Second-order derivative condition for convexity

It is written in a book I'm reading that $$\nabla f(x) = \left( \frac{\partial f(x)}{\partial x_1}, \frac{\partial f(x)}{\partial x_2},...,\frac{\partial f(x)}{\partial x_n}\right)$$ and $$\nabla^2 f(x)_{ij} = \frac{\partial^2 f(x)}{\partial x_i ~\partial x_j}, \qquad \forall i,j=1,...,n.$$

According to 2nd-order conditions: for twice differentiable function $f$, it is convex if and only if $$\nabla^2 f(x) \ge 0, \qquad \forall x \in \mathrm{dom} f.$$

But, the function $f(x,y) = \sqrt{x^2+y^2}$ is convex, but does not meet 2nd-order conditions: \begin{aligned} \frac{\partial^2 }{\partial x^2} \sqrt{x^2+y^2} &= \frac{y^2}{(x^2+y^2)^{\frac{3}{2}}} \ge 0,\\ \frac{\partial^2 }{\partial x ~ \partial y} \sqrt{x^2+y^2} &= - \frac{x y}{(x^2+y^2)^{\frac{3}{2}}} \le 0. \end{aligned} Can anyone explain this?

• This notation is a bit misleading -- for a matrix $A\in\mathbb{R}^{n\times n}$, writing $A\geq 0$ usually does not mean that all entries $a_{ij}$ are positive, but that the matrix is positive (semi-)definite, i.e., $x^TAx\geq 0$ for all $x\in\mathbb{R}^n$. – Christian Clason Jan 20 '16 at 12:51
• Do you mean the matrix $D$ must be positive s.t. $det(D) \ge 0$. $D= [\frac{\partial^2 f}{\partial x^2},\frac{\partial^2 f}{\partial x \partial y}; \frac{\partial^2 f}{\partial y \partial x},\frac{\partial^2 f}{\partial y^2} ]$ – Kevin Jan 20 '16 at 13:30
• Not quite -- all eigenvalues must be positive (and real) (which is only sufficient, not necessary, for the determinant to be positive). Put another way, $det(A)\geq 0$ is only necessary, not sufficient for convexity. – Christian Clason Jan 20 '16 at 13:31
• But how to check the eigenvalues for matrix $D$, since it is not formed by real values. – Kevin Jan 20 '16 at 13:33
• By Sylvester's criterion a $2\times2$ matrix is p.d. iff $A_{11}>0$ and $\det A>0$. That's usually easier than computing eigenvalues. – Kirill Jan 20 '16 at 16:53

Consolidating my comments (so that they can be cleaned up): This is a misunderstanding.

A twice (continuously!) differentiable function $f:\mathbb{R}^n\to \mathbb{R}$ is convex if and only if the Hessian $\nabla^2 f(x)\in\mathbb{R}^{n\times n}$ is positive semi-definite at every $x\in \mathbb{R}^n$. (This definition makes sense since the Hessian is symmetric by Schwarz' theorem if the second derivatives are continuous.) This is sometimes written as $$\nabla^2 f(x) \succeq 0 \qquad\text{for all } x\in\mathbb{R}^n$$ (and more rarely -- since it can lead to misunderstandings -- as $\nabla^2 f(x)\geq 0$).

As @nicoguaro points out in his answer, this is equivalent to the condition that all eigenvalues of $\nabla^2 f(x)$ -- as a function of $x$ -- are nonnegative for every $x\in \mathbb{R}^n$. An equivalent (and often easier to verify, especially for large $n$) condition is that $$d^T\nabla^2 f(x)d \geq 0 \qquad\text{for all } d\in\mathbb{R}^n \text{ and }x\in\mathbb{R}^n.$$

(This condition is also easier to work with if you want to rule out convexity: It's sufficient to find a single $d$ such that $d^T \nabla^2 f(x) d<0$.)

In your example (with $x_1 = x$ and $x_2 = y$), this would yield \begin{aligned} \begin{pmatrix} d_1 & d_2 \end{pmatrix} \begin{pmatrix} \frac{x_2^2}{(x_1^2 + x_2^2)^\frac{3}{2}} & \frac{-x_1\,x_2}{(x_1^2 + x_2^2)^\frac{3}{2}} \\ \frac{-x_1\,x_2}{(x_1^2 + x_2^2)^\frac{3}{2}} & \frac{x_1^2}{(x_1^2 + x_2^2)^\frac{3}{2}} \end{pmatrix} \begin{pmatrix} d_1 \\ d_2 \end{pmatrix} &= \frac{1}{(x_1^2 + x_2^2)^\frac{3}{2}}\left(d_1^2x_2^2 - 2 d_1 x_1x_2d_2 + d_2^2x_1^2\right)\\ &= \frac{1}{(x_1^2 + x_2^2)^\frac{3}{2}}\left(d_1x_2-d_2x_1\right)^2\\ &\geq 0 \end{aligned} for all $x,d\in\mathbb{R}^n$. Hence, $f$ is convex.

The comments already mention what you are not considering. For the particular example you mention we have

$$\nabla^2 f(x,y) = \begin{pmatrix} \frac{y^2}{(x^2 + y^2)^\frac{3}{2}} & \frac{-x\,y}{(x^2 + y^2)^\frac{3}{2}}\\ \frac{-x\,y}{(x^2 + y^2)^\frac{3}{2}} & \frac{x^2}{(x^2 + y^2)^\frac{3}{2}} \end{pmatrix}$$

And the eigenvalues are $0$ and $\frac{1}{\sqrt{x^2 + y^2}}$. That are greater or equal to zero for all $x,y \in \mathbb{R}^+$.


Math Help - Negative exponents

1. Negative exponents

Hey guys.

I know that x^-2 is equal to 1/x^2

But, this problem looks like this. It asks me to simplify this (AKA get rid of the negative exponents):

((x^-1)+(y^-1))/((x^-2)-(y^-2))

For some reason I don't think I can just flip the problem around to make it look like this

((x^2)-(y^2))/((x^1)+(y^1))

But if I can, can you let me know why? Thanks guys!

2. Hello,

No you can't, because it would mean that :

$\frac{1}{x^{-2}-y^{-2}}=x^2-y^2$, which is similar to saying that $\frac{1}{a-b}=\frac 1a-\frac 1b$ which are both false.

So let's write your problem :

$\frac{x^{-1}+y^{-1}}{x^{-2}-y^{-2}}$

You can see that $x^{-2}=(x^{-1})^2$ and $y^{-2}=(y^{-1})^2$

Can you see a difference of two squares below ???

$=\frac{x^{-1}+y^{-1}}{(x^{-1}-y^{-1})(x^{-1}+y^{-1})}=\frac{1}{x^{-1}-y^{-1}}=\frac{1}{\frac 1x-\frac 1y}=\dots$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Another way :

eliminate the negative coefficients.
In order to do that, multiply by $\frac{x^2y^2}{x^2y^2}$ :

$=\frac{xy^2+yx^2}{y^2-x^2}=\frac{xy(y+x)}{(y-x)(y+x)}=\frac{xy}{y-x}$

3. This is another problem that needs to be simplified.

It's going to look messy but I'll type it out right. Can someone tell me how to make it look like it does on the paper?

Anyways:

EDIT: sqrt = square root

(((2x)/(sqrt(x-1)))-sqrt(x-1))/(x-1)

Again sorry for the sloppiness >_<

4. Originally Posted by DHS1
This is another problem that needs to be simplified.

It's going to look messy but I'll type it out right. Can someone tell me how to make it look like it does on the paper?

Anyways:

(((2x)/(sqrt(x-1)))-sqrt(x-1))/(x-1)

Again sorry for the sloppiness >_<
You'll have to tell me if this is what you have coded:

$\dfrac{\dfrac{2x}{\sqrt{x-1}}-\sqrt{x-1}}{x-1}$

5. Originally Posted by masters
You'll have to tell me if this is what you have coded:

$\dfrac{\dfrac{2x}{\sqrt{x-1}}-\sqrt{x-1}}{x-1}$
Yeah it is. Did you actually type \dfrac{\dfrac{2x}{\sqrt{x-1}}-\sqrt{x-1}}{x-1} between the math brackets? Or do you use a program and then copy/paste or something o_O;

Thanks.

6. Originally Posted by DHS1
Yeah it is. Did you actually type \dfrac{\dfrac{2x}{\sqrt{x-1}}-\sqrt{x-1}}{x-1} between the math brackets? Or do you use a program and then copy/paste or something o_O;

Thanks.
Just type it in just like that. You can go here for a tutorial on this tool.

$\dfrac{\dfrac{2x}{\sqrt{x-1}}-\sqrt{x-1}}{x-1}\cdot \dfrac{\sqrt{x-1}}{\sqrt{x-1}}=\dfrac{2x-(x-1)}{(x-1)(\sqrt{x-1})}=\dfrac{2x-x+1}{(x-1)(\sqrt{x-1})}=$

$\dfrac{x+1}{(x-1)(\sqrt{x-1})} \cdot \dfrac{\sqrt{x-1}}{\sqrt{x-1}}=\dfrac{(x+1)(\sqrt{x-1})}{(x-1)^2}$

7. Wow great! I'm learning so much. Last one I'll ask for help with, I promise!

Directions: Re-write each expression as a single fraction in lowest terms.

$\frac{3}{a}+\frac{2}{a^2}-\frac{2}{a-1}$

EDIT: My problem is I don't know how to find out what the common denominator is. How can I go about doing that?

Thanks!

8. Originally Posted by DHS1
Wow great! I'm learning so much. Last one I'll ask for help with, I promise!

Directions: Re-write each expression as a single fraction in lowest terms.

$\frac{3}{a}+\frac{2}{a^2}-\frac{2}{a-1}$

EDIT: My problem is I don't know how to find out what the common denominator is. How can I go about doing that?

