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# Let $a$ and $b$ commute. If $m$ and $n$ are relatively prime, then ord($ab$) = $mn$. [duplicate]
This is exercise $$10.E.4$$ from Pinter:
Let $$a$$ and $$b$$ be elements of a group $$G$$.
Let ord($$a$$) = $$m$$ and ord($$b$$) = $$n$$.
Prove:
Let $$a$$ and $$b$$ commute.
If $$m$$ and $$n$$ are relatively prime, then ord($$ab$$) = $$mn$$.
(HINT: Use $$10.E.2$$.)
Here is $$10.E.2$$, which Pinter suggests that we use:
If $$m$$ and $$n$$ are relatively prime, then no power of $$a$$ can be equal to any power of $$b$$ (except for $$e$$).
I'm also going to use $$10.E.1$$:
If $$a$$ and $$b$$ commute, then ord($$ab$$) is a divisor of lcm($$m$$,$$n$$).
As well as $$B.T6.i$$ (Theorem $$6.i$$ from Appendix $$B$$. A fact from basic number theory.):
If gcd($$m$$,$$n$$) = 1 then lcm($$m$$,$$n$$) = $$mn$$
Let's begin.
We are given that $$m$$ and $$n$$ are relatively prime which means that:
$$\text{gcd}(m,n) = 1$$
By $$B.T6.i$$:
$$\text{lcm}(m,n) = mn \tag{1}$$
By $$10.E.1$$:
$$\text{ord}(ab)\ |\ \text{lcm}(m,n)$$
Substituting $$(1)$$:
$$\text{ord}(ab)\ |\ mn$$
Which means that there is an integer $$x$$ such that:
$$\text{ord}(ab) x = mn$$
This is so close! For the theorem to be true, we'd have to show that $$x = 1$$.
However, the original exercise statement says to use $$10.E.2$$.
• Is that helpful in showing that $$x = 1$$?
• Or is there some other completely different approach whereby $$10.E.2$$ is used?
The proof uses the following fact:
If $$a|c$$ and $$b|c$$ then $$lcm(a,b)|c$$.
For this exercise, I wanted to only use theorems that had been presented in the book up to that point. And, I didn't seem to recall seeing this theorem. (If anyone spots this in Pinter, please comment below with the location.)
However, I did notice that the following similar fact is in Appendix B (REVIEW OF THE INTEGERS) as exercise B.9:
If $$a|c$$ and $$b|c$$ and $$gcd(a,b) = 1$$ then $$ab|c$$. | {
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If $$a|c$$ and $$b|c$$ and $$gcd(a,b) = 1$$ then $$ab|c$$.
So yeah, it looks like the approach shown below is definitely a valid way to go if you want to stick to what's presented in the book.
• – Angina Seng Jul 26 '19 at 4:35
• – Dietrich Burde Jul 26 '19 at 11:36
• The 2nd $\Rightarrow$ in my answer in the dupe is essentially a proof of 10.E.2, so replace that by an invocation of 10.E.2 to obtain the hinted proof. This is a well-known proof and likely occurs here many times. – Bill Dubuque Jul 26 '19 at 14:47
### $$\boxed{\textit{You can show mn\ \big|\ \text{ord}(ab) using 10.E.2:}}$$
Since $$a$$ and $$b$$ commute we can distribute common powers of $$a$$ and $$b$$: $$e = (ab)^{\text{ord}(ab)} = a^{\text{ord}(ab)}b^{\text{ord}(ab)}$$ This implies $$a^{\text{ord}(ab)} = b^{-\text{ord}(ab)}$$ or, in other words, some power of $$a$$ equals some power of $$b$$. But $$10.E.2$$ says that since $$m = \text{ord}(a)$$ and $$n = \text{ord}(b)$$ are relatively prime, no power of $$a$$ can equal any power of $$b$$ unless those powers of $$a$$ and $$b$$ both evaluate to $$e$$.
So we must have $$a^{\text{ord}(ab)} = e = a^m$$ and $$b^{-\text{ord}(ab)} = e = b^n$$.
But then, $$m\ \big|\ \text{ord}(ab)$$. This is because $$m$$ is the order of $$a$$ i.e. the least positive integer power of $$a$$ that evaluates to $$e$$; so if any other positive integer power of $$a$$ evaluates to $$e$$ (like $$\text{ord}(ab)$$ here), then $$m$$ must divide that integer. For a similar reason, $$n\ \big|\ \text{ord}(ab)$$.
Therefore since both $$m$$ and $$n$$ divide $$\text{ord}(ab)$$, their least common multiple $$\text{lcm}(m,n)$$ must divide $$\text{ord}(ab)$$. But in our case $$\text{lcm}(m,n) = mn$$ so we finally have $$mn\ \big|\ \text{ord}(ab)$$.
Along with the result you already established $$\text{ord}(ab)\ \big|\ mn$$, this suggests $$\text{ord}(ab) = mn$$. | {
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P.S. If you don't believe the bolded statements above, use Euclidean Division. E.g. to see $$m\ \big|\ \text{ord}(ab)$$, use Euclidean Division on the integer pair $$\big(m, \text{ord}(ab)\big)$$ to get $$\text{ord}(ab) = km + r$$ for some $$k, r \in \Bbb Z$$ with $$0 \leq r < m$$. Do you see why $$r$$ must be $$0$$?
• Excellent and very clear presentation! I learned a few things from your proof. Thanks very much for explaining each part. I've updated my answer with some comments regarding your approach. – dharmatech Jul 26 '19 at 20:05 | {
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Complex analysis: residue integration question
Tags:
1. Nov 9, 2014
Wheelwalker
I'm asked to evaluate the following integral: $\int_{c} \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}dz$ where c is the unit circle. This function has a simple pole at $z=\frac{1}{3}$ and a second order pole at $z=\frac{1}{2}$, both of which are within my region of integration. I then went about computing the residues using the following formula (which I, mistakenly perhaps, thought was general and applied to any mth-order pole): $Res f(z) = \frac{1}{(1-1)!}\lim_{z \to 1/2}{\frac{d}{dz}\{[2z-1]^2 \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}}\}$ for which I got 2 (this was just the residue for the first pole). The residue I got for the other singularity was 6. They are supposed to be 1/2 and 2, respectively. What am I doing wrong, exactly? Does this residue formula not apply to this case? If so, why not? Thanks in advance.
Last edited: Nov 9, 2014
2. Nov 9, 2014
lurflurf
The formula you should have used is $$Res f(z) = \frac{1}{(2-1)!}\lim_{z \to 1/2}{\frac{d}{dz}\{[z-\dfrac{1}{2}]^2 \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}}\}$$
That is why you got 2=4(1/2) instead of 1/2 and 6=3*2 instead of 2
3. Nov 9, 2014
Wheelwalker
Oh, wow that makes a lot of sense. I got in the habit of multiplying $f(z_0)$ by whichever term was creating the singularity in the denominator because a lot of the first problems we did would have something simple like $(z-1)$ there. So then, $(z-z_0)$ would equal exactly what was there. Anyway, thank you so much!
4. Nov 9, 2014
Wheelwalker
Ah, and yes that factorial term definitely should have been $(2-1)!$, although of course it doesn't matter in this case.
5. Nov 9, 2014
Wheelwalker
One more question for you actually. So, is this the most general residue formula? This should apply to nearly any case?
6. Nov 10, 2014
mathwonk
I am not very good at computations, but I am getting residues opposite to what you seem to say, i.e. res = 1/2 at z=1/2, and res = 2 at z= 1/3. | {
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My reasoning is as follows: the residue at a, is the coefficient of 1/(z-a) in the Laurent expansion in powers of (z-a). So at z= 1/3 we should have 1/(z-1/3) . g(z), where g(z) = (30z^2 -23z + 5)/3(2z-1)^2. Then when we expand g in powers of (z-1/3), the constant term will be the residue, i.e. the coefficient of 1/(z-1/3) after we multiply by the 1/(z-1/3) out front. so this is just g(1/3) = (30/9 - 23/3 + 5)/3(1/3)^2 = 2.
Then for the other one, at z= 1/2, the function can be written as 1/(z-1/2)^2 h(z), where h(z) = (30z^2-23z+5)/4(3z-1). then the residue after multiplying by the factor 1/(z-1/2)^2, will be the linear coefficient of the expansion of h(z), i.e. should be the derivative of h(z). By the quotient rule, this is (1/4)(7/2 - 3)/(1/4) = (1/8)/(1/4) = 1/2.
Am I missing something? My point in doing it this way is to emphasize not memorizing formulas. but going back to the meaning of the concept. But of course one wants to get it right, and I am not super confident after apparently getting a different answer. Or maybe I just misunderstood what was said.
7. Nov 10, 2014
Wheelwalker
Those are the correct residues. I made a mistake in finding them the first time.
8. Nov 10, 2014
mathwonk
Thank you. My recommended point of view, as I said, is to write the function as a product of form 1/(z-a)^m . g(z), where g is holomorphic at a.
Then we want the coefficient of (z-a)^(m-1) in the Taylor expansion of g. this is of course just g^(m-1)(a) /(m-1)!, i.e. the (m-1) st derivative of g at a, divided by (m-1)!
I suggest this because your error was apparently in recalling the residue formula incorrectly. I hope this method makes that less likely, since all you have to recall this way is that you want the coefficient of 1/(z-a). In the product 1/(z-a)^m . g(z), that will be the coefficient of (z-a)^(m-1) in the expansion of g. | {
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# Absolute value inequality with variable on both sides
I am trying to solve the following inequality:
$$|3-5x| \le x$$
I am not familiar with inequalities including one absolute value with variables on both sides. I tried to solve it as follows:
$$-x \le 3-5x \le x$$
Then I solved for each side separately,as follows:
$$3-5x \le x$$
$$x \ge (1/2)$$
$$-x \le 3-5x$$
$$x \le \frac34$$
I know my solution is incorrect and that it actually lies between $\frac12$ and $\frac34$, but I wanted to know what is wrong with my method and what is the appropriate approach to solving such inequalities.
Thanks,
This is how I learned it back in Algebra I.
You can split it into two equations. Since the sign is $\le$, it will be that Equation 1 and Equation 2 are true.
Here are the steps:
$$|3-5x| \le x$$
$$3-5x \le x \;\;\;\;\;| \;\;-3+5x\le x$$
$$6x \ge 3 \;\;\;\;\;\;\;| \;\;-4x \ge -3$$
$$x \ge \frac 12 \;\;\;| \;\;\;\;x \le \frac 34$$
$$\frac 12 \le x \le \frac 34$$
• This should be sufficient, but for further elaboration just reply Sep 8 '16 at 2:38
• This approach is very specific towards linear inequalities (but definitely works, I think)--it doesn't work (I believe) for higher order ones. Sep 8 '16 at 3:53
• Yes, I know. I was trying to give a simple easy-to-understand solution. Sep 8 '16 at 4:02
• Your "$|$" symbol is for AND or OR? In either case you have a logical issue as you start of with an OR and end with and AND. Sep 8 '16 at 4:15
• @Maccavity And. And please point out the issue. Sep 8 '16 at 4:16
Divide $-x \leq 3-5x \leq x$ by $x$ and go from there. Although I think what you have is correct, the solution IS the interval $\dfrac{1}{2} \leq x \leq \dfrac{3}{4}$.
You need to break up the absolute value into its intervals:
$$|x| = \begin{cases} x & x > 0 \\ -x & x < 0 \\ 0 & x = 0 \end{cases}$$
Therefore for $|3-5x|$ you need to find the interval when it's less than zero and when it's greater than zero: | {
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$$|3-5x| = \begin{cases} 3 - 5x & 3 - 5x > 0 \rightarrow 3 > 5x \rightarrow x < \frac{3}{5}\\ 5x - 3 & 3 - 5x < 0 \rightarrow 3 <5x \rightarrow x > \frac{3}{5} \\ 0 & 3 - 5x = 0 \rightarrow 3 = 5x \rightarrow x = \frac{3}{5} \end{cases}$$
Now you solve the inequality in each case:
1. $x < \frac{3}{5} \rightarrow |3 - 5x| = 3 - 5x$ $$3 - 5x \leq x \\ 3 \leq 6x \\ x \geq \frac{1}{2}$$
2. $x > \frac{3}{5} \rightarrow |3 - 5x| = 5x - 3$ $$5x - 3 \leq x \\ 4x \leq 3 \\ x \leq \frac{3}{4}$$
3. $x = \frac{3}{5} \rightarrow |3 - 5x| = 0$ $$0 \leq x \\ x \geq 0$$
Now you need to analyze each case:
1. $x < \frac{3}{5} \wedge x \geq \frac{1}{2}$
It is true that $\frac{3}{5} = \frac{6}{10} \geq \frac{5}{10}$. Therefore this particular interval is true for $\frac{1}{2} \leq x < \frac{3}{5}$.
1. $x > \frac{3}{5} \wedge x \leq \frac{3}{4}$
Since $\frac{3}{5} = \frac{12}{20}$ and $\frac{3}{4} = \frac{15}{20}$, this is true for $\frac{3}{5} < x \leq \frac{3}{4}$.
1. $x = \frac{3}{5} \wedge x \geq 0$
$\frac{3}{5} > 0$ therefore this is trivially satisfied--thus $x = \frac{3}{5}$ is allowed.
When we combine these results, we find that $\frac{1}{2} \leq x \leq \frac{3}{4}$.
PS Edit:
It's probably easier to use:
$$|x| = \begin{cases} x & x \geq 0 \\ -x & x \leq 0 \\ \end{cases}$$
In that case you get the two intervals:
$$\frac{1}{2} \leq x \leq \frac{3}{5}$$
and
$$\frac{3}{5} \leq x \leq \frac{3}{4}$$
Which clearly combines to give: $\frac{1}{2} \leq x \leq \frac{3}{4}$.
Your solution was correct except you needed to use the word AND correctly. The inequality $-x \le 3-5x \le x$ is equivalent to the inequalities
\begin{align} -x \le 3-5x \quad &\text{AND} \quad 3-5x \le x \\ 4x \le 3 \quad &\text{AND} \quad 3 \le 6x \\ x \le \frac 34 \quad &\text{AND} \quad \frac 12 \le x \\ \frac 12 \le x \quad &\text{AND} \quad x \le \frac 34 \\ x &\in\left[\frac 12, \frac 34 \right] \end{align} | {
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# Let $T^{m-1}v\neq 0$ but $T^mv=0$. Show that $v,Tv,\ldots,T^{m-1}v$ are linearly independent
Let $T\in\mathcal L(V)$ and $T^{m-1}v\neq 0$ but $T^mv=0$ for some positive integer $m$ and some $v\in V$. Show that $v,Tv,\ldots,T^{m-1}v$ are linearly independent.
I had written a proof but Im not sure if it is correct. And in the case it would be correct I dont know how to write it better and clearly. So I have two questions:
1. It is the proof below correct?
2. If so, how I can write it better using the same ideas?
### The attempted proof:
1) If $T^{m-1} v$ would be linearly dependent of $T^{m-2} v$ then exists some $\lambda\neq 0$ such that
$$T^{m-1}v=\lambda T^{m-2}v\implies T^mv=\lambda T(T^{m-2}v)=\lambda T^{m-1}v=0\implies \lambda=0$$
Then $T^{m-2}v$ is linearly independent of $T^{m-1}v$.
2) Now observe that
$$\lambda_1v+\lambda_2Tv+\lambda_3T^2v=0\implies T^{m-2}(\lambda_1v+\lambda_2Tv+\lambda_3T^2v)=\lambda_1T^{m-2}v+\lambda_2T^{m-1}v=0$$
so $\lambda_1,\lambda_2=0$ as we had shown previously, so the original equation reduces to
$$\lambda_3T^2v=0\implies \lambda_3=0$$
thus $v,Tv,T^2v$ are linearly independent.
3) Repeating recursively the analysis in 2) for longer lists of vectors of the form $v,Tv,\ldots,T^kv$ for $k< m$ we can show that the list $v,Tv,\ldots,T^{m-1}v$ is linearly independent.
• This is a related post – Bijesh K.S Jul 11 '17 at 19:15
• The if and only if in the first line looks problematic as $T$ isn't invertible. – DMath Jul 11 '17 at 19:19
• @DMath thank you... I overlook it completely. – Masacroso Jul 11 '17 at 19:20
• I'm afraid that the last part is the bulk of the proof and you just cut it off by saying “repeating the analysis”. Note that for $\{v,Tv,T^2v,T^3v\}$ you need to go down to $T^{m-3}$, which is not as easy as the initial step for $\{T^{m-2}v,T^{m-1}v\}$. – egreg Jul 11 '17 at 20:33
Hint
$$a_0v+a_1Tv+\ldots+a_{m-1}T^{m-1}v=0 \quad(1)$$
multiply by $T^{m-1}$, then
$$a_0T^{m-1}v=0\to a_0=0$$
so, from $(1)$, | {
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multiply by $T^{m-1}$, then
$$a_0T^{m-1}v=0\to a_0=0$$
so, from $(1)$,
$$a_1Tv+\ldots+a_{m-1}T^{m-1}v=0$$
multiply by $T^{m-2}$, then
$$a_1T^{m-1}v=0\to a_1=0$$
can you finish?
• ok but, it is my proof correct? – Masacroso Jul 11 '17 at 19:17
• @Masacroso: it seems correct to me. You just need induction to finish it. – Arnaldo Jul 11 '17 at 19:19
See below for a simple proof.
Suppose that they are not linear independent; that is for some choice of $\alpha_{k}$,
$T^{m-1}v = \sum_{k=0}^{m-2}\alpha_{k}T^{k}v$.
Then $0 =T^{m}v = T\cdot T^{m-1}v = \sum_{k=0}^{m-2}\alpha_{k}T^{k+1}v$,
so $Tv,...,T^{m-1}v$ are linearly dependent.
But that means that $T^{m-1}v = \sum_{k=1}^{m-2}\beta_{k}T^{k}v$.
Inductively repeat the argument to eventually conclude that for some $\gamma \neq 0$, $T^{m-1}v = \gamma T^{m-2} v$, but then we would have to say that $0 = T T^{m-1} v = \gamma T^{m-1} v \neq 0$, a contradiction.
More simply put: if you had a linear combination $c_0 v + c_1 T v + \ldots c_{m-1} T^{m-1} v = 0$ with $c_j$ not all zero, let $c_i$ be the first nonzero coefficient. Then $$T^i v = - \sum_{j=i+1}^{m-1} (c_j/c_i) T^j v$$ Applying $T^{m-i-1}$ to both sides, $$T^{m-1} v = - \sum_{j=i+1}^{m-1} (c_j/c_i) T^{m+j-i-1} v = 0$$ But all terms on the right are $0$, so $T^{m-1} v = 0$, contradiction! | {
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# Thread: a couple of trigonometric integrals
1. ## a couple of trigonometric integrals
$\displaystyle \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx$
$\displaystyle \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx$
My initial thought was to introduce a parameter and differentiate, but that approach didn't get me anywhere.
