# Show that a connected graph on $n$ vertices is a tree if and only if it has $n-1$ edges.

Show that a connected graph on $n$ vertices is a tree if and only if it has $n-1$ edges.

$(\Rightarrow)$ If a tree $G$ has only $1$ vertex, it has $0$ edges. Now, assume that any tree with $k-1$ vertices has $k-2$ edges. Let $T$ be a tree with $k$ vertices. Remove a leaf $l$ to obtain a tree $T'$ with $k-1$ vertices. Then, $T'$ has $k-2$ edges, by the inductive hypothesis. The addition of $l$ to $T'$ produces a graph with $k-1$ edges, since $l$ has degree $1$.

$(\Leftarrow)$ Let $G$ be a connected graph on $n$ vertices, with $n-1$ edges. Suppose $G$ is not a tree. Then, there exists at least one cycle in $G$. Successively remove edges from cycles in $G$ to obtain a graph $G'$ with no cycles. Then, $G'$ is connected and has no cycles. Therefore, $G'$ is a tree, with $n$ vertices and $n-1-x$ edges, for some $x > 0$. However, this contradicts the previous derivation. Therefore, it must be the case that $G$ is a tree.

• it seems to be true. – mesel Jun 15 '14 at 12:40

The proofs are correct. Here's alternative proof that a connected graph with n vertices and n-1 edges must be a tree modified from yours but without having to rely on the first derivation:

Let $G$ be a connected graph on n vertices, with n−1 edges. Suppose $G$ is not a tree. Then, there exists at least one cycle in $G$. Remove one of the edges within a cycle. This leaves a connected graph on n vertices with n-2 edges which is impossible as a connected graph on n vertices must at least have n - 1 edges.

Yes, this seems like it is true, although you didn't prove a leaf must always exist, if you really want to be rigorous. And in the other part you didn't explain why if it contained a cycle you could remove an edge without the graph being disconnected, but I like the proof over all.

I have the following way of proving it. Let me know if you agree with it or not.

Suppose the graph on $n$ vertices with $n-1$ edges is not a tree. This implies it has at least $1$ polygon. Suppose in total there are $k$ edges involved in these polygons and we know that a polygon has as many edges as vertices, since polygons are of regular degree 2. Then, if we consider th egraph without the edges that form the polygons, then we have a connected graph with at least $n-k$ edges, since in a connected graph every vertex has degree at least 1. Thus, in total this would imply we have at least $n$ in all the graph. But this contradicts the fact that we have $n-1$ in the original graph. Thus it must be a tree.

The converse is not always true. Consider a disjoint graph with n vertices and n - 1 edges. Here we have n-1 edges but this is obviously not a tree. The assumption that G is connected is only for the forward implication.


# smooth functions or continuous

When wesay a function is smooth? Is there any difference between smooth function and continuous function? If they are the same, why sometimes we say f is smooth and sometimes f is continuous?

• "smooth" means (at least) "continously differentiable". Sometimes more (even infinite number of) derivatives are required to be continuous. Aug 20, 2013 at 16:59
• @njguliyev, not to nitpick but I think it's relatively common to call Lipschitz continuous ODEs "smooth" - being just smooth enough for existence and uniqueness of solutions. Aug 20, 2013 at 17:08

A function being smooth is actually a stronger case than a function being continuous. For a function to be continuous, the epsilon delta definition of continuity simply needs to hold, so there are no breaks or holes in the function (in the 2-d case). For a function to be smooth, it has to have continuous derivatives up to a certain order, say k. We say that function is $C^{k}$ smooth. An example of a continuous but not smooth function is the absolute value, which is continuous everywhere but not differentiable everywhere.

A smooth function is differentiable. Usually infinitely many times.

• ... or at least as often as we need it. Aug 20, 2013 at 17:14

Smooth implies continuous, but not the other way around. There are functions that are continuous everywhere, yet nowhere differentiable.

A smooth function can refer to a function that is infinitely differentiable. More generally, it refers to a function having continuous derivatives of up to a certain order specified in the text. This is a much stronger condition than a continuous function which may not even be once differentiable.

A smooth function is a function that has continuous derivatives up to some desired order over some domain. A function can therefore be said to be smooth over a restricted interval such as or . The number of continuous derivatives necessary for a function to be considered smooth depends on the problem at hand, and may vary from two to infinity. A function for which all orders of derivatives are continuous is called a C-infty-function.

Take $$f(x) = x|x|$$ it is smooth, and now consider $$g(x) = x^3$$, this other function is also smooth. However, $$g(x)$$ is much smoother than $$f(x)$$ because derivitive of $$f(x)$$. You can argue that $$g(x)$$ is infinitely many times smooth. All polynomials belong to $$C^\infty$$ meaning they are infinitely many times differentiable and are smooth.

However, $$h(x) = |x|$$ is not smooth, because it has corner. Please note that all three functions, $$f(x)$$, $$g(x)$$, and $$h(x)$$ are continous.