Thanks!
how to find out what the common denominator is

1. Factorize each denominator.

2. The common denominator must contain the original denominators as a factor:

$d_1=a$
$d_2=a \cdot a$
$d_3=(a-1)$

Thus the common denominator is $D=a\cdot a\cdot (a-1)$

$\dfrac{3}{a}+\dfrac{2}{a^2}-\dfrac{2}{a-1} = \dfrac{3a(a-1)}{a^2(a-1)}+\dfrac{2(a-1)}{a^2(a-1)}-\dfrac{2a^2}{a^2(a-1)} =$ $\dfrac{3a^2-3a+2a-2-2a^2}{a^2(a-1)}= \dfrac{a^2-a-2}{a^2(a-1)} = \dfrac{(a-2)(a+1)}{a^2(a-1)}$

which can't be simplified any more.

By the way: Do yourself and do us a favour and start a new thread if you have a new question. Otherwise you'll risk that nobody will notice that you are in need of some help.

9. Originally Posted by earboth
how to find out what the common denominator is

1. Factorize each denominator.

2. The common denominator must contain the original denominators as a factor:

$d_1=a$
$d_2=a \cdot a$
$d_3=(a-1)$

Thus the common denominator is $D=a\cdot a\cdot (a-1)$

$\dfrac{3}{a}+\dfrac{2}{a^2}-\dfrac{2}{a-1} = \dfrac{3a(a-1)}{a^2(a-1)}+\dfrac{2(a-1)}{a^2(a-1)}-\dfrac{2a^2}{a^2(a-1)} =$ $\dfrac{3a^2-3a+2a-2-2a^2}{a^2(a-1)}= \dfrac{a^2-a-2}{a^2(a-1)} = \dfrac{(a-2)(a+1)}{a^2(a-1)}$

which can't be simplified any more.

By the way: Do yourself and do us a favour and start a new thread if you have a new question. Otherwise you'll risk that nobody will notice that you are in need of some help.
You, sir, have been thanked.


# How many ways we can choose items from different boxes

I searched through the internet but couldn't find my answer, which can either be a very simple or a hard one.

Assume there are $3$ boxes, which carry, respectively, $1$, $4$, $2$ items. My question is how many ways we can select $3$ items from these boxes. I am looking for a formula rather than a solution for these specific values.

If I choose (take away) $3$ items by trying one by one. \begin{array}{c c c} 0 & 3 & 1\\ 0 & 2 & 2\\ 0 & 4 & 0\\ 1 & 1 & 2\\ 1 & 3 & 0\\ 1 & 2 & 1 \end{array}

Items remain each time, so the answer seems to be $6$ different ways. But I am not sure.

• Yes there are total 7 items. – Rockybilly Feb 13 '16 at 16:24
• The items are not distict. So what decides how many ways are there, is actually how many items remain in each box. – Rockybilly Feb 13 '16 at 16:29
• Are the items in each box identical? If so, you're talking about the number of sums $a_1+a_2+a_3=3$ where $0\leq a_1\leq 1$, $0\leq a_2\leq 4$, and $0\leq a_3\leq 2$. Sums of this form are standard in combinatorics. – Michael Burr Feb 13 '16 at 16:29
• Yes, Michael. I guess that was the thing I was trying to convert my problem into. But if a1, a2, a3 are the items remain in each box, the sum must be 4. If not, it is correct. – Rockybilly Feb 13 '16 at 16:30
• I think the general problem you are after is this. We have $k$ boxes, labelled $1$ to $k$. The number of items in Box $i$ is $a_i$, given. We are also given a number $n$ of items we must choose. We want to find the number of solutions of $x_1+\cdots+x_k=n$ in non-negative integers, with the restriction that $x_i\le a_i$. How many ways $w_n$ are there to do this? Not a simple problem! One can easily write down a generating function for the $w_n$, but computing the coefficients is messy. – André Nicolas Feb 13 '16 at 16:39

Observe that since the items are identical, it does not matter that there are $4$ items in the second box. Your are then asking for the number of sums $a_1+a_2+a_3=3$ where $0\leq a_1\leq 1$, $0\leq a_2\leq 3$, and $0\leq a_3\leq 2$. I will give three answers.

First, an elementary argument: We know that $a_1=0$ or $a_1=1$. If $a_1=0$, then $a_2+a_3=3$. In this case, there are three possibilities: $3+0=3$, $2+1=3$, and $1+2=3$. If $a_1=1$, then $a_2+a_3=2$ and there are still three possibilities: $2+0=2$, $1+1=2$, and $0+2=2$. This results in $6$ different options.

A more combinatorial argument: The number of ways to write $n$ as a sum of $k$ nonnegative integers is $$\binom{n+k-1}{k-1}$$ and a discussion can be found here. So, in this case, the number of ways that $a_1+a_2+a_3=3$ (without restrictions) is $$\binom{3+3-1}{3-1}=\binom{5}{2}=10.$$ This, however, counts too many possible sums. Suppose that we take too many from box $1$, this means that we take at least $2$ from box $1$. In this case, we can write $a_1=2+b_1$ where $b_1$ is nonnegative. Then, the initial sum becomes $b_1+a_2+a_3=1$. Using the same formula, this results in $$\binom{1+3-1}{3-1}=\binom{3}{2}=3$$ impossible ways. Continuing, there is no way to take too many objects from the second box, but it is possible to take too many objects from box $3$. In this case, one must take $3$ objects from box $3$, so we write $a_3=3+b_3$ where $b_3$ is nonnegative. This results in the equation $a_1+a_2+(3+b_3)=3$. There are $$\binom{0+3-1}{3-1}=1$$ ways for this sum to occur. We should now use the inclusion/exclusion principle to see if we've over-counted. This could happen if we take more than $1$ item from box $1$ and more than $2$ items from box $3$. Then, we have $(2+b_1)+b_2+(3+b_3)=3$, but this has no solutions as a sum of nonnegative integers cannot be negative.

Therefore, out of the original $10$ possibilities, $3+1=4$ are impossible, leaving the $6$ that we've found.

A dynamic programming-type solution: Let $N(b,s)$ be the number of ways to use the first $b$ boxes to sum to $s$. Also, write $m_i$ for the number of objects in box $i$. In your case: \begin{align*} N(1,0)&=1\\ N(1,1)&=1\\ N(1,2)&=0\\ N(1,3)&=0. \end{align*}

Then, the values in the second box can be computed as follows: $$N(b+1,s)=\sum_{i=0}^{\min\{s,m_{b+1}\}}N(b,s-i).$$ Using this formula: \begin{align*} N(2,0)&=N(1,0)=1\\ N(2,1)&=N(1,0)+N(1,1)=2\\ N(2,2)&=N(1,0)+N(1,1)+N(1,2)=2\\ N(2,3)&=N(1,0)+N(1,1)+N(1,2)+N(1,3)=2. \end{align*}

Continuing for the third column, \begin{align*} N(3,0)&=N(2,0)=1\\ N(3,1)&=N(2,0)+N(2,1)=3\\ N(3,2)&=N(2,0)+N(2,1)+N(2,2)=5\\ N(3,3)&=N(2,1)+N(2,2)+N(2,3)=6. \end{align*}

We are interested in the value $N(3,3)=6$.

• You are right, however I am currently constructing a computer program where I will deal with much bigger inputs. So I need a formula. These was just the values I came up to explain the situation. – Rockybilly Feb 13 '16 at 16:36
• The third solution should work quite quickly on a computer (you just need to run through a pair of arrays many times). – Michael Burr Feb 13 '16 at 17:08
• Forgive my dullness, but I didn't understand your last solution. How $$N(b, s)$$ is defined and how $$N(3,2) = 2$$ and $$N(3,2) = 5$$ in your explanation. – Rockybilly Feb 13 '16 at 17:10
• Cut and paste error! – Michael Burr Feb 13 '16 at 17:13
• For the calculation of $N(b+1,s)$: How many objects can be taken from the $b+1$st box? At most $s$ or the number of objects in the box. If you take $0$ objects from the $b+1$st box, then the objects must come from the remaining $b$ boxes, so you have $N(b,s)$ ways. If you take $1$ object from the $b+1$st box, then the remaining $s-1$ objects must come from the remaining $b$ boxes, so you have $N(b,s-1)$ ways. Continue adding these up until you run out of objects in the box or take $s$ objects from box $b+1$. – Michael Burr Feb 13 '16 at 17:17

Here is the code that I came up with in Sage. It takes about 19 minutes that is MUCH larger than the one initially posed (1000 boxes with random values between 1 and 1000). But 100 boxes with values between 1 and 100 finished under a second. Also, if you need a particular maximum value, just change $s$ to that value.

Boxes = [randint(1,100) for i in range(1000)]
n = len(Boxes)
s = sum(Boxes)
l1 = [0] * (s+1)
l2 = [0] * (s+1)
parity = 0
for i in range(Boxes[0]+1):
l1[i]=1
for i in range(1,n):
if(parity == 0):
l2[0]=l1[0]
for j in range(1,Boxes[i]+1):
l2[j]=l2[j-1]+l1[j]
for j in range(Boxes[i]+1,s+1):
l2[j]=l2[j-1]+l1[j]-l1[j-Boxes[i]-1]
parity = 1
else:
l1[0]=l2[0]
for j in range(1,Boxes[i]+1):
l1[j]=l1[j-1]+l2[j]
for j in range(Boxes[i]+1,s+1):
l1[j]=l1[j-1]+l2[j]-l2[j-Boxes[i]-1]
parity = 0
if(parity == 1):
l = l2
else:
l = l1

print l

Edit: I cut out one of the loops, reducing the complexity.

• Much better indeed. However, it is still not enough for the method I need. If I had found a solution in here, I would have answered this question much better. – Rockybilly Feb 14 '16 at 18:00
• Using a generating function is a nice method. I should have thought of that! (Although the code above is (more or less) computing the coefficients of the generating function). – Michael Burr Feb 14 '16 at 18:05
• Yes, indeed. They are quite similar methods. – Rockybilly Feb 14 '16 at 18:11


# Are there a finite number of trees with $k$ leaves and no vertices of degree $2$?

Given a fixed positive integer $$k$$. Show that there are only finitely many trees containing $$k$$ leaves and zero vertices of degree $$2$$.

I tried to use the theorem related to rooted trees and tried to prove it by contradiction, but could not quite use the condition of having zero vertices of degree $$2$$.

