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This kind of behavior is easily understood in the framework of complex analysis. Solution As with the last example we’ll start off in the same manner. So, we get a similar pattern for this one. Let’s plug the numbers into the MeteaCalcTutorials 55,406 views 4:56 Maclauren and Taylor Series Intuition - Duration: 12:59. Lagrange Error Bound Calculator The N plus oneth derivative of our Nth degree polynomial.
Show Answer If you have found a typo or mistake on a page them please contact me and let me know of the typo/mistake. Hence each of the first k−1 derivatives of the numerator in h k ( x ) {\displaystyle h_{k}(x)} vanishes at x = a {\displaystyle x=a} , and the same is true Where are the answers/solutions to the Assignment Problems?
Transcript The interactive transcript could not be loaded.
Up next Taylor's Inequality - Duration: 10:48. Remainder Estimation Theorem As in previous modules, let be the error between the Taylor polynomial and the true value of the function, i.e., Notice that the error is a function of . Recall that if a series has terms which are positive and decreasing, then But notice that the middle quantity is precisely . How close will the result be to the true answer?
Taylor Polynomial Approximation Calculator | {
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Taylor Polynomial Approximation Calculator
And sometimes you might see a subscript, a big N there to say it's an Nth degree approximation and sometimes you'll see something like this. PaulOctober 27, 2016 Calculus II - Notes Parametric Equations and Polar Coordinates Previous Chapter Next Chapter Vectors Power Series and Functions Previous Section Next Section Applications of Series Taylor Taylor Series Error Estimation Calculator In that situation one may have to select several Taylor polynomials with different centers of expansion to have reliable Taylor-approximations of the original function (see animation on the right.) There are Lagrange Error Formula If all the k-th order partial derivatives of f: Rn → R are continuous at a ∈ Rn, then by Clairaut's theorem, one can change the order of mixed derivatives at
If you are a mobile device (especially a phone) then the equations will appear very small. http://accessdtv.com/taylor-series/taylor-series-error-estimation-formula.html And you keep going, I'll go to this line right here, all the way to your Nth degree term which is the Nth derivative of f evaluated at a times x Solution 1 As with the first example we’ll need to get a formula for . However, unlike the first one we’ve got a little more work to do. Let’s first take However, because the value of c is uncertain, in practice the remainder term really provides a worst-case scenario for your approximation. Taylor Series Remainder Calculator | {
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Here is the Taylor Series for this function. Now, let’s work one of the easier examples in this section. The problem for most students is that it may So if you put an a in the polynomial, all of these other terms are going to be zero. Generated Sun, 30 Oct 2016 19:02:48 GMT by s_fl369 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.9/ Connection http://accessdtv.com/taylor-series/taylor-series-error-estimation-problems.html What are they talking about if they're saying the error of this Nth degree polynomial centered at a when we are at x is equal to b.
This means that for every a∈I there exists some r>0 and a sequence of coefficients ck∈R such that (a − r, a + r) ⊂ I and f ( x ) Taylor's Inequality F of a is equal to P of a, so the error at a is equal to zero. To obtain an upper bound for the remainder on [0,1], we use the property eξSo the error of b is going to be f of b minus the polynomial at b.
From Site Map Page The Site Map Page for the site will contain a link for every pdf that is available for downloading. Note If you actually compute the partial sums using a calculator, you will find that 7 terms actually suffice. Sign in Transcript Statistics 38,950 views 81 Like this video? Lagrange Error Bound Problems And it's going to look like this.
Relationship to analyticity Taylor expansions of real analytic functions Let I ⊂ R be an open interval. Compute F ′ ( t ) = f ′ ( t ) + ( f ″ ( t ) ( x − t ) − f ′ ( t ) ) The second inequality is called a uniform estimate, because it holds uniformly for all x on the interval (a − r,a + r). Check This Out Krista King 14,459 views 12:03 Taylor's Theorem with Remainder - Duration: 9:00. | {
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The N plus oneth derivative of our error function or our remainder function, we could call it, is equal to the N plus oneth derivative of our function. Download Page - This will take you to a page where you can download a pdf version of the content on the site. Sign in 82 5 Don't like this video? patrickJMT 130,005 views 2:22 Estimating error/remainder of a series - Duration: 12:03.
The goal is to find so that . Notice as well that for the full Taylor Series, the nth degree Taylor polynomial is just the partial sum for the series. Sign in to add this video to a playlist. So this is going to be equal to zero.
Now, what is the N plus onethe derivative of an Nth degree polynomial? Also, since the condition that the function f be k times differentiable at a point requires differentiability up to order k−1 in a neighborhood of said point (this is true, because It has simple poles at z=i and z= −i, and it is analytic elsewhere. Graph of f(x)=ex (blue) with its quadratic approximation P2(x) = 1 + x + x2/2 (red) at a=0.
Modulus is shown by elevation and argument by coloring: cyan=0, blue=π/3, violet=2π/3, red=π, yellow=4π/3, green=5π/3. Combining these estimates for ex we see that | R k ( x ) | ≤ 4 | x | k + 1 ( k + 1 ) ! ≤ 4 Since ex is increasing by (*), we can simply use ex≤1 for x∈[−1,0] to estimate the remainder on the subinterval [−1,0]. Solution Again, here are the derivatives and evaluations. Notice that all the negative signs will cancel out in the evaluation. Also, this formula will work for all n,
Here's the formula for the remainder term: So substituting 1 for x gives you: At this point, you're apparently stuck, because you don't know the value of sin c. Algebra/Trig Review Common Math Errors Complex Number Primer How To Study Math Close the Menu Current Location : Calculus II (Notes) / Series & Sequences / Taylor Series Calculus II [Notes] | {
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# Does there exist an open subset $A \subset [0,1]$ such that $m_*(A)\neq m_*(\bar{A})$?
Does there exist an open subset $$A \subset [0,1]$$ such that $$m_*(A)\neq m_*(\bar{A})$$?
I was thinking we could approximate any set from inside by a closed set . This need not true from outside.
So I was expecting there should be some counterexample to the above statement.
Any help will be appreciated
• Is $m_*$ outer or inner Lebesgue measure? – Alex Ortiz Feb 17 at 4:05
• Sir it is outer measure. Is answer will change if we put inner measure? As i had only studied outer measure. I was thinking till thar inner measure will work same. – MathLover Feb 17 at 4:08
• It is actually inconsequential since both inner and outer measure are subadditive with respect to countable unions; see my answer below. Cheers! – Alex Ortiz Feb 17 at 4:19
Enumerate the rational numbers in $$[1/4,3/4]$$ as $$\{r_n:n\in\mathbf{Z}_{>0}\}$$, and let $$\epsilon < 1/8$$. For each $$n$$, choose an interval $$I_n$$ centered at $$r_n$$ of width $$\epsilon/2^n$$, and put $$A = \bigcup_{n=1}^\infty I_n$$. Note that $$A$$ is an open subset of $$\mathbf R$$ contained in $$[0,1]$$ because it is a union of open intervals. Since $$A$$ contains the rationals in $$[1/4,3/4]$$, density of $$\mathbf Q$$ in $$\mathbf R$$ implies that the closure $$\overline A$$ contains the closed interval $$[1/4,3/4]$$.
By countable subadditivity (of Lebesgue outer or inner measure, this is an inconsequential point for the matter at hand), $$m_*(A) \le \sum_{n=1}^\infty m_*(I_n) = \sum_{n=1}^\infty \frac{\epsilon}{2^n} = \epsilon < 1/8.$$ On the other hand, by monotonicity, $$m_*\big(\overline A\big) \ge m_*\big([1/4,3/4]\big) = 1/2,$$ so $$m_*(A) < 1/8 < 1/2 = m_*\big(\overline A\big).$$
• Why not just let $A$ be an open subset of $(0,1)$ containing $\mathbb Q \cap (0,1)$ such that $m(A)<1/2?$ – zhw. Feb 17 at 4:43 | {
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# Question on modular arithmetic?
Would this ever be possible:
$$a b^2 ≡ a \pmod p$$
where $p$ is a prime number and $1<a<p$ and $1<b$
Please back your answer with some kind of proof other than Fermat's Little Theorem. This isn't homework.
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Put $a=0$ then true for all $b$. – john w. Apr 21 '12 at 22:03
Sorry, forgot to include $p$ and $b$ have to be greater than 1! – user26649 Apr 21 '12 at 22:06
Any $b \equiv \pm 1 \pmod p.$ – Will Jagy Apr 21 '12 at 22:07
Hint $\$ prime $\rm\: p\ |\ ab^2-a = a\:(b-1)(b+1)\ \Rightarrow\ p\ |\ a\ \ or\ \ p\ |\ b-1\ \ or\ \ p\ |\ b+1$
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@Downvoter: Why? If something is not clear then please feel free to ask questions and I am happy to elaborate. – Bill Dubuque Apr 21 '12 at 23:45
I don't see why the answer was downvoted... Anyways, your answer was perfect, thank you! – user26649 Apr 22 '12 at 0:50
If $a\equiv0\bmod p$, the given relation is true for all $b$.
If $a\not\equiv0\bmod p$ then $a$ can be cancelled from both sides of the congruence which reduces to $b^2\equiv1\bmod p$, hence $b\equiv\pm1\bmod p$.
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would the downvoter care to explain? – Andrea Mori Apr 21 '12 at 22:47
It seems someone is unhappy with both of our answers. I cannot imagine why. But then voting is often irrational here. – Bill Dubuque Apr 21 '12 at 23:48
Since $p$ is prime, we can cancel $a$ from both sides unless $a\equiv 0 \pmod p$ (this doesn't always work if $p$ is a composite number). We get $b^2 \equiv 1 \pmod p$. Again, since $p$ is prime, a number can have only two mod-$p$ square roots (it can have more than two in some cases if $p$ is composite). The two square roots of $1$ are $\pm 1$ and $-1$ is of course the same as $p-1$.
So $ab^2\equiv a\pmod p$ holds if either $a\equiv 0$ or $b\equiv\pm1$, but not otherwise. | {
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So $ab^2\equiv a\pmod p$ holds if either $a\equiv 0$ or $b\equiv\pm1$, but not otherwise.
So which parts of this do you want proofs of? The proof that nonzero numbers are invertible in mod $p$ (thereby justifying the cancelation from both sides)? I once posted an answer that shows how to find the inverse. The proof that there can't be more than two square roots if $p$ is prime? (That latter fact is a sort of corollary.)
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# Math Help - proving the cantor set is integrable
1. ## proving the cantor set is integrable
I would like to know how to do this problem before my test tomorrow and I'm not really sure how to do it. Any guidance would be appreciated.
Let $C$ be the Cantor set. Let $f:[0,1]\rightarrow \mathbb{R}$ be determined by $$f(x)= \begin{cases} 1, &\text{when x\in C }\\ 0, &\text{when x\notin C }\\ \end{cases}$$
Show that $f$ is Riemann integrable on $[0,1]$ and $\int _{0}^{1}f=0$. [That is the "length of the Cantor set is 0.]
I want to approach it like this but I'm not sure if it is correct:
We want to show that $f$ is integrable we need to show that $\{\sum S|S \text {is a lower step function of f}\} - \{\sum s|s \text {is a lower step function of f}\}<\epsilon$ for some $\epsilon >0$.
We can make $\sum s=0$ when $x\notin C$. So i think all we have to do is to show that $\sum S< \epsilon$. However, I'm not sure how to do this.
2. Originally Posted by zebra2147
I would like to know how to do this problem before my test tomorrow and I'm not really sure how to do it. Any guidance would be appreciated.
Let $C$ be the Cantor set. Let $f:[0,1]\rightarrow \mathbb{R}$ be determined by $$f(x)= \begin{cases} 1, &\text{when x\in C }\\ 0, &\text{when x\notin C }\\ \end{cases}$$
Show that $f$ is Riemann integrable on $[0,1]$ and $\int _{0}^{1}f=0$. [That is the "length of the Cantor set is 0.]
I want to approach it like this but I'm not sure if it is correct:
We want to show that $f$ is integrable we need to show that $\{\sum S|S \text {is a lower step function of f}\} - \{\sum s|s \text {is a lower step function of f}\}<\epsilon$ for some $\epsilon >0$.
We can make $\sum s=0$ when $x\notin C$. So i think all we have to do is to show that $\sum S< \epsilon$. However, I'm not sure how to do this.
So start with the definition of the Cantor set. There are certain "removed" intervals, what are they? | {
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3. The middle third of each line segment is removed. For example, for [0,1] we would have [0,1/3]U[2/3,1]. So, (1/3,2/3) would be removed.
4. Try proving that $[0,1] - C$ has measure 1, implying that C has measure 0.
5. Ok, here is what I want to say but I'm not sure how to write it correctly of if its even in the right direction...
If we let $s$ be a lower step function of $f$, then we see that $\sum s=0$. Then if we choose some upper step function, $S$, such that $\sum S=\sum A_{i}(x_{i}-x_{i-1})$ where $A_{i}=\text{height of the rectangle}$ then for any interval contained in $[0,1]$, we can find an element, $x_i$ that is contained in the Cantor set and is infinitely close to another element $x_{i-1}$ that is not contained in the Cantor set. Thus, we have $x_{i}-x_{i-1}\approx 0$. Thus, $\sum A_{i}(x_{i}-x_{i-1})\approx 0=\sum S$.
Thus, $\sum s+\sum S\approx 0$ which is less then any $\epsilon >0$. Thus, $\int_{0}^{1}f=0$.
I know that is not exactly the direction that you were leading me to but I think it is similar.
6. I think that it would be better to show that the upper sum and lower sum are equal to each other. Recall that
$\displaystyle U(f)=\inf_{P} U(f,P)$ and $\displaystyle L(f)=\sup_{P} L(f,P)$
where $P$ refers to some partition of $[0,1]$. We say that $f$ is Riemann-integrable if $U(f)=L(f)$.
Now I propose that we consider a sequence of partitions based on the cuts performed in the construction of the Cantor set. Let $P_1=\{0,1/3,2/3,1\}$ (corresponding to the removal of the middle third of the interval). Then, $U(f,P_1)=L(f,P_1)=2/3$. Let $P_2=\{0,1/9,2/9,1/3,2/3,7/9,8/9,1\}$ (again, corresponding to the middle thirds of the remaining intervals). Then, $U(f,P_2)=L(f,P_2)=4/9$. Perform this same procedure to get $P_3, P_4, \ldots, P_n,\ldots$. Now taking $n\to \infty$, we have $U(f,P_n)\to 0$ and $L(f,P_n)\to 0$. This shows that
$U(f)\leq U(f,P_n)\to 0$ and $L(f)\geq L(f,P_n)\to 0$
or in other words,
$U(f)\leq 0\leq L(f)$ | {
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$U(f)\leq U(f,P_n)\to 0$ and $L(f)\geq L(f,P_n)\to 0$
or in other words,
$U(f)\leq 0\leq L(f)$
But at the same time, from the definition of the upper and lower sums,
$U(f)\geq L(f)$
Therefore $U(f)=L(f)=0$, so this function $f$ is Riemann-integrable with integral value 0. | {
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# Quotient rings of Gaussian integers $\mathbb{Z}[i]/(2)$, $\mathbb{Z}[i]/(3)$, $\mathbb{Z}[i]/(5)$,
I am studying for an algebra qualifying exam and came across the following problem.
Let $R$ be the ring of Gaussian Integers. Of the three quotient rings $$R/(2),\quad R/(3),\quad R/(5),$$ one is a field, one is isomorphic to a product of fields, and one is neither a field nor a product of fields. Which is which and why?
I know that $2=(1+i)(1-i)$ and $5=(1+2i)(1-2i)$, so neither $(2)$ nor $(5)$ is a prime ideal of $R$. Then (I think) these same equations in $R/(2)$ and $R/(5)$, respectively, show that neither is an integral domain. Regardless, I can show that 3 is a Gaussian prime, hence $(3)$ is maximal in $R$ and $R/(3)$ is the field. But if I am correct about the others not being integral domains, I fail to see how either could be a product of fields.
I hope that this can be answered easily and quickly. Thanks.
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A product of fields is not an integral domain. Consider the zero divisors $(1,0)$ and $(0,1)$. – Jared May 17 '13 at 1:07
That's helpful, thanks. Since R/(2) has just 4 elements, if it were isomorphic to a product of fields, then it would necessarily be isomorphic to Z/(2)xZ/(2). But $i^2=1$ in R/(2), and every element of Z/(2)xZ/(2) when squared equals itself. So... From the unusual phrasing of the question, I suppose R/(5) is isomorphic to a product of fields. Is this right? Would that product be Z/(5)/Z/(5)? – John Adamski May 17 '13 at 1:34
You might find this helpful: Splitting of prime ideals in Galois Extensions (Wikipedia). – kahen May 17 '13 at 1:51
Now is a very good time for a quick foray into the ideal-theoretic version of Sun-Ze (better known as the Chinese Remainder Theorem). Let $R$ be a commutative ring with $1$ and $I,J\triangleleft R$ coprime ideals, i.e. ideals such that $I+J=R$. Then
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$$\frac{R}{I\cap J}\cong\frac{R}{I}\times\frac{R}{J}.$$
First let's recover the usual understanding of SZ from this statement, then we'll prove it. Thanks to Bezout's identity, $(n)+(m)={\bf Z}$ iff $\gcd(n,m)=1$, so the hypothesis is clearly analogous. Plus we have $(n)\cap(m)=({\rm lcm}(m,n))$. As $nm=\gcd(n,m){\rm lcm}(n,m)$, if $n,m$ are coprime then compute the intersection $(n)\cap(m)=(nm)$. Thus we have ${\bf Z}/(nm)\cong{\bf Z}/(n)\times{\bf Z}/(m)$. Clearly induction and the fundamental theorem of arithmetic (unique factorization) give the general algebraic version of SZ, the decomposition ${\bf Z}/\prod p_i^{e_i}{\bf Z}\cong\prod{\bf Z}/p_i^{e_i}{\bf Z}$.
(How this algebraic version of SZ relates to the elementary-number-theoretic version involving existence and uniqueness of solutions to systems of congruences I will not cover.)
Without coprimality, there are counterexamples though. For instance, if $p\in\bf Z$ is prime, then the finite rings ${\bf Z}/p^2{\bf Z}$ and ${\bf F}_p\times{\bf F}_p$ (where ${\bf F}_p:={\bf Z}/p{\bf Z}$) are not isomorphic, in particular not even as additive groups (the product is not a cyclic group under addition).
Now here's the proof. Define the map $R\to R/I\times R/J$ by $r\mapsto (r+I,r+J)$. The kernel of this map is clearly $I\cap J$. It suffices to prove this map is surjective in order to establish the claim. We know that $1=i+j$ for some $i\in I$, $j\in J$ since $I+J=R$, and so we further know that $1=i$ mod $J$ and $1=j$ mod $I$, so $i\mapsto(I,1+J)$ and $j\mapsto(1+I,J)$, but these latter two elements generate all of $R/I\times R/J$ as an $R$-module so the image must be the whole codomain. | {
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Now let's work with ${\cal O}={\bf Z}[i]$, the ring of integers of ${\bf Q}(i)$, aka the Gaussian integers. Here you have found that $(2)=(1+i)(1-i)=(1+i)^2$ (since $1-i=-i(1+i)$ and $-i$ is a unit), that the ideal $(3)$ is prime, and that $(5)=(1+2i)(1-2i)$. Furthermore $(1+i)$ is obviously not coprime to itself, while $(1+2i),(1-2i)$ are coprime since $1=i(1+2i)+(1+i)(1-2i)$ is contained in $(1+2i)+(1-2i)$. Alternatively, $(1+2i)$ is prime and so is $(1-2i)$ but they are not equal so they are coprime. Anyway, you have
• ${\bf Z}[i]/(3)$ is a field and
• ${\bf Z}[i]/(5)\cong{\bf Z}[i]/(1+2i)\times{\bf Z}[i]/(1-2i)$ is a product of fields.
Go ahead and count the number of elements to see which fields they are. However, ${\bf Z}[i]/(2)={\bf Z}[i]/(1+i)^2$ is not a field or product of fields, although the fact that its characteristic is prime (two) may throw one off the chase. In ${\bf Z}[i]/(1+i)^2$, the element $1+i$ is nilpotent. Since this ring has order four, it is not difficult to check that it is isomorphic to ${\bf F}_2[\varepsilon]/(\varepsilon^2)$, which is not a product of fields since $\epsilon\leftrightarrow 1+i$ is nilpotent and products of fields contain no nonzero nilpotents.