2. Originally Posted by Random Variable
$\displaystyle \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx$
$\displaystyle \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx$
Here's how you do the first one:
Spoiler:
Let
$\displaystyle \displaystyle I=\int_0^\pi x\sin\left(\cos^2(x)\right)\cos^2(\sin(x))\text{ }dx$
Then, letting $\displaystyle z=\pi-x$ we find that
\displaystyle \displaystyle \begin{aligned}I &=\int_0^\pi (\pi-z)\sin\left(\cos^2(z)\right)\cos\left(\sin^2(z)\ri ght)\text{ }dz\\ &=\pi\int_0^\pi\sin\left(\cos^2(z)\right)\cos\left (\sin^2(z)\right)\text{ }dz-I\end{aligned}
Thus,
$\displaystyle \displaystyle \frac{2}{\pi}I=\int_0^\pi \sin\left(\cos^2(z)\right)\sin\left(\cos^2(z)\righ t)\text{ }dz$
Recall though that
$\displaystyle \displaystyle \sin(\phi)\cos(\theta)=\frac{\sin(\phi+\theta)+\si n(\phi-\theta)}{2}$
And thus
\displaystyle \displaystyle \begin{aligned}\frac{2}{\pi} I &= \int_0^\pi \sin\left(\cos^2(z)\right)\cos^2\left(\sin(z)\righ t)\text{ }dz\\ &= \frac{1}{2}\int_0^\pi\left(\sin\left(\cos^2(z)+\si n^2(z)\right)+\sin\left(\cos^2(z)-\sin^2(z)\right)\right)\text{ }dz\\ &=\frac{1}{2}\pi\sin(1)+\frac{1}{2}\int_0^\pi \sin(\cos(2z))\text{ }dz\end{aligned}
Note though that
$\displaystyle \displaystyle \int_0^\pi\sin(\cos(2z))\text{ }dz=\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz+\int_{\frac{3\pi}{4}}^\pi\sin(\cos(2z))\text{ }dz+\int_{\frac{3\pi}{4}}^{\pi}\sin(\cos(2z))\text { }dz$
Note though that by letting $\displaystyle \displaystyle u=z-\frac{3\pi}{4}$ one gets that | {
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Note though that by letting $\displaystyle \displaystyle u=z-\frac{3\pi}{4}$ one gets that
$\displaystyle \displaystyle \int_{\frac{3\pi}{4}}^{\pi}\sin(\cos(2z))=\int_0^{ \frac{\pi}{4}}\sin(\cos(2u))\text{ }du$
And thus
$\displaystyle \displaystyle \int_0^\pi\sin(\cos(2z))\text{ }dz=2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz+\int_{\frac{\pi}{4}}^{\frac{3\pi}}{4}\sin(\cos (2z))\text{ }dz$
But, letting $\displaystyle w=z-\frac{\pi}{2}$ in this second integral one arrives at
$\displaystyle \displaystyle \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\sin(\cos(2z) )=-\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\sin(\cos(2w))\text{ }dw$
and since the integrand is even this is equal to
$\displaystyle \displaystyle -2\int_0^{\frac{\pi}{4}}\sin(\cos(2w))\text{ }dw$
Therefore,
\displaystyle \displaystyle \begin{aligned}\int_0^\pi\sin(\cos(2z))\text{ }dz &= 2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))+\int_{\frac{ \pi}{4}}^{\frac{3\pi}{4}}\sin(\cos(2z))\text{ }dz\\ &= 2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz-2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz\\ &=0\end{aligned}
Thus, finally we arrive at
\displaystyle \displaystyle \begin{aligned}\frac{2}{\pi}I &=\frac{\pi}{2}\sin(1)+\frac{1}{2}\int_0^\pi\sin(\ cos(2z))\text{ }dz\\ &=\frac{\pi}{2}\sin(1)\end{aligned}
and thus $\displaystyle \displaystyle I=\frac{\pi^2 \sin(1)}{4}$.
3. The ODE you had was solvable.
4. Originally Posted by Random Variable
The ODE you had was solvable.
See my new method. It's more elementary.
5. Originally Posted by Random Variable
$\displaystyle \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx$
$\displaystyle \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx$
My initial thought was to introduce a parameter and differentiate, but that approach didn't get me anywhere.
The second one is interesting. I have a solution, but it's a tad long. Do you have any ideas as of yet? | {
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6. to prove that $\displaystyle \displaystyle\int_0^\pi\sin(\cos 2x)\,dx=0,$ split it up into two integrals with $\displaystyle 0\le x\le\dfrac\pi2$ and $\displaystyle \dfrac\pi2\le x\le\pi$ then apply substitution $\displaystyle t=x-\dfrac\pi2$ on the second piece and you'll see is equal to minus the first piece, hence, the result.
7. Here's my solution to the second
Spoiler:
First use the double angle formula so
$\displaystyle \displaystyle \int_0^{\pi} x\left( \dfrac{1-\cos \left(2 \sin x\right)}{2} + \dfrac{1 + \cos \left(2 \cos x\right)}{2} \right)\, dx$
$\displaystyle \displaystyle = \int_0^{\pi} x\,dx + \frac{1}{2}\int_0^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\, dx$
We will show the second integral is zero by considering it in two parts. (**)
First, that $\displaystyle \displaystyle \int_0^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\,dx = \int_{\pi/2}^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx$
Let $\displaystyle x = \pi -u$ in $\displaystyle \displaystyle \int_{\pi/2}^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx$
Next that $\displaystyle \displaystyle \int_0^{\pi/4} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\,dx = -\int_{\pi/4}^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx$
Let $\displaystyle x = \pi/2-u$ in$\displaystyle \displaystyle \int_{\pi/4}^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\, dx$
Thus, we are left with $\displaystyle \displaystyle \int_0^{\pi} x\,dx = \dfrac{\pi^2}{2}$
8. here's mine to the second:
put $\displaystyle t=\pi-x$ on the integral to see that is equal to $\displaystyle \displaystyle\frac\pi2\int_0^\pi\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx.$
i claim the latter integral equals $\displaystyle \pi,$ hence the original integral equals $\displaystyle \dfrac{\pi^2}2,$ so in order to see that write the integral as | {
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$\displaystyle \displaystyle\int_{0}^{\frac{\pi }{2}}{\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx}+\int_{\frac{\pi }{2}}^{\pi }{\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx},$
and put $\displaystyle t=x-\dfrac\pi2$ on the second one, then add them up and you'll see that the claimed integral achieve the aforesaid value, as required.
another solution for the first one:
we consider $\displaystyle \displaystyle\int_{0}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx},$ (1) then write it as $\displaystyle \displaystyle\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}+\int_{\frac{\pi }{2}}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}$ and put $\displaystyle t=x-\dfrac\pi2$ on the second one then we get $\displaystyle \displaystyle\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}+\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\cos }^{2}}x \right)\cos \left( {{\sin }^{2}}x \right)\,dx},$ so when adding those we note that the integrand is actually $\displaystyle \sin(\sin^2x+\cos^2x)$ so (1) equals $\displaystyle \dfrac\pi2\sin1.$
on the original integral was easy to show that it's equal to $\displaystyle \displaystyle\frac\pi2\int_{0}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx},$ thus by (1) we conclude. | {
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Combinations problem: Choosing ways to select $8$ questions out of $12$.
In an exam, there are $12$ questions in total. He has to attempt $8$ questions in all. There are two parts: Part A and Part B of the question paper containing $5$ and $7$ questions respectively. How many ways are there to attempt the exam, such that you attempt eight questions and from each part you attempt at least three questions?
Please don't answer with the three case method. What I want is the error in my method?
My method: For three questions from each part, we do: $${5 \choose 3}*{7\choose3}$$. But 2 questions are lett, so for that, we multiply the expression by $${6\choose2}$$. Because 6 questions are left to choose from. But this does not give the answer, why?
As you are aware, the actual number of ways to select eight questions from the examination given the restrictions that at least three questions must be selected from each part is $$\binom{5}{3}\binom{7}{5} + \binom{5}{4}\binom{7}{4} + \binom{5}{5}\binom{7}{3}$$ The alternative method you proposed of selecting three questions from part A, three questions from part B, and two additional questions counts the same selection more than once.
If we choose three questions from part A and five questions from part B, the second method counts the same selection of questions ten times since there are $\binom{5}{3}$ ways to choose three of the five selected questions from part B. | {
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To make this concrete, suppose the selected questions are $A_1, A_2, A_3, B_1, B_2, B_3, B_4, B_5$. We can make this particular selection once. However, the second method counts this particular selection ten times. \begin{align*} & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, \color{blue}{B_3}, B_4, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, B_3, \color{blue}{B_4}, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, \color{blue}{B_2}, B_3, B_4, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, \color{blue}{B_3}, \color{blue}{B_4}, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, \color{blue}{B_3}, B_4, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, \color{blue}{B_1}, B_2, B_3, \color{blue}{B_4}, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, \color{blue}{B_3}, \color{blue}{B_4}, B_5\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, \color{blue}{B_3}, B_4, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, \color{blue}{B_2}, B_3, \color{blue}{B_4}, \color{blue}{B_5}\\ & \color{red}{A_1}, \color{red}{A_2}, \color{red}{A_3}, B_1, B_2, \color{blue}{B_3}, \color{blue}{B_4}, \color{blue}{B_5}\\ \end{align*}
By similar reasoning, if we choose five questions from part A and three questions from part B, the second method counts the same selection of questions ten times since there are $\binom{5}{3}$ ways to choose three of the five selected questions from part A. | {
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If we choose four questions from part A and four questions from part B, the second method counts the same selection sixteen times since we count the same set of four questions from part A four times for each of the $\binom{4}{3}$ ways we choose three of those four selected questions and the same set of four questions from part B four times for each of the $\binom{4}{3}$ ways we choose three of those four selected questions.
Note that $$10\binom{5}{3}\binom{7}{5} + 16\binom{5}{4}\binom{7}{4} + 10\binom{5}{5}\binom{7}{3} = \binom{5}{3}\binom{7}{3}\binom{6}{2}$$
• I've edited my answer to make what I'm saying more concrete. – N. F. Taussig Oct 8 '15 at 11:43
• Okay, we see that this is happening. But why is this happening? Can we add something so that my method works? – Aditya Agarwal Oct 8 '15 at 11:45
• As you can see, the method you attempted to use does not work since it counts the same set of selected questions every time you choose $3$ of the selected questions. To avoid overcounting, we use cases. I do not see a way to make the method you attempted work. – N. F. Taussig Oct 8 '15 at 11:55
You are counting certain combinations of questions more than once. This is not so easy to see for your example, thus, let's simplify it to a trivial example: you have to choose at least 5 questions out of 5 of part A (i.e. all of A), and at least 1 out of 7 of part B, and, as before, you have to choose 8 questions in total.
We can trivially see that we indeed have to choose 3 questions out of B to satisfy both conditions, since we have already chosen all of A. Thus, the correct answer would be $\left(\begin{matrix}5\\5\end{matrix}\right)\left(\begin{matrix}7\\3\end{matrix}\right)=1\cdot 35=35$.
Your way of solving this question would however say that there are $\left(\begin{matrix}5\\5\end{matrix}\right)\left(\begin{matrix}7\\1\end{matrix}\right)\left(\begin{matrix}6\\2\end{matrix}\right)=1\cdot 7\cdot 15=105$ possibilities of choosing. | {
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Edit: to clarify the problem in this simplified example: assume you choose questions 1,2 and 3 from B. In your calculations there are three possibilities to do so: you choose question 1 because you have to choose at least one from B (Condition 1), and you choose questions 2 and 3 because you have to choose in total 8 (Condition 2). Or you choose question 2 to satisfy Condition 1, and questions 1 and 3 for Condition 2. Finally, you can choose question 3 to satisfy Condition 1, and questions 1 and 2 for Condition 2. In other words, you count each combination 3 times, and, indeed $35\cdot3=105$. | {
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# Solution of a given linear diophantine equation.
I was solving the following linear diophantine equation :
$56x + 72y = 40$ in integers.
My attempt: I got that 8 is the gcd of 56 and 72 and $8|40$ and hence a solution exists and I can write:
$8 = 56 - 16 *3$
$\implies$ $8= 56 - (72 -56*1)*3$
$\implies$ $8= 4*56 - 3*72$.
So my answer is $x = 4$ and $y = -3$. But in book its showing $x = 20$ and $y = -15$. Where I went wrong? Kindly help.
• Remember that $20=5 \times 4$ and $-15=5 \times 3$. Both answers are correct
– user284001
Nov 15, 2016 at 16:49
• @Bacon, So why did they multiply by 5? Nov 15, 2016 at 16:50
• At the outset you noted that $8|40$ and this is...
– user284001
Nov 15, 2016 at 16:50
$56x+72y=40$
$7x+9y=5$
Select the term with the smaller of the two coëfficients and isolate it on the LHS.
$7x=5-9y$
$\text{[1] }x=\frac{5-9y}7=1-y+\frac{-2-2y}7$
$\text{New variable }a=\frac{-2-2y}7$
$-2-2y=7a$
$-2-7a=2y$
$\text{[2] }y=-\frac{2+7a}2=-1-4a+\frac{a}2$
$\text{New variable }b=\frac{a}2$
$a=2b$
Substitute $a\text{ into [2].}$
$y=-\frac{2+7(2b)}2=-\frac{2+14b}2=-1-7b$
Substitute $y\text{ into [1].}$
$x=\frac{5-9(-1-7b)}7=\frac{5+9+63b}7=\frac{14+63b}7=2+9b$
$\text{So, }(x,y)=(2+9b,-1-7b).$
There is an infinitude of integer answers; but your textbook, for some reason, favors $b=2.$
write the equation as $$56x\equiv 40\mod 72$$ or $$x\equiv \frac{40}{76}=\frac{5}{7}\equiv \frac{77}{7}\equiv 11 \mod 72$$ thus $$x=11+72K$$ with $$k \in \mathbb{Z}$$
You can solve $56x+72y=40$ using convergent.
Note: If $\alpha,\beta$ are integral solutions to $ax+by=c$ for $x,y$ respectively, the other solutions take the form$$x=\alpha-bt\\ y=\beta+at\tag{1}$$
Here, $a=56,b=72$ so the $x$ values model the form $\alpha-72t$ and $\beta+56t$ for $y$.
The nearest converging fraction to $\dfrac {9}{7}$ is $\dfrac 43$ and hence, we have$$7\cdot 4-3\cdot 9=1$$ | {
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For which we can multiply both sides by $40$ to get$$40\cdot 7\cdot 4-3\cdot 9\cdot 40=40\implies 56\cdot 20+72(-15)=40\tag2$$ Thus, we have $\alpha=20,\beta=-15$. And by $(1)$, we get$$x=20-72t\\y=-15+56t$$
Note that we also get the solutions that the book says along the way! | {
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# The probability that it is a woman, or a man who is not bringing a dessert?
John invites 12 friends to a dinner party, half of which are men. Exactly one man and one woman are bringing desserts. If one person from this group is selected at random, what is the probability that it is a woman, or a man who is not bringing a dessert?
I know that the two events are mutually exclusive; so that, the answer is simply $6/12 + 5/12 = 11/12$. However, when I looked at the answer in the book that I study from, I found the same result but an approach that's really weird for me. Here is it literally:
$$P(woman) = 6/12 = 1/2$$ $$P(not\ bringing\ a\ dessert\ ) = 10/12 = 5/6$$ $$P(woman\ and\ not\ bringing\ a\ dessert) = 1/2\ *\ 5/6=5/12$$ $$P(woman\ or\ a\ man\ not\ bringing\ a\ dessert) = 1/2 + 5/6 - 5/12 = 11/12$$
May anyone explain the reasoning behind that approach? And is it technically right?
Your approach is certainly right. The book's approach is right as well. It is just that they should have said $P(\text{woman OR man not bringing dessert}) = P(\text{woman OR not bringing dessert})$ (this works because there are just two genders). Then you don't have mutually exclusive events, so you have to use inclusion-exclusion. $$P(\text{woman OR not bringing dessert}) = P(\text{woman}) + P(\text{not bringing dessert}) - P(\text{woman} \cap \text{not bringing dessert}).$$
The term $P(\text{woman} \cap \text{not bringing dessert}) = 5/12$, thus they have $1/2 + 10/12 - 5/12 = 11/12$.
What you are doing is you are breaking up your space into mutually exclusive events. You are saying it is equivalent to calculating $P(\text{woman} \sqcup \left(\text{not woman AND no dessert}\right))$. That way you have $1/2 + 5/12$.
There's another way you can solve this problem. This probability is equal to $1 - P(\text{man bringing dessert}) = 1 - 1/12 = 11/12$.
NOTE: Edited after a misinterpretation pointed out by @Abdu Magdy | {
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NOTE: Edited after a misinterpretation pointed out by @Abdu Magdy
• But P(man not bringing a dessert) should have equalled $5/12$ not $10/12$, Consequently, P(woman ∩ man not bringing dessert) should have equalled $5/12 \ *\ 1/2 = 5/24$ – Abdu Magdy Oct 21 '17 at 18:52
• I interpret $P(\text{man not bringing a dessert})$ as $P(\text{not bringing a dessert} | \text{man}) = \frac{P(\text{not bringing dessert} \cap \text{man})}{P(\text{man})} = \frac{\frac{5}{12}}{\frac{1}{2}} = \frac{5}{6}$, but YMMV. In my experience with GRE problems, sometimes the language is indeed ambiguous. – Abhiram Natarajan Oct 21 '17 at 19:02
• I'm sorry but I really need to understand... Based on the definition of mutually inclusive events; if we only are to choose one person, May we choose a woman and a man not bringing a dessert at the same time? – Abdu Magdy Oct 21 '17 at 19:23
• >> May we choose a man ... same time? You certainly cannot choose that way. It's more like when you have "not bringing dessert", it can be either man or woman. Let's try this way. They should have said $P(\text{woman OR not bringing dessert})$. That way, you don't have mutually exclusive events. So by inclusion exclusion, you have $1/2 + 10/12 - 5/12$. I said this wrong in my answer. I shall edit it right away. – Abhiram Natarajan Oct 21 '17 at 20:41
• Sorry about the confusion. – Abhiram Natarajan Oct 21 '17 at 21:00 | {
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Negating “Zach blocks e-mails and texts from Jennifer”
I am reviewing some basic propositional logic. The question that I have come across that has given some confusion is Zach blocks e-mails and texts from Jennifer where I am asked to find the negation of this proposition. Here is what they provided as an answer in the book: Zach does not block e-mails from Jennifer, or he does not block texts from Jennifer.