Here is how $$f(x)$$ looks like:

Here is how the derivative of $$f(x)$$ looks like:

Here is the second derivate of $$f(x)$$, as you can see its second derivative is not even continuous:

Here is the graph of $$g(x)$$:

Here is graph of $$\frac{d f(x)}{dx} = 3x^2$$:

Here is graph of $$\frac{d^2 g(x)}{dx^2} = 6x$$:

• If a function $f$ is smooth, then can I suppose that $f$ is increasing ou decreasing? At least in some interval? Oct 9, 2020 at 0:52

Consider a sequence in $\mathbb{R}$ say $\{x_n\}_{n \in \mathbb{N}}$, which is continuous in $\mathbb{R}$. Usually we do not say it a smooth function.


# Math Help - Optimization

1. ## Optimization

An isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??

Any other good notes or questions for Optimization problems would be welcomed as well

2. Originally Posted by someone21
An isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??

Any other good notes or questions for Optimization problems would be welcomed as well
Minimum or maximum? The minimum area is trivially zero.

3. Originally Posted by mr fantastic
Minimum or maximum? The minimum area is trivially zero.
Minimum possible area but why is it zero??

And if maximum possible area then how to do it???

4. Originally Posted by someone21
Minimum possible area but why is it zero??

And if maximum possible area then how to do it???
Draw an isosceles triangle with non-equal side the diameter of the circle. Now move the non-equal side up so that the two equal sides get smaller and smaller. The triangle degenerates to a point - of area zero.

5. Originally Posted by someone21
An isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??

Any other good notes or questions for Optimization problems would be welcomed as well
Scroll down to the bottom of this: Triangles, Inscribed and Circumscribed Circles

Your circle has a radius r = 2. Let a = c and re-arrange:

$K = \frac{a^2 b}{8}$ .... (1)

where K is the area of the triangle.

You need to find a relationship between a and b and use it to get one variable in terms of the other, b in terms of a say. Substitute this into (1) to get K in terms of a single variable, a say. Now solve dK/da = 0, test nature etc ....

I haven't done it but I wouldn't be surprised if you find b = a gives maximum area, that is, an equilateral triangle has maximum area.

6. Originally Posted by mr fantastic
Scroll down to the bottom of this: Triangles, Inscribed and Circumscribed Circles

Your circle has a radius r = 2. Let a = c and re-arrange:

$K = \frac{a^2 b}{8}$ .... (1)

where K is the area of the triangle.

You need to find a relationship between a and b and use it to get one variable in terms of the other, b in terms of a say. Substitute this into (1) to get K in terms of a single variable, a say. Now solve dK/da = 0, test nature etc ....

I haven't done it but I wouldn't be surprised if you find b = a gives maximum area, that is, an equilateral triangle has maximum area.
Alternatively, you could work with finding the size of the two equal angles at the base of the triangle:

Let the angle at the top vertex of the triangle be of size $\phi$.

Let the two equal angles at the base of the triangle be of size $\theta$.

Let a be the two equal side lengths of the triangle. Let b be the length of the base of the triangle. Let h be the height of the triangle. Drop a perpendicular from the top vertex - half of the top vertex angle is $\frac{\phi}{2}$ and the perpendicular is the height of the triangle.

Area of any triangle: $A = \frac{bh}{2}$.

To find the angles that give a maximum triangle area, express the above area in terms $\phi$ and radius (= 2).

You should be able to show that $\frac{a}{2} = 2 \cos(\phi/2)$ (use the isosceles triangle with sides r = 2, r = 2, a and equal angles $\phi/2$). Then:

$a = 4 \cos(\phi/2)$.

$b = 8 \cos(\phi/2) \sin(\phi/2)$.

$h = 4 \cos(\phi/2) \cos(\phi/2) = 4 \cos^2 (\phi/2)$.

Therefore:

$A = \frac{bh}{2} = \frac{1}{2} [8 \cos(\phi/2) \sin(\phi/2)] [4 \cos^2(\phi/2)] = 16 \cos^3 (\phi/2) \sin(\phi/2)$.

Now solve $\frac{d A}{d \phi} = 0$ etc. I get $\phi = \frac{\pi}{3}$, that is, the triangle is equilateral. Substitute $\phi = \frac{\pi}{3}$ into A to get the maximum area.

Note: I reserve the right for this reply to contain algebraic errors.

7. Originally Posted by mr fantastic
Alternatively, you could work with finding the size of the two equal angles at the base of the triangle:

Let the angle at the top vertex of the triangle be of size $\phi$.

Let the two equal angles at the base of the triangle be of size $\theta$.

Let a be the two equal side lengths of the triangle. Let b be the length of the base of the triangle. Let h be the height of the triangle. Drop a perpendicular from the top vertex - half of the top vertex angle is $\frac{\phi}{2}$ and the perpendicular is the height of the triangle.

Area of any triangle: $A = \frac{bh}{2}$.

To find the angles that give a maximum triangle area, express the above area in terms $\phi$ and radius (= 2).