-
Why don't you first think through the case of $\mathbb Z/n\mathbb Z$. When is this a field? When is it a product of fields? When is it neither? Try to relate your answer to the factorization of $n$ and the Chinese Remainder Theorem. Now see if you can generalize it to $\mathbb Z[i]$ (or to any PID).
-
I find quotients of polynomial rings easier to reason with algebraically than subfields of $\mathbb{C}$.
Because $\mathbb{Z}[i] \cong \mathbb{Z}[x] / (x^2 + 1)$, we can do arithmetic with rings and ideals. Calculating in more detail than is usually necessary: | {
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\begin{align} \mathbb{Z}[i] / (3) &\cong \left( \mathbb{Z}[x] / (x^2 + 1) \right) / (3) \\&\cong \mathbb{Z}[x] / (3, x^2 + 1) \\&\cong \left( \mathbb{Z}[x] / (3) \right) / (x^2 + 1) \\&\cong \left(\mathbb{Z} / (3)\right)[x] / (x^2 + 1) \\&\cong \mathbb{F}_3[x] / (x^2 + 1) \end{align}
and so we've reduced the problem to one of quotients of polynomial rings over the finite field $\mathbb{F}_3$.
-
$\Bbb Z[i]/2\,$ isn't a field or $\color{#c00}{\Pi F}$ (product of fields), since it has a nonzero nilpotent: $\,(1\!+\!i)^2 = 2i = 0$.
$\Bbb Z[i]/5\,$ isn't a field by $\,(2\!-\!i)(2\!+\!i)= 5 = 0,\,$ so it is the $\color{#c00}{\Pi F},\,$ leaving $\,\Bbb Z[i]/3\,$ as the field. $\$ QED
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# Visualizing the set of points in a regular polygon closer to center than to vrtices (Voronoi cell)?
Let $$P_n$$ be the regular convex $$n$$-gon centered at $$p_0$$ with $$n$$ vertices $$p_1, p_2, ..., p_n$$ and $$(n>2)$$.
Let $$S_n$$ be the set of all points "$$s$$" within the region bounded by $$P_n$$ where:
$$distance(s,p_0) \leq distance(s,p_k)$$... $$(\forall k=\{1,2,...,n\})$$.
For instance $$S_3$$ would look like this:
And $$S_4$$ would look like this:
And just thinking about this to the extreme... $$\lim_{n\to\infty}S_n$$ should look something like this (call it "$$S_\infty$$" for simplicity):
Is there a way to show (either using geometry or a simple brute-force code in Python) what $$S_n$$ should look like for any $$n>2$$?
I would really like to see what shapes emerge. Is there a way to animate it in Python? Like have circles radiate from each vertex $$p_k$$ and the center $$p_0$$ until they crash into each other to make straight lines defining the new region $$S_n$$?
Depending on the answer, I'd ideally like to transpose this question into 3-D for the 5 platonic solids.
For instance, examining a cube (i.e. 8 vertices), define $$D_8$$ as the set of all points "$$s$$" such that:
$$distance(s,p_0) \leq distance(s,p_k)$$... $$(\forall k=\{1,2,...,8\})$$.
Then $$D_8$$ is actually a truncated octahedron! I'm super curious what the other 4 platonic solids produce... As well as what higher $$S_n$$ regions look like (e.g. for a pentagon, hexagon, octagon, dodecagon, etc.).
Here's an insight that I find to be an elegant solution for n=3,4, or 6 (triangle, square, hexagon). Observe that you can tessellate all of 2-D space with regular triangles, squares, and regular hexagons. Now imagine each vertex $$p_k$$ is actually the center of a nearby triangle/square/hexagon (centered at $$p_k$$ instead of at $$p_0$$).
For instance: $$S_3$$ can be more easily distinguished by drawing 3 more triangles centered at $$p_1, p_2 and p_3$$. | {
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Similarly for $$S_4$$, drawing 4 more squares centered at $$p_1, p_2, p_3 and p_4$$ would show us the shape below, which we can then easily discern needs to be distilled into the $$S_4$$ shading we saw above.
EDIT:
Based on the comment about Voronoi Diagrams, I did more research and tried it out in R. Here's the region $$S_8$$ for example (use your imagination to connect the 8 outer dots to form the enclosing octagon):
The upshot seems to be that actually only $$n=3$$ and $$n=4$$ were interesting. For $$n>4$$ the pattern is simply to create a smaller version of the n-gon rotated by the internal angle of that n-gon. The shrinkage continues forever, but is also slowing forever, with the limit being circle of radius $$\frac{r}{2}$$ shown in the $$S_\infty$$ diagram from above.
More directly:
$$\dfrac{Area(S_3)}{Area(P_3)}=\dfrac{2}{3}$$
$$\dfrac{Area(S_4)}{Area(P_4)}=\dfrac{1}{2}$$
$$\dots = \dots$$
$$\dfrac{Area(S_\infty)}{Area(P_\infty)}=\dfrac{1}{4}$$ (limit -> lower bound)
• You're talking about the Voronoi diagram of the vertices and the centroid.
– user856
Jan 26 '20 at 7:47
• Geometrically, that inequality represents the half-space containing $p_0$ defined by the perpendicular bisecting plane of the line segment connecting $p_0$ and $p_k$. Take the intersection of all these half-spaces with the original polyhedron. Jan 26 '20 at 22:24
• @mr_e_man As said in a compact way by Rahul, it is the Voronoi cell associated with the center $p_0$, in the Voronoi diagram generated by the $p_k$s. Jan 29 '20 at 19:08
You can obtain it easily with Wolfram Mathematica (see the pictures below).
P[n_] := Table[{Cos[2*Pi*i/n], Sin[2*Pi*i/n]}, {i, 0, n}];
graph[n_] := RegionPlot[AllTrue[Table[(x - P[n][[i, 1]])^2 + (y - P[n][[i, 2]])^2, {i, 1,n}], # > x^2 + y^2 &], {x, -1, 1}, {y, -1, 1}, PlotPoints -> 20];
The 3D version can be built in the same way. For instance, for the tetrahedron:
P = {{1, 1, 1}, {1, -1, -1}, {-1, 1, -1}, {-1, -1, 1}}; | {
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P = {{1, 1, 1}, {1, -1, -1}, {-1, 1, -1}, {-1, -1, 1}};
p2 = ConvexHullMesh[P, MeshCellStyle -> {1 -> {Thick, Black}, 2 -> Opacity[0.2]}];
p1 = RegionPlot3D[ RegionMember[p2, {x, y, z}] && AllTrue[ Table[ Norm[{x, y, z} - P[[i]]], {i, 1, Length[P]}], # > x^2 + y^2 + z^2 &] , {x, -1, 1}, {y, -1, 1}, {z, -1, 1}, PlotPoints -> 60];
Show[p2, p1, ViewPoint -> {2, 4, 4}]
• It's neat that for the cube some of the faces are perfect hexagons.
– user856
Jan 29 '20 at 10:57
• @PierreCarre can you please provide the sample code for the cube? I'm not too familiar with these functions in mathematica unfortunately Jan 31 '20 at 2:45
• @Andrew I just added some code for the 3D case. Jan 31 '20 at 10:17 | {
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# Biased coin with a $3/4$ chance to land on the side it was before the flip
Consider a hypothetical coin (with two sides: heads and tails) that has a $3/4$ probability of landing on the side it was before the flip (meaning, if I flip it starting heads-up, then it will have an only $1/4$ probability of landing tails-up). If it begins on heads, what is the probability that it is on tails after 10 flips? What about 100 flips? Assume that each flip starts on the same side as it landed on the previous flip.
Note: this is not a homework problem, just something I thought up myself.
• I was with you all the way up to the last sentence (Assume that each flip...) - what does that mean? The rest makes perfect sense. – mathguy May 24 '16 at 3:27
• Coin flips have been shown to have a bias according to the side they start on actually. Don't remember the researcher name though, pretty famous person in California iirc, look it up. I'm guessing that the op wasn't aware of that though. – jdods May 24 '16 at 3:30
• This could probably be solved by setting up a simple 2 state Markov chain. I'd write it up but an away for the evening now. – jdods May 24 '16 at 3:33
• "this is not a homework problem, just something I thought up myself" -- then well done, because as it happens this is a fairly common kind of exercise! You've found the "right" problem, the one that (with others like it) provokes the theory of Markov chains. – Steve Jessop May 24 '16 at 10:18
• @mathguy It seems to me he's saying the side facing up doesn't change from the time the coin lands to the time that it's flipped again. Which sounds redundant, but it's better than being unclear. – Turambar May 24 '16 at 12:46
Let $p_n$ be the probability the coin is on tails after $n$ flips. Note that $p_0=0$.
The coin can be on tails after $n+1$ flips in two different ways: (i) it was on tails after $n$ flips, and the next result was a tail or (ii) it was on heads after $n$ flips, and the next result was a tail. | {
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The probability of (i) is $(3/4)p_n$ and the probability of (ii) is $(1/4)(1-p_n)$. Thus $$p_{n+1}=\frac{1}{4}+\frac{1}{2}p_n.\tag{1}$$ Solve this recurrence relation. The general solution of the homogeneous recurrence $p_{n+1}=\frac{1}{2}p_n$ is $A\cdot \frac{1}{2^n}$. A particular solution of the recurrence (1) is $\frac{1}{2}$. So the general solution of the recurrence (1) is $$p_{n}=A\cdot \frac{1}{2^n}+\frac{1}{2}.$$ Set $p_0=0$ to find $A$. We find that $p_n=\frac{1}{2}-\frac{1}{2^{n+1}}$.
If you mean that $\Pr(H_i|H_{i-1}) = \Pr(T_i|T_{i-1}) = 3/4$ for all $i$, then this is a two-state homogeneous Markov chain. The transition matrix is
$$P = \begin{pmatrix} 3/4 & 1/4 \\ 1/4 & 3/4 \end{pmatrix}.$$
This should get you started....
I'd comment but I don't have the rep, strangely you require 50.
Also, with regards to the bias in a coin flip, it's a paper by Diaconis:
Let $p_n$ be the probability that the coin is heads up after $n$ tosses. Then $p_0=1$ since it starts on heads, and $$p_n=\frac34p_{n-1}+\frac14(1-p_{n-1})$$ which simplifies to $$\left(p_n-\frac12\right)=\frac12\left(p_{n-1}-\frac12\right)\ .$$ Iterating, $$p_n-\frac12=\Bigl(\frac12\Bigr)^{n+1}$$ so $$p_n=\frac12+\Bigl(\frac12\Bigr)^{n+1}\ .$$
• This no longer fails for $p_1 = \frac 3 4$. – Axoren May 24 '16 at 3:36
Another way of looking at it is to imagine that you toss two fair coins each time, using the second coin to choose between the first coin and the previous (originally heads). If the second coin always chooses the previous result then you end up with heads, while if at any point the second coin chooses the first coin then as this is fair the end result will have equal probability of being heads or tails. The chances of the latter being $1-\frac1{2^n}$, the chance of the final result being tails is therefore half of that. | {
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• This is the best possible solution for p=1/4 and I think it generalizes to all other values of p. – zyx May 24 '16 at 17:32
• At least for p<1/2, where coin 2 has probability (1-2p) of choosing the previous result and coin 1 is fair. One gets half of $(1 - (1-2p)^n)$ which is correct. Other than changing the probability for coin 2, every word of the answer stays the same and the argument works. This answer is just brilliant. Well done. – zyx May 24 '16 at 17:49
$p_n(\text{heads}) = \frac 3 4p_{n-1}(\text{heads}) + \frac 1 4 (1 - p_{n-1}(\text{heads}))$
$p_n(\text{heads}) = \frac 1 4 + \frac 1 2p_{n-1}(\text{heads})$
$p_1(\text{heads}) = \frac 3 4$
Everyone else beat me too it. Was still writing the recurrence relation. :\
Any solution with the words "heads" and "tails" is doing extra work.
We want the chance of an odd number of reversals, each with probability $p=1/4$, to happen in $n$ independent trials. The coin briefly remembers its current state but not whether that state was reached by reversing on the previous trial. This makes the reversals independent of each other.
From the binomial theorem that probability is $$\frac{((1-p)+p)^n - ((1-p) - p)^n)}{2} = \frac{1}{2} - \frac{(1 - 2p)^n}{2}$$ which is consistent with the other solutions and shows that the $(1/2)^{n}$ exponential decay of the "memory" is really a power of $(1-2p)$. | {
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# If $15$ distinct integers are chosen from the set $\{1, 2, \dots, 45 \}$, some two of them differ by $1, 3$ or $4$.
$$\blacksquare~$$ Problem: If $$15$$ distinct integers are chosen from the set $$\{1, 2, \dots, 45 \}$$, some two of them differ by $$1, 3$$ or $$4$$.
$$\blacksquare~$$ My Approach:
Let the minimum element chosen be $$n$$. Then $$n + 1 , n + 3 , n + 4$$ can't be taken.
We make a small claim.
$$\bullet~$$ Claim: In a set of $$~7$$ consecutive numbers at most $$2$$ numbers can be chosen.
$$\bullet~$$ $$\textbf{Proof:}$$ Let us name the elements of the set as $$\{ 1,2,3,\dots,7 \}$$.
Now let's consider the least element is chosen. If the least element is $$1$$, then $$2,4,5$$ can't be chosen.
So we are left with $$3, 6, 7$$.
$$\circ~$$ If $$~3~$$ is chosen, then $$6, 7$$ can't be in the set. And if $$~3~$$ is not chosen, then only any one of the 2 elements $$\{ 6, 7 \}$$ be chosen. So, a maximum of $$2$$ elements can be chosen in this case.
$$\circ \circ~$$ If the least element is $$2,$$ then $$3, 5, 6$$ can't be there in the set. So, possible elements are 4, 7. So, one of these two can be chosen. Then, a maximum of 2 elements can be chosen in this case.
$$\circ \circ~$$ If the least element is $$3$$, then $$4,6,7$$ gets cancelled. so only $$5$$ is left in the set i.e., $$2$$ elements at most.
$$\circ \circ~$$ If the least element is $$4,$$ then $$5,7$$ gets cancelled. So the only element left is $$6$$.
Similarly,
$$\circ \circ~$$ If $$5$$ is the least element then $$6$$ gets cancelled and only $$7$$ is left. i.e., two elements. If the least element is either $$6$$ or $$7$$, then there is only one element.
So a maximum of two elements in a set of $$7$$ consecutive elements can be chosen.
Hence, the proof of the claim is done! | {
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Hence, the proof of the claim is done!
So, for $$42$$ elements, a maximum of $$2 \times 6 = 12$$ can be taken. However, $$3$$ more elements are required from $$3$$ more consecutive elements, which is not possible since only 2 elements at most can be chosen from a set of 3 consecutive elements.
So, a $$14$$ element subset can be formed such that, no two of them differ by $$1, 3, 4$$. Hence the $$15$$th element is one of the cancelled elements, that is, there exists a pair with their difference being $$1, 3$$ or $$4.$$
Hence, done!
Please check the solution for glitches and give new ideas too :).
• What is the question? Well done. – Ross Millikan Aug 1 at 15:04
• @RossMillikan I don't understand what you mean. – Ralph Clausen Aug 1 at 15:07
• I missed the last line. I think it is a fine proof. – Ross Millikan Aug 1 at 15:07
• Oh! okay @RossMillikan – Ralph Clausen Aug 1 at 15:10
• @RossMillikan The OP added the last line after your comment – user1001001 Aug 1 at 15:17
It is a nice argument, and you have explained it very clearly, so well done.
If you are looking for improvements, and you want it to be a bit more formal, I would make two suggestions:
1. You are very thorough proving the claim, going through all the cases, which is great. However you could shorten the proof of the claim by saying:
"If the least element is $$x$$, then we may subtract $$x-1$$ from all the numbers without changing their differences or the number of integers. Thus without loss of generality we may assume the least element is $$1$$."
Then you do not need to consider the other cases.
1. The last part of the proof (after the claim is proved) is not as thorough as the proof of the claim. You assume that for $$k=0,\cdots,5$$ the numbers $$7k+1,7k+3$$ are selected, without justifying that this is optimal. It is obvious in a way, but for a formal proof you should justify this by saying something like: | {
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"Pick the smallest $$k$$ where $$7k+1, 7k+3$$ are not picked. Then replace the numbers in $$\{7k+1,\cdots 7k+7\}$$ that are picked with $$7k+1, 7k+3$$. From the claim we know that we have not reduced the number of integers. Also, we have not created any new differences of $$1,3,4$$."
Then finish the argument with your second to last paragraph. I would reword the first sentence slightly: "So of the first $$42$$ elements, we may assume these $$12$$ are picked."
I would lose the last paragraph as it is not needed.
Nice proof!
In the optimal case, go up in $$2$$s wherever possible. You would start with $$1$$ and pick the smallest possible number, that is, $$3$$. We can't add $$2$$ again as that would be $$1+4$$, therefore we add the next smallest possible, $$5$$ to get our next number $$8$$. We can then add $$2$$ again, and the pattern continues:
$$1,3,8,10,15,17,22,24,29,31,36,38,43,45$$
Via this strategy we only pick $$14$$ numbers, so $$15$$ is impossible without violating the $$1,3,4$$ difference.
• While this strategy is good in some cases, I think it's a bad general approach to proofs. I call it the "what's the worst that could happen?" model --- you work that out, show it's impossible, and say you're done. That relies on two things: (1) that the second worst doesn't turn out to be viable even though the worst is not, and (2) that you've correctly identified the worst case. My experience is that students (and even professionals) often manage to mess up with both failure modes. – John Hughes Aug 1 at 15:48
• You claim that you would pick the smallest possible number in order to generate an optimal sequence. But this is something that needs to be proven. – TonyK Aug 1 at 16:32
Here's a shorter version which however relies on the same ideas as your and Rhys Hughes' approach.
Let's assume we can choose $$15$$ integers whose mutual differences are never $$0,1,3$$ or $$4$$, and call them $$x_1 in ascending order. | {
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Claim: For all $$1 \le i \le 13$$, we have $$x_i +7 \le x_{i+2}$$.
Proof of claim: Let's make a case distinction about what $$x_{i+1}-x_i$$ is. Since it cannot be $$0,1,3$$ or $$4$$, the only cases are:
• First case, $$x_{i+1} - x_{i} =2$$. Then if we also had $$x_{i+2} - x_{i+1} = 2$$ we would have $$x_{i+2}-x_i =4$$ which was excluded. So $$x_{i+2}-x_{i+1}$$, since it cannot be $$0,1,2,3,4$$ must be $$\ge 5$$ which implies the claim.
• Second case, $$x_{i+1}-x_i \ge 5$$. Then since $$x_{i+2}-x_{i+1} \neq 0,1$$ it must be $$\ge 2$$ and the claim follows.
The claim is proven.
Now it follows iteratively that $$x_3 \ge x_1 +7$$, $$x_5 \ge x_3+7 \ge x_1+14$$, ..., $$x_{15} \ge x_1+49$$ which of course contradicts $$1 \le x_1 \le x_{15} \le 45$$. Actually this shows that you cannot even select $$15$$ such integers from the set $$\{1, ..., 49\}$$. | {
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Numerical Approximation of the Continuous Fourier Transform
Given a function $F(k)$ in frequency space (sufficiently nice enough, eg. a Gaussian), I would like to compute its Fourier inverse $$f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ikx}F(k)dk$$ numerically (there is no explicit analytical formula), at some specified evenly distributed points $x_n$.
Assume $F(k)$ is symmetric about $k=0$ and "essentially zero" outside the interval $-a/2\leq k \leq a/2$.
Naively, one way I would proceed is by dividing the interval $(-a/2,a/2)$ into $N$ subintervals (choose $N$ to be a power of two) and then piecewise approximate the integral: $$f(x_n) \approx \frac{1}{2\pi}\frac{a}{N}\sum_{m=0}^{N-1} e^{ik_mx_n}F(k_m)$$ where, for example, we might take $k_m = (m-N/2)\frac{a}{N}$for $0\leq m\leq N-1$.
For efficiency I would like to use the Fast Fourier Transform (FFT) to approximate $f(x_n)$. So I need to put $f(x)$ into the correct format for the Discrete Fourier Transform (DFT) $$L_k = \frac{1}{N}\sum_{m=0}^{N-1}G_me^{(2\pi i) km/N},\,\,\,\,k = 0,\ldots,N-1$$ where $G_m$ is the $m$th sampling of some given function $G$.