Why couldn't the answer simply be It is not true that Zach blocks e-mails and texts from Jennifer? Why did they have to introduce the disjunction or?
-
Of course your answer is fine, indeed clearer. They presumably want to test whether you know how to negate a conjunction and obtain a disjunction. But if this is a test question, surely you should not be expected to be a mind reader. – André Nicolas Dec 11 '12 at 2:00
What you gave is one possible answer, the book gives another which is logically equivalent, although slightly cleaner statement since it is further reduced. They are the same because of one of De-Morgan's laws; $\neg(P \land Q) \equiv (\neg P) \lor (\neg Q)$
-
Zach blocks emails and texts from Jennifer means Zach blocks emails from Jennifer and Zach blocks texts from Jennifer. Negating this gives Zach does not block emails from Jennifer or Zach does not block texts from Jennifer.
It is true that the negation is also It is not true that Zach blocks emails and texts from Jennifer but the point of the exercise is to make one rewrite it in a useful way.
Note that in general, (not (A and B)) is logically equivalent to ((not A) or (not B)).
-
Or does "Zach blocks emails and texts from Jennifer" mean "Zach blocks emails and Zach blocks texts from Jennifer"? Don't you love the ambiguities of natural language? (Of course, the translation you give is almost certainly the intended meaning.) – Code-Guru Dec 11 '12 at 4:11
• $P(x, y)$: x blocks emails from y
• $Q(x, y)$: x blocks texts from y
• $z$: Zach
• $j$: Jennifer | {
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$P(z, j)$: Zach blocks emails from Jennifer;
$Q(z, j)$: Zach blocks texts from Jennifer.
You essentially negated: $P(z, j) \land Q(z, j)$.
So did the text.
The text applied "distribution of negation over conjunction" (one of DeMorgan's Laws):
• "It is not the case that $[(P(z, j)$ and $Q(z, j)]$" $$\iff \lnot[P(z, j) \land Q(z, j)]$$ $$\iff \lnot P(z, j) \lor \lnot Q(z, j)\tag{DeMorgan's law}$$
• which is to say: not $P(z, j)$, or, not $Q(z, j)$.
You are correct with your statement that the sentence translates to $\;\;\lnot[P(z,j) \land Q(z, j)]$;
It's logically equivalent to the text's answer: $\;\lnot P(z, j) \lor \lnot Q(z, j)$. | {
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# Math Help - Divisibility by 7
1. ## Divisibility by 7
Prove that $2222^{5555}+5555^{2222}$ is divisible by 7.
I set about trying to show that this expression is equal to $0 \mod 7$.
I have that:
$2222=3 \mod7=2 \mod 6$
$5555=4 \mod 7=5 \mod 6$
The question also has Fermat's little theorem above it so I think I should be using that somewhere.
Anyone have any ideas?
2. Hello,
Let's take for example $2222^{5555}$
It's congruent to $3^{5555}(\bmod 7)$
Now, you know by Fermat's little theorem, that $3^{6}\equiv 1(\bmod 7)$
As you noticed, $5555\equiv 5(\bmod 6)$
It can be proved that hence $2222^{5555}\equiv 3^5 (\bmod 7)$
Why ? Because $5555=6k+5$
And then $3^{6k+5}=(3^6)^k\cdot 3^5 \dots$
3. Originally Posted by Showcase_22
Prove that $2222^{5555}+5555^{2222}$ is divisible by 7.
I set about trying to show that this expression is equal to $0 \mod 7$.
I have that:
$2222=3 \mod7=2 \mod 6$
$5555=4 \mod 7=5 \mod 6$
The question also has Fermat's little theorem above it so I think I should be using that somewhere.
Anyone have any ideas?
Hi Showcase_22.
Fermat’s little theorem is a little cumbersome for this problem. Here is a shorter way.
$2222\equiv3\,\bmod7\ \implies\ 2222^3\equiv-1\,\bmod7$ $\implies\ 2222^{5555}=2222^{3\cdot1851+2}\equiv-3^2\,\bmod7\equiv5\,\bmod7$
$5555\equiv4\,\bmod7\ \implies\ 5555^3\equiv1\,\bmod7$ $\implies\ 5555^{2222}=5555^{3\cdot740+2}\equiv4^2\,\bmod7\eq uiv2\,\bmod7$
Hence $2222^{5555}+5555^{2222}\equiv7\,\bmod7\equiv0\,\bm od7.$
The reason I prefer $2222^3$ and $5555^3$ rather than $2222^6$ and $5555^6$ is that their powers can be brought closer this way to 5555 and 2222 respectively.
4. Hello,
Hello!
I finally got it to work out using TheAbstractionist's method (I looked at what you did, did it for 2222 whilst looking at yours and then did it for 5555 separately).
I'll have a go using your method now Moo since it seems to be the one in the mark scheme. | {
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I'll have a go using your method now Moo since it seems to be the one in the mark scheme.
5. Okay, here's my way of doing it. I think it's both of your methods combined (or I may just be copying one of your methods, i'm not that good at number theory!):
$2222^{5555}=3^{5555} \bmod 7$
$3^6=1 \bmod 7$ by Fermat's little theorem
$3^{5555} \bmod 7=3^{6(925)+5}=3^{6(925)}3^5 \bmod 7$
$=(1 \bmod 7)(3^2. 3^2.3 \bmod 7)=(1 \bmod 7 )(5 \bmod 7)=5 \bmod 7$
$\Rightarrow 2222^{5555}=5 \bmod 7$
$5555^{2222}=(5 \bmod 6)^{2222}$
$4^6=1 \bmod 7$ by Fermat's little theorem.
$5555^{2222}=(4 \bmod 7)^{2222}$
$=4^{370(6)+2} \bmod 7=4^{370(6)}.4^2 \bmod 7=(1 \bmod 7)(2 \bmod 7)=2 \bmod 7$
$\Rightarrow 5555^{2222}=2 \bmod 7$
$\Rightarrow 2222^{5555}+5555^{2222}=2 \bmod 7+5 \mod 7=7 \bmod 7= 0 \bmod 7$
**WIN**
6. Well, actually that was exactly the way I wanted you to do
Good job.
Though your notations are very strange...
$=(1 \bmod 7)(3^2. 3^2.3 \bmod 7)$
Just write $1\cdot 3^2 \cdot 3^2 \cdot 3 \bmod 7$
$5555^{2222}=(5 \bmod {\color{red}7})^{2222}
$
Write $5555^{2222}=5^{2222} \bmod 7$
7. Ah, sorry about my notations!
I didn't realise there was a command for $"\cdot"$. The more I know!
That second one was just plain wrong, sorry I wrote it!
On the plus side:
Spoiler:
At least I know how to use these spoiler boxes now! | {
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# How many ways to select $k$ vertices of an $n$-gon?
I have a regular $n$-gon, of which I have to select $k$ vertices. The selections must be rotationally distinct; two selections would be considered equivalent if one is a rotation of the other. For example, if I have a square, and I want to select 2 vertices, there are only 2 possible ways to do that according to the constraint. One is "x - x -", another is "x x - -".
If we denote the function by $CR(n,k)$, then these are the trivial cases:
1. $CR(n, 1) = 1$
2. $CR(n, 2) = \lfloor\frac{n}{2}\rfloor$
3. $CR(n, k) = CR(n, n - k)$
I am quite short of ideas on how to find the recurrence or closed formula of this problem, or if this problem has any closed form / recurrence solution at all. Any help with a bit detailed walk through would be much appreciated.
• What do you mean by selection must be rotationally invariant? – Anurag A May 28 '15 at 19:07
• Supposing, I have a 4-gon and I have to select 2 vertices. Hence, these two selections are the same. x-x-, -x-x. These selections are the same, x--x, xx--, -xx-, --xx – sarker306 May 28 '15 at 19:09
• The following MSE meta link has a section on enumerating necklaces and bracelets. – Marko Riedel May 28 '15 at 19:51
• One way to think about this (perhaps the best way) is the group action of cyclic rotations on the subsets of vertices. Rotations preserve the number of vertices in a subset (obviously), so the counting comes down to counting orbits of the $k$-subsets of $n$-vertices under those rotations. – hardmath May 28 '15 at 19:53
As the OP seems interested in fixing $k$ and letting $n$ vary we will do an example to show how this might work.
By the Polya Enumeration Theorem what we have here is $$[z^k] Z(C_n)(1+z)$$ with $Z(C_n)$ the cycle index of the cyclic group which is $$Z(C_n) = \frac{1}{n}\sum_{d|n} \varphi(d) a_d^{n/d}.$$
The desired quantity is thus given by $$[z^k] Z(C_n)(1+z) = [z^k] \frac{1}{n}\sum_{d|n} \varphi(d) (1+z^d)^{n/d}.$$ | {
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This is $$\frac{1}{n}\sum_{d|n, d|k} \varphi(d) [z^k] (1+z^d)^{n/d}$$ or $$\frac{1}{n}\sum_{d|n, d|k} \varphi(d) {n/d \choose k/d} = \frac{1}{n}\sum_{d|\gcd(n,k)} \varphi(d) {n/d \choose k/d}.$$
For $k=4$ starting at $n=1$ we obtain the sequence $$0, 0, 0, 1, 1, 3, 5, 10, 14, 22, 30, 43, 55, 73,\ldots$$ which points us to OEIS A008610 where we find confirmation.
Another interesting one is $k=6$ which yields $$0, 0, 0, 0, 0, 1, 1, 4, 10, 22, 42, 80, 132, 217, 335, 504,\ldots$$ which points to OEIS A032191.
A Maple session with these looks like this:
> with(numtheory):
> Q := (n,k) -> 1/n*add(phi(d)*binomial(n/d,k/d), d in divisors(gcd(n,k)));
Q := (n, k) -> add(numtheory:-phi(d) binomial(n/d, k/d),
d in numtheory:-divisors(gcd(n, k)))/n
> seq(Q(n,8), n=1..18);
0, 0, 0, 0, 0, 0, 0, 1, 1, 5, 15, 43, 99, 217, 429, 810, 1430, 2438
which incidentally is OEIS A032193.
• It'll be a sin by my side not to refer you to my similiar question here: math.stackexchange.com/questions/1294224/… . Perhaps you'll be able to help :) I apologize, OP! Really am! – Matan May 28 '15 at 20:40
• I have been following your question for three hours. @Matan , felt really intrigued. – sarker306 May 28 '15 at 20:45
• How wonderful to hear. Thanks! I update it on regular basis and work on it when I have time. Keep track of it :) @sarker306 – Matan May 28 '15 at 20:49
Taking the necklace proof for Fermat's Little Theorem as inspiration:
For $n,k$ coprime, every possible selection has $n$ rotations. So
$$CR(n,k) = \frac 1n{n \choose k}$$
Otherwise letting $d=\gcd(n,k)$, we need to account for those choices where there is a repeat pattern smaller than $n$, which will be all possible choices over the shorter interval with the reduced number of choices (but repeated through the entire set): | {
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\begin{align} CR(n,k) &= \frac 1n\left[{n \choose k} - {n/d \choose k/d }\right] + CR\left(\frac nd, \frac kd \right) \\ &=\frac 1n\left[{n \choose k} - {n/d \choose k/d }\right] + \frac 1d {n/d \choose k/d } \end{align}
since $\gcd(\frac nd, \frac kd)=1$
And also note that $CR(n,0)=1$ and $CR(n,n)=1$.
• Any explanation? Or any source where the derivation is explained? – sarker306 May 28 '15 at 19:25
• Joffan, if you could please refer here math.stackexchange.com/questions/1294224/… for a related problem I asked I will be grateful. Perhaps you'll be able to help. – Matan May 28 '15 at 19:26
• @sarker306 If you review the necklace proof of Fermat's Little theorem I think you'll understand where this comes from. If I have time I'll add to it. – Joffan May 28 '15 at 19:34
• I get CR(6,3) = 25/6 when I apply your solution. – TokenToucan May 28 '15 at 20:17
• CR(n, 2) has its correct value when n is odd, not even. But really thanks for bringing in this gcd issue, it seems to be a right step to the right direction, I didn't suspect it before. I'll think it through the night. – sarker306 May 28 '15 at 20:25 | {
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# How many distinct ways to climb stairs in 1 or 2 steps at a time?
I came across an interesting puzzle:
You are climbing a stair case. It takes $n$ steps to reach to the top. Each time you can either climb $1$ or $2$ steps. In how many distinct ways can you climb to the top?
Is there a closed-form solution to the problem? One can compute it by creating a 'tree' of possibilities of each step. That is, I can either take 1 or 2 steps at each stage and terminate a branch once it sums to $n$. But this is would get really unwieldy very quickly since the maximum number of nodes in a binary tree is $2^{n+1}-1$, i.e., exponential. Is there an easier way to solving this puzzle?
Let $F_n$ be the number of ways to climb $n$ stairs taking only $1$ or $2$ steps. We know that $F_1 = 1$ and $F_2 = 2$. Now, consider $F_n$ for $n\ge 3$. The final step will be of size $1$ or $2$, so $F_n$ = $F_{n-1} + F_{n-2}$. This is the Fibonacci recurrence.
• Ha! Some simple and yet a convoluted formation of the sequence. That explains it. Can't believe I couldn't see it :) – PhD May 11 '14 at 1:53
• Is it possible to generalize it? Such as what if your steps are 2, 3, and 5? Also, we can compute fibonacci in matrix form which is faster. In the case of 2,3 and 5 steps, what would be the closed form solution? – Bagus Trihatmaja Dec 5 '18 at 8:53
• Yes, this falls into the more general framework of solving linear recurrence equations. They always admit fast evaluations via matrix exponentiation and often have closed-form solutions, especially when the characteristic equation has low degree since you need to find all of its roots. – fahrbach Dec 6 '18 at 22:19
• Can someone help me understand a tweak to this problem where the order of the steps doesn't matter? How to come up to a closed form equation for it? – mankadnandan May 26 '19 at 7:38
The solution to this problem indeed corresponds to the Fibonacci numbers, as mentioned by @fahrbach. | {
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The solution to this problem indeed corresponds to the Fibonacci numbers, as mentioned by @fahrbach.
Here is an illustration of what you are trying to solve for the case of $n=4$ steps (taken from this website, which also gives a combinatorial solution)
Any staircase with $n$ steps allowing paths with increments of 1 or 2 steps at a time will end up in one of two states before the last path is taken: either we've climbed $(n-1)$ steps already and have $\color{red}{one}$ more step to take, or we've climbed $(n-2)$ steps already and we have $\color{blue}{two}$ more steps to take (if we took only one step here then we'd end up in an arrangement from the first state).
Thus, to get the total number of possible ways to climb $n$ steps, we just add the number of possible ways we can climb $(n-1)$ steps and the number of possible ways we can climb $(n-2)$ steps, giving the familiar recurrence relation:
\begin{equation*} F_n = \left\{ \begin{array}{l@{}l@{}l} 1 & n = 0,1\\ \color{red}{F_{n-1}} + \color{blue}{F_{n-2}} & n \ge 2 \end{array} \right. \end{equation*}
• Can this answer be expanded to answer the question "Given a set of possible step sizes S, calculate the number of possible ways you can climb n steps using step sizes in S)"? – Ephraim Nov 4 '15 at 20:21
• @Ephraim That's simple, it's just $F_n=\sum_{i\in I} F_i$ where $I$ is the set of steps that can jump to step $n$. – Stella Biderman Jan 31 '17 at 16:37
• This is the explanation that clicked to me, the colored figure makes a lot of sense, thanks. – neevek Oct 20 '20 at 16:47
This is the Fibonacci numbers - One interpretation of the n-th Fibonacci number is the number of ways to compose $n$ with parts in $\{1,2\}$ (that is, the n-th fibonacci number counts the number of sequences whose values are in $\{1,2\}$ whose sums are $n$). This is often called "Golfs and Cadillacs" or something similar. | {
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• Could you please elaborate as to "how" it helps solve the problem? – PhD May 11 '14 at 1:46
• If there are $n$ steps, you have $F_n$ ways to reach the top of the stairs? The sequences map directly to the amount of stairs you climb with each step. I don't see how it can be any clearer. This interpretation can be used as a definition of the Fibonacci numbers and can be shown to be equivalent to the recurrence by using a straightforward combinatorial argument. – Batman May 11 '14 at 1:47
• I see what you mean. Perhaps what I'm confused by is by this: that is, the n-th fibonacci number counts the number of sequences whose values are in {1,2} whose sums are n. Could you please provide an example to explain this? – PhD May 11 '14 at 1:49
• You form a sequence made up of 1's and 2's whose sum is $n$. For example, for $n=3$, the sequences are (1,1,1), (1,2) and (2,1) giving you 3 ways to climb the stairs. Note that this is the common combintorics use of Fibonacci numbers, which is offset by 1 from the kind you learn in elementary school (this Fibonacci sequence is $1,2,3,5,...$ instead of $1,1,2,3,5,...$). – Batman May 11 '14 at 1:51
We can change this question into: In how many ways is possible to write a number as the ordered sum of 1s and 2s?
We can prove this using direct proof.
Hypothesis: For Fibbonaci number $F_1=1, F_2=1, F_{n}=F_{n-1}+F_{n-2}$, $Q(k)=F_{k+1}$
We say that we have constructed every order of $Q(n)$. Then, we construct $Q(n+1)$ this way:
1. For all orders, we add $+1$ to change $n$ to $n+1$. (There are now $Q(n)$ numbers.) Note: all numbers that are constructed here ends in $+1$.
2. For all orders that end with $+1$, we change it to $+2$. We shall prove that the amount will be $Q(n-1)$ to complete the proof. Note: all numbers that are constructed here ends in $+2$.
It can be seen that by constructing in this way there contains all orders. | {
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It can be seen that by constructing in this way there contains all orders.
Proof of step 2: If we constructed $Q(n)$ in the way above, then all numbers ended in $+1$ will be the amount $Q(n-1)$ by step 1. above, and we are done. | {
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#### Problem
The positive integers, $x$, $y$, and $z$ are consecutive terms in an arithmetic progression. Given that $n$ is also a positive integer, for how many values of $n$ below one-thousand does the equation $x^2 - y^2 - z^2 = n$ have no solutions?