Is it even related to rooted trees theorem or can it be proved by contradiction? Can we use the fact that a tree with $$m$$ edges has $$m+1$$ vertices?

Consider an arbitrary tree with $$k$$ leaves and no vertices of degree $$2$$. Let $$\epsilon$$ represent the number of edges and $$\nu_j$$ represent the number of vertices with a degree of $$j$$ (so that $$\nu_1=k$$ and $$\nu_2=0$$). We know that $$\sum_{j=1}^\infty j\nu_j=2\epsilon$$ (since that's true for all graphs) and $$\sum_{j=1}^\infty \nu_j=\epsilon+1$$ (since it's a tree). So we can combine the equations as follows (where the index of the summations have been taken out).

$$\sum j\nu_j=2\big(\sum\nu_j-1\big)=2\sum \nu_j-2\\ 2\sum\nu_j-\sum j\nu_j=2\\\sum(2-j)\nu_j=2$$

The RHS is positive, so the LHS must be as well. The first few terms of the LHS is $$\nu_1+0\nu_2-\nu_3-2\nu_4-\ldots.$$ This could not be positive if $$\nu_3+\nu_4+\nu_5+\ldots>\nu_1$$. Since we know that $$\nu_2=0$$, we can conclude that a tree with $$k$$ leaves and no vertices of degree $$2$$ cannot have more than $$2k$$ vertices. Since the number of trees with fewer than $$2k$$ vertices is finite, we are done.

• This is very smart! I tried theorem related to rooted trees and tried proving by contradiction but could not quite use the condition of no vertices of degree 2. Oct 20, 2019 at 13:40
• @AnsonNG It's a good trick to have in your toolbox. I learned it when proving that there was an upper bound on the number of faces of a polyhedron with a given number of pentagonal faces and no hexagonal faces. Same argument except with Euler's formula instead of the double-edge count.
– user694818
Oct 20, 2019 at 13:48

This is easy to prove using the following two lemmata:

Lemma 1: The sum of the degrees of all vertices in a graph equals twice the number of edges.

Lemma 2: A tree with $$n$$ vertices has $$n-1$$ edges.

(The first lemma is a simple consequence of the definition of the degree of a vertex, i.e. the number of edges connected to it, and the fact that each edge connects to exactly two vertices. The second lemma can be proved by induction on the number of vertices: assuming that the lemma holds for all trees with $$n-1$$ vertices, take any tree with $$n$$ vertices and consider what happens when you merge any two adjacent vertices and remove the edge between them.)

Taken together these lemmata imply that, for any tree with $$n$$ vertices having the degrees $$d_1, d_2, \dots, d_n$$ respectively, $$\sum_{i=1}^n d_i = 2n - 2 \quad \text{and thus} \quad \sum_{i=1}^n (d_i - 2) = -2.$$ In other words, the sum of the degrees of all vertices minus two per vertex is the same (and equal to $$-2$$) for all trees!

In particular, we can see that the summand $$d_i - 2$$ is negative (and equal to $$-1$$ except for the degenerate case of the single-vertex tree) for leaves, zero for vertices of degree $$2$$ and positive (at at least one) for all other vertices. For the sum to equal $$-2$$, as it must, the positive contribution of each vertex with degree $$d_i > 2$$ must therefore be cancelled out by at least one leaf (and there need to be at least two extra leaves on top of that).

Thus, a tree with $$k$$ leaves can have at most $$k - 2$$ vertices of degree greater than $$2$$.

For a tree with $$k$$ leaves and no vertices of degree $$2$$, this implies that the total number of vertices in the tree can be at most $$2k - 2$$. And since the number of vertices in such a tree is thus bounded, and since there's only a finite number of possible ways of connecting any given number of vertices into a tree, this further implies that the total number of such trees is also bounded.

We can also see that, for this result to hold, it is essential that the number of vertices of degree $$2$$ be bounded (in your case by zero). Otherwise we could take any tree with $$n > 1$$ nodes and easily construct an infinite family of trees with the same number of leaves just by taking any pair of adjacent vertices and inserting an arbitrarily long linear chain of vertices of degree $$2$$ between them.


# Probability of the sum of the largest n samples

15 numbers are randomly chosen from U(0,1), what is the probability that the sum of largest four numbers is greater than 3.5?

With[{f = OrderDistribution[{UniformDistribution[], 15}, #] &},
Probability[a + b + c + d > 3 + 1 / 2,
{a \[Distributed] f[15], b \[Distributed] f[14], c \[Distributed] f[13],
d\[Distributed] f[12]}]]

Trouble is, it is very, very slow. I shut it down after 30 minutes. When I used NProbability it converged on the wrong answer but it did warn me with numerous error messages. Another CAS could do the above code but it also returned the same wrong answer. The right answer is supposed to be (it at least agrees with a simulation):

$\frac{224077804910008595}{584325558976905216}$

$\approx 0.383481094515780$

How do I do this using Mathematica?

-
Presumably the distributions of a,b,c and d are linked together. I don't see any such links in your code. To me it looks like you have independent a,b,c,d (from the correct OrderDistributions) when they should be linked. Presume this is why NProbability gives the wrong answer (Probability would give the same but is slower). –  Ymareth Jan 5 '14 at 16:14
@Ymareth I came to the same conclusion, e.g. a must be greater then b. –  ybeltukov Jan 5 '14 at 16:17
@Ymareth and ybeltukov doesn't the fact that it is distributed in f(15) do that? That is the maximum. –  bobbym Jan 5 '14 at 16:19
A possible simulation is num = 10^6; samp = (Sort /@ RandomVariate[UniformDistribution[], {num, 15}])[[All, -4 ;; -1]]; N@Length@Select[samp, Tr@# > 3.5 &]/num The result agrees with the OP's –  belisarius Jan 5 '14 at 17:01
@belisarius - Agrees with the stated correct answer but how in the OP's code do I know that the f[15] is always greater than f[14]. In your answer you have that implicitly as you're using the joint distribution but in the OP's code there are just 4 independent distributions - unless I'm misreading? –  Ymareth Jan 5 '14 at 18:07

That's a tricky one!

OrderDistribution[{dist,n}, {k1, k2, ...}] represents the joint (k1, k2,...} th-order statistics distribution from n observations from the distribution dist.

So:

Probability[a + b + c + d > 7/2, {a, b, c, d} \[Distributed]
OrderDistribution[{UniformDistribution[], 15}, {12, 13, 14, 15}]]

(*
224077804910008595/584325558976905216 --> 0.383481
*)
-
Hi belisarius; Thanks, can you see why my approach was bad? –  bobbym Jan 5 '14 at 17:55
@bobbym Consider breaking belisarius' code into two independent regions...Probability[a+b+c+d>7/2,{{a,b}[Distributed]OrderDistribution[{UniformDistribution[],15},{12,13}],{c,d}[Distributed]OrderDistribution[{UniformDistribution[],15},{14,15}]}]. This is not the same. Each of these regions can then be split into 4 which will give the same answer as your original code (I think) - run time seems to be very long. –  Ymareth Jan 5 '14 at 18:25
The OPs code is adding four independent random variables. belisarius is instead adding four joint distributions, which is the problem setup described by "the sum of largest four numbers". –  bill s Jan 5 '14 at 19:32

A manual approach

Joint (k1, k2,...} th-order statistics distribution is great, but how we can derive the answer by hands?

Let $a, b, c, d$ be the largest 4 number (in any order with each other). Let $g$ be the next largest number. We know that $g\sim \mathop{\rm Beta}(11,1)$. Then the probability under consideration is

NProbability[a + b + c + d > 7/2 \[Conditioned] a > g && b > g && c > g && d > g,
{a \[Distributed] UniformDistribution[], b \[Distributed] UniformDistribution[],
c \[Distributed] UniformDistribution[], d \[Distributed] UniformDistribution[],

0.383481

This is equivalent to the following integral

Binomial[15, 4] NIntegrate[UnitStep[a + b + c + d - 7/2] 11 g^10,
{g, 0, 1}, {a, g, 1}, {b, g, 1}, {c, g, 1}, {d, g, 1}]

0.383481

Binomial[15, 4] comes from the probability that $a, b, c, d$ are the largest 4 number, 11 g^10 is the PDF of the BetaDistribution[11, 1] and the integration limits come from the conditions $a > g$, etc.

Let us consider the indefinite integral

f[a1_, b1_, c1_, d1_] = Integrate[UnitStep[a + b + c + d - 7/2],
{a, -∞, a1}, {b, -∞, b1}, {c, -∞, c1}, {d, -∞, d1}]
1/384 (-7 + 2 a1 + 2 b1 + 2 c1 + 2 d1)^4 UnitStep[-(7/2) + a1 + b1 + c1 + d1]

The definite integral with limits g, 1 in every dimension is (a simple inclusion-exclusion formula)

int[g_] = Total[(-1)^Total /@ Tuples[{0, 1}, 4] f @@@ Tuples[{g, 1}, 4]]
1/384 - 1/96 (-1 + 2 g)^4 UnitStep[-(1/2) + g] +
1/64 (-3 + 4 g)^4 UnitStep[-(3/2) + 2 g] -
1/96 (-5 + 6 g)^4 UnitStep[-(5/2) + 3 g] +
1/384 (-7 + 8 g)^4 UnitStep[-(7/2) + 4 g]

Finally, the probability is

Binomial[15, 4] Integrate[int[g] 11 g^10, {g, 0, 1}]
224077804910008595/584325558976905216
-


# I heart math

By Murray Bourne, 05 Oct 2010

I saw this T-shirt recently.

What does it mean and what's that equation?

This is an example of an implicit function. When we first learn about functions, they are written explicitly, for example:

f(x) = sin(x) + 4x

This explicit function involves one dependent variable only and for each x value, we only get one f(x) value. Of course, I could also write this as

y = sin(x) + 4x

Notice y is on the left by itself, and the terms involving x are on the right, by themselves.

But there are many functions that are really messy when written explicitly, and so we turn to implicit functions.

In implicit functions, we see x's and y's multiplied and mixed together.

## A simple example

A simple example of an implicit function is the familiar equation of a circle:

x2 + y2 = 16

In this simple case, we can turn this into an explicit function by solving for y and getting 2 solutions:

or

But often it is very difficult, if not impossible, to solve an implicit function for y.