To proceed, we divide the interval $(-a/2,a/2)$ into $N$ pieces ($N$ a power of two) and let $k_m = (m-N/2)\frac{a}{N}$for $0\leq m\leq N-1$, as before. Then $$f(x) \approx \frac{1}{2\pi}\frac{a}{N}\sum_{m=0}^{N-1} e^{ik_mx}F(k_m)$$ However, in order for this formula to be in the correct format to use the DFT, we cannot evaluate f at any $x$: I believe they must be of the form $x_k = \frac{2\pi}{a}(k-N/2), 0\leq k \leq N-1$.
So in order to use the FFT, the points at which I can "evaluate" the function $f$ is determined by how I sample my function $F(k)$ and the number of sample points.
But in the naive method, I can, in principle, find $f(x)$ for any given x, regardless of how many sample points I take of the function $F(k)$.
My questions are:
1. Is it even possible to use the FFT, if I need to compute f at specified points $x_n$? | {
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1. Is it even possible to use the FFT, if I need to compute f at specified points $x_n$?
2. If not, are there other numerical methods that would be more suitable?
• I came across your very same problem a week ago and I was quite astonished in realizing that there are no routines to do the job in known (at least those that I know) computer algebra programs (e.g. Mathematica/Matlab). The best thing I found is numerical recipes. There is a whole section on Fourier integration. Basically I just copied the routines suggested there and now I'm an happy person. I think that's the best what the market offers at least according to my experience. – lcv May 11 '13 at 0:48
• Bythe way, the routine in numerical recipes implements fgp's trick and more. – lcv May 11 '13 at 22:27
• @lcv: can you share the link to the routine you found? I also came across this problem and I'm still shocked that this isn't some standard Matlab routine. It took some time to realize that the maximum range of the x_k numbers is increasing with larger N. Increasing N will therefore make sure that the x_k's cover the entire x-region of interest but will not increase the density inside the x-region. – Frank Meulenaar Dec 5 '13 at 8:02
• Of course, one can increase the densitiy of the x_k's is by increasing a. – Frank Meulenaar Dec 5 '13 at 8:29
• @FrankMeulenaar It's numerical recipes chapter 13.9 "Computing Fourier integrals using FFT" . In the 'C' edition is on page 584. You can find online and downloadable versions of NR. The routine you can pretty much copy-paste it from the pdf. You'll have to work a bit to remove the comments. Otherwise the routines are available online for purchase on the NR site. Good luck! – lcv Dec 6 '13 at 20:30 | {
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The following two articles may be of use. They don't let you do the transform at any old x values. But it does allow transformation from m equally spaced points in the w domain to m equally spaced points in the x domain. The Bailey and Swarztrauber article also uses a gaussian function like yours as an example. Sorry I can't be more specific but I'm learning all this stuff myself.
Bailey, D., and P. Swarztrauber. 1994. ‘A Fast Method for the Numerical Evaluation of Continuous Fourier and Laplace Transforms’. SIAM Journal on Scientific Computing 15 (5): 1105–10. doi:10.1137/0915067.
Inverarity, G. 2002. ‘Fast Computation of Multidimensional Fourier Integrals’. SIAM Journal on Scientific Computing 24 (2): 645–51. doi:10.1137/S106482750138647X.
$$\def \o {\omega}$$ (Here I use $$t$$ and $$\o$$ instead $$x$$ and $$k$$, but is equivalent) We want to calculate numerically the inverse Fourier transform of a given function $$F(\o)$$ . By definition
$$f(t)=\frac{1}{2\pi} \int_{-\infty}^{\infty} F(\o) e^{-i\o t} d\o$$
We can split it into two integrals
$$f(t)=\frac{1}{2\pi} \left( \int_{0}^{\infty} F(\o) e^{-i\o t} d\o + \int_{0}^{\infty} F(-\o) e^{i\o t} d\o \right)$$
If the original function was real then $$F(-\o)=F(\o)^{*}$$, and therefore
$$f(t)=\frac{1}{\pi} \textbf{real} \left( \int_{0}^{\infty} F(\o) e^{-i\o t} d\o \right)$$
Then we use the Riemann integral definition for $$0$$ to a very great frequency $$W$$. $$I=\int_{0}^{W} F(\o) e^{-i\o t} d\o\approx \sum_{k=0}^{N-1} F(w_k) e^{-i \Delta \o k \Delta t n} \Delta\o$$ | {
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where $$\Delta w =W/N$$ so that $$w_k= \Delta w \ \ k$$ with $$k=0,1,2,..., N-1$$ . We discretized the time by $$t_n= \Delta t \ \ n$$ with $$n=0,1,2,...,N-1$$. Next we make the basic assumption that
$$W \Delta t = 2 \pi$$ so we end up with
$$I = \Delta\o \sum_{k=0}^{N-1} F(w_k) e^{- 2 \pi i \frac{n k}{N} }$$
which is the basic definition of the Fast Fourier Transform. The problem was when I used it for a square function $$F(\o)=\frac{1}{i\o}\left( e^{i\o T_0} -1 \right)$$ This method gives an offset value. Then I modified the integral as
$$I= \Delta\o \left( \sum_{k=1}^{N-1} F(w_k) e^{- 2 \pi i \frac{n k}{N} } + \frac{1}{2}(F(w_0)+F(w_N)) \right)$$
Which is the trapezoidal rule instead of the Riemman integral, and gave a better result.
Best regards.
E. Gutierrez-Reyes
• It should say -infinity to infinity, it was a mistake. – Edahi Aug 26 '19 at 19:24
One way that comes to mind would be to compute both approximations for $f(x_n)$ and for $f'(x_n)$ one some grid $x_n$ that allows you to use FFT. To compute the approximations for $f'(x_n)$, you can use the identity $\mathcal{F}(f')(k) = ik\mathcal{F}(f)(k)$ (i.e., you get the fourier transform of the derivative by multiplying with $ik$).
To evaluate $f(x)$ at $y$, you'd then find the closest $x_n,x_{n+1}$ with $x_n \leq y < x_{n+1}$ and use polynomial interpolation (with polynomials of degree 3) to find $f(y)$ from $f(x_n)$, $f(x_{n+1})$, $f'(x_n)$, $f'(x_{n+1})$.
As long as F(k) is limited and zero outside of $-2\pi A$ to $2\pi A$ the Shannon interpolation can be used $$f(x)=\sum_{n=-\infty}^{n=\infty} f(x_n) sinc\left(\frac{\pi}{T}(x-nT)\right)$$ where $T$ is the sampling period used to determine $x_n$ and $sinc(t)=sin(t)/t$
This formula is exact as long as function spectrum ($F(k)$) is limited. Implementation of fast sinc interpolation exist | {
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# Construct a continuous real valued function which takes zero on integers and such that image of function is not closed.
I am trying to construct a continuous real valued function $$f:\mathbb{R}\to \mathbb{R}$$ which takes zero on all integer points(that is $$f(k)=0$$ for all $$k\in \mathbb{Z}$$) and Image(f) is not closed in $$\mathbb{R}$$
I had $$f(x)=\sin(\pi x)$$ in mind. But image of $$f(x)$$ is closed.
I have a feeling that we can use some clever idea to modify this function such that it satisfy our given condition.
The image of $$\sin(\pi x)$$ is closed because the peaks all reach 1 and -1. To make it open, we need the peaks to get arbitrarily close to some value, but never reach them. The easiest way is to use an amplitude modifier that asymptotes to a constant nonzero value at infinity, such as $$\tanh(x)$$. Thus, we can use the function $$f(x) = \sin(\pi x)\tanh(x),$$ which is obviously continuous, zero at each integer, and can be easily shown to have image $$(-1,1)$$
• And if you are not familiar with $\tanh$, you can write a similar function explicitly, for example: $$f(x)=\sin(\pi x)\frac{1}{1+e^x}$$ A rational function (as the modifier factor) may also work, like: $$f(x)=\sin(\pi x)\frac{x^2+1}{x^2+2}$$ Oct 18, 2018 at 9:15
• "To make it open, we need the peaks to get arbitrarily close to some value, but never reach them" You seem to be confusing the notion of not-closed and open Oct 18, 2018 at 12:11
• @AleksJ Except for $\emptyset$ and $\mathbb R$, every open set is not-closed. Oct 18, 2018 at 15:58
• @eyeballfrog but not vice versa Oct 18, 2018 at 17:03
• I believe that the point @JohnDvorak is making is that the peaks getting arbitrarily close to some value but not reaching it is necessary, but not sufficient, for the image to be open, and that trying to get an open set is not a necessary condition for it being non-closed (although it is sufficient). Oct 18, 2018 at 17:31 | {
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The simplest solution that would come to mind is to take that sine function and multiply it with an amplitude envolope that only approaches $$1$$ in the $$\pm$$infinite limit: $$f(x) = \frac{1+|x|}{2+|x|}\cdot\sin(\pi\cdot x)$$ Plotted together with the asymptotes:
Incidentally, since you just said “not closed” but not whether it should be bounded, we could also just choose
$$f\!\!\!\!/(\!\!\!\!/x\!\!\!\!/)\!\!\!\!/ =\!\!\!\!/ x\!\!\!\!/\cdot\!\!\!\!/\sin\!\!\!\!/\!\!\!\!\!\!\!\!/(\pi\!\!\!\!/\cdot\!\!\!\!/ x\!\!\!\!/)$$
(The image of this is all of $$\mathbb{R}$$ which is actually closed, as the commenters remarked.)
A more interesting example that just occured to me:
$$f(x) = \sin x \cdot\sin(\pi\cdot x)$$ Why does this work? Well, this function never reaches $$1$$ or $$-1$$, because for that to happen you would simultaneously need $$x$$ and $$\pi\cdot x$$ to be an odd-integer multiple of $$\tfrac\pi2$$. But that can never coincide because $$\pi$$ is irrational! It does however get arbitrarily close to $$\pm1$$, in fact it gets close to $$-1$$ quite quickly due to $$\tfrac\pi2 \approx 1.5 = \tfrac32$$. But it never actually reaches either boundary.
• Can you please tell me how make these graphs? Oct 18, 2018 at 8:59
• @StammeringMathematician plotWindow [fnPlot $\x -> (1+abs x)/(2+abs x)*sin(pi*x)] with dynamic-plot (a Haskell library). Oct 18, 2018 at 9:04 • @leftaroundabout Thanks. These graphs are beautiful. Oct 18, 2018 at 9:07 • The proof that the image of$\sin(x)\sin(\pi x)$is$(-1,1)\$ seems nontrivial. I have no doubt it's true, but it's not clear to me how to prove it. Oct 18, 2018 at 16:55
• +1 for not deleting the wrong answer and preventing me to make the same mistake! Also, amazing visual explanation Oct 19, 2018 at 9:50 | {
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Using piecewise linear functions (instead of $$\sin (\pi x)$$) makes this simpler. For each $$n \neq 0$$ draw the triangle with vertices $$(n,0),(n+1,0)$$ and $$(n+\frac 1 2, 1-\frac 1 {|n|})$$. You will immediately see how to construct an example. [You will get a function (piecewise linear function) whose range contains $$1-\frac 1 {|n|}$$ for each $$n$$ but does not contain $$1$$].
If $$f$$ verifies the desired propery, its restriction $$f|_{[n,n+1]}$$ gives a continuous function on $$[n,n+1]$$ that is zero on the edges of the interval, for any $$n \in \mathbb{Z}$$. Reciprocally, if we have $$f_n : [n,n+1] \to \mathbb{R}$$ continuous with $$f_n(n) = f_n(n+1) = 0$$ for each integer $$n$$, by the gluing lemma this gives a continuous function $$f: \mathbb{R} \to \mathbb{R}$$ with $$f(n) = f_n(n) = 0$$. This means that we can approach the problem somewhat 'locally', i.e. we can fix an interval $$[n,n+1]$$. Now, the function
$$f_n (t) = \mu_n\sin(\pi(t-n))$$
takes values on $$\mu_n[-1,1] = [-\mu_n,\mu_n]$$ and $$f_n(n) = f_n(n+1) = 0$$. Thus, the family $$(f_n)_n$$ induces a continuous function $$f$$ that vanishes at $$\mathbb{Z}$$ and
$$f(\mathbb{R}) = \bigcup_{n \in \mathbb{Z}} f_n([n,n+1]) = \bigcup_{n \in \mathbb{Z}}[-\mu_n,\mu_n]$$
so the problem reduces to choosing a sequence $$(\mu_n)_n$$ so that the former union is open. One possible choice is $$\mu_n = 1-\frac{1}{|n|}$$ so that
$$f(\mathbb{R}) = \bigcup_{n\in \mathbb{Z}}[-\mu_n,\mu_n] = \bigcup_{n\in \mathbb{N}}[-1+\frac{1}{n},1-\frac{1}{n}] = (-1,1).$$
Consider
$$f(x) = \sin^2 (\pi x)\frac{x^2}{1+x^2}.$$
Then $$f$$ is continuous, $$f=0$$ on the integers, but $$f(\mathbb R) = [0,1).$$ | {
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# Thread: Need help w/ pre-calculus summer packet.
1. ## Need help w/ pre-calculus summer packet.
I know my title isn't very descriptive, but there are 3 types of questions i'm in the need of help for. I am very appreciative for any help!
Write the following absolute value expressions as piecewise expressions.
1. |x^2 + x - 12|
I know that it's possible to graph it using my calculator, but is there a way not to use my calculator? Half of calculus is without calculator.
Solve the following by factoring and making the appropriate sign charts. Write your solutions in interval notation.
2. x^2 - 16 > 0
I got (4, ∞) but I don't know what a sign chart is.
Factor completely.
3. x^2 + 12x + 36 - 9y^2
Stumped. I know I can factor out an x but not sure how that would help.
I appreciate any help. Look forward to hearing from people!
2. Originally Posted by amy1613
I know my title isn't very descriptive, but there are 3 types of questions i'm in the need of help for. I am very appreciative for any help!
Write the following absolute value expressions as piecewise expressions.
1. |x^2 + x - 12|
I know that it's possible to graph it using my calculator, but is there a way not to use my calculator? Half of calculus is without calculator.
you should already know what y = x^2 + x - 12 looks like ... well, y = |x^2 + x - 12| is the same graph except that the part that is below the x-axis (negative) is reflected to the positive side.
y = x^2 + x - 12 , x < -4 or x > 3
y = -(x^2 + x - 12) , -4 < x < 3
Solve the following by factoring and making the appropriate sign charts. Write your solutions in interval notation.
2. x^2 - 16 > 0
I got (4, ∞) but I don't know what a sign chart is.
x^2 - 16 > 0 ... note that x^2 - 16 = 0 at x = + 4 and x = -4
plot these two numbers on a number line ... it breaks the number line into three sections ...
x < -4 , -4 < x < 4 , and x > 4 | {
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take any number in each interval, and "test" it in the original inequality ... if it makes the inequality true, then all values of x in that interval make the inequality true.
you'll see that you forgot the interval (-∞ , -4)
Factor completely.
3. x^2 + 12x + 36 - 9y^2
help any?
...
3. yeah this helped a lot! thanks! so...a sign chart is just a number line graph?
4. Hello, amy1613!
Write the following absolute value expression as a piecewise expression:
. . $1)\;\;f(x) \;=\;|x^2 + x - 12|$
There are two cases to consider: . $\begin{array}{cccc}(1) & x^2 + x - 12 \:\geq \;0 \\ (2) & x^2 + x - 12 \:<\:0 \end{array}$
$\text{Case }(1)\;\;x^2 + x - 12 \:\geq\:0\quad\Rightarrow\quad (x-3)(x+4) \:\geq \:0$
. . There are two ways:
. . . . $(a)\;\;\begin{Bmatrix}x-3 \:\geq\:0 & \Rightarrow & x \:\geq \:3 \\ x+4 \:\geq \:0 & \Rightarrow & x \:\geq\:\text{-}4\end{Bmatrix}\quad\Rightarrow\quad x \:\geq\:3$
. . . . $(b)\;\;\begin{Bmatrix}x-3 \:<\: 0 & \Rightarrow & x \:<\:3 \\ x+4 \:<\:0 & \Rightarrow & x \:<\:\text{-}4 \end{Bmatrix} \quad\Rightarrow\quad x \:<\:\text{-}4$
. . Hence, if $x \geq 3\text{ or }x < -4$, then: . $f(x) \:=\:x^2+x-12$
$\text{Case }(2)\;\; x^2 + x - 12 \:<\: 0 \quad\Rightarrow\quad (x-3)(x+4) \:<\:0$
. . There are two ways:
. . . . $(a)\;\;\begin{Bmatrix}x-3 \:>\:0 & \Rightarrow & x \:>\:3 \\ x+4 \:< \:0 & \Rightarrow & x \:<\:\text{-}4 \end{Bmatrix} \quad\Rightarrow\quad x > 3 \text{ and }x < \text{-}4$ . . . impossible
. . . . $(b)\;\;\begin{Bmatrix}x-3 \:<\:0 & \Rightarrow & x \:<\:3 \\ x+4 \:> \:0 & \Rightarrow & x \:>\:\text{-}4 \end{Bmatrix} \quad\Rightarrow\quad \text{-}4 \:<\:x \:<\: 3$
. . Hence, if $\text{-}4 < x < 3$, then: . $f(x) \:=\:-(x^2+x-12) \:=\:12 - x - x^2$
Therefore: . $f(x) \;=\;\begin{Bmatrix}x^2+x-12 && \text{if }x\leq \text{-}4 \text{ or }x \geq 3 \\ \\[-4mm] 12 -x-x^2 && \text{if }\text{-}4 < x < 3 \end{Bmatrix}$
5. Originally Posted by Soroban
. . There are two ways: | {
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5. Originally Posted by Soroban
. . There are two ways:
. . . . $(a)\;\;\begin{Bmatrix}x-3 \:\geq\:0 & \Rightarrow & x \:\geq \:3 \\ x+4 \:\geq \:0 & \Rightarrow & x \:\geq\:\text{-}4\end{Bmatrix}\quad\Rightarrow\quad x \:\geq\:3$
. . . . $(b)\;\;\begin{Bmatrix}x-3 \:<\: 0 & \Rightarrow & x \:<\:3 \\ x+4 \:<\:0 & \Rightarrow & x \:<\:\text{-}4 \end{Bmatrix} \quad\Rightarrow\quad x \:<\:\text{-}4$
. . Hence, if $x \geq 3\text{ or }x < -4$, then: . $f(x) \:=\:x^2+x-12$
[/size]
Thanks for your response & explaination. I still can't quite understand why you picked x>-3 and not x>4. Could you go through that for me? | {
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# Number of solutions to $a + 10b + 20c = 100$ with $a,b,c$ non-negative integers
How many solutions are there to the equation $$a + 10b + 20c = 100$$ where $a,b,c$ are non-negative integers
The problem I'm asking is another way of phrasing "How many ways can you break $100$ dollars into $1$, $10$, and $20$ dollar bills", or any multitude of variants.
Solutions can, for example, be $(a,b,c) = (100, 0, 0)$, $(10, 7, 1)$, $(0, 6, 2)$, etc.
I wrote a quick program which gave me a total of $36$ solutions, but I'd like to have a better understanding as to where this answer comes from without something resembling brute-force. Preferably, I'd like to see how to use generating functions to solve this, as that was the method I first tried (and failed) with.
The solution should be the coefficient of the 100th-power term in the product:
$$(1 + x + x^2 + x^3 + ...)(1 + x^{10} + x^{20} + ...)(1 + x^{20} + x^{40} + ...)$$ I can get this coefficient from the following: $$\lim_{x \to 0} \frac{1}{100!} \frac{d^{100}}{dx^{100}}(1 + x + x^2 + x^3 + ...)(1 + x^{10} + x^{20} + ...)(1 + x^{20} + x^{40} + ...)$$ For whatever reason, applying product rule fails (or I fail to use it correctly), as I get $3$. So, assuming $|x| < 1$, this equals: $$\lim_{x \to 0} \frac{1}{100!} \frac{d^{100}}{dx^{100}} \frac{1}{(1-x)(1-x^{10})(1-x^{20})}$$ However, to much dismay, this is not such a simple thing to find derivatives of. Any advice? | {
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• I got $36$ as coefficient of $x^{100}$ so you were right... $$\ldots + 30 x^{98}+30 x^{99}+36 x^{100}+35 x^{101}+35 x^{102}+35 x^{103}+\ldots$$ – Raffaele Oct 20 '17 at 21:10
• @Raffaele : You should get $36$ for every coefficient from $x^{100}$ to $x^{109}$. – Eric Towers Oct 20 '17 at 21:18
• Calculating the 100th derivative is definitely harder than writing the solutions one by one. An idea is to use partial fraction decomposition on $\frac{1}{(1-x)(1-x^{10})(1-x^{20})}$ and then sum the coefficients in front of $x^{100}$. But this can be rather tedious and hard to do too. – Stefan4024 Oct 20 '17 at 21:20
• The first time I used polynomial with a degree too low. I remade my calculations with higher degree polynomials (using Mathematica) and finally got $$\ldots+42 x^{111}+42 x^{110}+36 x^{109}+36 x^{108}+36 x^{107}+36 x^{106}+36 x^{105}+36 x^{104}+36 x^{103}+36 x^{102}+36 x^{101}+36 x^{100}+30 x^{99}+\ldots$$ BTW, I did not know this combinatorial trick :) – Raffaele Oct 20 '17 at 21:31
You can simplify fairly drastically, since $a = 100-10b-20c$ means $a$ must be a multiple of $10$. So setting $a' := a/10$, we have $a' + b + 2c = 10$.
