10:55
• Thank you for answer. I tried it before, but then the problem is that i have the same random number for each cosine in my sum. And i would like to generate a new one for each (but not for each x when plotting). – Maciek Kowalczyk Mar 29 '17 at 11:08
• Generate as many as you need: With[{r = RandomReal[{0, 2 Pi}, 10]}, Plot[Total[Cos[Range[10] x + r]] // Evaluate, {x, -10, 10}]] – J. M. is away Mar 29 '17 at 11:14
J.M. has answered the question in the comments. This is to get that answer (slightly edited) on record in a more visible form. For 10 terms, we get
SeedRandom[42];
With[{n = 10},
With[{r = RandomReal[{0, 2 Pi}, n]},
Plot[Total[Cos[Range[n] x + r]], {x, -10, 10}]]]
For a 100 terms, which is what the OP asked for, the plot must be stretched out to see the details better.
SeedRandom[42];
With[{n = 100},
With[{r = RandomReal[{0, 2 Pi}, n]},
Plot[Total[Cos[Range[n] x + r]], {x, -10, 10},
AspectRatio -> 1/5,
ImageSize -> Full]]] | 2019-07-18T05:23:33 | {
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https://math.stackexchange.com/questions/1590059/better-proof-for-frac1-cos-x-sin-x1-cos-x-sin-x-equiv-frac1?noredirect=1 | # Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$
It's required to prove that $$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$ I managed to go about out it two ways:
1. Assume it holds: $$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$ $$\Longleftrightarrow\sin x(1+\cos x+\sin x)\equiv(1+\cos x)(1-\cos x+\sin x)$$ $$\Longleftrightarrow\sin x+\cos x\sin x+\sin^2 x\equiv1-\cos x+\sin x+\cos x-\cos^2 x+\sin x \cos x$$ $$\Longleftrightarrow\sin^2 x\equiv1-\cos^2 x$$ $$\Longleftrightarrow\cos^2 x +\sin^2 x\equiv1$$ $$\Longleftrightarrow true$$
2. Multiplying through by the conjugate of the denominator: $$LHS\equiv\frac{1+\cos x + \sin x}{1 - \cos x + \sin x}$$ $$\equiv\frac{1+\cos x + \sin x}{1 - (\cos x - \sin x)} ~~\cdot ~~\frac{1+(\cos x - \sin x)}{1 +(\cos x - \sin x)}$$ $$\equiv\frac{(1+\cos x + \sin x)(1+\cos x - \sin x)}{1 - (\cos x - \sin x)^2}$$ $$\equiv\frac{1+\cos x - \sin x+\cos x + \cos^2 x - \sin x \cos x+\sin x + \sin x \cos x - \sin^2 x}{1 - \cos^2 x - \sin^2 x + 2\sin x \cos x}$$ $$\equiv\frac{1+ 2\cos x + \cos^2 x- \sin^2 x}{2\sin x \cos x}$$ $$\equiv\frac{1+ 2\cos x + \cos^2 x- 1 + \cos^2 x}{2\sin x \cos x}$$ $$\equiv\frac{2\cos x (1+\cos x)}{2\cos x(\sin x)}$$ $$\equiv\frac{1+\cos x}{\sin x}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\equiv RHS~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square$$ Both methods of proof feel either inelegant or unnecessarily complicated. Is there a simpler more intuitive way to go about this? Thanks.
• For (1), you are not assuming it holds, since you have all those $\iff$ steps. Not sure how it helps, but the right side is one of the half-angle formulae - $\cot(x/2)$. – Thomas Andrews Dec 26 '15 at 23:34
• @ThomasAndrews That's true; but it's not a deductive $LHS \equiv \dots \equiv RHS$ type thing, which is what I was looking for. Interesting point about the half-angle formulae, thanks. – Luke Collins Dec 26 '15 at 23:37
• Still, "assume" is wrong. You aren't assuming it, you are proving an equivalence and proving that each statement implies the other. Indeed, you don't even need the $\implies$ direction of that argument. – Thomas Andrews Dec 26 '15 at 23:42
• @ThomasAndrews Right you are. – Luke Collins Dec 26 '15 at 23:46
• – lab bhattacharjee Dec 27 '15 at 3:50
Since $1-\cos^2 x = \sin^2 x$, we have $f(x) := \dfrac{1+\cos x}{\sin x} = \dfrac{\sin x}{1-\cos x}$. Therefore,
\begin{align*}\dfrac{1+\cos x + \sin x}{1-\cos x + \sin x} &= \dfrac{f(x)\sin x + f(x)(1-\cos x)}{1-\cos x + \sin x} \\ &= \dfrac{f(x)[1-\cos x + \sin x]}{1-\cos x + \sin x} \\ &= f(x) \\ &= \dfrac{1+\cos x}{\sin x}.\end{align*}
• Well that's clever. Awesome :D – Luke Collins Dec 26 '15 at 23:39
For fun, I created a trigonograph:
$$\frac{1 + \cos\theta + \sin\theta}{1 + \sin\theta - \cos\theta} = \frac{1 + \cos\theta}{\sin\theta}$$
• That's beautiful! :) – Luke Collins Dec 27 '15 at 22:58
Observe $$(1 - \cos x + \sin x)(1 + \cos x) = (1 - \cos^2 x) + (1 + \cos x)\sin x = \sin^2 x + (1 + \cos x)\sin x = (1 + \cos x + \sin x)\sin x,$$ from which the result immediately follows.