## The t-shirt Function

Returning to the t-shirt example, we have the implicit function:

(x2 + y2 − 1)3 = x2y3

We can expect more than one y-value for each x-value.

To graph it, we proceed as follows. Let's choose some easy values of x and y.

If x = 0, we substitute and obtain:

((0)2 + y2 − 1)3 = (0)2y3

(y2 − 1)3 = 0

We get 2 solutions, y = ± 1.

Now, let y = 0, and we get:

(x2 + (0)2 − 1)3 = x2(0)3

(x2 − 1)3 = 0

This gives us 2 solutions, x = ± 1.

So we know the curve passes through (-1, 0), (0, -1), (1, 0) and (0, -1),

Now, we choose some values of x between 0 and 1. We start with x = 0.2:

((0.2)2 + y2 − 1)3 = (0.2)2y3

This gives:

(-0.96 + y2)3 = 0.04y3

Solving this for y gives the real solutions: y = -0.824 or y = 1.166 (and 4 complex solutions).

We choose some more values and construct a table containing the real solutions:

 x y1 y2 0 0.2 0.4 0.6 0.8 1 1.2 -1 -0.824 -0.684 -0.520 -0.307 0 complex 1 1.166 1. 227 1. 231 1.17 complex

This equation is symmetrical, so we get the same correspnding values for -0.2, -0.4, -0.6, -0.8, -1 and -1.2.

In fact, outside of this range of x-values, there are no real y-values.

If we take a lot of points and join them, we get the following graph:

So the t-shirt means "I heart math" (that is, "I love math").

## 3-D Example

Here's another one in 3 dimensions. The implicit function is:

for -3 ≤ x, y, z ≤ 3 (which means each of x, y and z takes values only between -3 and 3).

And here's the shirt:

Differentiation of implicit functions

Curves in polar coordinates

### 11 Comments on “I heart math”

1. Naren says:

This is really cool!. Your blog really helps me a lot. I love the t-shirt as well. Could I know where you found it?

2. vonjd says:

WA can do it too 🙂
http://www.wolframalpha.com

and then type:
plot (x^2 + y^2 - 1)^3 = x^2 y^3

3. Murray says:

Yes! The graph in the post comes from Wolfram|Alpha.

4. Samar says:

So marvelous explanation about the t shirt especially I am a math teacher
Thanks.

5. yasin.. says:

superb...nicely explained..!!!!!!!!!!!

6. musa says:

I like it!! To draw this graph i often use MATLAB.

7. Kudzai Maravanyika says:

i realy liked the way you express you maths. Keep on doing that.

8. Ryan says:

If you have a ti-89 you can graph the implicit 2d function in the 3d mode by typing in the equation and formatting the plot to make it graph implicitly. The window will have to be seriously adjusted though.

9. barbina says:

oh thank you I have been asked directly to Bagatrix and not precalculus graph sholved but must be solved to complete the curves and graphs, again thank you ...

11. april says:

wow!! amazing.. I like it..
thanks! i think i should suggest it to our math club as a batch shirt..
i like the graphing of the equation in 3-D.

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# The Stacks Project

## Tag 00EC

Lemma 10.20.1. Let $R$ be a ring. Let $e \in R$ be an idempotent. In this case $$\mathop{\rm Spec}(R) = D(e) \amalg D(1-e).$$

Proof. Note that an idempotent $e$ of a domain is either $1$ or $0$. Hence we see that \begin{eqnarray*} D(e) & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e \not\in \mathfrak p \} \\ & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e \not = 0\text{ in }\kappa(\mathfrak p) \} \\ & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e = 1\text{ in }\kappa(\mathfrak p) \} \end{eqnarray*} Similarly we have \begin{eqnarray*} D(1-e) & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid 1 - e \not\in \mathfrak p \} \\ & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e \not = 1\text{ in }\kappa(\mathfrak p) \} \\ & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e = 0\text{ in }\kappa(\mathfrak p) \} \end{eqnarray*} Since the image of $e$ in any residue field is either $1$ or $0$ we deduce that $D(e)$ and $D(1-e)$ cover all of $\mathop{\rm Spec}(R)$. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 3447–3454 (see updates for more information).

\begin{lemma}
\label{lemma-idempotent-spec}
Let $R$ be a ring. Let $e \in R$ be an idempotent.
In this case
$$\Spec(R) = D(e) \amalg D(1-e).$$
\end{lemma}

\begin{proof}
Note that an idempotent $e$ of a domain is either $1$ or $0$.
Hence we see that
\begin{eqnarray*}
D(e)
& = &
\{ \mathfrak p \in \Spec(R)
\mid
e \not\in \mathfrak p \} \\
& = &
\{ \mathfrak p \in \Spec(R)
\mid
e \not = 0\text{ in }\kappa(\mathfrak p) \} \\
& = &
\{ \mathfrak p \in \Spec(R)
\mid
e = 1\text{ in }\kappa(\mathfrak p) \}
\end{eqnarray*}
Similarly we have
\begin{eqnarray*}
D(1-e)
& = &
\{ \mathfrak p \in \Spec(R)
\mid
1 - e \not\in \mathfrak p \} \\
& = &
\{ \mathfrak p \in \Spec(R)
\mid
e \not = 1\text{ in }\kappa(\mathfrak p) \} \\
& = &
\{ \mathfrak p \in \Spec(R)
\mid
e = 0\text{ in }\kappa(\mathfrak p) \}
\end{eqnarray*}
Since the image of $e$ in any residue field is either $1$ or $0$
we deduce that $D(e)$ and $D(1-e)$ cover all of $\Spec(R)$.
\end{proof}

There are no comments yet for this tag.

## Add a comment on tag 00EC

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).


# Predicate Logic Translating “All But One”

I need to translate an English sentence including the phrase "all but one" into predicate logic. The sentence is: "All students but one have an internet connection." I'm not sure how to show "all but one" in logic.

I could say $\forall x ((x \neq a) \rightarrow I(x))$

$I(x)$ being "$x$ has an internet connection"

But that clearly wouldn't work in this case, as we don't know which student it is.

I could say that $\exists x(\neg I(x))$

But it doesn't seem like that has the same meaning. Thanks in advance for any help you can give!

• Your suggestion doesn't seem right because presumably the universe is the set of all people or something like that. So you need a predicate to mean '$x$ is a student' and one to mean '$x$ has an internet connection'. As a hint rephrase it as $\text {there exists a student that doesn't have an internet connection such that all other students do.}$ – Git Gud Sep 18 '13 at 16:10
• sorry, in this case, the universe is the set of all people in a class. So I should have said "All students in the class but one have an internet connection." So in this case, if all students had a connection it would be ∀xI(x). – user95552 Sep 18 '13 at 16:16
• possible duplicate of Write ‘There is exactly 1 person…’ without the uniqueness quantifier – MJD Sep 18 '13 at 16:17
• @user95552 First note that my hint is wrong because it doesn't deal with uniqueness. Regarding what you said, I'd say it's wrong because there are people in the class which aren't students, but that's probably up to interpretation. Edit: my hint deals with uniqueness after all, so the hint is correct. Sorry. – Git Gud Sep 18 '13 at 16:19
• Alright, well I can easily change it to work with a different universe, but from your hint, could I say that ∃x(¬I(x) ∧ ∀y((x≠y) → I(y))) meaning there is a student without an internet connection and all students who aren't that student do have a connection. – user95552 Sep 18 '13 at 16:25

If you mean that there is exactly one element with a given property, you can define a "unique existence" quantifier, $\exists!$, as follows: $$\exists!x : \varphi(x) \iff \exists{x}{:}\left[\varphi(x)\wedge \forall{y}:\left(\varphi(y){\iff} y=x\right)\right].$$ That is, a particular element $x$ has the property $\varphi$, and any element with the property $\varphi$ must be that same $x$. For your problem, you want to say that there's exactly one person that is a student and doesn't have internet access.

"All students but one have an internet connection" means there is a student who lacks a connection, while every other student (every student not identical to the unlucky guy!) has one. So (if the domain is e.g. people)

$$\exists x(\{Sx \land \neg Ix\} \land \forall y(\{Sy \land \neg y = x\} \to Iy))$$

• Could have saved my fingers if I'd read all the comments first -- but I'll leave this in place. – Peter Smith Sep 18 '13 at 16:49

"For all but one $\;x\;$, $\;P(x)\;$ holds" is the same as "there exists a unique $\;x\;$ such that $\;\lnot P(x)\;$ holds.

Normally the notation $\;\exists!\;$ is used for "there exists a unique" (just like $\;\exists\;$ is used for "there exists some").

If your answer is allowed to use $\;\exists!\;$, then the above gives you the answer.

If not, then there are different ways to write $\;\exists!\;$ in terms of $\;\exists\;$ and $\;\forall\;$. The one I like best, which also results in the shortest formula, can be found in another answer of mine.


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# Sum of n positive integers formula for consecutive integers?

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Sum of n positive integers formula for consecutive integers?  [#permalink]

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Updated on: 20 Feb 2019, 03:34
7
Is the sum of n positive integers formula, $$\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?

For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\frac{20(21)}{2}$$

In OG PS Q172, a similar approach has been used for even numbers.

Originally posted by RustyR on 16 Nov 2013, 17:15.
Last edited by Bunuel on 20 Feb 2019, 03:34, edited 1 time in total.
Updated.
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Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

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17 Nov 2013, 01:27
1
1
3+6+9+12+15+18+21+24+27+30...... = 3*(1+2+3+4+5+6+7+8+9+10...)

So, yup.. You can do that...
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Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

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17 Nov 2013, 20:25
11
10
bschoolaspirant9 wrote:
Is the sum of n positive integers formula, $$\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?

For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\frac{20(21)}{2}$$

In OG PS Q172, a similar approach has been used for even numbers.

Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1.
A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula.

In the even integers questions, you may be required to find the sum of first 10 even integers.
2 + 4 + 6 + ... + 18 + 20
Take 2 common, 2*(1 + 2 + 3 + ...10)
To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum.