You should be able to show that $\frac{a}{2} = 2 \cos(\phi/2)$ (use the isosceles triangle with sides r = 2, r = 2, a and equal angles $\phi/2$). Then:

$a = 4 \cos(\phi/2)$.

$b = 8 \cos(\phi/2) \sin(\phi/2)$.

$h = 4 \cos(\phi/2) \cos(\phi/2) = 4 \cos^2 (\phi/2)$.

Therefore:

$A = \frac{bh}{2} = \frac{1}{2} [8 \cos(\phi/2) \sin(\phi/2)] [4 \cos^2(\phi/2)] = 16 \cos^3 (\phi/2) \sin(\phi/2)$.

Now solve $\frac{d A}{d \phi} = 0$ etc. I get $\phi = \frac{\pi}{3}$, that is, the triangle is equilateral. Substitute $\phi = \frac{\pi}{3}$ into A to get the maximum area.

Note: I reserve the right for this reply to contain algebraic errors.
I think the question was the circle is inside the isoceles triangle?

8. Originally Posted by mr fantastic
Alternatively, you could work with finding the size of the two equal angles at the base of the triangle:

Let the angle at the top vertex of the triangle be of size $\phi$.

Let the two equal angles at the base of the triangle be of size $\theta$.

Let a be the two equal side lengths of the triangle. Let b be the length of the base of the triangle. Let h be the height of the triangle. Drop a perpendicular from the top vertex - half of the top vertex angle is $\frac{\phi}{2}$ and the perpendicular is the height of the triangle.

Area of any triangle: $A = \frac{bh}{2}$.

To find the angles that give a maximum triangle area, express the above area in terms $\phi$ and radius (= 2).

You should be able to show that $\frac{a}{2} = 2 \cos(\phi/2)$ (use the isosceles triangle with sides r = 2, r = 2, a and equal angles $\phi/2$). Then:

$a = 4 \cos(\phi/2)$.

$b = 8 \cos(\phi/2) \sin(\phi/2)$.

$h = 4 \cos(\phi/2) \cos(\phi/2) = 4 \cos^2 (\phi/2)$.

Therefore:

$A = \frac{bh}{2} = \frac{1}{2} [8 \cos(\phi/2) \sin(\phi/2)] [4 \cos^2(\phi/2)] = 16 \cos^3 (\phi/2) \sin(\phi/2)$.

Now solve $\frac{d A}{d \phi} = 0$ etc. I get $\phi = \frac{\pi}{3}$, that is, the triangle is equilateral. Substitute $\phi = \frac{\pi}{3}$ into A to get the maximum area.

Note: I reserve the right for this reply to contain algebraic errors.
I think the question was the circle is inside the isoceles triangle?

9. Originally Posted by someone21
I think the question was the circle is inside the isoceles triangle?
Ah yes. I misread circumscribed as inscribed. Oh well, I guess I answered the second part:

"Any other good notes or questions for Optimization problems would be welcomed as well"

10. Originally Posted by someone21
An isosceles triangle is circumscribed in a circle of a radius 2m.Determine the minimum poosible area the isosecles triangle can have??

Any other good notes or questions for Optimization problems would be welcomed as well
Let 2b be the length of the base of the triangle, a the length of the other two sides and h the height of triangle.

Then A = bh.

A relationship between b and h is need so that A can be expressed in terms of a single variable.

Draw a line from the middle of the base to the (top) vertex of the triangle. This line goes through the center of the circle and is the height of the triangle.

Draw a line from the center of the circle to one of the sides of the triangle. This line is perpendicular to that side (why?).

Apply Pythagoras Theorem to the two obvious right-triangles:

$h^2 + b^2 = a^2$ .... (1)

$2^2 + {\color{red}(a-b)}^2 = (h - 2)^2 \Rightarrow 4 + a^2 - 2ab + b^2 = h^2 - 4h + 4 \Rightarrow a^2 - 2ab + b^2 = h^2 - 4h$ .... (2)

You should consider the geometry that gives the red length .....

Solve equations (1) and (2) simultaneously for b in terms of h (a small bit of algebra for you to do):

$b = \frac{2h}{\sqrt{h^2 - 4h}}$.

Therefore A = .......

Solve dA/dh = 0 (I get h = 6). Hence the minimum area of the triangle is equal to .....

Note: As you might anticipate, if you do the calculations you get $2b = 4 \sqrt{3} = a$ and so the triangle is equilateral.

Note: I reserve the right for this reply to contain careless errors.

11. Originally Posted by someone21
Minimum possible area but why is it zero??
One way to think about it....

An equilateral triangle is a shape where all three sides are of equal length. It is considered a special case of an isosceles triangle. Let s be a side, and A be the area of the triangle.

$\lim_{s \rightarrow 0} A$

$\lim_{s \rightarrow 0} \frac{1}{2\sqrt{2}}s^2 \rightarrow 0$

Since the area of the circle, $A_c$ is always inscribed in the triangle, it is always less than the area of the triangle, so $A_c < A$

The limit of A converges to zero, so due to the Squeeze Theorem or Comparison test, the area of the circle converges to zero. The minimum area of the circle is therefore zero.