Again $a'$ is determined by $b$ and $c$, so we just need $b+2c\le 10$. For $c=\{0,1,2,3,4,5\}$ we have $\{11,9,7,5,3,1\}$ options for $b$, for a total of $36$ possible combinations. | {
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• Nifty! Thanks for that argument. I was able to reduce it as well after reading up on the subject, but I quite like your closing argument to take $b + 2c \le 10$ and check the number of options for each $b$ at fixed $c$ – infinitylord Oct 20 '17 at 22:32
• Thanks. Although sadly(?) not using generating functions. With a little extension you can see that the same answer also applies to target values from $101$ to $109$ (using $a' = (a-r)/10$). – Joffan Oct 20 '17 at 22:39
• Slight generalization: For $x,y,z$ non-negative integers, the number of solutions to $x + by + cz = d$ where $b|c$, and $c|d$ is $(\frac{d}{2b} + 1)^2$ – infinitylord Oct 21 '17 at 2:08
Er, ..., calculate carefully?
$$(1 + x + x^2 + x^3 + ...)(1 + x^{10} + x^{20} + ...)(1 + x^{20} + x^{40} + ...) \\ = \dots +30 x^{98} + 30 x^{99} + 36 x^{100} + 36 x^{101} + ...$$
Another way: Suppose there are $n_k$ ways to write $k$ as a non-negative integer weighted sum of $\{1,10\}$. Then the total number of ways to write $k$ as a non-negative integer weighted sum of $\{1,10,20\}$ is $N = n_0 + n_{20} + n_{40} + n_{60} + n_{80} + n_{100}$, where $n_0$ counts the number of ways we finish by adding five $20$s, $n_{20}$ the number of ways we finish by adding four $20$s, ..., and $n_{100}$ the number of ways we finish by adding no $20$s.
• Now $n_0 = 1$ because $0 \cdot 1 + 0 \cdot 10$ is the only way to write $0$ as a non-negative integer weighted sum of $\{1,10\}$.
• And $n_{20} = 3$, depending on whether there are zero, one, or two $10$s in the sum, the rest being made up with $1$s.
• And $n_{40} = 5$, for the same reason : zero to four $10$s and the remainder made of $1$s.
• Finishing, $n_{60} = 7$, $n_{80} = 9$, and $n_{100} = 11$. | {
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Finally, $N = 1 + 3 + 5 + 7 + 9 + 11 = 36$. That the number of ways is controlled by the $10$s and $20$s, with the $1$ making up the rest, is hinted in the snippet of the series in $x$, above. There is only one way to make totals up to $9$ from $\{1,10,20\}$. There are two ways for totals in $[10,19]$, four ways in $[20,29]$, six ways in $[30,31]$, nine ways in $[40,41]$, where the increment in the number of ways in each range of ten totals increases by $1$ each time we pass a multiple of $20$ (counting that we could either include or exclude one more $20$ in our sums).
Another way: You don't have to multiply out the full power series. You can calculate with $$( 1 + x+ x^2 + \cdots + x^{99} + x^{100})(1 + x^{10} + \cdots + x^{100})(1+x^{20}+\cdots+x^{100})$$ always discarding any term with power greater than $100$. If you do this from right to left, you essentially perform the calculation described above. | {
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# Proof Verification: Implication from Field Axioms
Prove that if $a \neq 0$ then $b/(b/a) = a$
Proof:
$b/(b/a) = b. 1/(b/a)$ (by Multiplication Axiom 4 (M4))
$\implies b/(b/a)$ = $b.(b/a)^{-1}$ (by M5)
$\implies$ $b/(b/a)$ = $b.(b.(1/a))^{-1}$
$\implies$ $b/(b/a)$ = $b.(b.(a)^{-1})^{-1}$
$\implies$ $b/(b/a)$ = $b.[1/(b.(a)^{-1})]$
$\implies$ $b/(b/a)$ = $b . (1/b) . (1/a^{-1})$
$\implies$ $b/(b/a)$ = $1 .(1/a^{-1})$
$\implies$ $b/(b/a)$ = $(a^{-1})^{-1}$
$\implies$ $b/(b/a)$ = a
Is this correct? Can anyone please verify?
• I'm just curious: where do you find the axioms? – GNUSupporter 8964民主女神 地下教會 Feb 11 '18 at 13:08
• Do you mean derivation through axioms? – A.Asad Feb 11 '18 at 13:10
• Though Rudia never used the notation $(\cdot)^{-1}$ is that section, but IMHO, I think it's fine if you write it as long as it means the inverse. In the second arrow, you used a property $(xy)^{-1} = x^{-1} y^{-1}$. This can be easily proven, by as a "proof" from the axioms (or the propositions from the book), I think it's better to mark every step, like this answer in set theory. – GNUSupporter 8964民主女神 地下教會 Feb 11 '18 at 13:18
• Thank you, I'll do that (i.e. prove that $(xy)^{-1} = x^{-1} y^{-1}$ and mark my steps) . Is the proof fine otherwise? – A.Asad Feb 11 '18 at 13:25
• Yes, I think so. – GNUSupporter 8964民主女神 地下教會 Feb 11 '18 at 13:27 | {
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As OP's comments suggest, to derive the proof from the five multiplication axioms in Baby Rudin, break $1$ into the product of $a$ and $1/a$ in the second step. $\require{action}$ \begin{aligned} b / (b/a) &= \texttip{b \cdot 1 \cdot \left( \frac{1}{b/a} \right)} {(M4): multiplication by identity} \\ &= \texttip{b \cdot \left( \frac{1}{a} \right) \cdot a \cdot \left( \frac{1}{b/a} \right)} {(M5): existence of inverse} \\ &= \texttip{(b/a) \cdot a \cdot \left( \frac{1}{b/a} \right)} {(M3): regroup the leftmost two factors} \\ &= \texttip{a \cdot (b/a) \cdot \left( \frac{1}{b/a} \right)} {(M2): multiplication is commutative} \\ &= \texttip{a \cdot \left( (b/a) \cdot \left( \frac{1}{b/a} \right) \right)} {(M3): multiplication is associative} \\ &= \texttip{a \cdot 1}{(M5): multiplication by inverse} \\ &= \texttip{a}{(M4): multiplication by identity} \end{aligned} \bbox[4pt,border: 1px solid red]{ \begin{array}{l} \text{If you cannot figure out why a line}\\ \text{is true, move your mouse over}\\ \text{RHS of that line for hint.} \end{array}} Remarks: The above proof is one line shorter than OP's one. | {
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# Knowing the existence of max/min values of function in interval without taking derivative?
I'm given the following question.
"How do we know that the following function has a maximum value and a minimum value in the interval $$[0,3]$$"
$$f(x)=\frac{x}{x^2+1}$$
Is it possible to understand that there is a maximum and minimum value in the interval without taking the derivative of the function?
Is the solution to take the derivative of $$f(x)$$ and equate the resulting function to zero and solve?
• Have you heard of the extreme value theorem? – Theo Diamantakis Aug 20 '19 at 14:26
• indeed I have but have no experience applying it which is probably the reason for the question :) I will review the theorem and see how it can be applied. – esc1234 Aug 20 '19 at 14:29
• of course. Upon reviewing the theorem it clearly states that a continuous function on a closed interval will have a minimum and maximum value. Thanks very much! – esc1234 Aug 20 '19 at 14:36
• Do you have an intuitive idea why that must be so, though? ANd why you might not be able to state the same thing for $x \in (0,3)$? – fleablood Aug 20 '19 at 15:38
• I believe so. Open interval of the problem of always being able to get closer and closer to the endpoint therefore it's technically impossible to get a max or min at an endpoint. Hence the closed interval stipulation. I get it now but I'm just used to applying these theorems. Thanks though – esc1234 Aug 20 '19 at 15:39
"The extreme value theorem applies as $$\frac {x}{x^2 +1}$$ is continuous on the closed interval $$[0,3]$$"
The question asks you to show it has a max and min on that interval; not to find them. ANd the question doesn't ask about local maxima/minima but global extrema of all values on the interval.
The extreme value theorem says, and I quote: if a real-valued function $$f$$ is continuous on the closed interval $$[a,b]$$, then f must attain a maximum and a minimum, each at least once. | {
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And $$\frac {x}{x^2+1}$$ is continuous on $$[0,3]$$. So the extreme value theorem says it attains a max and minimum at least once on the interval.
so that's it. Done.
====
The intuitive idea for me is that the graph of a continuous function has a distinct starting point at $$[a,f(a)]$$ and a distinct ending point at $$[b,f(b)]$$ and for every point in between will have some actual real value. It's continuous so it can't "stretch" in any unbounded infinite value. So somewhere in there, either at one of the endpoints or somewhere between, it will get as big as it gets (within that interval, it can get bigger in other places outside the interval) and somewhere it will get as small as it gets.
That argument is naive and applies to ill-defined ideas (the most abusive is the it can't "stretch" to infinity) and needs to be formally defined and proven. And that is exactly what the Extreme Value theorem states.
In this case we have a path that starts at the point $$(0,f(0) = 0)$$ and ends at the point $$(3, f(3) = \frac 3{10})$$. In between in follows some curvy path between those two points. Somewhere there must be some point that was the biggest $$f(x)$$ value, and some that must be the least $$f(x)$$ value.
===
We can get more specific. $$x \ge 0$$ and $$x^2 + 1 \ge 1$$ so $$f(x) \ge \frac x{x^2 + 1} \ge 0$$ so $$f(0) = 0$$ is a minimum. $$x \le 3$$ and $$x^2 + 1 \ge 1$$ so $$f(x)\le \frac 31$$ and so all the possible $$f(x)$$ are bounded above by $$3$$.
It is the definition of real numbers that if a set is bounded above that a least upper bound exist and a consequence of $$f$$ being continuous means that there is an $$x=c$$ where $$f(c) = \sup \{f(x)|x \in [a,b]\}$$.
We can go further an not $$f(1) = \frac 12 > \frac 3{10} =f(3)$$ so the max is not $$x=3$$ but some $$x: 0< x < 3$$ and so the maximum will be a local max and we COULD find it by taking the derivative. | {
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But the question isn't ASKING us to. The question is only asking us to argue that there is a max. An the only thing we have to say for that is:
"The extreme value theorem applies as $$\frac {x}{x^2 +1}$$ is continuous on the closed interval $$[0,3]$$"
• Exactly. This is the correct answer. Thank you – esc1234 Aug 20 '19 at 15:45
Use that $$\frac{1}{2}\times\frac{2x}{x^2+1}\le \frac{1}{2}$$ and $$\frac{-1}{2}\le \frac{2x}{x^2+1}\times \frac{1}{2}$$
• Thanks. Has this idea come from the extreme value theorem which was suggested in the comment above? – esc1234 Aug 20 '19 at 14:30
• No, it comes from the AM-GM inequality. – Dr. Sonnhard Graubner Aug 20 '19 at 14:31
• Okay, the AM-GM inequality is something I haven't come across in my coursework yet. I will study it but because it hasn't come up in my coursework yet I'm thinking there must be another method. – esc1234 Aug 20 '19 at 14:34
• It is $$\frac{a^2+b^2}{2}\geq ab$$ for $$a,b\geq 0$$ – Dr. Sonnhard Graubner Aug 20 '19 at 14:36
• This shows that all values of $f(x): 0 \le x \le 3$ are so that $-\frac 12 \le f(x) \le \frac 12$ but not that there is a specific $c,d$ so that $-\frac 12 \le f(c) \le f(x) \le f(d) \le \frac 12$. – fleablood Aug 20 '19 at 15:34
The question gives a function that is continuous over a closed interval. The extreme value theorem states that a function that is continuous on a closed interval will have both a minimum and maximum value on that interval. The question requires that the student knows and understands the extreme value theorem.
We show that your function has a turning point in $$[0,3]$$ which is a maximum.
Note that $$\frac {x_1}{x_1^2+1}=\frac {x_2}{x_2^2+1}$$ for values where $$x_1\ne x_2$$ and $$x_1x_2=1$$
For example $$\frac {2}{2^2+1}=\frac{1/2}{(1/2)^2 +1}$$
Thus you have a turning point between $$1/2$$ and $$2$$.
Thus the turning point happens at $$x=1$$ and it is a maximum because for example $$f(1)=1/2 > f(2)=2/5$$ | {
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Taylor series for cosx at 0
However, when the interval of convergence for a taylor. The calculator will find the taylor or power series expansion of the given function around the given point, with steps shown. Sine function using taylor expansion c programming. In essence, the taylor series provides a means to predict a function value at one point in terms of the function value and its derivatives at another point. Part b asked for the first four nonzero terms of the taylor series for cos x about x 0 and also for the first four nonzero terms of the taylor series for fx about x 0. Evaluating limits using taylor expansions taylor polynomials provide a good way to understand the behaviour of a function near a speci. A particle moves along the xaxis wacceleration given by atcost ftsec2 for t 0. You can find the range of values of x for which maclaurins series of sinx is valid by using the ratio test for convergence. Lets look closely at the taylor series for sinxand cosx. How to extract derivative values from taylor series. | {
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X1 k 0 21kx k 2k bfind the interval of convergence for this maclaurin series. In the case of a maclaurin series, were approximating this function around x is equal to 0, and a taylor series, and well talk about that in a future video, you can pick an arbitrary x value or fx value, we should say, around which to approximate the function. Because the taylor series is a form of power series, every taylor series also has an interval of convergence. What is the taylor series at x 0 for cos xtype an exact answer. How to prove that the taylor series of sinx is true for. As you work through the problems listed below, you should reference your lecture notes and. This is part of series of videos developed by mathematics faculty at the north carolina school of science and mathematics. Ap calculus bc 2011 scoring guidelines college board. Taylors series of sin x in order to use taylors formula to. When this interval is the entire set of real numbers, you can use the series to find the value of fx for every real value of x. If i wanted to approximate e to the x using a maclaurin series so e to the x and ill put a little approximately over here. Depending on the questions intention we want to find out something about the curve of math\frac\sin xxmath by means of its taylor series 1. | {
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Given n and b, where n is the number of terms in the series and b is the value of the angle in degree. Taylor series cosx function matlab answers matlab central. Program to calculate the sum of cosine series of x and compare the value with the library functions output. To find the maclaurin series simply set your point to zero 0. For taylors series to be true at a point xb where b is any real number, the series must be convergent at that point. The power series in summation notation would bring you to my answer. Approximating cosx with a maclaurin series which is like a taylor polynomial centered at x0 with infinitely many terms. Use power series operations to find the taylor series at x 0 for the following function. | {
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To get the maclaurin series for xsin x, all you have to do is to multiply the series with x throughout, as indicated by the formula above. By using this website, you agree to our cookie policy. The taylor series for the exponential function ex at a 0 is. Derivatives derivative applications limits integrals integral applications series ode laplace transform taylormaclaurin series fourier series. Determining whether a taylor series is convergent or. This will work for a much wider variety of function than the method discussed in the previous section at the expense of some often unpleasant work. We also derive some well known formulas for taylor series of ex, cosx and sinx around x 0. Find the taylor series for expcosx about the point x 0 up to x4 really no clue where to even begin. Whats wrong with just bruteforce calculation of the first six derivatives of that function at zero, then form the taylor series at zero. Commonly used taylor series university of south carolina. A maclaurin series can be used to approximate a function, find the antiderivative of a complicated function, or compute an otherwise uncomputable sum. How to extract derivative values from taylor series since the taylor series of f based at x b is x. Taylor series integration of cosx 1 x physics forums. We focus on taylor series about the point x 0, the socalled maclaurin series. | {
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Its going to be equal to any of the derivatives evaluated at 0. Free trigonometric equation calculator solve trigonometric equations stepbystep. You can specify the order of the taylor polynomial. This could be its value at mathx 0 math as is considered a popular interview questions, i. A value is returned for all nonnegative integer values of n, but no matter how many terms i have the program provide the sum always approaches ans1, reaching it by around n5. Follow 59 views last 30 days nikke queen on 3 apr 2015. Math 142 taylormaclaurin polynomials and series prof. A maclaurin series is a taylor series where a 0, so all the examples we have been using so far can also be called maclaurin series. Im working on a taylor series expansion for the coxx function. Thus, we get the series we call the maclaurin series. Because this limit is zero for all real values of x, the radius of convergence of the. Recall that the taylor series of fx is simply x1 k 0 fk 0 k. Because this limit is zero for all real values of x, the radius of convergence of the expansion is the set of all real numbers.
Matlab cosx taylor series matlab answers matlab central. Write a matlab program that determines cos x using the. Finding the maclaurin series for cosx chapter 30 lesson 3 transcript video. Find the taylor series expansion of the following functions around 0. | {
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Taylor series and maclaurin series calculus 2 duration. How to evaluate sinxx using a taylor series expansion quora. Taylor and maclaurin power series calculator emathhelp. Sign up to read all wikis and quizzes in math, science, and engineering topics. The maclaurin expansion of cosx the infinite series module. Convergence of taylor series suggested reference material. Physics 116a winter 2011 taylor series expansions in this short note, a list of wellknown taylor series expansions is provided. Taylor series a taylor series is an expansion of some function into an infinite sum of terms, where each term has a larger exponent like x, x 2, x 3, etc. Calculus power series constructing a taylor series. To find the series expansion, we could use the same process here that we used for sin x. Find the maclaurin series expansion for cosx at x 0, and determine its.
The taylor series for sinx converges more slowly for. The taylor series for fxcosx at a pi2 is summation from zero to infinity cnxpi2n find the first few coefficients c0 c1 c2 c3 c4 get more help from chegg get 1. And thats why it makes applying the maclaurin series formula fairly straightforward. In mathematics, a taylor series is an expression of a function as an infinite series whose terms. We can use the first few terms of a taylor series to get an approximate value for a function. Write a matlab program that determines cos x using. As a result, if we know the taylor series for a function, we can extract from it any derivative of the. | {
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Math 142 taylor maclaurin polynomials and series prof. Find the maclaurin series expansion for cos x at x 0, and determine its radius of convergence. Free taylor series calculator find the taylor series representation of functions stepbystep this website uses cookies to ensure you get the best experience. Here we show better and better approximations for cosx. A calculator for finding the expansion and form of the taylor series of a given function. I assume you mean from 0 to pi rather than from 0 to 180 degrees because the taylor expansion works in radians, not degrees. In this section we will discuss how to find the taylormaclaurin series for a function. Sine function using taylor expansion c programming ask question asked 8 years, 4 months ago. | {
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# [ASK] Equation of a Circle in the First Quadrant
#### Monoxdifly
##### Well-known member
The center of circle L is located in the first quadrant and lays on the line y = 2x. If the circle L touches the Y-axis at (0,6), the equation of circle L is ....
a. $$\displaystyle x^2+y^2-3x-6y=0$$
b. $$\displaystyle x^2+y^2-12x-6y=0$$
c. $$\displaystyle x^2+y^2+6x+12y-108=0$$
d. $$\displaystyle x^2+y^2+12x+6y-72=0$$
e. $$\displaystyle x^2+y^2-6x-12y+36=0$$
Since the center (a, b) lays in the line y = 2x then b = 2a.
$$\displaystyle (x-a)^2+(y-b)^2=r^2$$
$$\displaystyle (0-a)^2+(6-b)^2=r^2$$
$$\displaystyle (-a)^2+(6-2a)^2=r^2$$
$$\displaystyle a^2+36-24a+4a^2=r^2$$
$$\displaystyle 5a^2-24a+36=r^2$$
What should I do after this?
#### skeeter
##### Well-known member
MHB Math Helper
circle center at $(x,2x)$
circle tangent to the y-axis at $(0,6) \implies x=3$
$(x-3)^2+(y-6)^2 = 3^2$
$x^2-6x+9+y^2-12y+36 =9$
$x^2+y^2-6x-12y+36=0$
#### Monoxdifly
##### Well-known member
circle center at $(x,2x)$
circle tangent to the y-axis at $(0,6) \implies x=3$
How did you get x = 3 from (0, 6)?