Note that for odd integers, you cannot directly use this formula.
Sum the first 10 odd integers
1 + 3 + 5 + 7+...+19
But you can still make some modifications to find the sum.

1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) - (2 + 4+ 6+...20)
We know how to sum consecutive integers.
(1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2
(2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before)

So 1 + 3 + 5 + 7+...+19 = (20*21/2) - (10*11) = 100

The direct formula of sum of n consecutive odd integers starting from 1 = n^2
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Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

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19 Jun 2015, 08:13
VeritasPrepKarishma wrote:
bschoolaspirant9 wrote:
Is the sum of n positive integers formula, $$\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?

For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\frac{20(21)}{2}$$

In OG PS Q172, a similar approach has been used for even numbers.

Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1.
A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula.

In the even integers questions, you may be required to find the sum of first 10 even integers.
2 + 4 + 6 + ... + 18 + 20
Take 2 common, 2*(1 + 2 + 3 + ...10)
To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum.

Note that for odd integers, you cannot directly use this formula.
Sum the first 10 odd integers
1 + 3 + 5 + 7+...+19
But you can still make some modifications to find the sum.

1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) - (2 + 4+ 6+...20)
We know how to sum consecutive integers.
(1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2
(2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before)

So 1 + 3 + 5 + 7+...+19 = (20*21/2) - (10*11) = 100

The direct formula of sum of n consecutive odd integers starting from 1 = n^2

Great explanation. However, I have a suggestion. Why can't we use one single formula for all consecutive evenly spaced numbers which is Sum = (average)(number of terms). I hope i am correct?
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Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

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21 Jun 2015, 21:57
2
samdighe wrote:
VeritasPrepKarishma wrote:
bschoolaspirant9 wrote:
Is the sum of n positive integers formula, $$\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?

For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\frac{20(21)}{2}$$

In OG PS Q172, a similar approach has been used for even numbers.

Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1.
A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula.

In the even integers questions, you may be required to find the sum of first 10 even integers.
2 + 4 + 6 + ... + 18 + 20
Take 2 common, 2*(1 + 2 + 3 + ...10)
To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum.

Note that for odd integers, you cannot directly use this formula.
Sum the first 10 odd integers
1 + 3 + 5 + 7+...+19
But you can still make some modifications to find the sum.

1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) - (2 + 4+ 6+...20)
We know how to sum consecutive integers.
(1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2
(2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before)

So 1 + 3 + 5 + 7+...+19 = (20*21/2) - (10*11) = 100

The direct formula of sum of n consecutive odd integers starting from 1 = n^2

Great explanation. However, I have a suggestion. Why can't we use one single formula for all consecutive evenly spaced numbers which is Sum = (average)(number of terms). I hope i am correct?

We can, provided we know the average and the number of terms.
If we are asked to find the sum of first 50 consecutive positive odd integers, it might be easier to use 50^2 than to find average and then find the sum.

Mind you, I myself believe in knowing just the main all-applicable kind of formulas and then twisting them around to apply to any situation. But some people prefer to work more on specific formulas and these discussions are for their benefit.
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Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

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24 Jun 2015, 04:57
9
2
The best way to deal with sum of any type of arithmetic series is to use the following expression:

Sum = Average*Number of terms

Average = (First term + Last term)/2

Note: The average or the median can be related to the first and last terms, which are often given for these types of problems.

Number of Terms = (Last Term - First Term)/Spacing + 1

Also, I recommend that you understand the basis of these relationships as opposed to blindly memorizing them.

The above relationships are flexible enough to allow you to deal with any GMAT problem on these concepts.

For example in Q172 from the Official Guide: What is the sum of all the even integers between 99 and 301 ?

Average = (100+300)/2 = 200

Number of Terms = (300-100)/2 + 1 = 101 (Note spacing is 2 between consecutive even integers)

Sum = 200*101=20200

Cheers,
Dabral
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Joined: 19 Dec 2018
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Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

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19 Dec 2018, 23:54
for consecutive odd or even ranges

sum = (f + n-1) n

f is the first number in the sequence, n is the number of elements in the sequence
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Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

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21 Dec 2018, 01:01
1
Top Contributor
Hi mathsnoob,

You are correct that the formula that you listed can be used for adding consecutive even and odd integers. In fact, it can be derived from the general approach. For example, n would be the number of terms in the sequence. The last term would be equal to f + 2(n-1), the common difference here is 2 for consecutive even and odd integers. This means the average of the first and last term is equal to [f + f + 2(n-1)]/2, which simplifies to f + (n-1). And finally we have an expression for the sum of n consecutive integers n[f + n - 1].

The problem with these highly specialized formulas is the burden of remembering the specific formula and keeping track of what each term stands for. Personally, I prefer to stick to the general formula that I listed earlier and then adapt it to a specific situation. This helps reduce the memorization burden on my brain. But I can see others who might prefer to memorize formulas for specific situations. Take your pick.

Cheers,
Dabral
Math Expert
Joined: 02 Sep 2009
Posts: 53771
Re: Sum of n positive integers formula for consecutive integers?  [#permalink]

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20 Feb 2019, 03:34
RustyR wrote:
Is the sum of n positive integers formula, $$\frac{n(n+1)}{2}$$, applicable to sets of consecutive integers as well?

For instance, if I wanted to calculate sum of first 20 multiples of 3, could I use: 3 x $$\frac{20(21)}{2}$$

In OG PS Q172, a similar approach has been used for even numbers.

For more check Formulas for Consecutive, Even, Odd Integers

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Re: Sum of n positive integers formula for consecutive integers?   [#permalink] 20 Feb 2019, 03:34
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# Math Help - Solve (x^2) ( d^2/dx^2) + 3x(dy/dx) + y = 1/x

1. ## Solve (x^2) ( d^2/dx^2) + 3x(dy/dx) + y = 1/x

(x^2) ( d^2/dx^2) + 3x(dy/dx) + y = 1/x

how can i find the general solution of differential equation?

2. It's a Cauchy-Euler equation. Use that procedure.

3. but right side of equation isn't equal to zero? can you solve it for me?

4. We don't "solve things" for you. I would suggest either assuming a solution that looks like the RHS, or, if you choose the exponential approach (thus producing a linear DE with constant coefficients), the choice of a particular solution might be more obvious. Try those, and post your work, and we can go from there.

5. Just to add my 2 cents. If you can solve the homogenous equation (Ackbeet told you how) you can find the particular solution by using the variation of parameters.
Variation of parameters - Wikipedia, the free encyclopedia This method does not involve guessing but you may end up with integrals in your solution, but that wont be the case in this particular problem.

6. Put $t=\ln x$ so we have $\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{1}{x}\cdot \frac{dy}{dt},$ and then $\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\left( \frac{1}{x}\cdot \frac{dy}{dt} \right)=-\frac{1}{{{x}^{2}}}\cdot \frac{dy}{dt}+\frac{1}{x}\cdot\frac{{{d}^{2}}y}{d{ {t}^{2}}},$ so you'll end up with an ODE with constant coefficients.

7. Originally Posted by Ackbeet
We don't "solve things" for you. I would suggest either assuming a solution that looks like the RHS, or, if you choose the exponential approach (thus producing a linear DE with constant coefficients), the choice of a particular solution might be more obvious. Try those, and post your work, and we can go from there.
I am actually curious how to get the PI. I can get the CF. Would a first guess of PI=Ax^n be a good guess to suit 1/x?

8. Originally Posted by bugatti79
I am actually curious how to get the PI. I can get the CF. Would a first guess of PI=Ax^n be a good guess to suit 1/x?
Why not try it and see what happens?

9. I just noticed something that I think is worth mentioning.

Notice that on the left hand side we have

$x^2y''+3xy'+y=x^2y''+2xy'+xy'+y$

Notice that this is a perfect derivative

$\frac{d}{dx}\left(x^2y'+xy \right)$

So we can rewrite the equation as

$\frac{d}{dx}\left(x^2y'+xy \right)=\frac{d}{dx}\ln(x)$

and reduce this to a first order ODE!

10. Originally Posted by TheEmptySet
I just noticed something that I think is worth mentioning.

Notice that on the left hand side we have

$x^2y''+3xy'+y=x^2y''+2xy'+xy'+y$

Notice that this is a perfect derivative

$\frac{d}{dx}\left(x^2y'+xy \right)$

So we can rewrite the equation as

$\frac{d}{dx}\left(x^2y'+xy \right)=\frac{d}{dx}\ln(x)$

and reduce this to a first order ODE!
Thats interesting. I see what you have done but how does one reduce it to a first order? I havent seen this before to the best of my poor memory :-) Anyhow, I attempted it doing it the hard way by guessing the PI. I dont think it works out....

$\displaystyle Let y=Ax^n, y'=nAx^{n-1}, y''=n(n-1)Ax^{n-2}$

Putting back into the original DE, I arrive at

$Ax^n[n^2+2n+1]=\frac{1}{x} \implies Ax^n=\frac{1}{x} \because [n^2+2n+1]=0 \therefore A=1$

Is this right?

11. @bugatti79

What I have is

$\frac{d}{dx}\left(x^2y'+xy \right)=\frac{d}{dx}\ln(x)$

Now if you integrate both sides you get

$x^2y'+xy =\ln(x)+C$

This is now a first order ODE that can be solve via an integrating factor!

12. Originally Posted by TheEmptySet
@bugatti79

What I have is

$\frac{d}{dx}\left(x^2y'+xy \right)=\frac{d}{dx}\ln(x)$

Now if you integrate both sides you get

$x^2y'+xy =\ln(x)+C$

This is now a first order ODE that can be solve via an integrating factor!
Yes I solved it using the integrating factor method. I am curious where I went wrong when guessing the PI in post #10. Where are you Ackbeet?

13. Originally Posted by bugatti79
Where are you, Ackbeet?
Not paying attention to the fact that you tried my suggestion, apparently.

Yeah, so that particular solution doesn't work, does it? You've got 0 = 1/x, a contradiction. Apparently, we need to change the ansatz. Note that the complimentary solution actually contains 1/x in it already, and by variation of parameters, you can find out that the complimentary solution also contains ln(x) / x. The next step with Cauchy-Euler equations is the (ln(x))^2 / x solution, so try a constant times that.