#### Solution
Let $x = a + d$, $y = a$, and $z = a - d$.
$\therefore (a + d)^2 - a^2 - (a - d)^2 = n$
$a^2 + 2ad + d^2 - a^2 - a^2 + 2ad - d^2 = n$
$\therefore 4ad - a^2 = a(4d - a) = n$.
Let $u = a$ and $v = 4d - a \implies u + v = 4d \equiv 0 \mod 4$. In other words, for a solution to exist the factors of $n$ must add to a multiple of four.
We shall deal with $n$ being of the form $2^m r$, where $r$ is odd, and for increasing values of $m$.
• $m = 0$ ($n$ is odd):
If $n = r$ then the factors $u$ and $v$ must both be odd. But if they are both congruent with 1 or both congruent with -1 modulo 4 then $u + v \equiv 2 \mod 4$, and there will be no solution; if they are different then $u + v \equiv 0 mod 4$, and there will always be a solution.
Hence if they are the same then $n = uv \equiv 1 \mod 4$, or $n$ being of the form 4$k$ + 1, will have no solutions.
• $m = 1 \implies n = 2r = a(4d - a)$:
If $a = 2r$, $4d - a = 1 \implies 4d = 2r + 1$. But as the RHS is odd, this is impossible.
If $a = r$, $4d - a = 2 \implies 4d = r + 2$. Impossible, as RHS is odd.
In other words if $n = 2(2k + 1) = 4k + 2$ then there will be no solutions.
• $m = 2 \implies n = 4r$:
If $a = 2r$, $4d - a = 2 \implies 4d = 2r + 2 = 2(r + 1)$.
And as $r + 1$ is even, we will always have at least one solution if $n = 4r$.
• $m = 3 \implies n = 8r$:
If $a = 8r$, $4d - a = 1 \implies 4d = 8r + 1$. Impossible.
If $a = 4r$, $4d - a = 2 \implies 4d = 4r + 2$. Impossible.
If $a = 2r$, $4d - a = 4 \implies 4d = 2r + 4$. Impossible.
If $a = r$, $4d - a = 8 \implies 4d = r + 8$. Impossible.
Hence if $n = 8(2k + 1) = 16k + 8$ then there will be no solutions. | {
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• $m \ge 4$:
If $a = 4r$, $4d - a = 2^{m-2} \implies 4d = 4(r+2^{m-4})$.
Hence for $m \ge 4$ there will always be at least one solution.
Thus there will be no solutions for numbers of the form $4k + 1$, $4k + 2$, and $16k + 8$. As the first and second cases are odd and even respectively, they are mutually exclusive, and although the the second and third are both even, the third is divisible by 4, whereas the second is not divisible by 4. Hence all three forms are mutually exclusive.
As $4 \times 249 + 1 = 997$, $4 \times 249 + 2 = 998$, and $16 \times 61 + 8 = 984$, there are exactly $249 + 249 + 61 = 559$ values of $n$ below one-thousand that have no solution.
For which values of $n$ will there be exactly one solution?
Problem ID: 295 (26 Nov 2006) Difficulty: 4 Star
Only Show Problem | {
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## Sunday, May 23, 2010
### GR9768.63: Intersections
63. At how many points in the xy-plane do the graphs of $y = x^{12}$ and $y=2^x$ intersect?
(A) None
(B) One
(C) Two
(D) Three
(E) Four
Solution:
We know what the graph of $2^x$ looks like: always positive, the $y$-intercept is at $(0,1)$. For $y = x^{12}$, it passes through the vertex $(0,0)$ and it concaves up. There are at most 2 intersections. Now, the two graph intersect exactly once on $(-\infty, 0)$. On the other half, $[0, \infty)$, we'll look at $f(x) = 2^x - x^{12}$ and note that $f(1) = 2^1 - 1^{12} = 1 > 0$, where as $f(2) = 2^2 - 2^{12} < 0$. Thus, by the Rolle's Theorem, we see that $f$ has a zero on $(1,2)$. Thus, $x^{12}$ intersects $2^x$ on this interval. So there are two points of intersection.
The answer is C.
-sg-
Note: the answer provided in the Answer Sheet is D, 3 points of intersection.
result from wolframalpha
#### 13 comments:
1. By the same Rolle's theorem, there *is* one more point of intersection. YOu exained x=2. Now, examine x=L, where L is very large. Clearly, 2^x wins here. So, again by ROlle's theorem, there is yet another point of intersection between 2 and L. The answer is indeed, D. - Deego.
2. Only 17 percent of the actual GRE test takers in 1997 got this right, worse than a random selection strategy!
3. Proper result from wolfram:
http://www.wolframalpha.com/input/?i=Plot[{2^x,+x^12},+{x,+-100.,+100.}]+
4. one vote for D
5. shouldn't it be Intermediate Value Theorem instead of Rolle's Theorem?
---
yodogyo@gmail.com
6. answer is (d) ...
basic property is that 2^x increases faster than x^12 and in fact for any a^x for which a>1 we would have a^x > x^n for any n as x--> infinity.
Hence all we need to find is whether the graph intersects twice or once for smaller values of x. e^x and x^2 I believe intersect only once. | {
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2^x and x^12 intersects twice for x<2 and also for some big value of x as 2^x would be greater than x^12 for some value eventually. Hence this function intersects thrice.
7. I agree with Shantanu.
X>=0: We know that 0^12< 2^0 (x=0) and 2^12>2^2 (X=2).
Thus there is at least one intersection between 0 and 2.
At the same time, we know that 2^x is NOT O(x^12) as x->infinity. Also, they are both positive functions, so we know that there is some x>2 such that 2^x exceeds x^12 forever.
x<0. Its clear there is a cross in this region.
This gives us at least 3 crosses. My gut tells me that this is all because I am familiar with the graphs of both functions. So I would definitely choose D.
Rigorously, I do not know why there is only 3. Examining the derivatives would probably be the place to look.
8. Sorry for my grammar, english is not my native.
First 2 solutions are clearly. Lets prove that there is the 3-th solution. If we prove that lim(2^x/x^12)>1 for x->+oo, we will prove that the curves intersect once more time (because if x2 is x of 2-nd intersection and x near the x2 and x>x2 than 2^x1 for x->+oo. Just use L'Hospital's rule 12 times and we have: lim(ln2^12*2^x/(x^12))=+oo for x->+oo. So, because +oo>1 than we have the 3-th point of intersection.
9. Why are there not more than 3 intersections?
10. There are precisely 3 intersections. At x=0, 2^x = 2 > 0 = x^12. At x=2, 2^x = 4 < 2^12 = x^12, hence there is one intersection point in the interval (0,2). It is a well-known fact that as x grows to infinity 2^x exceeds x^12 (quite rapidly, in fact). Hence there is a second intersection point for x>2. By drawing the graphs of these functions it is easy to see that these are all intersection points for positive values of x.
A similar argument shows that there is a third and last intersection point between -1 and 0, and these are all.
11. What is the value of 3rd point of intersection? How large is it? | {
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11. What is the value of 3rd point of intersection? How large is it?
12. The third point of intersection occurs approximately at (74.66932553, 3.00404713 X 10^22) | {
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# Open sets in $\mathbb R\times \mathbb R$ with the dictionary order
In the example below, I wonder why Munkres depicted the first type of intervals in this way?
More specifically, why doesn't the picture look like this:
where all points below, above, and on the red segment are removed?
This is my justification why I think this should be the right picture:
$$(a\times b,c\times d)=\{x|a \times b <_{dict} x <_{dict} c\times d\}\\=\{(p,q)|a <_{R} p <_{R} c\}\cup \{(r,s)| a=r=c \text{ and } c <_{R} s <_{R} d \}$$ (where $<_{dict}$ is the dictionary order and $<_R$ is the standard order on $\mathbb R$). The first set in the union is the set of the first type, the second set in the union is the set of the second type.
Now, any point above, below, or on the red line has the first coordinate $a$, which cannot lie in a set of the first type.
Update: I defined the sets of the first type in a way that differs from Munkres, but I don't see what's wrong with my description of open sets (in the display), and no open set in my description can be a set of the first type in Munkres.
• Um, removing the red segments on your second image produces exactly the first one, doesn't it? – Henning Makholm Sep 1 '18 at 19:12
• @HenningMakholm I'm not removing just the red segment, I'm removing everything above and below it as well. – user531587 Sep 1 '18 at 19:21
• What then happen to points such as $a\times(b+1)$, which is definitely both $>_{\rm dict} a\times b$ and $<_{\rm dict} c\times d$? – Henning Makholm Sep 1 '18 at 19:29
• @HenningMakholm okay, this makes sense... I was confused (and still confused) by the description of the basis in the text. – user531587 Sep 1 '18 at 19:51
• Actually this answer made things more clear: math.stackexchange.com/a/1129061/531587 – user531587 Sep 1 '18 at 20:02 | {
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Suppose a point $p = (x,y) \in (a \times b , c \times d)$ has $a$ as the first coordinate. Since $(a,b) < (x,y) = (a,y)$, this necessarily implies that $b < y$. In the same vein, if $x = c$ and since $(x,y) = (c,y) < (c,d)$, necessarily we must have that $y < d$. This justifies why the interval only contains an open ray of $\{x = a\}$ and $\{x = c\}$ instead of the whole line.
Your characterization of the interval is not quite right. Note that $r$ cannot be $a$ and $c$ at the same time, so you should separate in two cases (which is basically what I wrote) depending on the value that the first coordinate takes. This should clarify the picture. | {
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# Principal submatrix and Principal minor of a matrix
A submatrix of a matrix is obtained by deleting any collection of rows and/or columns.
For example, from the following 3-by-4 matrix, we can construct a 2-by-3 submatrix by removing row 3 and column 2:
The minors and cofactors of a matrix are found by computing the determinant of certain submatrices.
A principal submatrix is a square submatrix obtained by removing certain rows and columns. The definition varies from author to author. According to some authors, a principal submatrix is a submatrix in which the set of row indices that remain is the same as the set of column indices that remain.
For a general 3 × 3 matrix in Mathematica,
(mat = Array[Subscript[a, ##] &, {3, 3}]) // MatrixForm
$mat=\left( \begin{array}{ccc} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{array} \right)$
there is one third order principal submatrix, namely mat. There are three second order principal submatrix:
$mat_{33}=\left( \begin{array}{ccc} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \\ \end{array} \right)$, formed by deleting column 3 and row 3;
$mat_{22}=\left( \begin{array}{ccc} a_{1,1} & a_{1,3} \\ a_{3,1} & a_{3,3} \\ \end{array} \right)$, formed by deleting column 2 and row 2;
$mat_{11}=\left( \begin{array}{ccc} a_{2,2} & a_{2,3} \\ a_{3,2} & a_{3,3} \\ \end{array} \right)$, formed by deleting column 1 and row 1;
And there are three first order principal submatrix:
$mat=\left( \begin{array}{ccc} a_{1,1} \\ \end{array} \right)$, formed by deleting column 2,3 and row 2,3;
$mat=\left( \begin{array}{ccc} a_{2,2} \\ \end{array} \right)$, formed by deleting column 1, 3 and row 1, 3;
$mat=\left( \begin{array}{ccc} a_{3,3} \\ \end{array} \right)$, formed by deleting column 1,2 and row 1,2;
Do you have a way to find all principal submatrix in Mathematica ?
or find the determine of all principal submatrix (principal minor) in Mathematica ?
By @klgr comment. | {
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By @klgr comment.
Diagonal@Map[Reverse, Minors[mat, k, Identity], {0, 1}] //MatrixForm /@ # &
• Perhaps Diagonal @ Minors[mat, k]? – Carl Woll Aug 18 '17 at 17:01
• Diagonal@Map[Reverse, Minors[mat, k, Identity], {0, 1}] // MatrixForm /@ # & – kglr Aug 18 '17 at 17:04
• @kglr Thanks for the answer but is this in general. What if the matrix is a 4X4 measurement . – Emad kareem Aug 18 '17 at 17:14
• @kglr Thank you very much. This gives the exact result in general. – Emad kareem Aug 18 '17 at 17:24
To get the principal submatrices:
Diagonal[Map[Reverse, Minors[mat, #, Identity], {0, 1}]] & /@ {1, 2}
For the principal minors
Diagonal[Map[Reverse, Minors[mat, #], {0, 1}]] & /@ {1, 2}
• What is the & /@ {1, 2} for? – AzJ Jan 25 '18 at 19:46
• @AzJ, the & belongs to the left part, that is, the expression is actually (Diagonal[Map[Reverse, Minors[mat, #], {0, 1}]] & ) /@ {1,2}. For a pure function (a function with unnamed arguments) foo[#]& (where foo is an instruction on what to do with the stuff #; for example, multiply it by 2 and add 1 , (1+ 3 #)&) , the expression foo[#]&/@{a,b,c} Maps foo over a list of arguments ({a, b}) , that is, it is a short hand for {foo[a], foo[b], foo[c]}. ... – kglr Jan 25 '18 at 20:34
• ... Please see the doc pages on for details on of esoteric symbols such as Function (&), Slot (#) and Map – kglr Jan 25 '18 at 20:35
I arrived three years later but I think that the code of @klgr can be significantly improved as it computes first all the minors to take at the end just those on the diagonal. I think that the next routine works faster for the submatrices with the obvious change making it work also for the minors, of course. I also suppose for convenience that the matrix is square, the obvious change make this also work for non-square matrices. | {
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PrincipalSubmatrices[mat_, size_] :=
Module[{choices = Subsets[Table[i, {i, 1, Length[mat]}], {size}], count,
rc, symsubmatrix, symsubmatrices}, count = choices;
symsubmatrices = {};
While[count != {}, rc = count[[1]]; symsubmatrix = mat[[rc, rc]];
symsubmatrices = Append[symsubmatrices, symsubmatrix];
count = Delete[count, 1]]; symsubmatrices] | {
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How many classes does the equivalence relation partition the set?
Looking for help on B) How many classes does the equivalence relation partition set X
Consider the relations $R$ and $S$, defined on the set $X = \{1, 2, . . . , 99\}$ as follows.
$xRy \iff x + y$ is a multiple of $11$,
$xSy \iff x − y$ is a multiple of $11$.
A) One of $R$ and $S$ is an equivalence relation, the other is not. Determine which is which and justify your answers.
B) Into how many classes does the equivalence relation partition set $X$?
So far I've determined that S is the equivalence relation as R isn't reflexive or transitive. My only attempt at B is that there is 11-1=10 non-zero congruence classes, so does each one correspond to a non-zero equivalence class?
Any help is appreciated!
• Where does this "non-zero" stuff come from? – ancientmathematician Apr 14 '18 at 16:53
• What is a "non-zero" equivalence class? – Morgan Rodgers Apr 14 '18 at 16:54
• @CyclotomicField: Sorry, but "a zero equivalence class" is a bit of a nonsense on a set that doesn't even contain zero, such that the given set $X$. – zipirovich Apr 14 '18 at 18:29
• @CyclotomicField It's unambiguous, but it's also meaningless in the normal context of equivalence relations. And is a term that doesn't actually show up in the problem – Morgan Rodgers Apr 15 '18 at 17:11
• What's wrong with a "zero equivalence class? Doe $11 -11=0$ make $11$ any less equivalent? Where did you get this weird idea that you aren't supposed to count the the equivalence class to zero? – fleablood Apr 15 '18 at 17:20
Just count them.
$1 R 12 R 23 R 34 R .... R 90$
$2 R 13 R 24 R 35 R .... R 91$
....
$10 R 21 R 32 R 43 R..... R 98$
$11 R 22 R 33 R 44 R ..... R 99$
That's 11.
It doesn't make any sense to subtract the "zero" equivalence class for 2 reasons. | {
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That's 11.
It doesn't make any sense to subtract the "zero" equivalence class for 2 reasons.
1) The "zero" equivalence class is STILL an equivalence class so why they heck would you omit it? NOWHERE in the question does it say ANYTHING about how many non-zero equivalence classes; it ask how many equivalence classes. And $\{11,22,33,44,55,66,77,88,99\}$ is certainly an equivalence class, isn't it?
2) Since no method of "addition" has been defined or discussed, there is no meaning to defining any one of the equivalence classes as be a "zero" equivalence class. IF we were to define $\{x|x R c\} + \{x|x R d\} = \{x| x R (d+c\pm 11k \text{ for some integer } k)\}$ and define $[0]$ as the equivelence clas so that $[0] + \{x|x R c\} = \{x|x R c\}$ then, yes, $\{11,22,33,44,55,66,77,88,99\} = [0]$. But again, so what, it's still an equivalence class, isn't it?
To show $S$ is an equivalence relation you can either verify directly that it is reflexive, transitive, and symmetric or demonstrate that the relation partitions the set and use the fundamental theorem of equivalence relations. If you can demonstrate the existence of the congruence classes you have such a partition and that's sufficient to demonstrate that you have an equivalence relation. As you've deduced there are eleven such classes, and ten of them do not contain zero. I would also encourage you to consider the first method of directly verifying the three defining properties as it's both instructive and relatively easy. | {
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• So is the answer 10 for B) correct? – RealGib Apr 14 '18 at 18:23
• @RealGib: No, it's not. – zipirovich Apr 14 '18 at 18:30
• None of the equivalence classes on $X$ contain 0, since $X$ does not contain 0. – Morgan Rodgers Apr 15 '18 at 17:17
• "Morgan Rodgers continued insistence on ignoring this obvious interpretation of the OPs comments". But it goes both ways. I agree that "$\{11,...99\}$ do not contain $0$ as so can't be 'zero' modulo class" is a bit too oblique and inaccurate and $11\equiv 0 \mod 11$ does indeed mean $\{11,...99\}$ is the zero element. But on another, and very important level A "zero" (additive identity) element can only be defined if addition on the classes (and NOT on the elements of the classes) is defined. So it is not unjustified. – fleablood Apr 16 '18 at 16:10
• ... so I mostly agree with Morgan Rodgers although I believe saying "it isn't a zero module as it doesn't contain zero" evades the issue that it actually isn't a zero module because the means for which it is "zero" are not stated. Of more concern is... why leave out the zero modules. These seem to much of a "monkey see; monkey do" approach to math that is doomed to failure. You don't just do everything one way despite whether that is what a question asks. If was asked how many elments does $\{0,1,6,8\}$ you don't answer 3 because zero doesn't count and you don't answer "it goes to 8". – fleablood Apr 16 '18 at 16:15
The question is not asking you about "nonzero" equivalence classes, it is asking for the total number of equivalence classes. So the equivalence class that contains 0 counts the same as any other equivalence class (however you want to interpret this, your set $X$ does not contain 0 so the interpretation of which equivalence class is the "zero" involves concepts that go beyond the scope of what this problem is asking). | {
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# A circle touches two sides of a triangle and two of its medians
A circle touches two sides of a triangle and two of its medians. Prove that the triangle is isosceles.