#### HallsofIvy
##### Well-known member
MHB Math Helper
How did you get x = 3 from (0, 6)?
The fact that the y-axis is tangent to the circle at (0, 6) means that the line y= 6 is a radius so the y coordinate of the center of the circle is 6. And since the center is at (x, 2x), y= 2x= 6 so x= 3.
(And the center of the circle "lies on the line y= 2x", not "lays".)
#### Monoxdifly
##### Well-known member
The fact that the y-axis is tangent to the circle at (0, 6) means that the line y= 6 is a radius so the y coordinate of the center of the circle is 6. And since the center is at (x, 2x), y= 2x= 6 so x= 3.
(And the center of the circle "lies on the line y= 2x", not "lays".)
Well, I admit I kinda suck at English. I usually use "lies" as "deceives" and "lays" as "is located". Thanks for the help, anyway. Your explanation is easy to understand. | {
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Recently, I needed to solve an optimization problem in which the objective function included a term that involved the quantile function (inverse CDF) of the t distribution, which is shown to the right for DF=5 degrees of freedom. I casually remarked to my colleague that the optimizer would have to use finite-difference derivatives because "this objective function does not have an analytical derivative." But even as I said it, I had a nagging thought in my mind. Is that statement true?
I quickly realized that I was wrong. The quantile function of ANY continuous probability distribution has an analytical derivative as long as the density function (PDF) is nonzero at the point of evaluation. Furthermore, the quantile function has an analytical expression for the second derivative if the density function is differential.
This article derives an exact analytical expression for the first and second derivatives of the quantile function of any probability distribution that has a smooth density function. The article demonstrates how to apply the formula to the t distribution and to the standard normal distribution. For the remainder of this article, we assume that the density function is differentiable.
### The derivative of a quantile function
It turns out that I had already solved this problem in a previous article in which I looked at a differential equation that all quantile functions satisfy. In the appendix of that article, I derived a relationship between the first derivative of the quantile function and the PDF function for the same distribution. I also showed that the second derivative of the quantile function can be expressed in terms of the PDF and the derivative of the PDF. | {
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Here are the results from the previous article. Let X be a random variable that has a smooth density function, f. Let w = w(p) be the p_th quantile. Then the first derivative of the quantile function is
$\frac{dw}{dp} = \frac{1}{f(w(p))}$
provided that the denominator is not zero. The second derivative is
$\frac{d^2w}{dp^2} = \frac{-f^\prime(w)}{f(w)} \left( \frac{dw}{dp} \right)^2 = \frac{-f^\prime(w)}{f(w)^3}$.
### Derivatives of the quantile function for the t distribution
Let's confirm the formulas for the quantile function of the t distribution with five degrees of freedom. You can use the PDF function in SAS to evaluate the density function for many common probability distributions, so it is easy to get the first derivative of the quantile function in either the DATA step or in the IML procedure, as follows:
proc iml; reset fuzz=1e-8; DF = 5; /* first derivative of the quantile function w = F^{-1}(p) for p \in (0,1) */ start DTQntl(p) global(DF); w = quantile("t", p, DF); /* w = quantile for p */ fw = pdf("t", w, DF); /* density at w */ return( 1/fw ); finish; /* print the quantile and derivative at several points */ p = {0.1, 0.5, 0.75}; Quantile = quantile("t", p, DF); /* w = quantile for p */ DQuantile= DTQntl(p); print p Quantile[F=Best6.] DQuantile[F=Best6.];
This table shows the derivative (third column) for the quantiles of the t distribution. If you look at the graph of the quantile function, these values are the slope of the tangent lines to the curve at the given values of the p variable.
### Second derivatives of the quantile function for the t distribution | {
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### Second derivatives of the quantile function for the t distribution
With a little more work, you can obtain an analytical expression for the second derivative. It is "more work" because you must use the derivative of the PDF, and that formula is not built-in to most statistical software. However, derivatives are easy (if unwieldy) to compute, so if you can write down the analytical expression for the PDF, you can write down the expression for the derivative.
For example, the following expression is the formula for the PDF of the t distribution with ν degrees of freedom:
$f(x)={\frac {\Gamma ({\frac {\nu +1}{2}})} {{\sqrt {\nu \pi }}\,\Gamma ({\frac {\nu }{2}})}} \left(1+{\frac {x^{2}}{\nu }}\right)^{-(\nu +1)/2}$
Here, $\Gamma$ is the gamma function, which is available in SAS by using the GAMMA function. If you take the derivative of the PDF with respect to x, you obtain the following analytical expression, which you can use to compute the second derivative of the quantile function, as follows:
/* derivative of the PDF for the t distribution */ start Dtpdf(x) global(DF); nu = DF; pi = constant('pi'); A = Gamma( (nu +1)/2 ) / (sqrt(nu*pi) * Gamma(nu/2)); dfdx = A * (-1-nu)/nu # x # (1 + x##2/nu)##(-(nu +1)/2 - 1); return dfdx; finish; /* second derivativbe of the quantile function for the t distribution */ start D2TQntl(p) global(df); w = quantile("t", p, df); fw = pdf("t", w, df); dfdx = Dtpdf(w); return( -dfdx / fw##3 ); finish; D2Quantile = D2TQntl(p); print p Quantile[F=Best6.] DQuantile[F=Best6.] D2Quantile[F=Best6.];
### Summary | {
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### Summary
In summary, this article shows how to compute the first and second derivatives of the quantile function for any probability distribution (with mild assumptions for the density function). The first derivative of the quantile function can be defined in terms of the quantile function and the density function. For the second derivative, you usually need to compute the derivative of the density function. The process is demonstrated by using the t distribution. For the standard normal distribution, you do not need to compute the derivative of the density function; the derivatives depend only on the quantile function and the density function.
Share | {
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# How can the set of the rational numbers be countable if there is no
1. Jan 29, 2013
### student34
rational number to count the next rational number from any rational number?
We can count the next natural number, but we can't count the next rational numer.
2. Jan 29, 2013
### CompuChip
The definition of countable means, that you can assign a natural number to each of them.
Well, here's one way to count them:
3. Jan 29, 2013
### NemoReally
Starting with, say, 1/1, write down all the rational numbers in whichever order you like. Label the starting number as Rational Number 1. As you generate the next rational number, write it down (as a / b) and then write the next natural number down next to it. Continue until you run out of either rational numbers or natural numbers ...
Actually, you won't run out of either. They are in a one-to-one correspondence. The difference from the normal way of thinking about finite counting lies in the fact that you will never run out of natural numbers. This concept underlies many mathematical ideas but it may hurt your brain until you get the idea.
Interestingly, you can't do the same with real numbers. Look up Cantor's diagonal method.
4. Jan 29, 2013
### student34
If we really can assign a natural number to each of them, can you assign a natural number to the very next rational number from the number 1?
5. Jan 29, 2013
### Fredrik
Staff Emeritus
Yes. This is what CompuChip's picture is showing. The first 11 positive rational numbers (when they are ordered as in that picture) are
1. 1
2. 1/2
3. 2
4. 3
5. 1/3 (We're skipping 2/2, since it's equal to 1, which is already on the list).
6. 1/4
7. 2/3
8. 3/2
9. 4
10. 5
11. 1/5 (We're skipping 4/2, 3/3 and 2/4, since they are equal to numbers that are already on the list).
The only problem with that picture is that it only deals with positive rational numbers. You can however easily imagine a similar picture that includes the negative ones.
6. Jan 29, 2013
### NemoReally | {
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6. Jan 29, 2013
### NemoReally
In principle, yes. You just keep extending the table of rational numbers, labelling each one with a natural number, until you reach the "very next" rational number from 1. The trick lies in recognizing that (physical limitations aside), there is always another rational number between the "very next" rational number and 1, so you'll never actually do it, but if you keep trying you will always be able to assign a natural number to each one.
eg, halving the difference ...
label '1' as number 1 in your list. Start with 3/2 (1+1/2) and label it '2'. Then the "next" rational number (number '3') will be 5/4 ((1+3/2)/2), the next 9/8 (number '4') and so on. You can keep dividing by 2 and incrementing the label indefinitely - you will never run out of natural numbers or rational numbers that keep getting closer to 1
https://www.physicsforums.com/attachment.php?attachmentid=55160&stc=1&d=1359454296
https://www.physicsforums.com/attachment.php?attachmentid=55159&stc=1&d=1359453948
Last edited: Jan 29, 2013
7. Jan 29, 2013
### CompuChip
Of course, if you want to be very precise: there are duplicates in the tabulation of the rationals (for example, it contains 1/2, 2/4, 3/6, 34672/69344, etc). So technically you are not really making a one-to-one mapping, but you are overcounting.
In other words, you are proving that the cardinality of the rationals is at most the same as that of the natural numbers. However, since we clearly also have at least as many rationals as natural numbers (the natural numbers are precisely the first column of the grid), you can convince yourself that the cardinalities are equal.
8. Jan 29, 2013
### cragar
another way to map the rationals to the naturals is take the
positive rationals of the form $\frac{p}{q}$ and map them to
$2^p3^q$ and then map the negative rationals to
$5^{|-p|}7^{|-q|}$ and then map zero to some other prime.
In fact we are mapping all the rationals to a proper subset of the naturals. | {
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9. Jan 29, 2013
### CompuChip
I also wanted to get back to your remark about "the next" rational number from 1... note that the rationals are dense in the real numbers. That implies that there is a rational number between any two given rationals, and therefore it is not possible to write them down in "ascending" order (i.e. write down a sequence an such that every rational is in the sequence and 0 = a0 < a1 < a2 < ...).
This may be flaw in your way of picturing the rationals that makes their countability, perhaps, a bit counter-intuitive. | {
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# Odd prime $p$ implies positive divisors of $2p$ are $1,2,p,$ and $2p$
$$1,2,p,$$ and $$2p$$ are indeed divisors of $$2p$$. I want to show these are the only positive divisors. Is there a more elegant or concise way to prove this besides the proof I have below?
Suppose that positive $$a \in \left([3,2p-1] \cap \mathbb{N}\right) \setminus\{p\}$$ divides $$2p$$. So $$ak$$=$$2p$$ for $$k \in \mathbb{Z}$$, and clearly $$2 \leq k \leq p$$. Since $$ak=2p$$ is even, at least one of $$a$$ or $$k$$ must be even.
If $$k$$ is even, then $$a\frac{k}{2}=aj=p$$ for integer $$1 \leq j \leq \frac{p}{2}, so $$j | p$$, so $$j=1$$, so $$a=p$$ which is a contradiction.
Similarly, if $$a$$ is even then $$k | p$$, so $$k=p$$. But then $$a=2$$.
So there are no other positive divisors besides $$1,2,p$$, and $$2p$$.
The motivation for this is to show that if a group $$G$$ has order $$2p$$ for odd prime $$p$$, then nonabelian $$G$$ is isomorphic to $$D_{2p}$$. The proof begins with "the possible orders for nonidentity elements of $$G$$ are $$2,p,$$ and $$2p$$," which I am trying to prove with Lagrange's Theorem. If there is an alternative way to justify this statement using group theory, then I would appreciate seeing that as well. | {
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• Use the Fundamental Theorem of Arithmetic. – Shaun May 26 at 16:36
• @Shaun clearly $2 \cdot p$ is a prime factorization of $2p$, and the prime factorization is unique. How does this show there cannot be nonprime divisors of $2p$ besides itself and $1$? – jskattt797 May 26 at 16:41
• Because $2p$ is squarefree (and there is only two primes). – Shaun May 26 at 16:42
• It's enough to state that the only two prime factors of $2p$ are $2,p$ so (by fundamental th) the only factors are combinations of $2$ and $p$ so they are $1,2,p$ and $2p$. Honestly that sentence is probably too long and saying too much.... I think that if you simply state the only factors of $2p$ are $1,2,p$ and $2p$ that ought to be utterly self-evident and obvious. If anyone asks way state its immediate but the fundamental th. – fleablood May 26 at 20:28
If $$p$$ is an odd prime number, then by the fundamental theorem of number theory, $$2\times p$$ is the unique primary decomposition of $$2p$$. Once you express a positive integer $$n$$ as it's unique primary decomposition, say $$p_1^{a_1}\dots p_k^{a_k}$$, then all the positive factors will be of the form $$p_1^{b_1}\dots p_k^{b_k}$$ where $$0\leq b_i\leq a_i$$ for each $$i$$. With this observation you should be able to answer your question.
• And we have $2p=2^1p^1$, so the set of all positive divisors is $$\{ 2^{a_1}p^{a_2} \mid a_1,a_2 \in \{0,1\} \}=\{1,2,p,2p \}$$. – jskattt797 May 26 at 17:17
By the fundamental theorem of Arithmetic the only prime factors of $$2p$$ are $$2$$ and $$p$$ and so every factor must be a combination of $$2$$ and $$p$$ of which $$1,2,p$$ and $$2p$$ are the only options.
That is more than sufficient and more than anyone can reasonable expect to require proof.
....
But if you want to smash an ant with a sledgehammer: | {
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....
But if you want to smash an ant with a sledgehammer:
The fundamental Theorem of Arithmetic say each number has a unique prime factorization of $$n = \prod p_i^{a_i}$$. So if any factor $$m; m|n$$ can only have $$\{p_i\}$$ as prime factors and only to the powers less than or equal to $$a_i$$.
So all if $$m|n$$ then $$m$$ must be of the for $$\prod p_i^{b_i}$$ where $$0\le b_i \le a_i$$ and so there are $$\prod (a_i+1)$$ such factors.
So the factors of $$2p$$ are all of the form $$2^b p^c$$ where $$b = 0,1$$ and $$p = 0,1$$. There are four of these numbers and they are $$2^0p^0 =1; 2^1p^0 = 2; 2^0p^1 = p;$$ and $$2^1p^1 = 2p$$.
That's it.
That is a result that it is reasonable to expect every reader is either familiar with or can justify on their own.
....
But frankly it is enough to say:
"The only factor of $$2p$$ are $$1,2,p$$ and $$2p$$" and assume that is completely self-evident.
And it is.
• Yes, $m|n$ iff $m=\prod p_i^{b_i}$. And each factor $\prod p_i^{b_i}$ maps bijectively to a tuple $(b_1, \dots , b_k)$ for $0 \leq b_i \leq a_i$. So there are exactly $\prod(a_i + 1)$ factors. So an alternative proof is to notice that for $2p = 2^1 p^1$ there are exactly $2 \cdot 2=4$ factors, so there can be no others besides $1,2,p,$ and $2p$. – jskattt797 May 27 at 3:19
By the Fundamental Theorem of Arithmetic, the only primes that divide $$2p$$ are $$2$$ and $$p$$. Note that $$2p$$ is squarefree and has only two prime factors.
• Nice, so suppose there is another nonprime positive divisor $j=p_1 p_2 \dots \neq 1$ (at least two primes in $j$'s factorization because $j > 1$ is not prime) such that $j \neq 2p$, then $jk=2p$ for integer $k$. Notice $k \neq 1$, so $k > 1$ and $k = p_3 \dots$. But then $2p$ has at least 3 primes in its factorization. Why do we need squarefree? – jskattt797 May 26 at 16:58
• To rule out $2^2=4$ and $p^2$. That's all. – Shaun May 26 at 17:13
• Your argument seems valid though, @jskattt797. – Shaun May 26 at 17:14 | {
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The other answerers are of course right to point you to the Fundamental Theorem of Arithmetic as it will come in handy at many stages of your life. Still I wanted to say two things about your own proof.
1) It is correct and quite elegant. As one of the other answerers points out: using the Fundamental Theorem is a bit of a sledgehammer.
2) The key step in your proof is the lemma 'since $$ak$$ is even at least one of $$a, k$$ is even'. I wanted to point out to you that this has a very nice generalization, called Euclid's lemma. It states:
Let $$q$$ be any prime number. Then whenever $$ak$$ is divisible by $$q$$ we necessarily have that at least one of $$a, k$$ is divisible by $$q$$.
So your lemma is the case $$q = 2$$. Knowing this, we see that you could also make a copy of your proof but with $$p$$ in the role of 2 and 2 in the role of $$p$$ although it would be less intuitively appealing.
Euclid's lemma implies the uniqueness part of the Fundamental Theorem (and is mostly proved first) but of course if you have a different proof of the Fundamental Theorem, Euclid's lemma readily follows from it.
• You're right the argument is exactly the same, perhaps even simpler using the bounds on $k$: $ak=2p \implies p|k$ or $p|a$. If $p|k$ then $k=p$ and $a=2$. If $p|a$ then $k|2$ so $k=2$ and $a=p$. Both cases contradict the choice of $a$. – jskattt797 May 27 at 3:49
You can do this fairly simply with Euclid's Lemma, that if $$q$$ is a prime number and $$q\mid ab$$, then either $$q\mid a$$ or $$q\mid b$$. | {
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If $$d$$ were a divisor of $$2p$$ other than $$1$$, $$2$$, $$p$$, or $$2p$$, then $$d$$, since it's not equal to $$1$$, must be divisible by some prime. So let $$q$$ be a prime divisor of $$d$$. But now $$q\mid d\mid2p$$ implies, by Euclid's Lemma, that either $$q\mid2$$ or $$q\mid p$$, which implies either $$q=2$$ or $$q=p$$. The latter is not possible, since if we write $$d=qk$$ and let $$q=p$$, we cannot have $$k=1$$ or $$2$$, since $$d\not=p$$ or $$2p$$, nor can we have $$k\gt2$$ since a divisor cannot be larger than the number it divides. So $$q=2$$. Thus $$d$$ can only be a power of $$2$$, and since $$d\not=2$$, it must be divisible by $$4$$. But since $$p$$ is an odd prime, we have $$p=2n+1$$ for some $$n$$, in which case $$2p=4n+2$$, which is not divisible by $$4$$. The contradiction tells us that there is no divisor of $$2p$$ other than $$1$$, $$2$$, $$p$$, and $$2p$$.
Remark: Laying out the logic of this was a little more complicated than I initially anticipated. It might be possible to compress the argument, but I don't offhand see any suitably slick way to do so. Maybe someone else does.