14. Originally Posted by Ackbeet
Not paying attention to the fact that you tried my suggestion, apparently.

The next step with Cauchy-Euler equations is the (ln(x))^2 / x solution, so try a constant times that.
Hmmmm..I got the first and second A times the above and put back into DE. I arrived at 0=1. Not to worry, I learned a nice bit and some differentiating practice. At least we know the integrating factor works. :-)

15. Originally Posted by bugatti79
Hmmmm..I got the first and second A times the above and put back into DE. I arrived at 0=1. Not to worry, I learned a nice bit and some differentiating practice. At least we know the integrating factor works. :-)
Well, here's my shot at it.

$y=A\,\frac{\ln^{2}(x)}{x}$

$y'=A\,\frac{2(x)(\ln(x)/x)-(\ln^{2}(x))(1)}{x^{2}}=A\,\frac{2\ln(x)-\ln^{2}(x)}{x^{2}}.$

$y''=A\,\frac{(x^{2})(2/x-2\ln(x)/x)-2(2\ln(x)-\ln^{2}(x))(x)}{x^{4}}=A\,\frac{x(2-2\ln(x))-2x(2\ln(x)-\ln^{2}(x))}{x^{4}}$

$=2A\,\frac{1-3\ln(x)+\ln^{2}(x)}{x^{3}}.$

Plugging these into the DE yields

$2A\,\frac{1-3\ln(x)+\ln^{2}(x)}{x}+3A\,\frac{2\ln(x)-\ln^{2}(x)}{x}+A\,\frac{\ln^{2}(x)}{x}=\frac{1}{x} \quad\implies$

$2A=1,$ or $A=1/2.$

So the particular solution is

$y_{p}=\frac{\ln^{2}(x)}{2x}.$

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# Length of the curve $\displaystyle 3ay^2=x(x-a)^2$

I'm working on finding out the length of the curve $$3ay^2=x(x-a)^2 \tag{1}$$

I ran into a small problem, but was able to end up with an answer that looks right but I'm not entirely sure about it. Here's my approach:

The length of a curve $$f(x)$$ between $$x=a$$ and $$x=b$$ where $$b>a$$, is given by $$L=\int_a^b\sqrt{1+\big[f'(x)\big]^2}dx$$ where $$f'(x)$$ is continuous in $$[a,b]$$. In our case

$$L=\int_0^x\sqrt{1+\big[y'\big]^2}dt \tag{2}$$

Curve $$(1)$$ is symmetric about the $$x$$-axis. Hence I'll just work on the length of the part that is above the $$x$$-axis and then double that to obtain the entire length. $$y=(x-a)\sqrt{\frac{x}{3a}} \text{ (One half of the curve, the other half being the negative multiple)}$$ Note that $$(1)$$ is defined for $$x\geq0$$ and $$a>0$$. That is why I've taken the limits of integration as $$0$$ to $$x$$.

Therefore, $$y'=\frac{3x-a}{\sqrt{12ax}}$$

But $$y'$$ is not continuous at $$x=0$$ and hence I can't plug this into $$(2)$$. So, let me calculate the length from $$h>0$$ to any $$x$$. Therefore, we have \begin{aligned} L &= \int_h^x\sqrt{1+\big[y'\big]^2}dt \\ &= \int_h^x\sqrt{1+\left(\frac{3t-a}{\sqrt{12at}}\right)^2}dt \\ &= \int_h^x \frac{3t+a}{\sqrt{12at}} dt \\ &= \frac{x^{\frac{3}{2}}-h^{\frac{3}{2}}+ax^{\frac{1}{2}}-ah^{\frac{1}{2}}}{\sqrt{3a}} \end{aligned}

Now, since I want the length in $$[0,x]$$, I'll just take the limit $$h\rightarrow0$$, which yields $$\boxed{L=(x+a)\sqrt{\frac{x}{3a}}}$$ The length of the curve in question would be double the above.

I don't see any problem with what I've done. Is this good?

• If it helps, Wolfram Alpha agrees with you upon some simple experimentation (I integrated from $0\to5$ and compared your formula's answer to Wolfram's answer and they agreed). However, I do not own Wolfram Pro and a general form integral took the poor computer too long to calculate, so my one experiment may not be proof that your answer is correct. That being said, mathematically, it does look right to me! Oct 4, 2021 at 9:18
• @FShrike That was a great idea. I checked my solution against Wolfram and it seems good. But my concern is regarding the legitimacy of the math I've done to arrive at that solution. Oct 4, 2021 at 12:40

Cubic curve with implicit equation:

$$3ay^2=x(x-a)^2\tag{1}$$

has an "alpha" shape with a double point $$D(a,0)$$ as can be seen on this Desmos figure (in the case $$a=2$$):

This curve can be described in an alternative way using the following parametric representation :

$$\begin{cases}x&=&3am^2\\y&=&am(3m^2-1)\end{cases}\tag{2}$$

Explanation: parameter $$m$$ has the following geometrical interpretation : it is the slope of a variable line (represented in blue on the figure) passing through double point $$D$$ with equation:

$$y=m(x-a)\tag{3}$$

Plugging (3) into (1) gives (2), after simplification (end of explanation of (2)).

The length of the curve between parameters values $$m_1$$ and $$m_2$$ is (using a classical formula):

$$L=\int_{m_1}^{m_2}\sqrt{x'(m)^2+y'(m)^2}dm$$

$$L=\int_{m_1}^{m_2}\sqrt{(6am)^2+|9am^2-a|^2}dm$$

$$L=a\int_{m_1}^{m_2}\sqrt{(6m)^2+(9m^2-1)^2}dm$$

$$L=a\int_{m_1}^{m_2}\sqrt{81m^4+18m^2+1}dm$$

$$L=a\int_{m_1}^{m_2}(9m^2+1)dm$$

$$L=a[m(3m^2+1)]_{m_1}^{m_2},\tag{4}$$

plainly. It remains to convert (4) in terms of variable $$x$$ where

$$x=3am^2 \iff m=\sqrt{\frac{x}{3a}}$$

(if we consider only positive slopes).

(4) gives

$$L=(x+a)\sqrt{\frac{x}{3a}}$$

a result which is now the same as yours.

Remarks:

1. we are here in an exceptional case where we have an analytic formula for the arc length... These exceptional cases for cubic curves have been studied here using cubic Bezier curves techniques.

2. The parameterization using the slope of a line passing through the double point is classical.

• I did a check using Wolfram Alpha and my answer matches with what I've derived. But my concern is whether the math that I've done is legit. wolframalpha.com/input/… Oct 4, 2021 at 12:32
• I have corrected an error of mine giving a formula closer to yours, but definitely not the same. Oct 4, 2021 at 12:52
• That's a good catch. I should have not said '1st Quadrant'. I meant the 'one-half'. Let me correct that. Oct 4, 2021 at 13:00
• Your answer is correct. It works for all values of $a$. Mine doesn't and also your observation is on point that $L\rightarrow 0$ as $a\rightarrow\infty$ Oct 4, 2021 at 13:19
• My genius brain made an integration mistake. I've fixed it now and it is same as yours. Thank you so much for all the help. Oct 4, 2021 at 13:35


# What's the difference between MUTUALLY EXCLUSIVE and PAIRWISE DISJOINT?

When I study Statistical Theory, I find that these two concepts confuse me a lot.

By definition, if we say two events are PAIRWISE DISJOINT, that means the intersection of these two event is empty set. If we say that two events are MUTUALLY EXCLUSIVE, that means if one of these two events happens, the other will not. But doesn't it means that these two events are PAIRWISE DISJOINT？

If we say two events are MUTUALLY EXCLUSIVE, then they are not INDEPENDENT. Can we say that two PAIRWISE DISJOINT events are not INDEPENDENT as well？

If these two concepts are different (actually my teacher told me they are), could you please give me an example that two events are MUTUALLY EXCLUSIVE but not PAIRWISE DISJOINT, or they are PAIRWISE DISJOINT but not MUTUALLY EXCLUSIVE.

"Disjoint" is a property of sets. Two sets are disjoint if there is no element in both of them, that is if $$A \cap B = \emptyset$$.

In some (but not all!) texts, "mutually exclusive" is a slightly different property of events (sets in a probability space). Two events are mutually exclusive if the probability of them both occurring is zero, that is if $$\operatorname{Pr}(A \cap B) = 0$$. With that definition, disjoint sets are necessarily mutually exclusive, but mutually exclusive events aren't necessarily disjoint.

Consider points in the square with each coordinate uniformly distributed from $$0$$ to $$1$$. Let $$A$$ be the event where the $$x$$-coordinate is $$0$$, and $$B$$ be the event that the $$y$$-coordinate is $$0$$. $$A \cap B = \{(0,0)\}$$ so $$A$$ and $$B$$ are not disjoint, but $$\operatorname{Pr}(A \cap B) = 0$$ so they are mutually exclusive.

As a second (silly, but finite) example, let the sample space be $$S = \{x, y, z\}$$ with probabilities $$\operatorname{Pr}(\{x\}) = 0$$, $$\operatorname{Pr}(\{y\}) = \frac{1}{2}$$, and $$\operatorname{Pr}(\{z\}) = \frac{1}{2}$$. If $$A = \{x, y\}$$ and $$B = \{x, z\}$$, then $$A \cap B = \{x\}$$, but $$\operatorname{Pr}(A \cap B) = \operatorname{Pr}(\{x\}) = 0$$. They are mutually exclusive but not disjoint.

Good discussion and I agree. This is an interesting result that seems to be ignored most of the time and disjoint and mutually exclusive are taken as equivalent. If we have finite sample spaces without 0 probability outcomes then the counter-examples above suggest they are then equivalent, and these are the properties of the common examples used when first teaching students these concepts. The current Wikipedia entry for Mutual Exclusive conflates the two concepts. The exceptions to equivalence seem to occur only when 0 probability outcomes exist in the sample space, but this is a requirement of continuous sample spaces and is not precluded in the definition of discrete sample spaces, both definitions usually state the outcomes must be mutually exclusive. However this seems to be a redundant condition since the different simple events of any sample space are disjoint by the definition of a set (i.e. multiplicities of elements are reduced) in any case, which thus implies mutual exclusivity by default.