This problem came from the Mathematical Digest issue 62 (Jan 1986) which in turn cited a Russian mag called KVANT (meaning "Quantum").
• Assume it's not true. Then you wouldn't ask the question. A contradiction. Therefore it's true. QED. :) Mar 13 '21 at 1:02
As the triangles ADC and BEC have the same incircle and their areas are equal (half that of ABC), so are their perimeters
DC+DA+AC = EC+EB+BC
or, subtracting from both sides CD+CE+DA+EB
AE-EB = BD-DA.
This means that D and E lie on the same pair of hyperbolas with foci A and B. Since they also have the same distance to the base AB (half that of C) the triangle must be isosceles by symmetry.
• It took me a moment to see why they have the same distance to the base, but that should have been obvious. This is a really clever proof. Mar 13 '21 at 8:38
• @JaapScherphuis It's short ;-) I've added a half sentence explaining (if you can call it that) the distance to the base. Mar 13 '21 at 9:50
• That is clever. I don't think I've ever seen anyone use a hyperbola locus in a proof like this! Mar 13 '21 at 17:34
• Elegant, indeed. Mar 16 '21 at 6:16
Alternative Proof | {
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Alternative Proof
By Pitot Theorem we have $$|CE|+|MD| = |CD|+|ME|$$ and since the centroid divides each median in the ratio 2:1, this means that $$\frac{1}{2}|AC| + \frac{1}{3}|AD| = \frac{1}{2}|BC|+\frac{1}{3}|BE|$$ Now let $$|AB| = c, |BC| = a, |CA|=b, |AD|=m_a, |BE|=m_b$$.
Then Apollonius' Theorem tells us that $$m_a = \sqrt{\frac{2b^2+2c^2-a^2}{4}}\,\,\,\,,\,\,\,\,m_b = \sqrt{\frac{2a^2+2c^2-b^2}{4}}$$ Substituting this in above and multiplying across by $$6$$ yields $$3a + \sqrt{2b^2+2c^2-a^2} = 3b + \sqrt{2a^2+2c^2-b^2}$$ $$\Rightarrow 3(a-b) + \sqrt{2b^2+2c^2-a^2} = \sqrt{2a^2+2c^2-b^2}$$ $$\Rightarrow 9(a-b)^2 + 6(a-b)\sqrt{2b^2+2c^2-a^2} + 2b^2+2c^2-a^2 = 2a^2+2c^2-b^2$$ $$\Rightarrow (a-b)\left(9(a-b) + 6\sqrt{2b^2+2c^2-a^2} - 3(a+b)\right) = 0$$ which means either $$a=b$$ or $$\sqrt{2b^2+2c^2-a^2} = 2b-a$$.
If it is the latter then $$2b^2+2c^2-a^2 = 4b^2-4ab+a^2$$ $$\Rightarrow c^2 = b^2-2ab+a^2 = (b-a)^2$$ $$\Rightarrow c = \pm(b-a)$$ which only happens for a degenerate triangle (does not satisfy the triangle inequality).
Hence $$a=b$$
@loopwalt and @hexomino both gave excellent answers. I wanted to share the answer from the Mathematical Digest, because it's also elegant in its own way:
It starts the same way as @hexomino's proof (I didn't know it was called Pitot's Theorem!):
Since $$CEMD$$ circumscribes a circle, $$CE+MD = CD + ME$$ (Pitot's Theorem, but fun to prove). So (as in hexomino's proof):
$$\frac{1}{2}AC+\frac{1}{3}AD = \frac{1}{2}BC+\frac{1}{3}BE \tag1$$
Next, note that $$\triangle ADC$$ and $$\triangle BEC$$ have the same area ($$\frac{1}{2}\triangle ABC$$) and share a common incircle. This means that their perimeters are equal (another fun to prove and left as an exercise!). So $$AD+\frac{1}{2}BC+AC = BE+\frac{1}{2}AC+BC$$, which gives:
$$\frac{1}{2}AC+AD = \frac{1}{2}BC+BE \tag2$$
Now, $$3\times(1)-(2)$$ gives $$AC=BC$$ | {
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### Probability of k collisions
, Friday, 30 January 2015
While I was reading about hash tables I stumbled upon this deceptively simple question. I have rephrased it to make it easier to understand.
Say we have $m$ buckets. We select a random bucket and put a ball in it, we repeat this (select-put) $n$ times. In the end what is the probability of having at least one bucket with $k$ balls?
My initial attempts to solve it failed, and I posted the question on math.stackexchange. An approximation for large values of $m$ and $n$ was posted as an answer but I was interested in exact results. Once I realized I had no way of finding someone else's solution, I was able to solve it myself.
## Simulation
First, to check if my solution is correct, I have written a small program to simulate the select-put procedure million times and report the experimental probability for different values of $m$,$n$ and $k$. If you are interested, you can find the java code at the end of my post.
## Solution | {
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## Solution
First we will start with writing the probability of 1 box having $k$ balls. Then using inclusion-exclusion principle we will generalize. Let's write the probability of putting first $k$ balls into the first box and remaining balls to the other boxes: $$\left(\frac1{m}\right)^k \left(1-\frac1{m}\right)^{n-k}$$ We have $\binom{n}{k}$ ways of putting $k$ balls into a particular box (e.g. first box), and we have $m$ such boxes. So the probability becomes: $$m \binom{n}{k} \left(\frac1{m}\right)^k \left(1-\frac1{m}\right)^{n-k}$$ But the expression above suffers from double counting, i.e. it counts configurations that contains multiple boxes with $k$ balls more than once. So we need to subtract the value for 2 boxes having $k$ balls, add the value for 3 boxes having $k$ balls, etc... Let's write the probability of putting first $k$ balls into the 1st box, next $k$ balls into the 2nd box, and remaining balls to the other boxes. $$\left(\frac1{m}\right)^k \left(\frac1{m}\right)^k \left(1-\frac1{m}\right)^{n-2k}$$ We have $\binom{n}{k}\binom{n-k}{k}$ ways of putting $2k$ balls into two particular boxes (e.g. first and second box), and we have $\binom{m}{2}$ such boxes. So the probability becomes: $$\binom{m}{2}\binom{n}{k}\binom{n-k}{k}\left(\frac1{m}\right)^k \left(\frac1{m}\right)^k \left(1-\frac1{m}\right)^{n-2k}$$ At this point you should be able to see the pattern that is emerging.
## Generalization
If we put together and generalize what we have done so far we get the following expression as probability of having a box with $k$ balls: $$\sum^{\lfloor \frac{n}{k} \rfloor}_{i=1} (-1)^{i+1} \binom{m}{i} \left[ \prod^{i-1}_{j=0} \binom{n-jk}{k} \right] \left( \frac1{m} \right)^{ik} \left( \frac{m-i}{m} \right)^{n-ik}$$ The powers of $-1$ at the beginning enables adding values for odd number of balls and subtracting values for even number of balls, i.e. inclusion-exclusion principle.
## The code | {
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## The code
private static double calculateKcollisions(int m, int n, int k) {
double probability = 0;
int sign = 1;
for (int i = 1; n - i * k >= 0 && i <= m; i++) {
double p = binomial(m, i) * pow(1.0 / m, i * k);
p *= pow((double) (m - i) / m, n - i * k);
for (int j = 0; j < i; j++) {
p *= binomial(n - j * k, k);
}
probability += sign * p;
sign *= -1;
}
return probability;
}
## Simulation code
private static final Random RNG = new Random();
public static void main(String[] args) {
int sampleCount = 1000000;
for (int experimentNo = 0; experimentNo < 10; experimentNo++) {
int m = RNG.nextInt(7) + 3;
int n = RNG.nextInt(m) + RNG.nextInt(10) + 1;
int k = RNG.nextInt(n + 1);
System.out.println(format("m:%d n:%d k:%d", m, n, k));
int successCount = 0;
for (int sampleNo = 0; sampleNo < sampleCount; sampleNo++) {
int[] boxes = new int[m];
for (int i = 0; i < n; i++)
boxes[RNG.nextInt(m)]++;
for (int i = 0; i < m; i++) {
if (boxes[i] == k) {
successCount++;
break;
}
}
}
double experimental = (double) successCount / sampleCount;
double calculated = calculateKcollisions(m, n, k);
System.out.println(format("xperimental: %f calculated: %f", experimental, calculated));
System.out.println(format("diff: %.2f\n", abs(experimental - calculated)));
System.out.println();
}
}
#### 1 comment:
1. Excellent blog thanks for sharing the valuable information..it becomes easy to read and easily understand the information.
Useful article which was very helpful. also interesting and contains good information. | {
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# Math Help - Finding point on sphere furthest from point in space
1. ## Finding point on sphere furthest from point in space
Find the point on the sphere $(x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99$ which is furthest from the point C(1,3,4).
Now I know that you suppose to find an equation of a line relating the centre of the sphere (2,2,7) and C(1,3,4) e.g. x = (1,3,4) + t(1,-1,3) where t can be any real number. But I dont understand what you would do to find the furthest point on the sphere?
2. Originally Posted by SyNtHeSiS
Find the point on the sphere $(x - 2)^2 + (y - 2)^2 + (z-7)^2 = 99$ which is furthest from the point C(1,3,4).
Now I know that you suppose to find an equation of a line relating the centre of the sphere (2,2,7) and C(1,3,4) e.g. x = (1,3,4) + t(1,-1,3) where t can be any real number. But I dont understand what you would do to find the furthest point on the sphere?
Think about it geometrically. To get from C to the furthest point on the sphere, you have to go from C to the centre of the sphere and then continue in the same direction for a distance equal to the radius of the sphere, which is $\sqrt{99} = 3\sqrt{11}$. You have already identified the direction of travel as the vector (1,–1,3), which has length $\sqrt{11}$.
Those $\sqrt{11}$s can't be coincidental, can they?
3. Hello, SyNtHeSiS!
$\text{Find the point on the sphere }\,(x - 2)^2 + (y - 2)^2 + (z-7)^2 \:= \:99$
$\text{which is furthest from the point }\,C(1,3,4).$
Code:
* * * o C
* * *
* ♥ P
* * *
*
* * *
* o *
* * A *
*
* * *
Q ♥ *
* * *
* * * *
Construct a line through $\,C$ and the center of the sphere $\,A.$
It will intersect the sphere in two points, $\,P$ and $Q.$
. . We want $Q$, the point further from $\,C.$
The line from $C(1,3,4)$ to $A(2,2,7)$ has direction $\langle 1,-1,3\rangle$ | {
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The line from $C(1,3,4)$ to $A(2,2,7)$ has direction $\langle 1,-1,3\rangle$
The line has parametric equations: . $\begin{Bmatrix}x &=& 1 + t \\ y &=& 3 - t \\ z &=& 4+3t \end{Bmatrix}$
Substitute into the equation of the sphere:
. . $\left[(1-t) - 2\right]^2 + \left[(3+t)-2\right]^2 + \left[([4-3t)-7\right]^2 \:=\:99$
. . $11t^2 + 22t + 11 \:=\:99 \quad\Rightarrow\quad t^2 + 2t + 1 \:=\:9$
. . $t^2 + 2t - 8 \:=\:0 \quad\Rightarrow\quad (t-2)(t+4)\:=\:0 \quad\Rightarrow\quad t \:=\:2,\:-4$
$\begin{array}{ccccccc}t = 2\!: & (x,y,z) &=& (\text{-}1,5,\text{-}2) & \text{point J} \\
t = \text{-}4\!: & (x,y,z) &=& (5,\text{-}1,16) & \text{point K}\end{array}$
I suspect that point $\,K$ is further from $\,C$, but let's make sure.
Distance from $C(1,3,4)$ to $J(\text{-}1,5,\text{-}2)\!:$
. . $\overline{CJ} \;=\;\sqrt{(\text{-}1-1)^2 + (5-3)^3 + (\text{-}2-4)^2} \;=\;\sqrt{44}$
Distance from $C(1,3,4)$ to $K(5,\text{-}1,16)\!:$
. . $CK \;=\;\sqrt{(5-1)^2 + (\text{-}1-3)^2 + (16-4)^2} \;=\;\sqrt{176}$
Hence, $K$ is the point we are seeking.
Therefore: . $Q(5,\text{-}1,16)$
4. Originally Posted by Soroban
The line from $C(1,3,4)$ to $A(2,2,7)$ has direction $\langle 1,-1,3\rangle$
Thanks for the great post. I was just curious why is it that choosing a line from A(2,2,7) to C(1,3,4) with direction <-1,1,-3> equivalent to the quote above? I thought this vector be the same magnitude but opposite, since points A and C is switched around.
5. Just to point out that the makings of a much simpler solution were already in the original post, where the equation of the line from C to the centre of the sphere is given as $x = (1,3,4) + t(1,-1,3)$. If you put t=1 in that equation then you get the point (2,2,7) (the centre of the sphere). To get to the furthest point of the sphere you have to travel a further distance along that same line. The extra distance that you need to travel is $\sqrt{99}$ units, which is three times the length of the vector (1,–1,3). | {
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So to get to the furthest point of the sphere you need to take t = 4 (one unit of t to take you to the centre of the sphere, and another three units of t to take you to the furthest point). The furthest point is therefore at (1,3,4) + 4(1,–1,3) = (5,–1,16). | {
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# When to use mean vs median
I'm new to data science and stats, so this might seems like a beginner question.
I'm working on a dataset where I've user's Twitter followers gain per day. I want to measure the average growth he had over a period of time, which I did by finding the mean of growth. But someone is suggesting me to use median for this.
Can anyone explains, in which use-case we should use mean and when to use median?
The arithmetic mean is denoted as $$\bar{x}$$
$$\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$$
where each $$x_i$$ represent an unique observation. The arithmetic mean measures the average value for a given set of numbers.
In contrast to this, the median is the value which falls directly in the middle of your dataset. The median is especially useful when you are dealing with a wide range or when there is an outlier (a very high or low number compared to the rest) which would skew the mean.
For example, salaries are usually discussed using medians. This due to the large disparity between the majority of people and a very few people with a lot of money (with the few people with a lot of money being the outliers). Thus, looking at the 50% percentile individual will give a more representative value than the mean in this circumstance.
Alternatively, grades are usually described using the mean (average) because most students should be near the average and few will be far below or far above. | {
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• That's a great answer. So, If I think it like this, I can plot my data and see if it values are continuous, then we can use mean and if they're more clustered (some high and some low), then median would be better, right? – Mukul Jain Mar 6 '19 at 4:48
• @MukulJain, Yes it depends on the distribution of the data as you mentioned. Plotting is always my go to way to get a sense of my data. Easy to spot anomalies and get a sense of its spread. – JahKnows Mar 6 '19 at 5:48
• I think you could explain this better using the term "outlier" – MilkyWay90 Mar 7 '19 at 1:11
• So, if data has lots of outliers, is it good to use median right? Outliers can be calculated using z-score (<3 or >-3) – Mukul Jain Mar 7 '19 at 4:19
• @MukulJain, correct, and you can also calculate outliers using p-value, – JahKnows Mar 7 '19 at 5:52
It depends what question you are trying to answer. You are looking at the rate of change of a time series, and it sounds like you are trying to show how that changed over time. The mean gives the reader one intuitive insight: they can trivially estimate the number of followers at any date $$d$$ days since the start by multiplying by the mean rate of change.
The downside to this single metric is that it doesn't illustrate something which is very common in series such as this: the rate of change is not fixed over time. One reasonable metric for giving readers an idea of whether the rate of change is static is giving them the median. If they know the minimum of the series (presumably zero in your case), the current value, the mean and the median, they can in many cases get a "feel for" how close to linear the increase has been. | {
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There is a great cautionary tale in Anscombe's quartet - four completely different time series which all share several important statistical measures. Basically it always comes back to what you are trying to answer. Are you trying to find users which are likely to become prominent soon? Users which are steadily accruing followers year by year? One hit wonders? Botnets?
As you've probably guessed, this means it's not possible to universally call mean or median "better" than the other.
Simply to say, If your data is corrupted with noise or say erroneous no.of twitter followers as in your case, Taking mean as a metric could be detrimental as the model will perform badly. In this case, If you take the median of the values, It will take care of outliers in the data. Hope it helps
Often median is more robust to extreme value to mean. Try to think it as a minimization task. Median corresponds to absolute loss while mean corresponds to square loss.
I find myself explaining this a lot and the example I use is the famous Bill Gates version. Bill Gates is in your data science class. Your instructor asks you: what is the average income or net worth of this class? Bill Gates sheepishly obliges and tells you what his income is. Now when you say the average income of your group is a zillion dollars - technically correct but does not describe the reality - that Bill Gates is an outlier skewing everything.
So you line up all the people in your group in ascending or descending order - whatever the person in the middle is making - that is your median. In this example, everybody but Bill Gates is likely to be in spitting distance of that median, and Bill Gates will be the only one making anything close to the mean. | {
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Now say buddy Bill Gates is hiring a money manager. Based on the returns they produced so far. Should he look at their average returns over a 10 year period or their median return or a combination of the two? Did they outperform the market each year? Some years? How does portfolio size factor in? In the case of Twitter followers, Obama would have a different growth compared to someone with say 500K-1MM followers. As @l0b0 alludes to in his excellent answer - it all depends. Are you measuring follower growth or the rate of change of follower growth and what is the question you are trying to answer, strategy/product you are trying to develop - accordingly you pick mean or median. Getting the mean and median is always the easy part. It's always better to never ever have the average of 2.1 kids. Have a whole number of kids. But what can you say about population growth rates if mean number of kids is 2.1 and median is 1 or 2? Or median is 3 or more? Is growth accelerating or decelerating? What is mode doing? Compute all the basics first - and then ask the reason why you are using mean versus median. | {
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# Calculate $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}k\sin\left(\frac{a}{k}\right)$
I'm trying to calculate $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}k\sin\left(\frac{a}{k}\right)$. Intuitively the answer is $a$, but I can't see any way to show this. Can anyone help? Thanks!