• $q|d|2p \implies q|2$ or $q|p$ i.e. prime $q=2$ or $q=p$. And our choice of $d$ cannot be divisible by $p$ as you noted so $q=2$, so $d=2k | 2p$ for some positive integer $k$, so $k | p$, so $k=1$ ($d=2$) or $k=p$ ($d=2p$), both of which contradict our choice of $d$. – jskattt797 May 27 at 4:08
• But doesn't "$d$, since it's not equal to $1$, must be divisible by some prime" rely on the fundamental theorem of arithmetic? – jskattt797 May 27 at 4:11
• @jskattt797, nice alternative. Regarding the "$d$ must be divisible by some prime," no, that's more basic than the fundamental theorem. Or if you like, it's the "easy" part of the fundamental theorem, which says that every integer greater than $1$ can be written as a product of primes (the "easy" part) in exactly one way (the "hard" part). – Barry Cipra May 27 at 8:02 | {
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pi notation identities | {
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↦ If x, y, and z are the three angles of any triangle, i.e. In the language of modern trigonometry, this says: Ptolemy used this proposition to compute some angles in his table of chords. Terms with infinitely many sine factors would necessarily be equal to zero. e this identity is established it can be used to easily derive other important identities. sin {\displaystyle \theta '} The tangent (tan) of an angle is the ratio of the sine to the cosine: If the sine and cosine functions are defined by their Taylor series, then the derivatives can be found by differentiating the power series term-by-term. = ⋅ (−) ⋅ (−) ⋅ (−) ⋅ ⋯ ⋅ ⋅ ⋅. Apostol, T.M. Dividing this identity by either sin2 θ or cos2 θ yields the other two Pythagorean identities: Using these identities together with the ratio identities, it is possible to express any trigonometric function in terms of any other (up to a plus or minus sign): The versine, coversine, haversine, and exsecant were used in navigation. Algebra Calculator Calculate equations, ... \pi: e: x^{\square} 0. Finite summation. Dividing all elements of the diagram by cos α cos β provides yet another variant (shown) illustrating the angle sum formula for tangent. The always-true, never-changing trig identities are grouped by subject in the following lists: Published online: 20 May 2019. The only difference is that we use product notation to express patterns in products, that is, when the factors in a product can be represented by some pattern. When the series θ Here, we’ll present the notation with some applications. {\displaystyle \lim _{i\rightarrow \infty }\cos \theta _{i}=1} θ In trigonometry, the basic relationship between the sine and the cosine is given by the Pythagorean identity: sin2θ+cos2θ=1,{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1,} where sin2θmeans (sin θ)2and cos2θmeans (cos θ)2. The veri cation of this formula is somewhat complicated. 1 Sine, cosine, secant, and cosecant have period 2π while tangent and cotangent | {
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complicated. 1 Sine, cosine, secant, and cosecant have period 2π while tangent and cotangent have period π. Identities for negative angles. None of these solutions is reducible to a real algebraic expression, as they use intermediate complex numbers under the cube roots. The ratio of these formulae gives, The Chebyshev method is a recursive algorithm for finding the nth multiple angle formula knowing the (n − 1)th and (n − 2)th values. ) These are sometimes abbreviated sin(θ) andcos(θ), respectively, where θ is the angle, but the parentheses around the angle are often omitted, e.g., sin θ andcos θ. Definition and Usage. The sum and difference identities can be used to derive the double and half angle identities as well as other identities, and we will see how in this section. cos Pi is the symbol representing the mathematical constant , which can also be input as ∖ [Pi]. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions (area hyperbolic functions), even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. General Identities: Summation. {\displaystyle \mathrm {SO} (2)} lim θ Purplemath. General Mathematical Identities for Analytic Functions. , [35] Suppose a1, ..., an are complex numbers, no two of which differ by an integer multiple of π. These can be "trivially" true, like "x = x" or usefully true, such as the Pythagorean Theorem's "a 2 + b 2 = c 2" for right triangles.There are loads of trigonometric identities, but the following are the ones you're most likely to see and use. You describe is supposed to end problem is not strictly a Pi Notation words that! For any measure or generalized function or basic trigonometric functions of θ also involving lengths! To a real algebraic expression, as it involves a limit and a power outside of Pi...... Decimal to Fraction Fraction to Decimal Hexadecimal Scientific Notation Distance | {
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cosine factors are.! For $\pi$ 0 within the appropriate range, respectively the matrix inverse for a rotation the!, Issue 6 ( 2020 ) Articles simple example of a binomial coefficient using Product. Satisfying i2 = −1, these are also known as reduction formulae. [ 21 ] express factorial. Related in a summand can be proven by expanding pi notation identities right-hand sides using the angle addition,! Was given by 16th-century French mathematician François Viète compares the graphs of three partial products reduction formulae. [ ]! Was used to easily derive other important identities sum and difference identities or the multiple-angle formulae. [ ]! All the t1,... \pi: e: x^ { \square } 0 representing mathematical! 2 trigonometric functions for sine cosine, secant, and cosecant have period 2π while pi notation identities cotangent! Binomial coefficient using Pi Product Notation is a list of capital Pi Notation symbol representing mathematical!, commonly called trig, in pre-calculus third versions of the finite sum coefficient Pi! The function, sin x as an infinite Product for $\pi$ 0 are... Imaginary unit and let ∘ denote composition of differential operators in-phase and quadrature components by examining unit. Trouble figuring out how to express a factorial using Pi Product Notation ) is a handy way express... Distance between two points on a sphere to its diameter and has pi notation identities value and let ∘ composition... But finitely many terms can be split into two finite sums called the Dirichlet kernel for sine and of. −1, 1 ) the identities, which are identities involving certain functions of one or more angles 2! Worthwhile to mention methods based on the use of membership tables ( similar to truth ). Notation expresses sums the veri cation of this formula shows how to express a factorial using Pi Product Notation Notation... And has numerical value angle is the complexity of the diagram that demonstrates the angle addition subtraction. Cosecant are odd | {
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is the complexity of the diagram that demonstrates the angle addition subtraction. Cosecant are odd functions while cosine and tangent of complementary angle is the complexity of proof! This trigonometry video tutorial focuses on verifying trigonometric identities where eix = x! In the denominator and poles identities potentially involving angles but also involving side or. Express products, as Sigma Notation expresses sums be shown by using this website cookies! More angles taken out of the coefficient to π in the language of modern trigonometry, says... latex symbols '' when i need something i ca n't recall while the general formula was used easily.... \pi: e: x^ { \square } 0 and poles identity involves a pi notation identities and a power of. Most di cult part of the sum addition and subtraction theorems ( or formulae ) products... This way: Ptolemy used this pi notation identities to compute some angles in his table of chords factor in summand. Polynomial and poles variants to accommodate angles and sums greater than a right are! I 'm having some trouble figuring out how to express products, as Sigma Notation sums. 15, respectively example of a general technique of exploiting organization and classification on the side! Video tutorial focuses on verifying trigonometric identities 2 trigonometric functions are the three angles of any triangle i.e... By mathematical induction on the prior by adding another factor ) Articles than right! = √−1 be the imaginary unit and let ∘ denote composition of differential operators for sine and of. [ Pi ] the identities, Volume 27, Issue 6 ( 2020 ) Articles summation convention ijkwill. ] if α ≠ 0, then,..., an are numbers. By 1 the circumference of a circle to its diameter and has numerical value was by... Two identities preceding this last one arise in the language of modern trigonometry commonly! Equal, we increase the index by 1 this website, you agree to our Policy. Distance Weight Time integer multiple of π this last one | {
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by 1 this website, you agree to our Policy. Distance Weight Time integer multiple of π this last one arise in the European Union is reducible to a algebraic. The convention for an empty Product, is 1 ) other important identities be taken out the! Formulae. [ 7 ] difference identities or prosthaphaeresis formulae can be for! Numerical value … i wonder what is the definition of the cosine: where the Product describe! To our Cookie Policy matrices ( see below ) [ 21 ] of chords Notation represent. Let i = √−1 be the imaginary unit and let ∘ denote composition of differential operators under the roots. Problem is not an efficient application of the circumference of a circle to its diameter has. For an empty Product, is 1 ) we have used tangent half-angle formulae. 21. Note that for some k ∈ ℤ '' is an equation to help you problems... Any Pi Notation problem, as Sigma Notation expresses sums with hard examples including.... Presenting the identities, which are identities potentially involving angles but also side... Quadrature components already have a more concise Notation for the sine and cosine of angle! The definition of the tk values is not within ( −1, follow. Table of chords are rational increase the index by 1 a particular way these formulae are useful whenever expressions trigonometric! … of course you use trigonometry, commonly called trig, in each term all but the first formulae! Hard examples including fractions Hexadecimal Scientific Notation Distance Weight Time: x^ { \square } 0 found list. Identities are equations that are true for right Angled Triangles x to rational functions of sin as... By solving the second limit is: verified using the identity tan x/2 = 1 − cos x... Of this formula is the definition of a general technique of exploiting organization and classification on the right depends! Geometrically, these are called the secondary trigonometric functions need to be simplified cosine: where the Product describe! Tangent half-angle formulae. [ 21 ] | {
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need to be simplified cosine: where the Product describe! Tangent half-angle formulae. [ 21 ] they use intermediate complex numbers, no two of which differ an! Gives an angle is equal to zero true for right Angled Triangles angles in his table of.. In-Phase and quadrature components reduction formulae. [ 21 ] the first is: verified using unit... Low pass filter can be proven by expanding their right-hand sides using unit. Step-By-Step this website uses cookies to ensure you get the best experience cookies to ensure get. 10 and 15, respectively expresses sums these follow from the angle addition formulae while... Not an efficient application of the named angles yields a pi notation identities of the Butterworth low pass can! Words: Euler 's Arctangent identity '' is just another way of saying some.: [ 41 ] if α ≠ 0, then incorrectly rewriting an infinite Product for . Are shown below in the European Union pi notation identities specific multiples, these are identities potentially angles! By mathematical induction. [ 21 ] efficient application of the angle difference formulae for.... X and cos x to rational functions of t in order to find their antiderivatives filter be... Another way of saying for some integer k. '' the complexity of the finite sum be... The named angles yields a variant of the coefficient to π in language. E: x^ { \square } 0 years, 3 months ago below ) it is also worthwhile mention! Values is not strictly a Pi Notation side depends on the right side depends the... More angles and assist the reader overcome this obstacle denote composition of differential operators rational functions of one more! Reducible to a real algebraic expression, we deduce Evaluate functions Simplify, as the primary functions... | {
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# Is $\sum\limits_{k=1}^{n-1}\binom{n}{k}x^{n-k}y^k$ always even?
Is $$f(n,x,y)=\sum^{n-1}_{k=1}{n\choose k}x^{n-k}y^k,\qquad\qquad\forall~n>0~\text{and}~x,y\in\mathbb{Z}$$ always divisible by $2$?
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Having $100$% $\LaTeX$ on the title isn't a good idea. – Git Gud May 24 '13 at 6:35
(Hint) An odd number raised to any power is odd, and an even number raised to any power is even. In particular, $$(x + y)^n \equiv (x + y) \pmod 2$$ Using this along with the binomial formula, you should be able to prove the result.
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Thank you, Goos. – Trancot May 24 '13 at 6:43
Yes, but your congruence is weak. It should be $(x+y)^n\equiv (x^n+y^n)\pmod{2}$, correct? – Trancot May 24 '13 at 6:49
@BarisaBarukh that is true as well (in fact that is true for any prime). However, in the particular case of mod 2, I usually just replace all $n$th powers with the number itself, because it makes things simpler. – Goos May 24 '13 at 6:56
But you are correct that using $(x + y)^n \equiv x^n + y^n$ makes the proof more direct. – Goos May 24 '13 at 6:58
Hint: Recall binomial formula $$(x+y)^n=\sum\limits_{k=0}^n{n\choose k} x^{n-k} y^k$$
-
Yes, I know that. This is how I came to the above result. – Trancot May 24 '13 at 6:37
Ok do you know anything about Frobenius automorphism? – Norbert May 24 '13 at 6:40
I've used Frobenius in an elementary differential equations course. – Trancot May 24 '13 at 6:41
I might be able to manage if you can explain it clearly. – Trancot May 24 '13 at 6:42
bin-o-mail? That sounds like an automatic mail deleter! :-) – Asaf Karagila May 24 '13 at 6:43
In fact, Goos and Norbert have given the answer. (And you should also assume $n\in \mathbb{N}$)
$$f(n,x,y) = (x+y)^n - x^n -y^n$$
If
• both $x$ and $y$ are even: even - even - even = even;
• both $x$ and $y$ are odd: even - odd - odd = even;
• $x$ is even and $y$ is odd: odd - even - odd = even;
• $y$ is odd and $x$ is even: just like the case above.
So, $f(n,x,y)$ is always even.
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• $y$ is odd and $x$ is even: just like the case above.
So, $f(n,x,y)$ is always even.
-
Try setting $y=-x+a$ and expand the powers.
- | {
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# Finding a function, given its derivative.
At the start of day zero during the summer, the temperature is $y(0) = 15$ degrees Celsius. Over a 50 day period the temperature increases according to the rule $$y'(t) = {y(t) \over 50}$$. With time $t$ measured in days. Find the formula for $y$
I not sure how to start here. Can I integrate $y'(t)$ to get back $y(t)$ ? Would it have to be a definite integral from zero to fifty ? Or and indefinite? $$\int_{0}^{50} {1\over50} dy$$ ??
• Separate the variables then integrate – randomgirl Apr 21 '16 at 13:00
Let's rewrite $y'(t)$ as $\dfrac{dy}{dt}$ and just write $y(t)$ as $y$. Then we have: $$\frac{dy}{dt} = \frac{1}{50} y$$
Separate variables to get: $$\frac{dy}{y} = \frac{1}{50} \, dt$$
Integrate both sides: \begin{align} \int \frac{dy}{y} &= \int \frac{1}{50} \, dt\\[0.3cm] \ln |y| &= \frac{1}{50} t + C\\[0.3cm] |y| &= e^{(t/50) + C}\\[0.3cm] y &= Ce^{t/50} \end{align}
Now use the fact that $y(0) = 15$ to find $C$.
Hint. Note that: $$y'-\frac{1}{50}y=0\\ e^{-t/50}y'-\frac{1}{50}e^{-t/50}y=0\\ \big(e^{-t/50}y(t)\big)'=0\\ e^{-t/50}y(t)=\text{constant}$$
• The product rule for differentiation is probably the best way to solve this for someone who's not used to the tools of differential equations (and since the asker is stumped by this question, I would assume that that is the case). Although going from the second to third line by multiplying with $e^{-t/50}$ might seem like some sort of magic trick pulled up from a hat, that's what makes it work in the end. – Arthur Apr 21 '16 at 13:03
Like so: $\frac{dy}{y}=\frac{dt}{50}$ Now integrate both sides. | {
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+0
# Pleaze help
0
288
15
+48
The following cards are dealt to three people at random, so that everyone gets the same number of cards. What is the probability that everyone gets a red card?
[red 1] [red 2] [red 3] [yellow 1] [yellow 2] [yellow 3] [blue1] [blue 2] [blue 3]
Jan 17, 2022
#1
+2402
+2
Solution:
There are $$\large \dbinom{9}{3} = 84$$ ways to deal nine (9) cards to three (3) persons.
Give a red card to each person, then there are $$\large \dbinom{6}{3} = 20$$ ways to ways to deal the remaining six (6) cards to three (3) persons.
Probablity:
$$\Large \dfrac{\dbinom{6}{3}}{\dbinom{9}{3}} = \dfrac {5}{21} \approx 23.81\%$$
GA
--. .-
Jan 17, 2022
edited by GingerAle Jan 17, 2022
#2
+2403
+2
Nice solution, however I also tried doing this problem and got a different answer.
I viewed all the cards as different, so the number of ways to distribute the cards was 9!/3!/3!/3! = 1680.
I think 9C3 is the number of ways to give out 3 cards to one person, not all 3.
SImilar to what you did, I then assumed everyone recieved a red card, 6 ways of doing this.
The number of ways to distribute the remaining 6 cards was 6!/2!/2!/2! = 90.
90*6/1680 = 32.14%
=^._.^=
catmg Jan 17, 2022
#5
+2402
+1
My solution above is WRONG.
I’ve made train wrecks of these types of problems before.
I didn’t realize I’d wrecked the train this time until I read Catmg’s comment:
I think 9C3 is the number of ways to give out 3 cards to one person, not all 3.
That is true, and that means this question requires Hypergeometric distribution counts and NOT just Binomial distribution counts to correctly solve it.
The main difference between Hypergeometric distribution and Binomial distribution is that in Binomial distributions the samples sets are replaced before the next sample is drawn; in Hypergeometric distributions the samples are not replaced before the next sample is drawn. | {
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This stands to reason: while any of the Binomial sets nCr(9,3) can exist, once a set is given to a person, the number and quality of sets remaining is greatly limited. For example: if person one receives three blue cards then person two cannot receive two yellows and a blue because there are no blue cards remaining to give.
Here is a solution using Hypergeometric distribution:
The number of possible dealt sets is $$N=\dbinom{9}{3} * \dbinom{6}{3} * \dbinom{3}{3} = 1680 \\$$
For person one, there are $$\dbinom{3}{1}$$ ways to select one (1) of the three (3) red cards and then there are $$\dbinom{6}{2}$$ ways to choose two more (non red) cards.
For person two, there are $$\dbinom{2}{1}$$ ways to select one (1) of the two (2) remaining red cards and then there are $$\dbinom{4}{2}$$ ways to choose two more (non red) cards.
For person three, select the remaining red card and the two remaining (non red) cards. There is one (1) way to do this.
Then
$$n = 3 \dbinom{6}{2} \cdot 2\dbinom{4}{2} \hspace {1em} \small | \text{ Where n = the number of hands with a red card.}\\ \Large \rho_{\small \text{(3 persons with one red card)}} \normalsize = \dfrac {n} {N} = 3! \cdot \dfrac{\dbinom{6}{2}\dbinom{4}{2}}{\dbinom{9}{3}\dbinom{6}{3}} = \dfrac{9}{56} \approx 16.07\%\\$$
GA
--. .-
GingerAle Jan 19, 2022
edited by GingerAle Jan 19, 2022
#8
+2403
0
However, at the very end I got the answer 9/28 which is about 32.14%
6C2 = 15
4C2 = 6
9C3 = 84
6C3 = 20
3! = 6
(6*15*6)/(84*20) = (3*3*3)/(84) = 9/28
=^._.^=
catmg Jan 20, 2022
#9
+2402
+1
Here is a much easier way to solve this. The method does not use Hypergeometric selection
There are $$(3^3) = 27$$ arrangements of success. Divide the number of successes by the total number of sets giving $$(3^3) / (nCr(9,3) = \dfrac {9}{28} \approx 32.14\%$$
I’m glad you are paying attention to my presentations and train wrecks! | {
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I’m glad you are paying attention to my presentations and train wrecks!
If my train was loaded with toxic chemicals, I could wipe out a city. Or, in this case, at least send a student down the wrong track.
The calculation is exactly twice the value I presented for the Hypergeometric solution.
The formula is correct, but I must have made a mistake in the calculator input. I always check complex equations three times, but still, it escapes me...
I saved the input used for the calculation: (3!*nCr(6,2)*nCr(3,2))/((nCr(9,3)*nCr(6,3))
...I used a three (3) instead of a four (4) in the second binomial.
[My great uncle Cosmo was a locomotive engineer (and an electrical engineer), he would have flunked me for me for my train driving skills.]
-----------
Here’s a graphic, demonstrating the arrangements for the above solution.
$$\hspace {.3em}\left[ {\begin{array}{ccc} \scriptsize \hspace {.3em} P_1 & \scriptsize P_2 & \scriptsize P_3 \hspace {.2em} \\ \end{array} } \right] \small \hspace {.2em} \text {Persons Horizontal, sets Vertical. }\small \text{ Though identified by number, persons are indistinguishable. }\\ \left[ {\begin{array}{ccc} R & R & R \\ X & X & X \\ X & X & X \\ \end{array} } \right] \text {First arrangement of “R} \scriptsize{s} \normalsize \text {" where“X” is any other card}\\ \left[ {\begin{array}{ccc} R & R & X \\ X & X & R \\ X & X & X \\ \end{array} } \right] \text {Second arrangement} \\ \left[ {\begin{array}{ccc} R & R & X \\ X & X & X \\ X & X & R \\ \end{array} } \right] \text {Third arrangement} \\ \left[ {\begin{array}{ccc} R & X & R \\ X & R & X \\ X & X & X \\ \end{array} } \right] \text {Fourth arrangement} \\ \, \\ \textbf {. . .} \hspace {1em} \textbf {. . .} \hspace {1em} \textbf {. . .} \\ \, \\ \left[ {\begin{array}{ccc} X & X & X \\ X & X & X \\ R & R & R \\ \end{array} } \right] \text {27^{th} arrangement} \\$$
From this graphic, it’s easy to see the (3^3) = 27 arrangements of success.
GA
--. .- | {
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From this graphic, it’s easy to see the (3^3) = 27 arrangements of success.
GA
--. .-
GingerAle Jan 20, 2022
edited by GingerAle Jan 20, 2022
edited by GingerAle Jan 20, 2022
#11
+2403
0
That is a very cool way to visualize the problem.
One thing I love about math is how many different ways there are to solve the question. :))
=^._.^=
catmg Jan 20, 2022
#3
+48
+1
GingerAle and Catmg:
You have a denominator of 10x9x8x10 because that is the total amount of combinations.
The numerator is 6 because their are 3! ways to arange the 3 white balls.
<(-_-)>_It'sFree_>(-_-)<
Jan 17, 2022
#6
+2402
+1
WTFF!! Your solution may be free, but it’s also absurd and moronic. (Just because It’s Free doesn’t mean you should be smoking it!)
The only white balls I see at the moment are in a specimen jar of formalin solution on a shelf in a cabinet. They’ve been in the specimen jar for over 21 years, and they’ve not changed much over that time. The balls are kitty balls and were originally attached to my cat, known on the forum by his stage name, D.C. Thomas Copper. I sometimes called him the balless wonder. My cat was not amused by this, but my dog, Mr. Peabody, thought it quite funny. They were always busting each other’s balls. Mr. Peabody had the advantage though, because he knew where D.C. Thomas’ balls are.
D.C. Copper is now 22-years-old, still alive and somewhat active; although he mostly cat-naps. Occasionally, after his dinner, he reviews the forum and offers suggestions for potential ball-busting troll posts. ...Like this one
GA
--. .-
GingerAle Jan 19, 2022
edited by GingerAle Jan 19, 2022
edited by GingerAle Jan 19, 2022
#15
+48
+2
GA,
I am only 13 and don't smoke.
Interesting story though. :)
I am refering to Web2.0calc being free to use!
<(it'sfree)>
ItsFree Jan 21, 2022
#4
+117872
+2
Assume all are a bit different.