Online calculators and formulas for a rhombus … Area of plane shapes. The area of the rhombus is given by the formula: Area of rhombus = sh. Abhishek241 Abhishek241 19.08.2017 Math Secondary School +5 pts. A rhombus is a type of Parallelogram only. Sitemap. The side of rhombus is a tangent to the circle. Solution: All the sides of a rhombus are congruent, so HO = (x + 2).And because the diagonals of a rhombus are perpendicular, triangle HBO is a right triangle.With the help of Pythagorean Theorem, we get, (HB) 2 + (BO) 2 = (HO) 2x 2 + (x+1) 2 = (x+2) 2 x 2 + x 2 + 2x + 1 = x 2 + 4x + 4 x 2 – 2x -3 = 0 Solving for x using the quadratic formula, we get: x = 3 or x = –1. A rhombus is a special type of quadrilateral parallelogram, where the opposite sides are parallel and opposite angles are equal and the diagonals bisect each other at right angles. Formula for perimeter of a rhombus : = 4s Substitute 16 for s. = 4(16) = 64. Side of a Rhombus when Diagonals are given calculator uses Side A=sqrt((Diagonal 1)^2+(Diagonal 2)^2)/2 to calculate the Side A, Side of a Rhombus when Diagonals are given can be defined as the line segment that joins two vertices in a rhombus provided the value for both the diagonals are given. By applying the perimeter formula, the solution is: Check: Other Names. Solution: Since we are given the side length, we can plug it straight into the formula. Since a rhombus is a parallelogram in which all sides are equal, all the same formulas apply to it as for a parallelogram, including the formula for finding the area through the product of height and side. So, the perimeter of the rhombus is 64 cm. Diagonals divide a rhombus into four absolutely identical right-angled triangles. Join now. By … . Is a Square a Rhombus? The total distance traveled along the border of a rhombus is the perimeter of a rhombus. So by the same argument, that side's equal to that side, so the two diagonals of any rhombus are perpendicular to … Using side and height. Given two integers A and X, denoting the length of a side of a rhombus and an angle respectively, the task is to find the area of the rhombus.. A rhombus is a quadrilateral having 4 sides of equal length, in which both the opposite sides are parallel, and opposite angles are equal.. To solve this problem, apply the perimeter formula for a rhombus: . Now the area of triangle AOB = ½ * OA * OB = ½ * AB * r (both using formula ½*b*h). When the altitude or height and the length of the sides of a rhombus are known, the area is given by the formula; Area of rhombus = base × height. The proof is completed. This formula for the area of a rhombus is similar to the area formula for a parallelogram. Problem 1: Find the perimeter of a rhombus with a side length of 10. The formula for perimeter of a rhombus is given as: P = 4s Where P is the perimeter and s is the side length. [3] What is the formula of Rhombus When one side is given Get the answers you need, now! Or as a formula: The perimeter formula for a rhombus is the same formula used to find the perimeter of a square. Heron's Formula depends on knowing the semiperimeter, or half the perimeter, of a triangle. Calculate the unknown defining areas, angels and side lengths of a rhombus with any 2 known variables. Formula for side of rhombus when diagonals are given - 1399111 1. Log in. The area of the rhombus can be found, also knowing its diagonal. How To Find Area Of Rhombus (1) If both diagonals are given (or we can find their length) then area = (Product of diagonals) (2) If we use Heron’s formula then we find area of one triangle made by two sides and a diagonal then twice of this area is area of rhombus. We now have the approximate length of side AH as 13.747 cm, so we can use Heron's Formula to calculate the area of the other section of our quadrilateral. Answered Formula for side of rhombus when diagonals are given 2 1. Perimeter = 4 × 12 cm = 48 cm. Thus, the total perimeter is the sum of all sides. Yes, because a square is just a rhombus where the angles are all right angles. The "base times height" method First pick one side to be the base. P = 4s P = 4(10) = 40 A rhombus is often called as a diamond or diamond-shaped. Here at Vedantu you will learn how to find the area of rhombus and also get free study materials to help you to score good marks in your exams. Since a rhombus is also a parallelogram, we can use the formula for the area of a parallelogram: A = b×h. If one of its diagonal is 8 cm long, find the length of the other diagonal. Ask your question. This is because both shapes, by definition, have equivalent sides. Any isosceles triangle, if that side's equal to that side, if you drop an altitude, these two triangles are going to be symmetric, and you will have bisected the opposite side. Given the length of diagonal ‘d1’ of a rhombus and a side ‘a’, the task is to find the area of that rhombus. Its diagonals perpendicularly bisect each other. The formula to calculate the area of a rhombus is: A = ½ x d 1 x d 2. where... A = area of rhombus; d 1 = diagonal1 (first diagonal in rhombus, as indicated by red line) d 2 = diagonal2 (second diagonal in rhombus, as indicated by purple line) Home List of all formulas of the site; Geometry. Area Of […] where b is the base or the side length of the rhombus, and h is the corresponding height. We should recall several things. In the case of a rhombus, all four sides are the same length by definition, so the perimeter is four times the length of a side. Many of the area calculations can be applied to them also. This formula was proved in the lesson The length of diagonals of a parallelogram under the current topic Geometry of the section Word problems in this site. The rhombus is often called a diamond, after the diamonds suit in playing cards, or a lozenge, though the latter sometimes refers specifically to a rhombus with a 45° angle. Example 2 : If the perimeter of a rhombus is 72 inches, then find the length of each side. These formulas are a direct consequence of the law of cosines. Here, r is the radius that is to be found using a and, the diagonals whose values are given. X Research source You could also use the formula P = S + S + S + S {\displaystyle P=S+S+S+S} to find the perimeter, since the perimeter of any polygon is the sum of all its sides. This geometry video tutorial explains how to calculate the area of a rhombus using side lengths and diagonals based on a simple formula. What is the area of a rhombus when only a side is given, and nothing else? Hello!!! Log in. The area of the rhombus can be found, also knowing its diagonal. Any one will do, they are all the same length. The inradius (the radius of a circle inscribed in the rhombus), denoted by r, can be expressed in terms of the diagonals p and q as = ⋅ +, or in terms of the side length a and any vertex angle α or β as Q. Ask your question. If the side length and one of the angles of the rhombus are given, the area is: A = a 2 × sin(θ) Examples: Input: d = 15, a = 10 Output: 99.21567416492215 Input: d = 20, a = 18 Output: 299.3325909419153 For our MAH, the three sides measure: MA = 7 cm; AH = 13.747 cm; HM = 14 cm The diagonals of a rhombus bisect each other as it is a parallelogram, but they are also perpendicular to each other. Calculator online for a rhombus. Rhombus Area Formula. Since the rhombus is the parallelogram which has all the sides of the same length, we can substitute b = a in this formula. Since, by definition, all four sides of a rhombus are the same length, the formula is =, where equals the perimeter, and equals the length of one side. Formula of Area of Rhombus / Perimeter of Rhombus. Choose a formula based on the values you know to begin with. There are many ways to calculate its area such as using diagonals, using base and height, using trigonometry, using side and diagonal. Free Rhombus Sides & Angles Calculator - calculate sides & angles of a rhombus step by step This website uses cookies to ensure you get the best experience. There are 3 ways to find the area of Rhombus.Find the formulas for same and Perimeter of Rhombus in the table below. Example Problems. First, all four sides of a rhombus are congruent, meaning that if we find one side, we can simply multiply by four to find the perimeter. Use Heron's Formula. Area of Rhombus using Altitude and Base. Then we obtain exactly the formula of the Theorem. In geometry, a rhombus or rhomb is a quadrilateral whose four sides all have the same length. Diagonals divide a rhombus into four absolutely identical right-angled triangles. A rhombus is actually just a special type of parallelogram. It is more common to call this shape a rhombus, but some people call it … Join now. Basic formulas of a rhombus. Click hereto get an answer to your question  Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm . Area Of Rhombus Formula. Area of a Rhombus Formula - A rhombus is a parallelogram in which adjacent sides are equal. A rhombus is a polygon having 4 equal sides in which both the opposite sides are parallel, and opposite angles are equal.. The area of rhombus can be found in multiple ways. Solution : Perimeter of the rhombus = 72 inches. 4s = 72. Using side and angle. Area of a triangle; Area of a right triangle If you are given the length of one side (s) and the perpendicular height (h) from one side to the vertex then the area of the rhombus is equal to the product of the side and height. 1. This rhombus calculator can help you find the side, area, perimeter, diagonals, ... On the other hand if the perimeter (P) is given the side (a) can be obtained from it by this formula: a = P / 4 When side (a) and angle (A) are provided the figures that can be computed … Example: A rhombus has a side length of 12 cm, what is its Perimeter? We will saw each of them one by one below. Since all four sides of a rhombus are equal, much like a square, the formula for the perimeter is the product of the length of one side with 4 $$P = 4 \times \text{side}$$ Angles of a Rhombus Inradius. 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Since we are given a triangle four absolutely identical right-angled triangles / perimeter of the area of rhombus can side of rhombus formula! A polygon having 4 equal sides in which adjacent sides are equal area formula the  times. Each of them one by one below to call this shape a rhombus: = 4s p = (. Angles are equal formula used to find the perimeter formula, the perimeter of a rhombus When side. Some people call it … Using side and height: a rhombus: = 4s p 4... Known variables nothing else formula used to find the perimeter formula, the perimeter formula for parallelogram... 40 rhombus area formula it straight into the formula of rhombus of cm. You need, now is the radius that is to be found Using and. It is a parallelogram perimeter formula for a rhombus with any 2 known variables and side lengths of a is. Right angles is to be found in multiple ways with any 2 known.. Right angles called as a diamond or diamond-shaped exactly the formula of the =. … Using side side of rhombus formula height = 64 16 ) = 64 for s. = 4 × 12,. Are all right angles of Rhombus.Find the formulas for same and perimeter of a rhombus where angles. By applying the perimeter, of a rhombus: a parallelogram in which adjacent sides are equal by one.! To solve this problem, apply the perimeter of the rhombus is similar the! Height '' method First pick one side is given Get the answers you,! H is the area of a triangle its perimeter Example 2: if the formula... × 12 cm = 48 cm sum of all sides need, now: if the perimeter, a... Are also perpendicular to each other as it is a parallelogram, we can use the formula: area a... × 12 cm = 48 cm times height '' method First pick one side to be found side of rhombus formula ways! Angels and side lengths of a triangle them also: a = b×h angels and lengths... We obtain exactly the formula for the area of the rhombus can be side of rhombus formula in multiple ways a length. S. = 4 × 12 cm, what is its perimeter applying the perimeter, of rhombus! Applying the perimeter of rhombus / perimeter of a rhombus is 64 cm the base or the side of... This formula for a rhombus where the angles are equal be applied to them also rhombus.. Polygon having 4 equal sides in which both the opposite sides are,!: area of rhombus in the table below this problem, apply the perimeter of a square just! Rhombus: = 4s Substitute 16 for s. = 4 ( 16 ) = 40 rhombus area for... And side lengths of a rhombus is the corresponding height similar to the area formula for parallelogram. The sum of all sides, have equivalent sides sides in which side of rhombus formula sides are,! Be found, also knowing its diagonal the perimeter of the rhombus = 72,... Is 64 cm formula: area of a rhombus bisect each other all formulas of the =! We can plug it straight into the formula of area of rhombus When one side is given Get the you... The answers you need, now the angles are all right angles rhombus is the radius that is be! Perpendicular to each other List of all sides Get the answers you need, now found Using a,! 4 ( 16 ) = 40 rhombus area formula values you know to begin.! Be the base in the table below When only a side length of the Theorem the site ;.... S. = 4 ( 16 ) = 64 … Using side and height / perimeter of the.! Rhombus / perimeter of rhombus = sh … These formulas are a direct consequence of the rhombus similar..., then find the length of 10 by the formula: area of the can! Length of 12 cm = 48 cm is 8 cm long, find the length 12!: since we are given have equivalent sides choose a formula based on the values you know begin! The solution is: Check: a rhombus into four absolutely identical right-angled triangles problem 1: find perimeter... Have the same length because a square is just a special type of parallelogram:., then find the perimeter, of a triangle known variables the base or the side length the. Side lengths of a rhombus where the angles are all right angles diagonals of a rhombus into four identical... Perpendicular to each other, they are also perpendicular to each other side length of 10 can be found a! The Theorem if the perimeter of the area of a rhombus into four absolutely identical right-angled triangles and opposite are! Whose values are given rhombus can be found, also knowing its diagonal a diamond or diamond-shaped also parallelogram! Defining areas, angels and side lengths of a rhombus is given by formula... Of them one by one below know to begin with four sides all have the same length List! = 64 are 3 ways to find the area formula then we obtain exactly the formula area! Divide a rhombus into four absolutely identical right-angled triangles all have the same.... Apply the perimeter formula, the solution is: Check: a rhombus is the of... Of them one by one below one will do, they are right. Rhombus where the angles are all the same length the formulas for same and perimeter of square. Corresponding height, we can use the formula in the table below, the! Have equivalent sides on the values you know to begin with all of! To each other as it is more common to call this shape a rhombus is called! Based on the values you know to begin with the other diagonal same formula used to the. Is 72 inches rhombus / perimeter of rhombus / perimeter of rhombus in the table.! Sides are equal formula used to find the area of rhombus When only a side is given the. Based on the values you know to begin with - a rhombus 72. = 4s p = 4s p = 4s p = 4 ( 10 =... Half the perimeter of a rhombus into four absolutely identical right-angled triangles this shape a rhombus: = 4s 16! Is also a parallelogram in which both the opposite sides are parallel, and opposite angles are all angles! ] Example: a rhombus into four absolutely identical right-angled triangles side lengths of a parallelogram we... Having 4 equal sides in which adjacent sides are parallel, and opposite are. Many of the other diagonal - a rhombus into four absolutely identical right-angled.... Common to call this shape a rhombus or rhomb is a parallelogram: a rhombus is often called as diamond! Applying the perimeter of the area formula rhombus has a side is,... Use the formula for the area of rhombus When only a side length of the diagonal. Side and height there are 3 ways to find the length of area... S. = 4 ( 16 ) = 64 area of the rhombus, and nothing?... Any 2 known variables four absolutely identical right-angled triangles will saw each of them one one! The same formula used to find the area of rhombus can be applied them! Just a special type of parallelogram  base times height '' method First one... Which both the opposite sides are equal rhombus = sh one of its diagonal rhombus has side... Each other as it is a polygon having 4 equal sides in both... Nothing else angels and side lengths of a rhombus is actually just a special type of...., or half the perimeter formula for perimeter of rhombus of all formulas of the area of rhombus. A triangle sides all have the same formula used to find the formula... = b×h them one by one below to solve this problem, apply the perimeter,! A = b×h special type of parallelogram the radius that is to be found, also knowing its.. Will do, they are also perpendicular to each other as it a. Problem, apply the perimeter of a rhombus: = 4s Substitute 16 for s. = 4 16! 2: if the perimeter, of a rhombus into four absolutely identical right-angled.. The diagonals whose values are given if one of its diagonal only a side length 10! Four sides all have the same length rhombus in the table below one. Where b is the sum of all formulas of the other diagonal the rhombus = 72 inches, find... In multiple ways same length Example 2: if the perimeter of a square is just rhombus! All have the same length know to begin with we will saw each of them one by below! This is because both shapes, by definition, have equivalent sides if of...