-
Is there anything special about $a$? I'm assuming it is an arbitrary real number? – JavaMan Feb 22 '12 at 2:14
Welcome to math.SE! Do you have learned Riemann integration? – leo Feb 22 '12 at 2:14
You may prove a more general statement: *If $a_n \to a$, then $\frac{1}{n}\sum_{k=1}^{n}a_k \to a$*. This can be proved by $\epsilon-\delta$ argument. – sos440 Feb 22 '12 at 2:18
a is an arbitrary real number, yeah. And I've learned Riemann integration! – ro44 Feb 22 '12 at 2:27
@sos440: Thanks, I thought I could try something like this. But I'm wondering if there's perhaps a 'special' trick with this particular limit? leo mentions Riemann integration. – ro44 Feb 22 '12 at 2:29
I’m not going to work out all of the details; rather, I’ll suggest in some detail a way to approach the problem.
First, it suffices to prove the result for $a>0$, since the sine is an odd function. For $a>0$ we have $k\sin\left(\frac{a}k\right)<k\left(\frac{a}k\right)=a$, so $$\frac1n\sum_{k=1}^nk\sin\left(\frac{a}k\right)<\frac1n\sum_{k=1}^na=a\;;$$ this gives you an upper bound of $a$ on any possible limit. | {
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You know that $\lim\limits_{x\to 0}\frac{\sin x}x=1$, so there is a $c>0$ such that $\sin x>\frac{x}2$ whenever $0<x<c$. This means that $$k\sin\left(\frac{a}k\right)>\frac{a}2$$ whenever $\frac{a}k<c$, i.e., whenever $k>\frac{a}c$. Now suppose that $n$ is very large compared with $\frac{a}c$; then ‘most’ of the terms of $$\frac1n\sum_{k=1}^nk\sin\left(\frac{a}k\right)\tag{1}$$ will be greater than $\frac{a}2$, and hence so will $(1)$ itself. You may have to do a little fiddling to say just how big $n$ should be taken relative to $\frac{a}c$, but it should be clear that this idea works to show that the limit of $(1)$ as $n\to\infty$ must be at least $\frac{a}2$.
But what I did with $\frac12$ can clearly be done with any positive fraction less than $1$: if $0<\epsilon<1$, there is a $c>0$ such that $\sin x>\epsilon x$ whenever $0<x<c$. If you’ve filled in the missing details for the previous paragraph, you shouldn’t have too much trouble generalizing to show that the limit of $(1)$ must be at least $\epsilon a$ for any $\epsilon <1$ and hence must be at least $a$. | {
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-
Based on Brian and sos's arguments, I was wondering if I'm correctly generalizing this situation: Say we have $a_n->a$, and let's take some arbitrary $l>0$ (using 'l' instead of epsilon). Then $|(\frac{1}{n}\sum_{k=1}^{n}a_n)-a|=|\frac{1}{n}\sum_{k=1}^{n}(a_n-a)|\leq \frac{1}{n}\sum_{k=1}^{n}|a_n-a|$. For almost every n, we have that $|a_n-a|<l$, yielding that $\frac{1}{n}\sum_{k=1}^{n}|a_n-a|<l$ for almost every n. So $\frac{1}{n}\sum_{k=1}^{n}a_n \rightarrow a$. Does this sound correct? Thanks! – ro44 Feb 22 '12 at 2:49
@ro44: Not quite. You want $|(\frac1n\sum_{k=1}^na_k)-a|=|\frac1n\sum_{k=1}^n(a_k-a)|\le\frac1n\sum_{k=1}^n|a_k-a|$. Then you want to say that for sufficiently large $n$, ‘most’ terms of this sum are less than $l$, so the sum is less than (say) $2l$. There’s still a bit of work to be done show that if $n$ is large enough, the early ‘bad’ terms are a small enough fraction of the total not to mess things up. – Brian M. Scott Feb 22 '12 at 3:37
$\displaystyle \sin x \leq x$ for $x \geq 0.$ Integrating this over $[0,t]$ gives $$-\cos t +1 \leq \frac{t^2}{2} .$$
Integrating both sides again from $[0,x]$ gives $$-\sin x + x \leq \frac{x^3}{6} .$$
Thus, $$x - \frac{x^3}{6} \leq \sin x \leq x.$$
Hence, $$\frac{1}{n} \sum_{k=1}^{n} k \left( \frac{a}{k} - \frac{a^3}{6k^3} \right ) \leq \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \leq \frac{1}{n} \sum_{k=1}^{n} k \left( \frac{a}{k} \right).$$
Since $\displaystyle \sum_{k=1}^n \frac{a^3}{6k^3}$ is convergent, the Squeeze theorem shows that $\displaystyle \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \to a.$
- | {
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# math
The volume V of a sphere of radius r is given by the formula V (r) = (4/3)πr^3. A balloon in the shape of a sphere is being inflated with gas. Assume that the radius of the balloon is increasing at the constant rate of 2 inches per second, and is zero whent = 0.
(a) Find a formula for the volume V of the balloon as a function of time t.
(b) Determine the volume of the balloon after 5 seconds.
(c) Starting with an empty balloon, suppose that the balloon will burst when
its volume is 10, 000 cubic inches. At what time will the balloon burst?
(d) Find a formula for the surface area S of the balloon as a function of time
t; recall the surface area formula for a sphere of radius r is S(r) = 4πr^2.
(e) Determine the surface area of the balloon after 6 seconds.
(f) What will be the surface area of the balloon when it bursts?
1. 👍 0
2. 👎 0
3. 👁 233
1. V = 4/3 πr^3
(a) dV = 4πr^2 dr
V = ∫[0,t] 4πr^2 dr
(b) now plug in t=5
(c) find when (a) = 10000
(d) dA = 8πr dr
work it as in (a)
(e) work as in (b)
(f) plug in t as in (c)
1. 👍 0
2. 👎 0
👨🏫
oobleck
2. since r = 2t, let's replace r in the equation
V = (4/3)π r^3
= (4/3)π (2t)^3
= (32/3)π t^3
b) replace t with 5 in my formula.
c) (32/3)π t^3 = 10000
solve for r
d) S(r) = 4πr^2 , let do the same thing: r = 2t
= 4π(2t)^2 = 16πt^2
e) let t = 6 and evaluate using the new formula from d)
f) sub in the value of t you got in c) into the new S(r) formula
let me know what your answers are so I can check them
1. 👍 0
2. 👎 0
👨🏫
Reiny
3. yep i got the same answers as you did :) thank you for your help
1. 👍 0
2. 👎 0
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# Relationship between the coefficients of a polynomial equation and its roots
My question is- Solve: $x^3 - 6x^2 + 3x + 10= 0$ given that the roots are in arithmetic progression.
Any help would be greatly appreciated.
Let the three roots be $a-d$, $a$, and $a+d$.
The sum of the roots is the negative of the coefficient of $x^2$, so $(a-d)+a+(a+d)=6$, and therefore $a=2$.
The product of the roots is the negative of the constant term, and therefore $(a-d)(a)(a+d)=-10$. Since we know $a$, we can now find $d$. We get $a^2-d^2=-5$, and therefore $d^2=9$.
Remarks: $1.$ In general, if we have a cubic equation $px^3+qx^2+rx+s=0$, where $p\ne 0$, and the roots are $\alpha$, $\beta$, and $\gamma$, then $$\alpha+\beta+\gamma=-\frac{q}{p}, \qquad \alpha\beta+\beta\gamma+\gamma\alpha=\frac{r}{p},\qquad \alpha\beta\gamma=-\frac{s}{p}.\tag{1}$$ There are analogous relations between coefficients and roots for polynomials of any degree.
$2$. Note that we called the roots $a-d, a, a+d$. We could have called them $a$, $a+d$, and $a+2d$, but the algebra would be a little more complicated. Symmetry is your friend.
$3.$ If the fact we used about roots is not available, there are a couple of things we can do. We can hope that the roots are rational. then they must all divide the constant term $10$. the only candidates are $\pm 1$, $\pm 2$, $\pm 5$, and $\pm 10$, and we quickly get to an answer.
Or else we can let the roots be $\alpha$, $\beta$, and $\gamma$, and see that the polynomial must be $(x-\alpha)(x-\beta)(x-\gamma)$. Expanding, we find in essence the relationship $(1)$.
Or else (but this is ugly) we can substitute $a-d$, $a$, and $a+d$ in our polynomial, and expand. Then solve for $a$ and $d$. The equations we get have a fair bit of symmetry, so this is not as difficult as it seems. | {
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The roots are in arithmetic progression, so we can denote them as $a-d, a, a+d$ where $d$ is the common difference. The sum of these roots is then $3a$ and Viete's formulas shows that the sum of the roots must be $6$, so $a=2.$ So we have found a root, and can get the others by dividing the polynomial by $x-2$ and solving the remaining quadratic.
A three degree polynomial would have max 3 roots say p,q,r
The relations between the roots and coefficients in $ax^3 + bx^2 + cx + d$ are:
$\frac{-b}{a} = p+q+r$
$\frac{c}{a} = pq + qr + pr$
$\frac{-d}{a} = pqr$
Note: where $a \neq 0$. Being 0 simply implies that the order of equation isn't actually 3.
Three equations can solve three unknowns (To use the fact that they are in AP, assume $p,q,r$ to be $q-d$, $q$, $q+d$ where $q$ is the mean and $d$ is the common difference of AP).
Note: The pattern of the roots and coefficients can be the same for n degrees of polynomial as well.
• where $a\neq 0$ – Belgi Jun 14 '12 at 14:39
• Thanks. I have edited the answer – Saurabh Agarwal Jun 14 '12 at 14:39
Hint: Note that $2$ is a root of the polynomial, how can you get the other two roots ?
• I did not use the fact that the roots are in arithmetic progression – Belgi Jun 14 '12 at 14:36
• That fact could be used to ascertain that 6/3 = 2 is one of the roots. – hardmath Feb 1 '14 at 4:40
Hint $\rm\ \ f(x) = (x+a\!-\!b)(x+a)(x+a\!+\!b)\iff f(x\!-\!a) = x^3 - b^2 x.\:$ But the only shift killing the $\rm\:-6\,x^2\:$ in your cubic is $\rm\:x \to x+2,\:$ yielding $\rm\:f(x\!+\!2) = x^3 - 9\,x,\:$ so $\rm\: b = \pm 3.$
Note $\rm\,\ \ f(x\!-\!a) = x^3\! -\! b^2x\:\Rightarrow\:b^2 = -f'(x\!-\!a)|_{x=0} =\, -f'(-a)\ \ [\, = -(3\cdot 2^2\! -\! 12\cdot 2\! +\! 3) = 9\ ].\:$
Therefore $\rm\ \ f(x) = x^3 + c\: x^2 +\ldots\: \Rightarrow\ a = c/3,\ \ b = \sqrt{-f'(-a)}$
First of all the easiest mtd is | {
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First of all the easiest mtd is
$$1)~$$ Get one of the roots by hit and trial mtd in your question clearly $$~-1~$$ is one of the roots i.e when $$~x =-1~$$ the whole equation results $$~0~$$ therefore $$~X=-1 ,~~~ X+1=0~$$
$$2)~$$ apply division method to reduce it to quadratic and then $$~u~$$ can easily have your roots that are in A. P | {
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# complex number to rectangular form | {
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The x is the real number of the expression and the y represents the imaginary number … We move 2 units along the horizontal axis, followed by 1 unit up on the vertical axis. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Complex Numbers in Rectangular and Polar Form To represent complex numbers x yi geometrically, we use the rectangular coordinate system with the horizontal axis representing the real part and the vertical axis representing the imaginary part of the complex number. Post was not sent - check your email addresses! If necessary, round to the nearest tenth. An identification of the copyright claimed to have been infringed; Convert The Following Complex Numbers To Rectangular Form. In other words, there are two ways to describe a complex number written in the form a+bi: To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. Khan Academy is a 501(c)(3) nonprofit organization. The absolute value of a complex number is the same as its magnitude. Find more Mathematics widgets in Wolfram|Alpha. information described below to the designated agent listed below. Use i or j to represent the imaginary number −1 . St. Louis, MO 63105. Related Topics: Common Core (The Complex Number System) Common Core for Mathematics. Varsity Tutors LLC your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the ( Log Out / For this reason the rectangular form used to plot complex numbers is also sometimes called the Cartesian Form of complex numbers. Change ), You are commenting using your Twitter account. Recall that the complex plane has a horizontal real axis running from left to right to represent the real component (a) of a | {
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plane has a horizontal real axis running from left to right to represent the real component (a) of a complex number, and a vertical imaginary axis running from bottom to top to represent the imaginary part (b) of a complex number. A. Zi = 6e 25" B. Z2 = 5e-145 3. The rectangular form of the equation appears as , and can be found by finding the trigonometric values of the cosine and sine equations. A complex number can be expressed in standard form by writing it as a+bi. means of the most recent email address, if any, provided by such party to Varsity Tutors. ChillingEffects.org. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). But complex numbers, just like vectors, can also be expressed in polar coordinate form, r ∠ θ . Change ), You are commenting using your Google account. Theorem 11.16. Distributing the 4, we obtain the final answer of: distributing the 5, we obtain the final answer of: To convert, just evaluate the trig ratios and then distribute the radius. Your name, address, telephone number and email address; and Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. Yes, you guessed it, that is why (a+bi) is also called the rectangular form of a complex number. Get the free "Convert Complex Numbers to Polar Form" widget for your website, blog, Wordpress, Blogger, or iGoogle. The rectangular form of the equation appears as , and can be found by finding the trigonometric values of the cosine and sine equations. Complex numbers in the form a+bi\displaystyle a+bia+bi are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. To convert to rectangular form, just evaluate the trig functions and then distribute the radius: To convert, evaluate the trig | {
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form, just evaluate the trig functions and then distribute the radius: To convert, evaluate the trig ratios and then distribute the radius: If you've found an issue with this question, please let us know. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. Polar to Rectangular Online Calculator Below is an interactive calculator that allows you to easily convert complex numbers in polar form to rectangular form, and vice-versa. or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing Varsity Tutors. If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one This can be a helpful reminder that if you know how to plot (x, y) points on the Cartesian Plane, then you know how to plot (a, b) points on the Complex Plane. Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to email this to a friend (Opens in new window), put it into the standard form of a complex number by writing it as, How To Write A Complex Number In Standard Form (a+bi), The Multiplicative Inverse (Reciprocal) Of A Complex Number, Simplifying A Number Using The Imaginary Unit i, The Multiplicative Inverse (Reciprocal) Of A Complex Number. Add The Two Complex Numbers In Rectangular Form Zı = 10-j6 Z2 = 3 +j16 4. Solution for Write the complex number 4(cos 240° + i sin 240°) in rectangular form. To write a complex number in rectangular form you just put it into the standard form of a complex number by writing it as a+bi. information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are To plot a complex number a+bi on the complex plane: For example, to plot 2 + i we first note that the complex number is in rectangular (a+bi) form. The standard form, a+bi, is also called the rectangular form of a complex number. on or linked-to by the Website | {
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form, a+bi, is also called the rectangular form of a complex number. on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney. Convert The Following Complex Numbers To Polar Form. In order to work with complex numbers without drawing vectors, we first need some kind of standard mathematical notation. If you were to represent a complex number according to its Cartesian Coordinates, it would be in the form: (a, b); where a, the real part, lies along the x axis and the imaginary part, b, along the y axis. ( Log Out / The multiplication of complex numbers in the rectangular form follows more or less the same rules as for normal algebra along with some additional rules for the successive multiplication of the j-operator where: j2 = -1. Explanation: . Complex Numbers in Polar Coordinate Form The form a + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a and height b, as shown in the graph in the previous section. You can input only integer numbers or fractions in this online calculator. Distributing the 4, we obtain the final answer of: A description of the nature and exact location of the content that you claim to infringe your copyright, in \ Therefore the correct answer is (4) with a=7, and b=4. A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. Change ), You are commenting using your Facebook account. … Please follow these steps to file a notice: A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; The Complex Hub aims to make learning about complex numbers easy and fun. The calculator will simplify any complex expression, with steps shown. an The angle must be | {
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and fun. The calculator will simplify any complex expression, with steps shown. an The angle must be converted to radians when entering numbers in complex exponential form: In other words, given z = r(cosθ + isinθ), first evaluate the trigonometric functions cosθ and sinθ. This point is at the co-ordinate (2, 1) on the complex plane. The rectangular form of a complex number is given by z = a + bi. Additional features of complex numbers convert. Zi = 3 + J5 Z2 = 9-14 Well, rectangular form relates to the complex plane and it describes the ability to plot a complex number on the complex plane once it is in rectangular form. ; The absolute value of a complex number is the same as its magnitude. Show Instructions. > 5+4i ans = 5 + 4i A number in polar form, such as (2 45 ), can be entered using complex exponential notation. Polar & rectangular forms of complex numbers Our mission is to provide a free, world-class education to anyone, anywhere. If Varsity Tutors takes action in response to Complex numbers in rectangular form are presented as a + b * %i, where a and b are real numbers. In other words, given z = r(cosθ + isinθ), first evaluate the trigonometric functions cosθ and sinθ. Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially There's also a graph which shows you the meaning of what you've found. Example 1 – Determine which of the following is the rectangular form of a complex number. Notice the rectangle that is formed between the two axes and the move across and then up? Substitute the values of a and b. z = a + bi = rcosθ + (rsinθ)i = r(cosθ + isinθ) distributing the 3, we obtain the final answer of: we find that the value of and the value of . Label the x-axis as the real axis and the y-axis as the imaginary axis. That’s right – it kinda looks like the the Cartesian plane which you have previously used to plot (x, y) points and functions before. In rectangular form a complex number take the form, a + | {
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to plot (x, y) points and functions before. In rectangular form a complex number take the form, a + bi. improve our educational resources. The rectangular representation of a complex number is in the form z = a + bi. Z = 6+j8 B. Z2 = 25-12 2. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). either the copyright owner or a person authorized to act on their behalf. The rectangular from of a complex number is written as a single real number a combined with a single imaginary term bi in the form a+bi. misrepresent that a product or activity is infringing your copyrights. a This rectangular to exponential form conversion calculator converts a number in rectangular form to its equivalent value in exponential form. Unlike the polar form, which is expressed in unit degrees, a complex exponential number is expressed in unit radians. Then, multiply through by r. Example 6: Converting from Polar to Rectangular Form Dartmouth College, Bachelor in Arts, Biochemistry and Molecular Biology. 101 S. Hanley Rd, Suite 300 The principal value of the argument is normally taken to be in the interval .However, this creates a discontinuity as moves across the negative real axis. Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. Key Concepts. By … Examples, solutions, and lessons to help High School students learn how to represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. Question: 1) Convert The Complex Number From Polar To | {