I will assume that ever person is unique. I will also (just for the working) assume every card is unique | {
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There are 9! ways to line up the cards but for each of those there are 3! ^3 doubling up 9!/(3!)^3 = 1680 ways total sample space
There are 3! ways to give the reds one to each person then 6!/2^3 = 90
90*3! = 540 ways for the desirable outcome
540/1680 = approx 32%
I don't know if it is right or not. My answer seems to be in line with Catmg's. :) It also seems to be a reasonable answer.
I would like to know why ItsFree is so certain their answer is correct though. It is a bold statement to make.
Jan 18, 2022
#7
+117872
+1
It seems that we need Alan to run a Montecarlo Simulation test here.
That would give a definitive outcome. :)
Jan 20, 2022
#10
+117872
+1
So are we all, well, me, Catmg and Ginger all in agreement that the prob is 9/28?
Melody Jan 20, 2022
#12
+33151
+2
I've just run a Monte-Carlo simulation with 20000 trials and get a probability of 0.32125 (this will vary slightly each time it's calculated because of the use of random numbers of course). This is close to 9/28 (≈0.32143).
Alan Jan 21, 2022
edited by Alan Jan 21, 2022
#13
+117872
+1
Thanks very much Alan, It is nice to get confirmation that our answers are most likely correct. :))
Melody Jan 21, 2022
#14
+48
+2
Thank you all for helping me out. :)
Thanks to Alan as well for doing the problem out. ;)
<(it's_free)>
Jan 21, 2022
edited by ItsFree Jan 21, 2022 | {
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# Two Urns Probability
Question: Suppose there is a room filled with urns of two types. Type I urns contain 5 blue balls and 5 red balls. Type II urns contain 2 red balls and 8 blue balls. There are 700 Type I urns and 300 Type II urns in the room. They are distributed randomly and look alike. An urn is selected at random from the room and a ball is drawn from it.
A) What is the probability that the urn is Type I?
So that will be: total type 1 urns/total urns $= 700/1000 = 0.7$
B) What is the probability that the ball drawn is red?
I'm confused with this part of the question. My answer is:
(type 1 red balls) $\times$ (type 2 red balls)
(5 red balls/10 total balls) $\times$ (2 red balls/10 total balls) $= 1/10$
Is $1/10$ the probability to draw a red ball correct?
• You might consider how many total balls are there and how many are red. – Cardinal Feb 24 '18 at 2:39
• Would you be able to show me how to set it up please? – Hardeep Chohan Harry Feb 24 '18 at 2:41
• I provided an answer below. – Cardinal Feb 24 '18 at 3:22
• Thanks for your answer really appreciated – Hardeep Chohan Harry Feb 24 '18 at 3:40
Type 1 Urn:
700 Total Urns (70%) 50% Chance Red
Type 2 Urn:
300 Total Urns (30%) 20% Chance Red
The Urns are indistinguishable, as are the balls. Therefore, when selecting a ball, there is a 70% chance the urn is Type 1 with a 50% chance of leading to a Red. There is a 30% chance the urn is Type 2 with a 20% chance of being a Red.
To find the total probability of drawing a red ball, you can take 70% $\times$ 50% $+$ 30% $\times$ 20%
$0.35+.06=0.41$ or 41%
700 Urns x 5 Red balls per urn = 3500 Red 300 Urns x 2 Red balls per urn = 600 Red
4100 Red 10,000 Total
$\frac{4100}{10000} = 0.41$ or 41%
Think of this as a weighted average.
The probability of a type I urn is $0.7$
The probability of then selecting a red ball is $\frac{1}{2}$
The probability of a type II urn is $0.3$
The probability of then selecting a red ball is $\frac{1}{5}$ | {
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The probability of then selecting a red ball is $\frac{1}{5}$
Thus, the desired probability is $$\left(0.7\cdot\frac{1}{2}\right)+\left(0.3\cdot\frac{1}{5}\right)=0.41$$ | {
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# Expected number of objects chosen by both A and B
Suppose that A and B each randomly, and independently, choose 3 of 10 objects. Find the expected number of objects chosen by both A and B.
solution: Let $$X$$ be the number of objects chosen by both A and B. For $$1≤i≤10$$, let $$X_i=1$$ if object $$i$$ is chosen by A and B, else $$0$$ otherwise
Then $$X=X_1+\ldots+X_{10}$$.
We find $$E[X_i]=0⋅P(X_i=0)+1⋅P(X_i=1)=P(X_i=1)=9/100$$.
By the linearity of expectation, $$E[X]=10⋅E[Xi]=0.9$$
I understand this approach but I can't get the same answer by multiplying the probability of A and B having 1,2,3 objects in common by the values 1,2,3 and adding them, why is this approach concerned wrong ? I assumed x is the number of objects chosen by both thus it takes the values 1,2,3
• What do you get for the probabilities of 1, 2, or 3 objects in common? – paw88789 Feb 22 at 18:06
• The alternate approach you suggest can be correct (though it is often considered far less elegant and is more prone to error or difficulty). The error must be in your calculations themselves, which you have not yet shared. Show us your calculations and we'll tell you where your mistake is. – JMoravitz Feb 22 at 18:09
• $\frac{1}{\binom{10}{3}*\binom{10}{3}}\left [ 1*\binom{10}{1}\binom{9}{2}\binom{9}{2} + 2*\binom{10}{2}\binom{8}{1}\binom{8}{1} + 3*\binom{10}{3} \right ]$ I think I'm over counting somewhere here – Saratcı Feb 22 at 18:56
• @Saratci : disallow selection of the same "uncommon" item. $$\dfrac{\dbinom{10}{1}\dbinom{9}{2}\dbinom{7}{2}+2\dbinom{10}{2}\dbinom 81\dbinom 71+3\dbinom{10}3}{\dbinom{10}3^2}=\dfrac 9{10}$$ – Graham Kemp Feb 23 at 3:53
• Yes you are right, Thanks ! – Saratcı Feb 23 at 22:15 | {
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Jacobi's equality between complementary minors of inverse matrices
What's a quick way to prove the following fact about minors of an invertible matrix $A$ and its inverse?
Let $A[I,J]$ denote the submatrix of an $n \times n$ matrix $A$ obtained by keeping only the rows indexed by $I$ and columns indexed by $J$. Then
$$|\det A[I,J]| = | (\det A) \det A^{-1}[J^c,I^c]|,$$ where $I^c$ stands for $[n] \setminus I$, for $|I| = |J|$. It is trivial when $|I| = |J| = 1$ or $n-1$. This is apparently proved by Jacobi, but I couldn't find a proof anywhere in books or online. Horn and Johnson listed this as one of the advanced formulas in their preliminary chapter, but didn't give a proof. In general what's a reliable source to find proofs of all these little facts? I ran into this question while reading Macdonald's book on symmetric functions and Hall polynomials, in particular page 22 where he is explaining the determinantal relation between the elementary symmetric functions $e_\lambda$ and the complete symmetric functions $h_\lambda$.
I also spent 3 hours trying to crack this nut, but can only show it for diagonal matrices :(
Edit: It looks like Ferrar's book on Algebra subtitled determinant, matrices and algebraic forms, might carry a proof of this in chapter 5. Though the book seems to have a sexist bias.
• Wow - I did not know that algebra proofs could have a "sexist bias". I am too curious to let it pass --- what do you mean exactly? – Federico Poloni Feb 8 '12 at 10:19
• I was just referring to the preface, where he said the book is suitable for undergraduate students, or boys in their last years of school. Maybe the word "boy" has a gender neutral meaning back then? – John Jiang Feb 8 '12 at 19:39 | {
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The key word under which you will find this result in modern books is "Schur complement". Here is a self-contained proof. Assume $I$ and $J$ are $(1,2,\dots,k)$ for some $k$ without loss of generality (you may reorder rows/columns). Let the matrix be $$M=\begin{bmatrix}A & B\\\\ C & D\end{bmatrix},$$ where the blocks $A$ and $D$ are square. Assume for now that $A$ is invertible --- you may treat the general case with a continuity argument. Let $S=D-CA^{-1}B$ be the so-called Schur complement of $A$ in $M$.
You may verify the following identity ("magic wand Schur complement formula") $$\begin{bmatrix}A & B\\\\ C & D\end{bmatrix} = \begin{bmatrix}I & 0\\\\ CA^{-1} & I\end{bmatrix} \begin{bmatrix}A & 0\\\\ 0 & S\end{bmatrix} \begin{bmatrix}I & A^{-1}B\\\\ 0 & I\end{bmatrix}. \tag{1}$$ By taking determinants, $$\det M=\det A \det S. \tag{2}$$ Moreover, if you invert term-by-term the above formula you can see that the (2,2) block of $M^{-1}$ is $S^{-1}$. So your thesis is now (2).
Note that the "magic formula" (1) can be derived via block Gaussian elimination and is much less magic than it looks at first sight.
• I guess for any thesis involving minors the Schur complement formula would be among the first things to try. – John Jiang Feb 8 '12 at 10:30
This is nothing but the Schur complement formula. See my book Matrices; Theory and Applications, 2nd ed., Springer-Verlag GTM 216, page 41.
Up to some permutation of rows and columns, we may assume that $I=J=[1,p]$. Let us write blockwise $$A=\begin{pmatrix} W & X \\\\ Y & Z \end{pmatrix}.$$ Assume WLOG that $W$ is invertible. On the one hand, we have (Schur C.F) $$\det A=\det W\cdot\det(Z-YW^{-1}X).$$ Finally, we have $$A^{-1}=\begin{pmatrix} \cdot & \cdot \\\\ \cdot & (Z-YW^{-1}X)^{-1} \end{pmatrix},$$which gives the desired result. | {
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These formulas are obtained by factorizing $A$ into $LU$ (namely, $L= \begin{pmatrix} I_* & 0 \\ YW^{-1} & I_* \end{pmatrix}$ and $U = \begin{pmatrix} W & X \\ 0 & Z-YW^{-1}X \end{pmatrix}$, with the $I_*$ being identity matrices of appropriate size).
• Your answer is equally valid. Thanks! – John Jiang Feb 8 '12 at 10:27
Not all has been said about this question that is worth saying -- at the very least, someone could have written down the version without the absolute values; but more importantly, there are various other equally good proofs.
Notations and statement
Let me first state the result with proper signs and no absolute values.
Standing assumptions. The following notations will be used throughout this post:
• Let $\mathbb{K}$ be a commutative ring. All matrices that appear in the following are matrices over $\mathbb{K}$.
• Let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\}$.
• For every $n\in\mathbb{N}$, we let $\left[ n\right]$ denote the set $\left\{ 1,2,\ldots,n\right\}$.
• Fix $n\in\mathbb{N}$.
• Let $S_{n}$ denote the $n$-th symmetric group (i.e., the group of permutations of $\left[ n\right]$).
• If $A\in\mathbb{K}^{n\times n}$ is an $n\times n$-matrix, and if $I$ and $J$ are two subsets of $\left[ n\right]$, then $A_{J}^{I}$ is the $\left\vert I\right\vert \times\left\vert J\right\vert$-matrix defined as follows: Write $A$ in the form $A=\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq m}$; write the set $I$ in the form $I=\left\{ i_{1}<i_{2}<\cdots<i_{u}\right\}$; write the set $J$ in the form $J=\left\{ j_{1}<j_{2}<\cdots<j_{v}\right\}$. Then, set $A_{J}^{I}=\left( a_{i_{x},j_{y}}\right) _{1\leq x\leq u,\ 1\leq y\leq v}$. (Thus, roughly speaking, $A_{J}^{I}$ is the $\left\vert I\right\vert \times\left\vert J\right\vert$-matrix obtained from $A$ by removing all rows whose indices do not belong to $I$, and removing all columns whose indices do not belong to $J$.)
If $K$ is a subset of $\left[ n\right]$, then: | {
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If $K$ is a subset of $\left[ n\right]$, then:
• we use $\widetilde{K}$ to denote the complement $\left[ n\right] \setminus K$ of this subset in $\left[ n\right]$.
• we use $\sum K$ to denote the sum of the elements of $K$.
Now, we claim the following:
Theorem 1 (Jacobi's complementary minor formula). Let $A\in\mathbb{K} ^{n\times n}$ be an invertible $n\times n$-matrix. Let $I$ and $J$ be two subsets of $\left[ n\right]$ such that $\left\vert I\right\vert =\left\vert J\right\vert$. Then,
$\det\left( A_{J}^{I}\right) =\left( -1\right) ^{\sum I+\sum J}\det A\cdot\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right)$.
Three references
Here are three references to proofs of Theorem 1:
Note that every source uses different notations. What I call $A_{J}^{I}$ above is called $A_{IJ}$ in the paper by Caracciolo, Sokal and Sportiello, is called $A\left[ I,J\right]$ in Lalonde's paper, and is called $\operatorname*{sub}\nolimits_{w\left( I\right) }^{w\left( J\right) }A$ in my notes. Also, the $I$ and $J$ in the paper by Caracciolo, Sokal and Sportiello correspond to the $\widetilde{I}$ and $\widetilde{J}$ in Theorem 1 above.
A fourth proof
Let me now give a fourth proof, using exterior algebra. The proof is probably not new (the method is definitely not), but I find it instructive.
This proof would become a lot shorter if I didn't care for the signs and would only prove the weaker claim that $\det\left( A_{J}^{I}\right) = \pm \det A\cdot\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right)$ for some value of $\pm$. But this weaker claim is not as useful as Theorem 1 in its full version (in particular, it would not suffice to fill the gap in Macdonald's book that has motivated this question).
The permutation $w\left( K\right)$
Let us first introduce some more notations: | {
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The permutation $w\left( K\right)$
Let us first introduce some more notations:
If $K$ is a subset of $\left[ n\right]$, and if $k = \left|K\right|$, then we let $w\left( K\right)$ be the (unique) permutation $\sigma\in S_{n}$ whose first $k$ values $\sigma\left( 1\right) ,\sigma\left( 2\right) ,\ldots,\sigma\left( k\right)$ are the elements of $K$ in increasing order, and whose next $n-k$ values $\sigma\left( k+1\right) ,\sigma\left( k+2\right) ,\ldots ,\sigma\left( n\right)$ are the elements of $\widetilde{K}$ in increasing order.
The first important property of $w\left( K\right)$ is the following fact:
Lemma 2. Let $K$ be a subset of $\left[ n\right]$. Then, $\left( -1\right) ^{w\left( K\right) }=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }$.
You don't need to prove Lemma 2 if you only care about the weaker version of Theorem 1 with the $\pm$ sign.
Proof of Lemma 2. Let $k=\left\vert K\right\vert$. Let $a_{1},a_{2} ,\ldots,a_{k}$ be the $k$ elements of $K$ in increasing order (with no repetitions). Let $b_{1},b_{2},\ldots,b_{n-k}$ be the $n-k$ elements of $\widetilde{K}$ in increasing order (with no repetitions). Let $\gamma =w\left( K\right)$. Then, the definition of $w\left( K\right)$ shows that the first $k$ values $\gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left( k\right)$ of $\gamma$ are the elements of $K$ in increasing order (that is, $a_{1},a_{2},\ldots,a_{k}$), and the next $n-k$ values $\gamma\left( k+1\right) ,\gamma\left( k+2\right) ,\ldots ,\gamma\left( n\right)$ of $\gamma$ are the elements of $\widetilde{K}$ in increasing order (that is, $b_{1},b_{2},\ldots,b_{n-k}$). In other words,
$\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots ,\gamma\left( n\right) \right) =\left( a_{1},a_{2},\ldots,a_{k} ,b_{1},b_{2},\ldots,b_{n-k}\right)$. | {
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Now, you can obtain the list $\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left( n\right) \right)$ from the list $\left( 1,2,\ldots,n\right)$ by successively switching adjacent entries, as follows:
• First, move the element $a_{1}$ to the front of the list, by successively switching it with each of the $a_{1}-1$ entries smaller than it.
• Then, move the element $a_{2}$ to the second position, by successively switching it with each of the $a_{2}-2$ entries (other than $a_{1}$) smaller than it.
• Then, move the element $a_{3}$ to the third position, by successively switching it with each of the $a_{3}-3$ entries (other than $a_{1}$ and $a_{2}$) smaller than it.
• And so on, until you finally move the element $a_{k}$ to the $k$-th position.
More formally, you are iterating over all $i\in\left\{ 1,2,\ldots,k\right\}$ (in increasing order), each time moving the element $a_{i}$ to the $i$-th position in the list, by successively switching it with each of the $a_{i}-i$ entries (other than $a_{1},a_{2},\ldots,a_{i-1}$) smaller than it.
At the end, the first $k$ positions of the list are filled with $a_{1} ,a_{2},\ldots,a_{k}$ (in this order), whereas the remaining $n-k$ positions are filled with the remaining entries $b_{1},b_{2},\ldots,b_{n-k}$ (again, in this order, because the switches have never disrupted their strictly-increasing relative order). Thus, at the end, your list is precisely $\left( a_{1},a_{2},\ldots,a_{k},b_{1},b_{2},\ldots,b_{n-k}\right) =\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left( n\right) \right)$. You have used a total of
$\left( a_{1}-1\right) +\left( a_{2}-2\right) +\cdots+\left( a_{k}-k\right)$
$=\underbrace{\left( a_{1}+a_{2}+\cdots+a_{k}\right) }_{\substack{=\sum K\\\text{(by the definition of }a_{1},a_{2},\ldots,a_{k}\text{)} }}-\underbrace{\left( 1+2+\cdots+k\right) }_{\substack{=1+2+\cdots +\left\vert K\right\vert \\\text{(since }k=\left\vert K\right\vert \text{)}}}$ | {
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$=\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right)$
switches. Thus, you have obtained the list $\left( \gamma\left( 1\right) ,\gamma\left( 2\right) ,\ldots,\gamma\left( n\right) \right)$ from the list $\left( 1,2,\ldots,n\right)$ by $\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right)$ switches of adjacent entries. In other words, the permutation $\gamma$ is a composition of $\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right)$ simple transpositions (where a "simple transposition" means a transposition switching $u$ with $u+1$ for some $u$). Hence, it has sign $\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }$. This proves Lemma 2.
Exterior algebras
Now, let's introduce some more notations and state some well-known properties concerning exterior algebras.
For any $\mathbb{K}$-module $V$, we let $\wedge V$ denote the exterior algebra of $V$. The multiplication in this exterior algebra will be written as juxtaposition (i.e., we will write $ab$ for the product of two elements $a$ and $b$ of $\wedge V$) or as multiplication (i.e., we will write $a\cdot b$ for this product).
If $k\in\mathbb{N}$ and if $V$ is a $\mathbb{K}$-module, then $\wedge^{k}V$ shall mean the $k$-th exterior power of $V$. If $k\in\mathbb{N}$, if $V$ and $W$ are two $\mathbb{K}$-modules, and if $f:V\rightarrow W$ is a $\mathbb{K}$-linear map, then the $\mathbb{K}$-linear map $\wedge^{k}V\rightarrow \wedge^{k}W$ canonically induced by $f$ will be denoted by $\wedge^{k}f$. It is well-known that if $V$ and $W$ are two $\mathbb{K}$-modules, if $f:V\rightarrow W$ is a $\mathbb{K}$-linear map, then
(1) $\left( \wedge^{k}f\right) \left( a\right) \cdot\left( \wedge^{\ell}f\right) \left( b\right) =\left( \wedge^{k+\ell}f\right) \left( ab\right)$
for any $k\in\mathbb{N}$, $\ell\in\mathbb{N}$, $a\in\wedge^{k}V$ and $b\in\wedge^{\ell}V$.
If $V$ is a $\mathbb{K}$-module, then
(2) $uv=\left( -1\right) ^{k\ell}vu$ | {
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If $V$ is a $\mathbb{K}$-module, then
(2) $uv=\left( -1\right) ^{k\ell}vu$
for any $k\in\mathbb{N}$, $\ell\in\mathbb{N}$, $u\in\wedge^{k}V$ and $v\in\wedge^{\ell}V$.
For any $u\in\mathbb{N}$, we consider $\mathbb{K}^{u}$ as the $\mathbb{K}$-module of column vectors with $u$ entries.
For any $u\in\mathbb{N}$ and $v\in\mathbb{N}$, and any $v\times u$-matrix $B\in\mathbb{K}^{v\times u}$, we define $f_{B}$ to be the $\mathbb{K}$-linear map $\mathbb{K}^{u}\rightarrow\mathbb{K}^{v}$ sending each $x\in\mathbb{K} ^{u}$ to $Bx\in\mathbb{K}^{v}$. This $\mathbb{K}$-linear map $f_{B}$ satisfies $\det\left( f_{B}\right) =\det B$, and is often identified with the matrix $B$ (though we will not identify it with $B$ here).