12. Originally Posted by colby2152
One way to think about it....

An equilateral triangle is a shape where all three sides are of equal length. It is considered a special case of an isosceles triangle. Let s be a side, and A be the area of the triangle.

$\lim_{s \rightarrow 0} A$

$\lim_{s \rightarrow 0} \frac{1}{2\sqrt{2}}s^2 \rightarrow 0$

Since the area of the circle, $A_c$ is always inscribed in the triangle, it is always less than the area of the triangle, so $A_c < A$

The limit of A converges to zero, so due to the Squeeze Theorem or Comparison test, the area of the circle converges to zero. The minimum area of the circle is therefore zero.
why is it s goes to zero and can it be explained more detaily please

13. Originally Posted by someone21
why is it s goes to zero and can it be explained more detaily please
You are looking at the area as a side goes to zero. The area, dependent upon the sides, then goes to zero as well.


# A Geometric Proof of $\zeta(2)=\frac{\pi^2}6$? (and other integer inputs for the Zeta)

Is there a known geometric proof for this famous problem? $$\zeta(2)=\sum_{n=1}^\infty n^{-2}=\frac16\pi^2$$

Moreover we can consider possibilities of geometric proofs of the following identity for positive even inputs of the Zeta function: $$\zeta(2n)=(-1)^{n+1} \frac{B_{2n}(2\pi)^{2n}}{2(2n)!}$$ and negative inputs: $$\zeta(-n)=-\frac{B_{n+1}}{n+1}$$

• There are many nice proofs at math.stackexchange.com/questions/8337/… . Is one of them geometric enough for you? (David Speyer's answer might fit the bill.) – Qiaochu Yuan Aug 27 '12 at 1:04
• Here's a proof by Prof. Greene at UCLA: math.ucla.edu/~greene/How%20Geometry.pdf – Elchanan Solomon Aug 27 '12 at 2:20
• thank you Qiaochu one of them is ;) – finnlim Aug 27 '12 at 2:44
• I wanted to ask this question some time ago. I'm glad to see it posted. (I thought of an elementary geometrical proof) – user 1591719 Aug 27 '12 at 5:25
• do you mean that you have an elementary geometric proof? Would you mind sharing it? – finnlim Aug 27 '12 at 5:42

Yes, there is a geometric proof for $\zeta(2)=\sum_{n=1}^\infty n^{-2}=\frac16\pi^2$, and, on top of that, it is a very unusual and, in my opinion, beautiful one.

In his gorgeous paper "How to compute $\sum \frac{1}{n^2}$ by solving triangles", Mikael Passare offers this idea for proving $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$:

Proof of equality of square and curved areas is based on another picture:

Recapitulation of Passare's proof using formulas is as follows: (for each step there is a geometric justification)

$$\sum_{n=1}^\infty \frac{1}{n^2} = \sum_{n=1}^\infty \int_0^\infty \frac{e^{-nx}}{n}\; dx\; = -\int_0^\infty log(1-e^{-x})\; dx\; = \frac{\pi^2}{6}$$

I'm not sure what you mean by a geometric proof, but the following should fit the bill, as here the identity is deduced from a comparison of two areas: the first area is

$\displaystyle\int_{[0,1]^2} \frac{1}{1 - xy} \frac{dx dy}{\sqrt{xy}}$

and the second is

$4 \displaystyle \int_{\substack{\xi, \eta>0 \\ \xi \eta \leq 1}} \frac{ d \xi \, d \eta}{(1+\xi^2)(1+\eta^2)}$;

They are equal by a change of variables. For the first quantity, expand $(1-xy)^{-1}$ as a geometric series and integrate term-wise to get $3 \cdot \zeta(2)$. The second can be computed to $\pi^2/2$. This can be found in detail in Kontsevich and Zagier's Periods (bottom of page 8), where they attribute the idea to Calabi. (You should easily be able to hunt down an identity of $\zeta (n)$ being equal to an integral over the unit square in $\mathbb{R}^n$, like the first above. It might be a fun exercise to see if this idea is can be adapted for even $n$.)

One geometric/probabilistic proof is by Eugenio Calabi.