• Its simpler to start from the RHS – lab bhattacharjee Dec 27 '15 at 2:39
if $$\frac{a}{b}=\frac{c}{d}=k$$ then $$\frac{a+c}{b+d} = \frac{kb+kd}{b+d} =k =\frac{a}{b}$$ since $$1-\cos^2 x =(1+\cos x)(1-\cos x) =\sin^2 x$$ we have $$\frac{1+\cos x}{\sin x} = \frac{\sin x}{1-\cos x} =\frac{1+\cos x +\sin x}{1 -\cos x +\sin x}$$
alternatively, using abbreviations $c=\cos x$ and $s=\sin x$ we have $$s(1+c+s)=s(1+c) + s^2 = s(1+c) + 1-c^2=s(1+c)+(1-c)(1+c)=(1+s-c)(1+c)$$ | 2019-07-19T10:40:49 | {
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https://math.stackexchange.com/questions/3387129/dimension-of-a-vector-space-of-functions | Dimension of a vector space of functions
Where $$S$$ is a set, the functions $$f: S \rightarrow \mathbb{R}$$ form a vector space under the natural operations: the sum $$f+g$$ is the function given by $$f +g(s)=f(s)+g(s)$$ and the scalar product is $$r\cdot f(s) = r\cdot f(s)$$. What is the dimension of the space resulting for each domain?
(a) $$S = \{1\}$$ $$\hspace{1mm}$$ (b) $$S = \{1,2\}$$ $$\hspace{1mm}$$ (c) $$S = \{1,...,n\}$$
The result of my attempt at the problem is that the dimension in each case should be infinitely large. My reasoning is that the set of functions, in either (a), (b), or (c), contains the space of $$n$$ degree polynomials functions $$\mathscr{P}_n$$, as any polynomial function takes values in all mentioned domains and returns a value in $$\mathbb{R}$$. Since $$\mathscr{P}_n$$ has an infinitely large dimension, as $$n \in \mathbb{N}$$ and $$\mathbb{N}$$ (the set of natural numbers) is infinitely large, the basis of the space containing $$\mathscr{P}_n$$ is infinitely large, and so must be the bases for the spaces in question. Therefore, the spaces in question have infinitely large dimensions.
However, the solution to the problem is
(a) dimension $$=1$$
(b) dimension $$=2$$
(c) dimension $$=n$$
I see how the solutions apply to each case but my question is why? Where in my reasoning am I making a mistake? Thank you in advance.
• Distinct polynomials can have the same values on $S$ and therefore induce the same function from $S \to \mathbb R$. For example, all polynomials $p$ satisfying $p(1) = 0$ result in the same function from $\{1\} \to \mathbb R$. – Bungo Oct 9 at 17:52
• Thank you for the response Bungo. Can't you just apply the same logic and say all polynomials $p(1) =1$, $p(1)=2$, and so forth and still have an infinitely large collection of functions? – Andy Oct 9 at 18:06
• Of course. The dimension of a vector space isn't the same as its cardinality. For example, the vector space $\mathbb R^2$ has dimension 2 over $\mathbb R$ (any basis contains two vectors), but there are infinitely many vectors in the space, namely all linear combinations of the two basis vectors. – Bungo Oct 9 at 18:08
• See if you can convince yourself that the set of all functions $S = \{1,2\} \to \mathbb R$ is actually the same thing as $\mathbb R^2$. – Bungo Oct 9 at 18:13
2 Answers
Hint: For the general case $$S=\{1,...,n\}$$, prove that the functions $$f_1,...,f_n$$ defined as $$f_i(i)=1$$ and $$f_i(j)=0$$ for $$i\neq j$$, forms a basis.