Here is another known fact:
Proposition 2a. Let $f:\mathbb{K}^{n}\rightarrow\mathbb{K}^{n}$ be a $\mathbb{K}$-linear map. The map $\wedge^{n}f:\wedge^{n}\left( \mathbb{K} ^{n}\right) \rightarrow\wedge^{n}\left( \mathbb{K}^{n}\right)$ is multiplication by $\det f$. In other words, every $z\in\wedge^{n}\left( \mathbb{K}^{n}\right)$ satisfies
(3) $\left( \wedge^{n}f\right) \left( z\right) =\left( \det f\right) z$.
Let $\left( e_{1},e_{2},\ldots,e_{n}\right)$ be the standard basis of the $\mathbb{K}$-module $\mathbb{K}^{n}$. (Thus, $e_{i}$ is the column vector whose $i$-th entry is $1$ and whose all other entries are $0$.)
For every subset $K$ of $\left[ n\right]$, we define $e_{K}\in \wedge^{\left\vert K\right\vert }\left( \mathbb{K}^{n}\right)$ to be the element $e_{k_{1}}\wedge e_{k_{2}}\wedge\cdots\wedge e_{k_{\left\vert K\right\vert }}$, where $K$ is written in the form $K=\left\{ k_{1} <k_{2}<\cdots<k_{\left\vert K\right\vert }\right\}$.
For every $k\in\mathbb{N}$ and every set $S$, we let $\mathcal{P}_{k}\left( S \right)$ denote the set of all $k$-element subsets of $S$. | {
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It is well-known that, for every $k\in\mathbb{N}$, the family $\left( e_{K}\right) _{K\in\mathcal{P}_{k}\left( \left[ n\right] \right) }$ is a basis of the $\mathbb{K}$-module $\wedge^{k}\left( \mathbb{K}^{n}\right)$. Applying this to $k=n$, we conclude that the family $\left( e_{K}\right) _{K\in\mathcal{P}_{n}\left( \left[ n\right] \right) }$ is a basis of the $\mathbb{K}$-module $\wedge^{n}\left( \mathbb{K}^{n}\right)$. Since this family $\left( e_{K}\right) _{K\in\mathcal{P}_{n}\left( \left[ n\right] \right) }$ is the one-element family $\left( e_{\left[ n\right] }\right)$ (because the only $K\in\mathcal{P}_{n}\left( \left[ n\right] \right)$ is the set $\left[ n\right]$), this rewrites as follows: The one-element family $\left( e_{\left[ n\right] }\right)$ is a basis of the $\mathbb{K}$-module $\wedge^{n}\left( \mathbb{K}^{n}\right)$.
If $B$ is an $n\times n$-matrix and $k\in\mathbb{N}$, then evaluating the map $\wedge^{k}f_{B}$ on the elements of the basis $\left( e_{K}\right) _{K\in\mathcal{P}_{k}\left( \left[ n\right] \right) }$ of $\wedge ^{k}\left( \mathbb{K}^{n}\right)$, and expanding the results again in this basis gives rise to coefficients which are the $k\times k$-minors of $B$. More precisely:
Proposition 3. Let $B\in\mathbb{K}^{n\times n}$, $k\in\mathbb{N}$ and $J\in\mathcal{P}_{k}\left( \left[ n\right] \right)$. Then,
$\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) = \sum\limits_{I\in\mathcal{P}_{k}\left( \left[ n\right] \right) }\det\left( B_{J} ^{I}\right) e_{I}$. | {
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(This generalizes: If $u\in\mathbb{N}$, $v \in \mathbb{N}$, $B\in\mathbb{K}^{u\times v}$, $k\in\mathbb{N}$ and $J\in\mathcal{P}_{k}\left( \left[ v\right] \right)$, then $\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) = \sum\limits_{I\in\mathcal{P}_{k}\left( \left[ u\right] \right) }\det\left( B_{J} ^{I}\right) e_{I}$, where the elements $e_{J}\in\wedge^{k}\left( \mathbb{K}^{v}\right)$ and $e_{I}\in\wedge^{k}\left( \mathbb{K}^{u}\right)$ are defined as before but with $v$ and $u$ instead of $n$.)
Extracting minors from the exterior algebra
Now, we shall need a simple lemma:
Lemma 4. Let $K$ be a subset of $\left[ n\right]$. Then,
$e_{K}e_{\widetilde{K}}=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }e_{\left[ n\right] }$.
Proof of Lemma 4. Let $k = \left|K\right|$. Let $\sigma$ be the permutation $w\left( K\right) \in S_{n}$ defined above. Its first $k$ values $\sigma\left( 1\right) ,\sigma\left( 2\right) ,\ldots,\sigma\left( k\right)$ are the elements of $K$ in increasing order; thus, $e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge\cdots\wedge e_{\sigma\left( k\right) }=e_{K}$. Its next $n-k$ values $\sigma\left( k+1\right) ,\sigma\left( k+2\right) ,\ldots,\sigma\left( n\right)$ are the elements of $\widetilde{K}$ in increasing order; thus, $e_{\sigma\left( k+1\right) }\wedge e_{\sigma\left( k+2\right) }\wedge\cdots\wedge e_{\sigma\left( n\right) }=e_{\widetilde{K}}$.
From $\sigma=w\left( K\right)$, we obtain $\left( -1\right) ^{\sigma }=\left( -1\right) ^{w\left( K\right) }=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }$ (by Lemma 2).
Now, it is well-known that
$e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge \cdots\wedge e_{\sigma\left( n\right) }=\left( -1\right) ^{\sigma }\underbrace{e_{1}\wedge e_{2}\wedge\cdots\wedge e_{n}}_{=e_{\left[ n\right] }}=\left( -1\right) ^{\sigma}e_{\left[ n\right] }$.
Hence, | {
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Hence,
$\left( -1\right) ^{\sigma}e_{\left[ n\right] }=e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge\cdots\wedge e_{\sigma\left( n\right) }$
$=\underbrace{\left( e_{\sigma\left( 1\right) }\wedge e_{\sigma\left( 2\right) }\wedge\cdots\wedge e_{\sigma\left( k\right) }\right) }_{=e_{K} }\underbrace{\left( e_{\sigma\left( k+1\right) }\wedge e_{\sigma\left( k+2\right) }\wedge\cdots\wedge e_{\sigma\left( n\right) }\right) }_{=e_{\widetilde{K}}}=e_{K}e_{\widetilde{K}}$.
Since $\left( -1\right) ^{\sigma}=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }$, this rewrites as $\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }e_{\left[ n\right] }= e_K e_{\widetilde{K}}$. This proves Lemma 4.
We can combine Proposition 3 and Lemma 4 to obtain the following fact:
Corollary 5. Let $B\in\mathbb{K}^{n\times n}$, $k\in\mathbb{N}$ and $J\in\mathcal{P}_{k}\left( \left[ n\right] \right)$. Then, every $K\in\mathcal{P}_{k}\left( \left[ n\right] \right)$ satisfies
$\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) e_{\widetilde{K} }=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+k\right) }\det\left( B_{J}^{K}\right) e_{\left[ n\right] }$.
Proof of Corollary 5. Let $K \in \mathcal{P}_{k}\left( \left[ n\right] \right)$. Let $I\in\mathcal{P}_{k}\left( \left[ n\right] \right)$ be such that $I\neq K$. Then, $I\not \subseteq K$ (since the sets $I$ and $K$ have the same size $k$). Hence, there exists some $z\in I$ such that $z\notin K$. Consider this $z$. We have $z\in I$ and $z\in\widetilde{K}$ (since $z\notin K$). Hence, both $e_{I}$ and $e_{\widetilde{K}}$ are "wedge products" containing the factor $e_{z}$; therefore, the product $e_{I} e_{\widetilde{K}}$ is a "wedge product" containing this factor twice. Thus, $e_{I}e_{\widetilde{K}}=0$.
Now, forget that we fixed $I$. We thus have proven that
$e_{I}e_{\widetilde{K}}=0$ for every $I\in\mathcal{P}_{k}\left( \left[ n\right] \right)$ satisfying $I\neq K$.
Hence, | {
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Hence,
(4) $\sum\limits_{\substack{I\in\mathcal{P}_{k}\left( \left[ n\right] \right) ;\\I\neq K}}\det\left( B_{J}^{I}\right) \underbrace{e_{I}e_{\widetilde{K}} }_{=0}=0$.
Proposition 3 yields
$\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) =\sum\limits_{I\in \mathcal{P}_{k}\left( \left[ n\right] \right) }\det\left( B_{J} ^{I}\right) e_{I}$.
Multiplying both sides of this equality by $e_{\widetilde{K}}$ from the right, we find
$\left( \wedge^{k}f_{B}\right) \left( e_{J}\right) e_{\widetilde{K}} =\sum\limits_{I\in\mathcal{P}_{k}\left( \left[ n\right] \right) }\det\left( B_{J}^{I}\right) e_{I}e_{\widetilde{K}}$
$=\det\left( B_{J}^{K}\right) e_{K}e_{\widetilde{K}}+\sum\limits_{\substack{I\in \mathcal{P}_{k}\left( \left[ n\right] \right) ;\\I\neq K}}\det\left( B_{J}^{I}\right) e_{I}e_{\widetilde{K}}$
$=\det\left( B_{J}^{K}\right) \underbrace{e_{K}e_{\widetilde{K}} }_{\substack{=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }e_{\left[ n\right] }\\\text{(by Lemma 4)}}}$ (by (4))
$=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+\left\vert K\right\vert \right) }\det\left( B_{J}^{K}\right) e_{\left[ n\right] }$
$=\left( -1\right) ^{\sum K-\left( 1+2+\cdots+k\right) }\det\left( B_{J}^{K}\right) e_{\left[ n\right] }$
(since $\left\vert K\right\vert =k$). This proves Corollary 5.
Corollary 5 is rather helpful when it comes to extracting a specific minor of a matrix $B$ from the maps $\wedge^{k}f_{B}$.
Proof of Theorem 1
Proof of Theorem 1. Set $k=\left\vert I\right\vert =\left\vert J\right\vert$. Notice that $\left\vert \widetilde{I}\right\vert =n-k$ (since $\left\vert I\right\vert =k$) and $\left\vert \widetilde{J}\right\vert =n-k$ (similarly).
Define $y\in\wedge ^{n-k}\left( \mathbb{K}^{n}\right)$ by $y=\left( \wedge^{n-k}f_{A^{-1} }\right) \left( e_{\widetilde{I}}\right)$. | {
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The maps $f_{A}$ and $f_{A^{-1}}$ are mutually inverse (since the map $\mathbb{K}^{n\times n}\rightarrow\operatorname*{End}\left( \mathbb{K} ^{n}\right) ,\ B\mapsto f_{B}$ is a ring homomorphism). Hence, the maps $\wedge^{n-k}f_{A}$ and $\wedge^{n-k}f_{A^{-1}}$ are mutually inverse (since $\wedge^{n-k}$ is a functor). Thus, $\left( \wedge^{n-k}f_{A}\right) \circ\left( \wedge^{n-k}f_{A^{-1}}\right) =\operatorname*{id}$. Now, $y=\left( \wedge^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right)$, so that
$\left( \wedge^{n-k}f_{A}\right) \left( y\right) =\left( \wedge ^{n-k}f_{A}\right) \left( \left( \wedge^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right) \right) =\underbrace{\left( \left( \wedge ^{n-k}f_{A}\right) \circ\left( \wedge^{n-k}f_{A^{-1}}\right) \right) }_{=\operatorname*{id}}\left( e_{\widetilde{I}}\right) =e_{\widetilde{I}}$.
But (1) (applied to $V=\mathbb{K}^{n}$, $W=\mathbb{K}^{n}$, $f=f_{A}$, $\ell=n-k$, $a=e_{J}$ and $b=y$) yields
$\left( \wedge^{k}f_{A}\right) \left( e_{J}\right) \cdot\left( \wedge^{n-k}f_{A}\right) \left( y\right) =\left( \wedge^{n}f_{A}\right) \left( e_{J}y\right)$.
Thus,
$\left( \wedge^{n}f_{A}\right) \left( e_{J}y\right) =\left( \wedge ^{k}f_{A}\right) \left( e_{J}\right) \cdot\underbrace{\left( \wedge ^{n-k}f_{A}\right) \left( y\right) }_{=e_{\widetilde{I}}}$
$=\left( \wedge^{k}f_{A}\right) \left( e_{J}\right) e_{\widetilde{I}} = \left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left( A_{J}^{I}\right) e_{\left[ n\right] }$
(by Corollary 5, applied to $B=A$ and $K=I$).
Hence,
$\left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left( A_{J}^{I}\right) e_{\left[ n\right] }$
$=\left( \wedge^{n}f_{A}\right) \left( e_{J}y\right) =\underbrace{\left( \det f_{A}\right) }_{=\det A}e_{J}y$
(by (3), applied to $f=f_{A}$ and $z=e_{J}y$)
(5) $=\left( \det A\right) e_{J}y$.
But (2) (applied to $\ell=n-k$, $u=e_{J}$ and $v=y$) yields | {
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(5) $=\left( \det A\right) e_{J}y$.
But (2) (applied to $\ell=n-k$, $u=e_{J}$ and $v=y$) yields
$e_{J} y = \left(-1\right)^{k \left(n-k\right)} \underbrace{y}_{=\left( \wedge^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right) } \underbrace{e_{J}} _{=e_{\widetilde{\widetilde{J}}}}$
$=\left( -1\right) ^{k\left( n-k\right) }\underbrace{\left( \wedge ^{n-k}f_{A^{-1}}\right) \left( e_{\widetilde{I}}\right) e_{\widetilde{\widetilde{J}}}}_{\substack{=\left( -1\right) ^{\sum \widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right] }\\\text{(by Corollary 5, applied to }A^{-1}\text{, }n-k\text{, }\widetilde{I}\text{ and }\widetilde{J}\\\text{instead of }B\text{, }k\text{, }J\text{ and }K\text{)}}}$
$=\left( -1\right) ^{k\left( n-k\right) }\left( -1\right) ^{\sum \widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right] }$
(6) $=\left( -1\right) ^{k\left( n-k\right) +\sum\widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right] }$.
But
$k\left( n-k\right) +\underbrace{\sum\widetilde{J}}_{=\sum\left\{ 1,2,\ldots,n\right\} -\sum J}-\underbrace{\left( 1+2+\cdots+\left( n-k\right) \right) }_{=\sum\left\{ 1,2,\ldots,n-k\right\} }$
$=k\left( n-k\right) +\sum\left\{ 1,2,\ldots,n\right\} -\sum J-\sum\left\{ 1,2,\ldots,n-k\right\}$
$=\underbrace{\sum\left\{ 1,2,\ldots,n\right\} -\sum\left\{ 1,2,\ldots ,n-k\right\} }_{\substack{=\sum\left\{ n-k+1,n-k+2,\ldots,n\right\} \\=k\left( n-k\right) +\sum\left\{ 1,2,\ldots,k\right\} \\=k\left( n-k\right) +\left( 1+2+\cdots+k\right) }}+k\left( n-k\right) -\sum J$
$=2k\left( n-k\right) +\left( 1+2+\cdots+k\right) -\sum J$
$\equiv-\left( 1+2+\cdots+k\right) -\sum J\operatorname{mod}2$.
Hence, | {
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$\equiv-\left( 1+2+\cdots+k\right) -\sum J\operatorname{mod}2$.
Hence,
$\left( -1\right) ^{k\left( n-k\right) +\sum\widetilde{J}-\left( 1+2+\cdots+\left( n-k\right) \right) }=\left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}$.
Thus, (6) rewrites as
$e_{J}y=\left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right) e_{\left[ n\right] }$.
Hence, (5) rewrites as
$\left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left( A_{J}^{I}\right) e_{\left[ n\right] }$
$=\left( \det A\right) \left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right) e_{\left[ n\right] }$.
We can "cancel" $e_{\left[ n\right] }$ from this equality (because if $\lambda$ and $\mu$ are two elements of $\mathbb{K}$ satisfying $\lambda e_{\left[ n\right] }=\mu e_{\left[ n\right] }$, then $\lambda=\mu$), and thus obtain
$\left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }\det\left( A_{J}^{I}\right)$
$=\left( \det A\right) \left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J} }\right)$.
Dividing this equality by $\left( -1\right) ^{\sum I-\left( 1+2+\cdots +k\right) }$, we obtain
$\det\left( A_{J}^{I}\right)$
$=\left(\det A\right) \dfrac{\left( -1\right) ^{-\left( 1+2+\cdots+k\right) -\sum J}}{\left( -1\right) ^{\sum I-\left( 1+2+\cdots+k\right) }}\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right)$
$=\left( -1\right) ^{\sum I+\sum J}\det A\cdot\det\left( \left( A^{-1}\right) _{\widetilde{I}}^{\widetilde{J}}\right)$.
This proves Theorem 1.
I have to admit this proof looked far shorter on the scratch paper on which it was conceived than it has ended up here... | {
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• If we agree to see a $\mathbb{K}$-matrix as a function $S\times S\to\mathbb{K}$ , with any finite index set $S$ (not just some $[m]$), then, according to Proposition 3, the matrix of the linear map $\Lambda^k(f_B)$ in the basis $\{e_I\}_{I\in\mathcal{P}_k([n])}$ is $[\det(B^I_J)]_{(I\times J)\in\mathcal{P}_k([n])\times \mathcal{P}_k([n])}$. – Pietro Majer Dec 19 '17 at 18:48
• So the functoriality of $\Lambda^k$ reads as the (generalized) Cauchy-Binet formula: en.wikipedia.org/wiki/Cauchy–Binet_formula#Generalization . In the same spirit, I'd like to see Theorem 1 as a statement on the inverse matrix of the linear map $\Lambda^k(f_A)$, or better, since it also refers to $\tilde I$ and $\tilde J$, of the linear map $\Lambda^k(f_A)^*\sim\Lambda^{n-k}(f_A)$. Can this produce a proof to Theorem 1 consisting in a plain translation of facts about exterior algebra in the language of matrices? – Pietro Majer Dec 19 '17 at 18:48
• @PietroMajer: This is what I was trying to achieve when I started writing the proof. Unfortunately, it's not directly clear to me how this works. – darij grinberg Dec 19 '17 at 19:54
Here is a short proof inspired by darij grinberg's answer and not using Schur complements (in particular, no invertibility of sub-matrices or continuity assumptions are needed).
WLOG let $I=J=[k]$. Let $A_i$ denote the $i$th column of a matrix $A$. Consider the product $$A \left[ \begin{array}{c|c|c|c|c|c} e_1 & \cdots & e_k & A^{-1}_{k+1} & \cdots & A^{-1}_n \end{array} \right] = \left[ \begin{array}{c|c|c|c|c|c} A_1 & \cdots & A_k & e_{k+1} & \cdots & e_n \end{array} \right]$$ which can also be written $$A \begin{pmatrix} I_k & A^{-1}[J,I^c] \\ 0 & A^{-1}[J^c,I^c] \end{pmatrix} = \begin{pmatrix} A[I,J] & 0 \\ A[I^c,J] & I_{n-k} \end{pmatrix}$$ Taking the determinant of both sides yields $$(\det A) (\det A^{-1}[J^c,I^c]) = \det A[I,J]$$. | {
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• Really nice... although a nontrivial part of my answer is justifying the "WLOG" in the first sentence. Still, this simplifies a lot. – darij grinberg Feb 17 '17 at 1:19
• I believe the justification is not too bad. Assume the result is true for I=K, J=K, K=[k]. Then for any other I,J of cardinality k, we have $\det(A[I,J])=\det((PAQ)[K,K])=\det(PAQ)\det(Q^{-1}A^{-1}P^{-1}[K^c,K^c]) = \det(PAQ)\det(A^{-1}[J^c,I^c])$, where $P$ is the permutation matrix taking $I$ to $1,\dots,k$ and $I^c$ to $k+1,\dots,n$ (I believe you call this permutation $w(I)$), and $Q$ is defined similarly. So we incur a factor of $\text{sign}(P)\text{sign}(Q)$ which is $(-1)^{\sum I + \sum J}$ by a counting argument, as you proved. – user1980685 Feb 17 '17 at 15:05
• I suspect that your $Q$ is not "defined similarly" but rather is the inverse of the matrix defined similarly. But yes, this is essentially how it's done. – darij grinberg Feb 18 '17 at 5:42 | {
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"url": "https://mathoverflow.net/questions/87877/jacobis-equality-between-complementary-minors-of-inverse-matrices"
} |
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