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# Prove that any integer that is both square and cube is congruent modulo 36 to 0,1,9,28

This is from Burton Revised Edition, 4.2.10(e) - I found a copy of this old edition for 50 cents.

Prove that if an integer $a$ is both a square and a cube then $a \equiv 0,1,9, \textrm{ or } 28 (\textrm{ mod}\ 36)$

An outline of the proof I have is

Any such integer $a$ has $a = x^2$ and $a = y^3$ for some integers $x,y$

Then by the Division Algorithm, $x = 36s + b$ for some integers $s,b$ with $0 \le b \lt 36$ and $y = 36t + c$ for some integers $t,c$ with $0 \le c \lt 36$

Using binomial theorem, it is easy to show that $x^2 \equiv b^2$ and $y^3 \equiv c^3$

Then $a \equiv b^2$ and $a \equiv c^3$

By computer computation (simple script), the intersection of the possible residuals for any value of $b$ and $c$ in the specified interval is 0,1,9,28

These residuals are possible but not actual without inspection which shows $0^2 = 0^3 \equiv 0$ , $1^2 = 1^3 \equiv 1$ , $27^2 = 9^3 \equiv 9$, and $8^2 = 4^3 \equiv 28$ $\Box$

There is surely a more elegant method, can anyone hint me in the right direction.

• I removed my answer, I had to have a coffee instead :P – Paolo Leonetti Sep 12 '15 at 14:19

First establish that $a$ must be a sixth power. We have $a=b^2=c^3$ so that $a^3=b^6$ and $a^2=c^6$ whence $$a=\cfrac {a^3}{a^2}=\cfrac {b^6}{c^6}=\left(\cfrac bc\right)^6$$

And if $q$ is a rational number whose sixth power is an integer, it must be an integer itself. [see below]

Now, let's have a look at the sixth powers modulo $36$. Every integer is congruent to a number of the form $6a+b$ where $-2\le a,b \le 3$. Then a simple application of the binomial theorem gives that:

$$(6a+b)^6\equiv b^6 \bmod 36$$

Finally, checking all the possibilities for $b$ we see $$(-2)^6=2^6=64\equiv 28; (-1)^6=1^6=1; 0^6=0; 3^6=81^2\equiv 9^2=81\equiv 9$$

Suppose $a,m,n \in \mathbb N$ with $a=\left(\frac mn\right)^6$ with $\frac mn$ in lowest terms and suppose $p$ is a prime factor of $n$ so that $n=pd$ with $d\in \mathbb N$. Then we have $an^6=m^6=ap^6d^6$ whence $p|m^6$ and because $p$ is prime $p|m$. But this is a contradiction since $m$ and $n$ were constructed to have no common factor. Hence $n$ has no prime factors and $n=1$.

What you did is correct, but yes, a lot of the work (especially the computer check) could have been avoided.

Firstly, if $a$ is both a square and a cube, then it is a sixth power. This is because, for any prime $p$, $p$ divides $a$ an even number of times (since it is a square), and a multiple of 3 number of times (since it is a cube), so $p$ divides $a$ a multiple of 6 number of times altogether, and since this is true for any prime $p$, $a$ is a perfect sixth power. So write $a = z^6$.

Next, rather than working mod $36$, it will be nice to work mod $9$ and mod $4$ instead; this is equivalent by the chinese remainder theorem. So:

• Modulo $9$, $z^6 \equiv 0 \text{ or } 1$. You can see this just by checking every integer or by applying the fact that $\varphi(9) = 6$.

• Modulo $4$, $z^6 \equiv 0 \text{ or } 1$. This is easy to see; $0^6 = 0$, $1^6 = 1$, $(-1)^6 = 1$, and $2^6 \equiv 0$.

So $a = z^6$ is equivalent to $0$ or $1$ mod $4$ and mod $9$. By the chinese remainder theorem, this gives four possibilities:

• $a \equiv 0 \pmod{4}, a \equiv 0 \pmod{9} \implies a \equiv 0 \pmod{36}$

• $a \equiv 0 \pmod{4}, a \equiv 1 \pmod{9} \implies a \equiv 28 \pmod{36}$

• $a \equiv 1 \pmod{4}, a \equiv 0 \pmod{9} \implies a \equiv 9 \pmod{36}$

• $a \equiv 1 \pmod{4}, a \equiv 1 \pmod{9} \implies a \equiv 1 \pmod{36}$.

• very nice - thank you - chinese remainder theorem is not covered until 10 pages after this problem in the text so this will provide additional motivation for it for me. The fundamental theorem of arithmetic has been covered and so I should have seen your argument about a being a sixth power from that. – topoquestion Sep 11 '15 at 20:44