In other words, show that $$f_1,...,f_n$$ are linearly independent and that for any function $$g:S\rightarrow \mathbb{R}$$, we can write $$g$$ as $$\alpha_1f_1 + \alpha_2f_2 + ... + \alpha_nf_n$$ for suitable $$\alpha_1,...,\alpha_n$$.
How can a polynomial belong to those spaces? If you are talking about polynomial functions, then you are making a mistake. Consider, for instance, the polynomial functions $$x^n$$ ($$n\in\mathbb N$$). They are all distinct as functions from $$\mathbb R$$ into $$\mathbb R$$. But they become all the same function if you restrict them to $$\{1\}$$!
• Thank you for the response José. I see your point, but can't you just take the polynomial functions $x^n + x^{n-1}$ and have another set of functions that are distinct on $\{1\}$? – Andy Oct 9 at 18:11
• @Andy When restricted to $S = \{1\}$, we have $x^n = x^{n-1}$ (they are the same function on the restricted domain), so $x^n + x^{n-1} = 2x^n$. In the same way, any function $\{1\} \to \mathbb R$ is a scalar multiple of $x^n$ restricted to $\{1\}$. In other words, the set consisting of the single function $x^n$ (or any other nonzero function $\{1\} \to \mathbb R$) constitutes a basis. – Bungo Oct 9 at 19:08 | 2019-10-22T17:03:22 | {
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http://cisautos.co.uk/minimum-volatility-meuttm/bc65ba-logarithmic-differentiation-problems | Use logarithmic differentiation to differentiate each function with respect to x. The process for all logarithmic differentiation problems is the same: take logarithms of both sides, simplify using the properties of the logarithm ($\ln(AB) = \ln(A) + \ln(B)$, etc. (3x 2 – 4) 7. Do 1-9 odd except 5 Logarithmic Differentiation Practice Problems Find the derivative of each of the Solution to these Calculus Logarithmic Differentiation practice problems is given in the video below! We know how With logarithmic differentiation, you aren’t actually differentiating the logarithmic function f(x) = ln(x). Now, as we are thorough with logarithmic differentiation rules let us take some logarithmic differentiation examples to know a little bit more about this. A logarithmic derivative is different from the logarithm function. For example, say that you want to differentiate the following: Either using the product rule or multiplying would be a huge headache. View Logarithmic_Differentiation_Practice.pdf from MATH AP at Mountain Vista High School. Problems. 11) y = (5x − 4)4 (3x2 + 5)5 ⋅ (5x4 − 3)3 dy dx = y(20 5x − 4 − 30 x 3x2 + 5 − 60 x3 5x4 − 3) 12) y = (x + 2)4 ⋅ (2x − 5)2 ⋅ (5x + 1)3 dy dx = … There are, however, functions for which logarithmic differentiation is the only method we can use. Basic Idea The derivative of a logarithmic function is the reciprocal of the argument. Logarithmic Differentiation example question. You do not need to simplify or substitute for y. Practice 5: Use logarithmic differentiation to find the derivative of f(x) = (2x+1) 3. (3) Solve the resulting equation for y′ . Instead, you’re applying logarithms to nonlogarithmic functions. (x+7) 4. ), differentiate both sides (making sure to use implicit differentiation where necessary), (2) Differentiate implicitly with respect to x. Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. It spares you the headache of using the product rule or of multiplying the whole thing out and then differentiating. In some cases, we could use the product and/or quotient rules to take a derivative but, using logarithmic differentiation, the derivative would be much easier to find. Using the properties of logarithms will sometimes make the differentiation process easier. For differentiating certain functions, logarithmic differentiation is a great shortcut. Instead, you do […] One of the practice problems is to take the derivative of $$\displaystyle{ y = \frac{(\sin(x))^2(x^3+1)^4}{(x+3)^8} }$$. Find the derivative of the following functions. Click HERE to return to the list of problems. Begin with y = x (e x). Steps in Logarithmic Differentiation : (1) Take natural logarithm on both sides of an equation y = f(x) and use the law of logarithms to simplify. The function must first be revised before a derivative can be taken. SOLUTION 2 : Because a variable is raised to a variable power in this function, the ordinary rules of differentiation DO NOT APPLY ! Apply the natural logarithm to both sides of this equation getting . (2) Differentiate implicitly with respect to x. We could have differentiated the functions in the example and practice problem without logarithmic differentiation. Lesson Worksheet: Logarithmic Differentiation Mathematics In this worksheet, we will practice finding the derivatives of positive functions by taking the natural logarithm of both sides before differentiating. (3) Solve the resulting equation for y′ . A variable is raised to a variable power in this function, ordinary! 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