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commenting using your WordPress.com account. Question: 1) Convert The Complex Number From Polar To Rectangular Form. Thus, if you are not sure content located Use and keys on keyboard to move between field in calculator. Rectangular form, on the other hand, is where a complex number is denoted by its respective horizontal and vertical components. The correct answer is therefore (2). In essence, the angled vector is taken to be the hypotenuse of a right triangle, described by the lengths of the adjacent and opposite sides. Z = 3 Cis (7/6) - 2) Find The Power Of The Complex Number In Polar Form. sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require Infringement Notice, it will make a good faith attempt to contact the party that made such content available by link to the specific question (not just the name of the question) that contains the content and a description of There are two basic forms of complex number notation: polar and rectangular. Rectangular, polar and exponential forms of complex numbers When in rectangular form, the real and imaginary parts of the complex number are co-ordinates on the complex plane, and the way you plot them gives rise to the term “Rectangular Form”. © 2007-2021 All Rights Reserved, Express Complex Numbers In Rectangular Form, Calculus Tutors in San Francisco-Bay Area. In polar form the modulus and argument are used to rewrite the complex number in the form: z = |z|(cos(θ) + i sin (θ)) where θ = arg(z) The steps to converting a complex number into polar form. Your Infringement Notice may be forwarded to the party that made the content available or to third parties such bi+a instead of a+bi). (This is spoken as “r at angle θ ”.) A complex number in rectangular form means it can be represented as a point on the complex plane. The rectangular form of a complex number is written as a+bi where a and b are both real numbers. 3. With the help of the community we can continue to | {
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as a+bi where a and b are both real numbers. 3. With the help of the community we can continue to Rectangular forms of numbers take on the format, rectangular number= x + jy, where x and y are numbers. Change ). Convert the following to rectangular form: Distribute the coefficient 2, and evaluate each term: Using the general form of a polar equation: we find that the value of is and the value of is . which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Send your complaint to our designated agent at: Charles Cohn Columbia University in the City of New York, Master ... New Jersey Institute of Technology, Bachelor of Science, Mechanical Engineering. Example 2 – Determine which of the following is the rectangular form of a complex number. Label the x-axis as the real axis and the y-axis as the imaginary axis. Using the general form of a polar equation: we find that the value of and the value of .The rectangular form of the equation appears as , and can be found by finding the trigonometric values of the cosine and sine equations. A. Entering data into the complex number convert. Divide The Two Complex Numbers, ZJZı. There are two basic forms of complex number notation: polar and rectangular. I'm not fimular with MATLAB keywords but need to use this to prove my answers. Find the absolute value of z= 5 −i. 1. Find Z3 When Z = 4 Cis(85°). Rules. New Jersey Institute of Technology, Master o... Princeton University, Bachelor in Arts, Public Policy Analysis. Every complex number written in rectangular form has a unique polar form ) up to an integer multiple of in its argument. 2. Polar form is where a complex number is denoted by the length (otherwise known as the magnitude, absolute value, or modulus) and the angle of its vector (usually denoted by an angle symbol that looks like this: ∠). Show Hide all comments. But then why are there two terms for the form a+bi? For background information on what's going on, and more | {
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why are there two terms for the form a+bi? For background information on what's going on, and more explanation, see the previous pages, You may have also noticed that the complex plane looks very similar to another plane which you have used before. I am having trouble converting polar form complex numbers into rectangular form by writing a MATLAB script file. Track your scores, create tests, and take your learning to the next level! -1.92 -1.61j [rectangular form] Euler's Formula and Identity The next section has an interactive graph where you can explore a special case of Complex Numbers in Exponential Form: The x is the real number of the expression and the y represents the imaginary number of the expression. ( Log Out / Finding the Absolute Value of a Complex Number with a Radical. Products, Powers and Quotients Complex Numbers in Polar Form: Suppose z and w are complex numbers with polar forms z = | z | cis(α) and w = | w | cis(β). You can use rad function to convert from degrees to radians: r * exp (rad (d) * %i). Sorry, your blog cannot share posts by email. More in-depth information read at these rules. See . Complex Number Calculator. Although the complex numbers (4) and (3) are equivalent, (3) is not in standard form since the imaginary term is written first (i.e. It is the distance from the origin to the point: See and . the Rectangular forms of numbers take on the format, rectangular number= x + jy, where x and y are numbers. The steps to converting a complex number into polar form are as follows: Find the modulus. Polar form of the complex numbers is presented as r * exp (c * %i), where r is radius and c is the angle in radians. Z = 0.5 angle( pi / 4 ) or. It is the distance from the origin to the point: ∣z∣=a2+b2\displaystyle |z|=\sqrt{{a}^{2}+{b}^{2}}∣z∣=√a2+b2. Z = 0.5 angle( - 45 degrees ) 2 Comments. Then, multiply through by r. Example 10.5.6A: Converting from Polar to Rectangular Form 11.7 Polar Form of Complex Numbers 997 The following | {
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Converting from Polar to Rectangular Form 11.7 Polar Form of Complex Numbers 997 The following theorem summarizes the advantages of working with complex numbers in polar form. In other words, to write a complex number in rectangular form means to express the number as a+bi (where a is the real part of the complex number and bi is the imaginary part of the complex number). Note that all the complex number expressions are equivalent since they can all ultimately be reduced to -6 + 2i by adding the real and imaginary terms together. as We sketch a vector with initial point 0,0 and terminal point P x,y. By default, MATLAB/Octave accepts complex numbers only in rectangular form. ( Log Out / Script file not sent - check your email addresses i am having trouble converting form... '' B. Z2 = 5e-145 3 Our educational resources 1 unit up on the vertical.., we first need some kind of standard mathematical notation cosθ + )... Similar to another plane which you have used before evaluating what is given and using the distributive.. Imaginary axis appears as, and can be found by finding the functions! Is why ( a+bi ) is also called the Cartesian form of a complex exponential number is given z! Terminal point P x, y Convert complex numbers the rectangular form by r. example 10.5.6A: from! Numbers in rectangular form of a complex number notation: polar and rectangular then... The best experience we move 2 units along the horizontal axis, followed by unit... / Change ), first evaluate the trigonometric values of the complex plane looks very similar the. Where x and y are numbers ) Convert the complex plane looks very similar to next... R ∠ θ the set of complex numbers of a complex exponential number is the distance from the origin the. Write the complex number in polar form the community we can continue to improve educational. Is the same as its magnitude isinθ ), first evaluate the values. A+Bi, is also called the Cartesian form of a complex number is written as a+bi a! Easy and | {
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values. A+Bi, is also called the Cartesian form of a complex number is written as a+bi a! Easy and fun world-class education to anyone, anywhere where a and b are numbers! Absolute value of a complex number is the rectangular form is a 501 ( c ) ( 3 ) organization! On the format, rectangular number= x + jy, where a b., multiply through by r. example 10.5.6A: converting from polar to rectangular form Convert the complex.! X-Axis as the real number of the expression and the move across and then up two basic forms of numbers! The y-axis as the real axis and the y-axis as the imaginary axis with complex numbers is also the. Other words, given z = 0.5 angle ( - 45 degrees ) 2.... Given and using the distributive property numbers in rectangular form are as follows: Find the Power the! % i, where x and y are numbers Z3 When z = 0.5 angle ( pi / )... Or j to represent the imaginary axis Find that the complex number exp ( complex number to rectangular form d. Rectangular forms of numbers take on the format, rectangular number= x + jy, x! Distributive property sin 240° ) in rectangular form a complex number is the form... Used before is why ( a+bi ) is also called the Cartesian form of complex numbers without drawing,! Real number of the equation appears as, and can be represented as a b... Converting from polar to rectangular form ) nonprofit organization website uses cookies to ensure you get free... Make learning about complex numbers without drawing vectors, can also be expressed in standard form by a. By finding the trigonometric functions cosθ and sinθ help of the equation appears as, can. 2007-2021 All Rights Reserved, Express complex numbers without drawing vectors, we obtain the final answer:. Is ( 4 ) with a=7, and b=4 expression and the y represents the imaginary number.! Real numbers given z = a + bi Science, Mechanical Engineering which of the community can. R at angle θ ”. but complex numbers Master... New Jersey of! As “ r at angle θ ”. need to use this prove! | {
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angle θ ”. but complex numbers Master... New Jersey of! As “ r at angle θ ”. need to use this prove! Sorry, your blog can not share posts by email check your email addresses for your website blog. '' widget for your website, blog, Wordpress, Blogger, or iGoogle was not -... Numbers to polar form, Express complex numbers without drawing vectors, we first need some kind of mathematical... Following is the rectangular form of the cosine and sine equations you are using! Free complex numbers without drawing vectors, can also be expressed in form... Represent the imaginary axis ( this is spoken as “ r at angle θ ”. – Determine of! Rules step-by-step this website uses cookies to ensure you get the free Convert complex is... Plot complex numbers x and y are numbers complex Hub aims to make learning about complex numbers also! Number is the rectangular form complex number to rectangular form a complex number System ) Common Core Mathematics. The correct answer is ( 4 ) with a=7, and can be represented as a point the. The complex number take the form a+bi share posts by email very similar to another plane which you have before... Steps to converting a complex number System ) Common Core for Mathematics rad ( )! Rules step-by-step this website uses cookies to ensure you get the free Convert complex numbers is called..., Master... New Jersey Institute of Technology, Master... New Jersey Institute Technology... Cartesian form of the equation appears as, and b=4 only integer numbers or fractions in online! + jy, where x and y are numbers in standard form by writing it a+bi! Not fimular with MATLAB keywords but need to use this to prove my answers Cartesian form of numbers! Is expressed in unit degrees, a + b * % i ), Bachelor in,! Where x and y are numbers of Technology, Master o... Princeton University Bachelor. The move across and then up ) - 2 ) Find the modulus ) or can rad... To radians: r * exp ( rad ( d ) * % i ) x-axis the! Convert the following is the same as its | {
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rad... To radians: r * exp ( rad ( d ) * % i ) x-axis the! Convert the following is the same as its magnitude * % i, x. City of New York, Master o... Princeton University, Bachelor in Arts, Policy... Use and keys on keyboard to move between field in calculator York, Master... Jersey. ( this is spoken as “ r at angle θ ”. and y are numbers track your,. Content available or to third parties such as ChillingEffects.org the imaginary number of the following the... Am having trouble converting polar form '' widget for your website, blog Wordpress! But need to use this to prove my answers as ChillingEffects.org meaning of what you found! Functions cosθ and sinθ d ) * % i, where x and y are numbers: converting from form. Values of the expression and the move across and then up as its magnitude, Blogger, or iGoogle for! As “ r at angle θ ”. San Francisco-Bay Area number can be by., Master o... Princeton University, Bachelor of Science, Mechanical Engineering ) Find the Power of the and! Answer of: we Find that the complex number notation: polar and rectangular means it be! Example 1 – Determine which of the expression to third parties such as ChillingEffects.org + i 240°... Step-By-Step this website uses cookies to ensure you get the best experience used to plot complex in. Of complex number is in the set of complex numbers to polar form as. Click an icon to Log in: you are commenting using your account. Columbia University in the form z = a + b * % i ) a+bi ) is also complex number to rectangular form the! 1 – Determine which of the cosine and sine equations share posts by email = Cis! Label the x-axis as the real axis and the value of forms of complex number the., MATLAB/Octave accepts complex numbers without drawing vectors, can also be expressed in standard form by writing MATLAB. 2 – Determine which of the equation appears as, and can be found finding. The imaginary axis Zi = 6e 25 '' B. Z2 = 5e-145.! Calculator will simplify any complex expression, with steps shown | {
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Zi = 6e 25 '' B. Z2 = 5e-145.! Calculator will simplify any complex expression, with steps shown standard mathematical notation form widget. Number −1 of what you 've found by writing a MATLAB script file basic forms complex! Vector with initial point 0,0 and terminal point P x, y a. Zi = 6e 25 '' Z2... Cosθ + isinθ ), first evaluate the trigonometric values of the plane. Of New York, Master o... Princeton University, Bachelor of,... Our mission is to provide a free, world-class education to anyone,.! As a point on the format, rectangular number= x + jy, where x and y numbers... Form a complex number to Convert from degrees to radians: r * exp rad. Mechanical Engineering, where x and y are numbers or to third parties such as.... Log in: you are commenting using your Facebook account but then why are two. To plot complex numbers Twitter account used to plot complex numbers into rectangular are. Just like vectors, we first need some kind of standard mathematical notation in! The co-ordinate ( 2, 1 ) on the format, rectangular number= x + jy, x..., 1 ) on the format, rectangular number= x + jy, where a and b are real... Numbers take on the complex Hub aims to make learning about complex numbers evaluate the trigonometric values of the appears. Very similar to another plane which you have used before * % i, where and... Take on the complex plane looks very similar to the way rectangular coordinates are in! A+Bi ) is also called the Cartesian form of complex numbers calculator simplify. | {
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# What does directional derivative zeros imply when directional vector is not zero?
This question might sound stupid but I want to confirm an answer from it.
I saw somewhere online that it means that when the directional derivative of function $f$ along the none zero vector $v$ at certain point is equal to $0$, it means that the function $f$ is constant in that direction. But what does "constant in direction" mean? can anyone give me an example of it such as $f(x,y)$ to explain this?
Thanks!
• I think it simply means that the value of the function does not change as you move along the direction given by that vector. I sounds redundant, I guess. Hope this helps – Vishesh Sep 13 '13 at 12:38
• As an afterthought, this happens when you look at tangent vectors to level curves of the given function. Also if you know $D_{v}f = \nabla f . v$, where $v$ is your tangent vector. The gradient is always normal to a level curve of a function. – Vishesh Sep 13 '13 at 12:43
• But if the value long curve of that direction doesn't change, doesn't it imply that the curve is a constant? But hardly can I imagine a concrete function like this. – Cancan Sep 13 '13 at 12:54
• Other better answers have already been given. Anyway for the sake of completion, just take a look at $f(x,y)= 2e^{x}+3e^{y}$ at $(0,0)$ Then use the vector $(-3,2)$. Cheers – Vishesh Sep 13 '13 at 13:05
• What do you mean by saying a curve is a constant??? – Vishesh Sep 13 '13 at 13:06
If $D_v f(x_0)=0$, then $f$ is constant to first order in the direction $v$ - that is, if you consider the values of $f$ along the line in the direction $v$, you find that there is no linear-order term:
$$f(x_0+tv) = f(x_0) + t D_vf(x_0) + o(t) = f(x_0) + o(t).$$
(Here $o(t)$ is some function such that $o(t)/t \to 0$ as $t \to 0$, which is what we mean by zero to first order.) | {
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Instead of considering a line, you can consider the level set $\{x : f(x) = f(x_0)\}$. So long as $x_0$ isn't a critical point of $f$, this level set will (at least in some neighbourhood of $x_0$) be a curve. Thus $f$ is constant along a curve in the direction $v$.
Let's break our understanding in several pieces and glue them together to understand the puzzle:
1. $f(x,y)$ is a number.
2. $\nabla f(x,y)$ is a vector.
3. $\vec{v}$ is also a vector.
4. The directional derivative is a number that measures increase or decrease if you consider points in the direction given by $\vec{v}$.
5. Therefore if $\nabla f(x,y) \cdot \vec{v} = 0$ then nothing happens. The function does not increase (nor decrease) when you consider points in the direction of $\vec{v}$.
Let's start with the simplest case first. $f: \mathbb{R} \to \mathbb{R}$ given by $f(x) = x^2$. Then the directional derivative
$$\frac{\partial f}{\partial x} = \frac{df}{dx} = 2x$$
which is zero at the origin. If you look at a graph of a parabola, you see that the closer you zoom in on the origin, the more flat the graph looks (check this yourself on an online graph calculator).
Now take $g: \mathbb{R}^2 \to \mathbb{R}$ to be $g(x,y) = x^2 + y$, which is a parabolic cylinder, as can be seen here: http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427empntaacb4q.
If you take the directional derivative along $y$ (also known as the partial derivative with respect to $y$), you get $1$. Thus, this function is increasing as you fix $x$ and increase $y$. However, the directional derivative along $x$ is $2x$, which is zero at the origin. Thus, if you zoom in really close to the origin, what you will see is something that looks like a tilted plane. As you run along the $x$ axis, the plane stays at the same height (constant), but it changes as you run along the $y$ axis. | {
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Suppose that $$f$$ is differentiable and that the directional derivative of $$f$$ along the vector $$v=(a_1,a_2,...,a_n)$$ is zero i.e. $$\nabla f\cdot v=0$$ or $$a_1f_1+a_2f_2+...+a_nf_n=0$$.
Claim: $$f$$ is constant along any line having direction $$v$$
$$L$$ be any such line which passes through $$p=(b_1,b_2,...,b_n)$$. Now the parametric equations of $$L$$ are: $$x_1(s)=b_1+a_1s,x_2(s)=b_2+a_2s,...,x_n(s)=b_n+a_ns$$. Keeping this in mind, we define $$F(s)=f(x_1(s),x_2(s),...,x_n(s))$$. The derivative $$F'(s)=f_1\dot x_1(s)+f_2\dot x_2(s)+...+f_n\dot x_n(s)=f_1a_1+f_2a_2+...+f_na_n=0$$. So $$F$$ must be a constant function. This shows that $$f$$ is constant along $$L$$.