Consider the double integral $$I=\int_{0}^{1}\int_{0}^{1} \frac{1}{1-x^2y^2} \ dy \ dx.$$ Expand the integrand into a geometric series as such $$\frac{1}{1-x^2y^2} =\sum_{n=0}^{\infty} (xy)^{2n},$$ which is justified by the region of integration, that is $0<x,y<1.$ Exchange summation and integration (application of Tonelli's theorem), and integrate term by term to get $$I=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}.$$ Now, we evaluate $I$ directly. Make the change of variables $$x=\frac{\sin(u)}{\cos(v)}, y=\frac{\sin(v)}{\cos(u)}.$$ This transformation has Jacobian Determinant $$\frac{\partial (x,y) }{\partial(u,v)}=1-x^2y^2,$$ which cancels with the integrand, and the region becomes the open triangle defined by the inequalities $$u+v<\frac{\pi}{2}, u, v>0$$ We know from geometry that this triangle has base $\frac{\pi}{2}$ and height $\frac{\pi}{2},$ hence the area is $\frac{\pi^2}{8}.$ Hence $I=\frac{\pi^2}{8}.$ Now to obtain $\zeta(2),$ we use the fact that $$\zeta(2)=\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=1}^{\infty} \frac{1}{(2n)^2}+\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2},$$ which implies by some algebraic manipulation that

$$\zeta(2)=\frac{4}{3}I=\frac{\pi^2}{6}.$$

We can extend this proof to the $\zeta(2k).$ Let

$$I_{2k}= \int_{0}^{1}... \int_{0}^{1} \frac{1}{1-x_1^2 \dots x_{2k}^2} \ dx_{2k} \ ... \ dx_1.$$

Converting this integrand into a geometric series and exchanging summation and integration gives the sum: $$I_{2k}=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2k}}.$$ We can then recover $$\zeta(2k)=\frac{2^{2k}}{2^{2k}-1} I_{2k},$$ by observing $$\zeta(2k)= \sum_{n=1}^{\infty} \frac{1}{(2n)^{2k}} + I_{2k},$$ and using some algebraic manipulation.

On the other hand, we make the change of variables $$x_i=\frac{\sin(u_i)}{\cos(u_{i+1})}, 1 \leq i \leq 2k,$$ and $u_{2k+1}:=u_{1}$ here. This transformation turns out to have a Jacobian Determinant which cancels with the denominator of the integrand in $I_{2k}.$ The region of integration becomes the open polytope: $$\Delta^{2k}= \left \lbrace (u_1, \dots, u_{2k}): u_{i}+u_{i+1} < \frac{\pi}{2}, u_i>0, 1 \leq i \leq 2k \right \rbrace$$ in which $u_{2k+1}:=u_1.$ Computing the volume of this polytope is much more difficult. Rescaling with the change of variables $u_i=\frac{\pi}{2} v_i,$ and letting $V_1, \dots, V_{2k}$ being $2k$ independent, uniformly distributed random variables on $(0,1),$ we get

$$I_{2k}=\text{Vol}(\Delta^{2k})=\left(\frac{\pi}{2}\right)^{2k} \text{Pr} \left( V_1+V_2<1, \dots, V_{2k-1}+V_{2k}<1, V_{2k}+V_{1}<1 \right),$$ the probability that $V_1, \dots, V_{2k}$ have cyclically pairwise consecutive sums less than $1.$

In the literature, the aforementioned probability has been computed in several ways (see: https://www.maa.org/sites/default/files/pdf/news/Elkies.pdf https://arxiv.org/pdf/1003.3602.pdf https://pdfs.semanticscholar.org/35be/01e63c0bfd32b82c97d58ccc9c35471c3617.pdf)

Joshua Seaton also mentioned Zagier's and Kontsevich's reproduced Calabi proof. The general version of this is to evaluate

$$J_{2k}= \int_{0}^{1}... \int_{0}^{1} \frac{1}{\sqrt{x_1 \dots x_{2k}}(1-x_1 \dots x_{2k})}\ dx_{2k} \ ... \ dx_1.$$

A quick substitution shows $J_{2k}=2^{2k}I_{2k}.$ The generalized version of the change of variables is

$$x_i=\frac{\xi_i^2(1+\xi_{i+1}^2)}{1+\xi_i^2}, \dots, x_{2k}=\frac{\xi_{2k}^2(1+\xi_{1}^2)}{1+\xi_{2k}^2},$$ and upon computing its Jacobian Determinant, we get that

$$J_{2k}=\int_{\mathbb{H}^{2k}} \frac{1}{\xi_1^2+1} \dots \frac{1}{\xi_{2k}^2+1} \ d \xi_{1} \dots d \xi_{2k},$$ where $\mathbb{H}^{2k}$ is the hyperbolic defined by the inequalities: $$\xi_{i}\xi_{i+1} <1, \xi_i>0,$$ where $1 \leq i \leq 2k$ and $\xi_{2k+1}:=\xi_1.$ Letting $\Xi_1, \dots \Xi_{2k}$ be $2k$ independent, nonnegative Cauchy random variables, we get

$$I_{2k}=\left(\frac{\pi}{2}\right)^{2k} \text{Pr} \left( \Xi_1\Xi_2<1, \dots, \Xi_{2k-1}\Xi_{2k}<1, \Xi_{2k}\Xi_{1}<1 \right),$$ the probability that $\Xi_1, \dots, \Xi_{2k}$ have cyclically pairwise consecutive products less than $1.$

In my paper, I show that the probabilities lead to a very gargantuan formula:

$$I_{2k}=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2k}}= \left(\frac{\pi}{4} \right)^{2k} +\left(\frac{\pi}{4} \right)^{2k} \sum_{n=1}^{k} \sum_{\substack{(r_1, \dots, r_n) \in [2k]^n: \\ |r_p-r_q| \notin \lbrace 0,1,2k-1 \rbrace, \\ p,q \in [n]} } \prod_{i=1}^{n} \frac{1}{i+\sum_{j=1}^{i} \alpha_j},$$ where $[m]:= \lbrace 1, \dots, m \rbrace$ and $$\alpha_j=2- \delta(2k,2) - \sum_{m=1}^{j-1} \delta(|r_m-r_j|,2)+\delta(|r_m-r_j|,2k-2)$$ and $\delta(a,b)$ is the Kronecker Delta Function. In particular, the inner sum is taken over all tuples $(r_1, \dots, r_n) \in [k]^ n$ having cyclically pairwise nonconsecutive entries.


# Linear Programming Word Problem: Theater

I'm having trouble understanding the second constraint.

A theater is presenting a program on drinking and driving for students and their parents[...] admission is 2.00 dollars for parents and 1.00 dollar for students. However, the situation has two constraints: 1) The theater can hold no more than 150 people and 2) every two parents must bring at least one student. How many parents and students should attend to raise the maximum amount of money?

Let $$x$$ = number of students

$$y$$ = number of parents

Since the question prompt wants the maximum amount of revenue, then the objective function is $$z = x + 2y$$

The theater can hold no more than 150 people so the first constraint is simply: $$x + y \leq 150$$

The second constraint is, "every two parents must bring at least one student," but I don't understand how to model this.

I've looked up a solution and it said $$y \leq 2x$$ is how to model this constraint, but I don't understand why. If there must be at least one student for every two parents then why wouldn't the inequality be $$2y \geq x$$?

Yes

y <= 2x

The easiest way to test this is with some data points and sketch a graph:

y_______x

2_______1,2,3,4,5…

4_______2,3,4,5,6…

6_______3,4,5,6,7…

Hence:

y <= 2x

2 <= 2x1 True

and

2y >= x

2x2 >= 5 False

Graph of y <= 2x


It is currently 23 Mar 2018, 02:07

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# If y is an integer, is y^2 divisible by 4?

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If y is an integer, is y^2 divisible by 4? [#permalink]

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07 Feb 2012, 20:41
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If y is an integer, is y^2 divisible by 4?

(1) y is divisible by 4
(2) y is divisible by 6
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Feb 2012, 02:16, edited 1 time in total.
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Re: Number Properties - Question 2 [#permalink]

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Hi, there. I'm happy to help with this.

Prompt: if y is an integer , is y^2 divisible by 4?

Fact #1: In order for an integer N to be divisible by 4, N must have at least two factors of 2 in its prime factorization.

Fact #2: When you square a number, say T^2, whatever prime factors are in the prime factorization of T are doubled in the prime factorization of T^2. Say a particular prime factor appears three times in T --- then it will appear six times in T^2.

Fact #3: An even number, by definition, has at least one factor of 2 in its prime factorization.

Therefore, the square of any even number will have at least two factors of 2, and therefore will be divisible by 4.

The question "is y^2 divisible by 4?" is entirely equivalent to the question "is y an even integer?"

All of that was the mathematical heavy-lifting for the question. Now, the statements will be a piece of cake.

Statement #1: y is divisible by 4. Therefore y is even. Sufficient.

Statement #2: y is divisible by 6. Therefore y is even. Sufficient.

Both statements sufficient. Answer = D

Does all that make sense?

Here's another odd/even DS question for practice.

http://gmat.magoosh.com/questions/868

Please let me know if you have any questions.

Mike
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Re: Number Properties - Question 2 [#permalink]

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08 Feb 2012, 02:16
If y is an integer, is y^2 divisible by 4?

(1) y is divisible by 4 --> as y itself is divisible by 4 then y^2 will also be divisible by 4. Sufficient.
(2) y is divisible by 6 --> y=6k, for some integer k --> y^2=36k^2=4*(9k^2) --> we can see that y^2 has 4 as a factor hence it is divisible by 4. Sufficient.

Hope it's clear.
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If y is an integer, is y^2 divisible by 4? [#permalink]

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26 Nov 2017, 11:28
nimc2012 wrote:
If y is an integer, is y^2 divisible by 4?

(1) y is divisible by 4
(2) y is divisible by 6

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variables and 0 equation, we need to consider each condition one by one.

Condition 1)
$$y = 4n$$ for some integer $$n$$.
$$y^2 = 16n^2$$
Since $$16$$ is divisible by $$4$$, this is sufficient.

Condition 2)
$$y = 4m$$ for some integer $$m$$.
$$y^2 = 36m^2$$
Since $$36$$ is divisible by $$4$$, this is sufficient.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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If y is an integer, is y^2 divisible by 4?   [#permalink] 26 Nov 2017, 11:28
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# Probability of random integer's digits summing to 12

What is the probability that a random integer between 1 and 9999 will have digits that sum to 12?