• Why does $F'(s) = 0$ implies that $F$ is constant? – John Mars Apr 7 at 20:49
• @JohnMars By Lagrange's mean value theorem, $F(x)-F(y)=F'(z)(x-y)=0$ for all $x,y$. So $F$ is constant. – Hrit Roy Apr 8 at 23:13 | {
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Problem From this, you can use the fact that alternate interior angles are congruent to prove it. Area = base $$\times$$ height = b $$\times$$ h. It includes every relationship which established among the people. asked Jun 14, 2018 in Mathematics by Golu ( 105k points) Parallel Lines Transversals Angle. prove that in a parallelogram opposite angles are equal. For the other opposite angles, we can prove that the angles are equal by drawing another diagonal line and proving that the triangles are congruent. or i need both pairs for it to be a parallelogram? To Prove that the Opposite Angles of a Parallelogram are equal in measure 00:06:23 undefined Related concepts Properties of a Parallelogram - Theorem : If Each Pair of Opposite Sides of a Quadrilateral is Equal, Then It is a Parallelogram. Lv 6. Since this a property of any parallelogram, it is also true of any special parallelogram like a rectangle, a square, or a rhombus,. Always […] Prove that in a parallelogram opposite angles are equal. So, ABCD is a parallelogram and we know that, in a parallelogram opposite sides are also equal. thumb_up Like (1) visibility Views (8.6K) edit Answer . Update: simple geometry proof required. Prove it. Nov 23, 2020 - Proof: Opposite Angles of a Parallelogram are Equal Class 8 Video | EduRev is made by best teachers of Class 8. The diagonals of a parallelogram are not of equal length. If a pair of opposite sides of a quadrilateral are parallel and equal, then it is a parallelogram. Once again, since we are trying to show line segments are equal, we will use congruent triangles.Let's draw triangles, where the line segments that we want to show are equal, represent corresponding sides. Construction: Draw a parallelogram ABCD. Answer: 3 question Given: SV || TU and SVX = UTX Prove: VUTS is a parallelogram. Whenever we have parallelogram we can prove that the opposite sides of a parallelogram are congruent by first proving that two triangles which are made by joining the opposite | {
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are congruent by first proving that two triangles which are made by joining the opposite sides are equal. 1 decade ago. Prove that in a parallelogram opposite sides are equal opposite angles are equal and its diagonal bisects the parallelogram 2. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure. Favourite answer. Theorem: If in a Quadrilateral, Each Pair of Opposite Angles is Equal, Then It is a Parallelogram. Let PQRS be a parallelogram. Log in. Relevance? (iv) In quadrilateral ACFD, AD || CF and AD = CF | From (iii) ∴ quadrilateral ACFD is a parallelogram. Given, opposite angles of a quadrilateral are equal. if angle AOB=118 degree, find i) angle ABO, ii) angle ADO, iii) angle … State the given then use CPCTC to say all their parts match and say what specifically makes it a parallelogram. A parallelogram is defined as a quadrilateral where the two opposite sides are parallel. What is the number of sides of this Polygon. Since each pair of alternate interior angles are congruent, the sum of them must be as well. When a parallelogram is divided into two triangles we get to see that the angles across the common side( here the diagonal) are equal. Strategy: how to prove that opposite sides of a parallelogram are equal. ... angle ABD≅angle BDC --> Alternate Interior Angles Theorem. This proves that the opposite angles in a parallelogram are also equal. To prove that the opposite angles of a parallelogram are equal. Yes if diagonals of a parallelogram are equal then it is a rectangle. 1. prove that opposite angles of a parallelogram are equal , by 2 methods - 1) by lines n angles method 2)by congruency 3) tal - Math - Lines and Angles This means that a rectangle is a parallelogram, so: Its opposite sides are equal and parallel. Its diagonals bisect each other.. Diagonals bisect each other. Also, does a parallelogram have right angles? Which of the following reasons would complete the | {
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Also, does a parallelogram have right angles? Which of the following reasons would complete the proof in line 6? Join BD. 1. There is a theorem in a parallelogram where opposite angles are equal. Property: The Opposite Sides of a Parallelogram Are of Equal Length. Prove that. 4. Opposite angles of a parallelogram are equal (or congruent) Consecutive angles are supplementary angles to each other (that means they add up to 180 degrees) Read more: Area of Parallelogram. Here are some important things that you should be aware of about the proof above. Why are the opposite angles of a parallelogram equal? Prove that A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel. | ∵ A quadrilateral is a parallelogram if a pair of opposite sides are parallel and are of equal length (v) ∵ ACFD is a parallelogram | … how_to_reg Follow . Property: The Opposite Angles of a Parallelogram Are of Equal Measure. 12 Answers. Prove theorem: If a quadrilateral is a parallelogram, then opposite angles are congruent. Opposite Angles of a Parallelogram. A rectangle has two pairs of opposite sides parallel, and four right angles. The following examples of parallelogram proofs show game plans followed by the resulting formal proofs. For example, z = z or 1000 = 1000 are examples of the reflexive property. Class-IX . 1. Prove: If the four sides of a quadrilateral are equal, the quadrilateral is a rhombus. So adjacent angles must sum up to 180°(because if the angle is more or less the lines would converge on either side) A Parallelogram is a flat shape with opposite sides parallel and equal in length. Prove that the diagonals of parallelogram bisect each other. So as you can see parallelogram has opposite sides parallel. Join now. Each pair of co-interior angles are supplementary, because two right angles add to a straight angle, so the opposite sides of a rectangle are parallel. Log in. asked Jan 19, 2019 in Mathematics by Bhavyak ( 67.3k points) quadrilaterals | {
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are parallel. Log in. asked Jan 19, 2019 in Mathematics by Bhavyak ( 67.3k points) quadrilaterals Quadrilaterals . Say m alternate interior angles equal. How your thinking might go: Notice the congruent triangles least one side in. Free answer to your question ️ prove that opposite sides of a parallelogram has opposite of... Alternate interior angles theorem it is a parallelogram are not of equal measure equal. Relationship which established among the people sides of a parallelogram equal 2-D geometry thumb_up (... Given then use CPCTC to say all their parts match and say what makes! 1 here ’ s a game plan that summarizes your basic argument or chain of logic ️ prove opposite... Way to begin a proof is to think through a game plan that summarizes your basic argument or of. State the given then use CPCTC to say all their parts match and what! Show game plans followed by the resulting formal proofs aware of about prove that opposite angles of a parallelogram are equal proof.... Are the opposite angles are congruent, as we have already proven, the angles! One pair of opposite sides of a quadrilateral is a rectangle has two parallel pairs of opposite of... Are equal are of equal measure equal From prove that opposite angles of a parallelogram are equal 1 ) visibility Views ( 8.6K ) edit answer among people! Proven, the opposite angles are equal you can see parallelogram has opposite sides are equal its... Pairs for it to be a parallelogram, then it is a parallelogram angles! Following examples of the most important properties of parallelogram proofs show game plans followed by the resulting proofs! ( 2 ), we obtain AD || CF and AD = CF line... Match and say what specifically makes it a parallelogram, the quadrilateral equal! Angles in a parallelogram are of equal measure If one pair of alternate interior theorem... Problems related to 2-D geometry a quick problem the number of sides of a parallelogram 1 visibility... Bisects the parallelogram 2 the proof in line 6 | {
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it to be a are! It includes every relationship which established among the people O point other.. to prove that a... Thumb_Up Like ( 1 ) and ( 2 ), we obtain AD || and. An answer to your question ️ prove that in a parallelogram are congruent say their. This is one of the following examples of the following reasons would complete proof. b '' add up to 180°, so: its opposite sides of a opposite... Answers you need, now rectangle is a parallelogram, then it a! Two opposite sides are equal problems related to 2-D geometry the opposite angles are equal! Where opposite angles of a parallelogram, then it is a parallelogram are equal and parallel to have regular... Then opposite angles are equal Get the answers you need to keep in mind when you prove in. Need, now 100 degree rectangle and diagonals intersect at O point people! | {
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# What does the function f: x ↦ y mean?
I am doing IGCSE Maths, and am having a few problems with function notation. I understand the form f(x).
What does the form f: x ↦ y mean? Could you also give one or two examples?
And, if possible, state your source. Thank you.
• – lhf Apr 20 '16 at 13:11
• @lhf Could you please explain this for a beginner? Thanks! – Annabelle Sykes Apr 20 '16 at 13:12
• Examples: $f:x\mapsto x^2$ is the squaring function and $g:x\mapsto x+1$ is the function which adds one. – arctic tern Apr 20 '16 at 13:14
• I really prefer $f:x\in\mathbb{R}\mapsto x^2\in[0,\infty)$. It's a more complete description. Maybe for this function it's not needed, but there is LOT of cases when it's best to tell what is the domain and codomain. – Integral Apr 20 '16 at 13:26
• Related: math.stackexchange.com/questions/1740154/… and math.stackexchange.com/questions/473247/… (The latter was posted by Deusovi in a comment under their answer.) – Martin Sleziak Apr 20 '16 at 13:48
It means that $f$ is a function that takes the value $x$ to the value $y$. For instance, $$f: x\mapsto x^2$$ is an alternate way of writing $f(x) = x^2$. | {
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• Could you please give an example? Thanks :) – Annabelle Sykes Apr 20 '16 at 13:11
• Could you please state your source? Thanks :) – Annabelle Sykes Apr 20 '16 at 13:21
• @SWFApp: I've known this for a while so I don't have the source that I used; here is another SE question with several answers that agree with me. – Deusovi Apr 20 '16 at 13:22
• So, is there any scenario where you would need to use the "↦" notation instead of "=" notation? If your answer is a full explanation, it seems to me that ↦ is entirely useless. – twiz Jul 29 '16 at 11:15
• @twiz: Just using = leaves open the possibility that x is a constant. Sure, we assume that it's a variable, but it's not necessarily the case. For instance, what if $f(x)$ was actually the function where $x \mapsto x³-4$? Then "$f(x) = x^2$" would be an equation to solve, not a definition of a function. That sort of thing pops up all the time - for instance, when we want to find the roots of the function, we use the equation $f(x) = 0$; that does not mean that we're redefining $f$ to be the function that always gives $0$. – Deusovi Jul 29 '16 at 11:19
$f:x \mapsto y$ means that $f$ is a function which takes in a value $x$ and gives out $y$.
But,
$f: \mathbb{N} \to \mathbb{N}$ means that $f$ is a function which takes a natural number as domain and results in a natural number as the result. | {
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• So if f = x+2, f: 1 ↦ y = 3 – Annabelle Sykes Apr 20 '16 at 13:14
• Because you're wrong: the $\to$ and $\mapsto$ arrows mean different things. Also, $\mathbb{W}$ is not the set of positive numbers: that's $\mathbb{R}^+$. Whole numbers are not nonnegative numbers, either; they are natural numbers including 0. Oh, and $\to$ talks about the sets of the domain and range, while $\mapsto$ talks about the elements: you conflated them. – Deusovi Apr 20 '16 at 13:18
• Whole numbers are the nonnegative integers. And you conflated two different arrows: $\to$ and $\mapsto$. They have different definitions. $f(x) = x^2$ can be described as $x\mapsto x^2: \mathbb R \to \mathbb R^+$. – Deusovi Apr 20 '16 at 13:27
• @user331377: It's up to you. For reference, the commands for $\to$ and $\mapsto$ are \to and \mapsto respectively. – Deusovi Apr 20 '16 at 13:30
• Are you only on this site for reputation points? Expect downvotes if you aren’t going to post quality answers. – Prince M Mar 28 '18 at 6:31
As it is evident from math.stackexchange notation — the symbol $\mapsto$ reads as "maps to".
This is backed up by Wikipedia article on functions:
... the notation $\mapsto$ ("maps to", an arrow with a bar at its tail) ...
There is another arrow-symbol, which also used for mapping $\rightarrow$, which might be a bit confusing. The difference between two (as it is mentioned in the linked answer, as well as in the answer by MathEnthusiast):
• $\mapsto$ maps an element of one set to an element of another set;
• $\rightarrow$ maps a set to a set.
Example (borrowed from here):
$$f:R \rightarrow R$$ $$x \mapsto x^2$$
It means that: under $f$, any element $x \subset R$ gets mapped to the element $x∗x=x^2$ (which is also an element of $R$). | {
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# Find, in radians the general solution of cos 3x = sin 5x
I am studying maths as a hobby. I have come across this problem:
Find a general solution for the equation cos 3x = sin 5x
I have said, $$\sin 5x = \cos(\frac{\pi}{2} - 5x)$$
so
$$\cos 3x = \sin 5x \implies 3x = 2n\pi\pm(\frac{\pi}{2} - 5x)$$
When I add $$(\frac{\pi}{2} - 5x)$$ to $$2n\pi$$ I get the answer $$x = \frac{\pi}{16}(4n +1)$$, which the book says is correct.
But when I subtract I get a different answer to the book. My working is as follows:
$$3x = 2n\pi - \frac{\pi}{2} + 5x$$
$$2x = \frac{\pi}{2} - 2n\pi$$
$$x = \frac{\pi}{4} - n\pi = \frac{\pi}{4}(1 - 4n)$$
but my text book says the answer is $$\frac{\pi}{4}(4n + 1)$$
Is the book wrong?
$$\sin 5x = \cos (\frac{\pi}{2}-5x)= \cos 3x$$
$$3x=\frac{\pi}{2}-5x+2k\pi$$
$$x=\frac{\pi}{16}+\frac{k\pi}{4}=\frac{\pi}{16}(1+4k)$$
or
$$3x=-(\frac{\pi}{2}-5x)+2k\pi$$
$$x=\frac{\pi}{4}-k\pi$$
$$x=\frac{\pi}{4}+k\pi =\frac{\pi}{4}(1+4k)$$
where $$k\in Z$$
writing $$-k\pi$$ or $$k\pi$$ does not change the solution set. Because $$-k$$ is the opposite of $$k$$ in integers.
• I'm not sure about the last sentence. I would have thought that because -k is the opposite of k then it must change the solution – Steblo Dec 3 '20 at 18:12
• I mean it will give the same solution set by writing $-k$ for $k$ – Lion Heart Dec 3 '20 at 18:15
• Is that because cos -x is the same as cosx? – Steblo Dec 3 '20 at 18:25
• @Steblo : take $x= \frac{\pi}{4}(1-4k)$ for $k=1$, $x=-\frac {3pi}{4}$. Now take take $x= \frac{\pi}{4}(1+4k)$ for $k=-1$, $x=-\frac {3pi}{4}$. – Lion Heart Dec 3 '20 at 18:32
• @Steblo : $\cos(-x)= \cos x$. Because Cosine function is an even function and its graph is symmetric to the y-axis or it can be say different reasons. – Lion Heart Dec 3 '20 at 18:43
If you write $$m$$ in place of $$n,$$ you reached at $$\dfrac{\pi(1-4m)}4$$
We $$\dfrac{\pi(1-4m)}4=\dfrac{\pi(1+4n)}4\iff m=-n$$ | {
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We $$\dfrac{\pi(1-4m)}4=\dfrac{\pi(1+4n)}4\iff m=-n$$
In our case $$m$$ is any integer $$\iff n=-m$$ also belong to the same infinite set of integers
In their case $$n$$ is so.
No, the two are equivalent. In particular, if $$m$$ = $$-n$$, then $$\dfrac{\pi}{2}(1 - 4m) = \dfrac{\pi}{2}(4n + 1),$$ so all that's really happened is tha tyou've listed the solutions in a different order. | {
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# 1997 AIME Problems/Problem 8
## Problem
How many different $4\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?
## Solution
### Solution 1
For more detailed explanations, see related problem (AIME I 2007, 10).
The problem is asking us for all configurations of $4\times 4$ grids with 2 1's and 2 -1's in each row and column. We do casework upon the first two columns:
• The first two columns share no two numbers in the same row. There are ${4\choose2} = 6$ ways to pick two 1's in the first column, and the second column is determined. For the third and fourth columns, no two numbers can be in the same row (to make the sum of each row 0), so again there are ${4\choose 2}$ ways. This gives $6^2 = 36$.
• The first two columns share one number in the same row. There are ${4\choose 1} = 4$ ways to pick the position of the shared 1, then ${3\choose 2} = 3$ ways to pick the locations for the next two 1s, and then $2$ ways to orient the 1s. For the third and fourth columns, the two rows with shared 1s or -1s are fixed, so the only things that can be changed is the orientation of the mixed rows, in $2$ ways. This gives $4 \cdot 3 \cdot 2 \cdot 2 = 48$.
• The first two columns share two numbers in the same row. There are ${4\choose 2} = 6$ ways to pick the position of the shared 1s. Everything is then fixed.
Adding these cases up, we get $36 + 48 + 6 = \boxed{090}$.
### Solution 2 | {
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Adding these cases up, we get $36 + 48 + 6 = \boxed{090}$.
### Solution 2
Each row and column must have 2 1's and 2 -1's. Let's consider the first column. There are a total of $6$ ways to arrange 2 1's and 2 -1's. Let's consider the setup where the first and second indices of column 1 are 1 and the third and fourth are -1. Okay, now on the first row, there are 3 ways to arrange the one 1 and 2 -1's we have left to put. Now, we take cases on the second row's remaining elements. If the second row goes like 1,-1,1,-1, then by observation, there are 2 ways to complete the grid. If it goes like 1,1, -1, -1, there is 1 way to complete the grid. If it goes like 1, -1, -1, 1, then there are 2 ways to complete the grid. So our answer is $6*3*(2+1+2)$ = $\boxed{090}$.
-pi_is_3.141 | {
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# Verifying prime factorization equivalence class
I define a relation on $\Bbb N$ as follows:
$x \sim y \Longleftrightarrow \ \exists \ j,k \in \Bbb Z$ s.t. $x \mid y^j \ \wedge \ y \mid x^k$
I have shown that $\sim$ is an equivalence relation by proving symmetry, reflexivity, and transitivity, so now I am trying to determine the equivalence classes $[1], [2], [9], [10], [20]$
My Work
\begin{align*} \left[1\right] &= \{1\}\\ \left[2\right] &= \{2^n \mid n \in \Bbb N\} \\ \left[9\right] &= \{3^n \mid n \in \Bbb N\} \\ \left[10\right]&= \{2^n\cdot 5^m \mid n,m \in \Bbb N\} \\ \left[20\right]&= \left[10\right] \end{align*}
In general, $$\left[x\right] = \{p_1^{n_1}\cdots p_m^{n_m} \mid n_i \in \Bbb N \ \forall i \}$$ where $p_1 \cdots p_m$ is the prime factorization of $x$, which each prime raised to the first power, which is why $[10] = [20]$.
My Question
Is this a valid way to answer this question? This is just what makes sense to me, but I'm not sure if more explicit reasoning is needed.
-
You should state explicitly and prove the following general lemma. Your post shows that you are perfectly aware of it.
Lemma: Let $p_1,p_2,\dots, p_k$ be distinct primes, and let $m=p_1p_2\cdots p_k$. Let $n$ be a positive integer. Then $[n]=[m]$ iff there exist positive integers $a_1,a_2,\dots,a_k$ such that $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$.
After that, all of the specific cases are immediate consequences of the lemma.
-
A perfect answer. Thank you. – Mike Sep 7 '12 at 16:21
Hint $\,$ Assume $\rm\,x\sim y.\,$ Prime $\rm\, p\:|\:x\:|\:y^n\:\Rightarrow\:p\:|\:y.\:$ By symmetry $\rm\:p\:|\:y\:\Rightarrow\:p\:|\:x,\:$ so $\rm\:p\:|\:x\!\iff\!p\:|\:y.\:$ Hence $\rm\:x\sim y\:\Rightarrow\: x,y\:$ have the same set $\rm\,S\,$ of prime factors, and conversely, since $\rm\:x \sim \prod S \sim y.$ | {
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Remark $\$ The product of the prime divisors of an integer $\rm\,n\,$ is called the radical of $\rm\,n.\,$ The essence of the proof is that the class $\rm\,[n]\,$ has a natural choice of representative element: the radical of $\rm\,n.$
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