As a user suggested, I could make a spreadsheet and count them, but is there a quicker way to do this?

• For a range like this, why not make a spreadsheet and count them? – Ross Millikan Apr 25 '13 at 16:52
• Your question is phrased as a stand-alone problem, without any further information or context. This does not match our quality standards, and hence is likely to attract downvotes, or be closed. It is impossible for us to assess your issues with the problem, and the level of answer appropriate for you. This is a guide to asking a good question. Concretely: please provide context, and include your work and thoughts on the problem. This helps attract more appropriate answers and will likely remove down- and close votes. – The Chaz 2.0 Apr 25 '13 at 16:54
• I vote against closing this question. – MJD Apr 25 '13 at 17:40
• I also vote against closing this question. – Zev Chonoles Apr 25 '13 at 22:02
• @TheChaz I am very disappointed to see valid questions like this being closed for strange reasons, e.g. "too localized". Ditto for analogous recent votes. – Math Gems Apr 26 '13 at 15:50

The problem does not require a spreadsheet. It does not even require paper.

The question is to count the number of integer tuples $\langle a,b,c,d\rangle$ with $a+b+c+d=12$ and $0\le a,b,c,d < 10$. We could enumerate this by choosing $a$ and then counting the tuples $\langle b,c,d\rangle$ with $b+c+d = 12-a$, and recursing, but an easier method is available.

First, note that if we drop the $a,b,c,d < 10$ restriction, the problem is easy. By the stars and bars method, there are $\binom{15}{12} = 455$ tuples that sum to 12.

From these 455 we need to eliminate the ones that contain $10, 11,$ or $12$. Let $t_i$ be the number of tuples where $a =i$ for $i\in\{10,11,12\}$. Clearly, $t_{12} = 1$: the only tuple is $\langle 12, 0,0,0\rangle$. For $a=11$ we need $b+c+d=1$, so exactly one of $b,c,d$ is 1 and the other two are 0, and thus $t_{11} = 3$.

For $a=10$ there are two possibilities. Either $\{b,c,d\} = \{2,0,0\}$ or $\{b,c,d\} = \{1,1,0\}$. In either case there are 3 tuples, so $t_{10} = 6$.

Since at most one of $a,b,c,d$ is greater than 9, the total number of tuples that contain 10, 11, or 12 is $4(t_{10}+t_{11}+t_{12}) = 40$.

Thus the total number of tuples of just 0 through 9, and the answer to the question, is 455 - 40 = 415; the probability is $\frac{415}{9999}$.

• Note that with the range starting at 1, leading zeros are explicitly OK. This makes it easier. – Ross Millikan Apr 25 '13 at 18:12
• There are several features of this problem that make it easier than it seems at first: leading zeroes are acceptable, so $a,b,c,$ and $d$ are symmetric. The required sum is close to 9, so only a few integer tuples need to be subtracted, and no inclusion-exclusion argument is needed. Recognizing such features when they appear can be an important aspect of solving this type of problem. – MJD Apr 25 '13 at 18:27
• Does not even require paper?? :) Neither did mine – wolfies Apr 25 '13 at 19:17
• The problem does not require an integer either. What is the probability that randomly chosen combination of four numbers in the range 0 to 9 adds up to to twelve. – Kaz Apr 25 '13 at 21:53
• @Kaz Probability is 0, because if you permit $(a,b,c,d) \in \mathbb{R}^4$, even restricting the possible range to between 0 and 9, $a+b+c+d=12$ is still a three-dimentional figure. That's like asking what the probability that a random point in a plane happens to lie on a given line, or the probability that a randomly chosen number just happens to be exactly the same as a given number. Without restricting the digits to integers, the problem becomes trivial. – AJMansfield Apr 26 '13 at 1:24

Use generating functions. The generating function for a single digit is:

$$1 + x + \cdots + x^9 = \frac{1-x^{10}}{1-x}.$$

The generating function for the sum of four digits is the fourth power:

$$\frac{(1-x^{10})^4}{(1-x)^4} = (1-x^{10})^4 (1-x)^{-4}.$$

To solve the problem, find the coefficient of $x^{12}$.

$$(1-x^{10})^4 = 1 -4 x^{10} + 6 x^{20} - \cdots$$

So, we only need the coefficients of $x^2$ and $x^{12}$ in $(1-x)^{-4}$ using the generalized binomial theorem. These are $\binom{5}{2} = 10$ and $\binom{15}{12} = 455$. The coefficient of $x^{12}$ is therefore

$$-4\cdot10 + 1\cdot455 = 415.$$

So the answer is $\frac{415}{9999}$.

• where do 5,2,15 and 12 come from in ncr(5, 2) and ncr(15, 12)? – michaelsnowden Sep 8 '15 at 17:46
• @michaelsnowden The coefficient of $x^2$ in $(1 - x)^{-4}$ is $(-1)^2\binom{-4}{2} = \binom{4 + 2 - 1}{2}$, because of the binomial theorem which says that the coefficient of $z^n$ in $(1+z)^n$ is $\binom{n}{r}$, and the fact that $\binom{-n}{r} = (-1)^r\binom{n + r - 1}{r}$. Similarly, the coefficient of $x^{12}$ in $(1-x)^{-4}$ is $(-1)^{12}\binom{-4}{12} = \binom{4 + 12 - 1}{12}$. – ShreevatsaR Jan 5 '19 at 21:16
• In general, if we're looking for $k$ random digits summing to $n$, by this answer we're looking for the coefficient of $x^n$ in $(1-x^{10})^k (1-x)^{-k}$. This is going to be $$\sum_{10r+s=n} (-1)^{r+s} \binom{k}{r} \binom{-k}{s} = \sum_{10r+s=n} (-1)^{r} \binom{k}{r} \binom{k+s-1}{s}$$ if I'm not mistaken. – ShreevatsaR Jan 5 '19 at 21:24

With Mathematica code (FROM 1 TO 9999):

Count[Map[Total, IntegerDigits[Range[9999]]], 12]

415

So: 415/9999

• As an alternative approach: Probability[x == 12, x \[Distributed] Total[IntegerDigits[Range[9999]], {2}]] directly yields 415/9999. – Sasha Apr 25 '13 at 19:05
• Wow, "FrankenLisp". – Kaz Apr 25 '13 at 21:54

Just so GAP doesn't get left out. Here's a GAP version to obtain the count:

Number([1..9999],n->Sum(ListOfDigits(n))=12);

which returns 415. So, the probability is $415/9999$.

• Go GAP! Good to see it's still around and alive. – Peter K. Apr 25 '13 at 20:57

With PARI/GP code

Q(n)=if(n<10,n,n%10 + Q(n\10))

sum(i=1,9999,Q(i)==12)/9999

I obtain $$\frac{415}{9999}.$$


# Thread: Normal distribution with population knee heights

1. ## Normal distribution with population knee heights

Given: Men have sitting knee heights that are normally distributed with mean of 22.0 in. and std dev of 1.1 in. Women have mean of 20.3 in and std dev of 1.0 in.

Question: Find min and max sitting knee heights that include at least 95% of all men and at at least 95% of all women.

What I did: I originally took the avg of the means given and the avg of std devs given and using invNorm found the min and max using .025 from either side of the curve. Min: 19.10 in and Max: 23.21 in. After testing these limits with the men's average I found that I was cutting out 20.9% of the men, so obviously I did this wrong. NOTE: I also found 95% max and min for each gender, but then didn't know what to do with this information.

What should I have done to find the answers?

Thanks.

2. Originally Posted by stats08
Given: Men have sitting knee heights that are normally distributed with mean of 22.0 in. and std dev of 1.1 in. Women have mean of 20.3 in and std dev of 1.0 in.

Question: Find min and max sitting knee heights that include at least 95% of all men and at at least 95% of all women.

What I did: I originally took the avg of the means given and the avg of std devs given and using invNorm found the min and max using .025 from either side of the curve. Min: 19.10 in and Max: 23.21 in. After testing these limits with the men's average I found that I was cutting out 20.9% of the men, so obviously I did this wrong. NOTE: I also found 95% max and min for each gender, but then didn't know what to do with this information.

What should I have done to find the answers?

Thanks.
To get the minimum for men you require the value of $x_{\min}$ such that $\Pr(X < x_{\min}) = 0.025$.

Note that $z^* = \frac{x_{\min} - 22.0}{1.1}$ where $\Pr(Z < z^*) = 0.025 \Rightarrow z^* = -1.96$.

Then $-1.96 = \frac{x_{\min} - 22.0}{1.1} \Rightarrow x_{\min} = 19.84$.

In a similar way I get $x_{\max} = 24.16$.

Do the same thing for the women.

3. Originally Posted by mr fantastic
To get the minimum for men you require the value of $x_{\min}$ such that $\Pr(X < x_{\min}) = 0.025$.

Note that $z^* = \frac{x_{\min} - 22.0}{1.1}$ where $\Pr(Z < z^*) = 0.025 \Rightarrow z^* = -1.96$.

Then $-1.96 = \frac{x_{\min} - 22.0}{1.1} \Rightarrow x_{\min} = 19.84$.

In a similar way I get $x_{\max} = 24.16$.

Do the same thing for the women.
Is there some confusion here?

Are we looking for two knee height intervals one containing at least 95% of men and the other 95% of women, or one interval which contains at least 95% of men and 95% of women simultaneously?

Working this numerically I find that the smallest interval symmetric about the mean knee height of men and women combined is approximately [18.48, 23.82] inches. This is the larger of the two symmetric intervals that contain 95% of womens knees centred at the grand mean and that contain 95% of mens knees centred at the grand mean.

The interval given contains about 95% of mens knees and 96.5% of womens knees.

(of course if we had chossen to centre the interval at another point we would have found a different result, now the question is what is the smallest interval that meets our requirements? I'm pretty sure that the interval found here is not the smallest that will do the job)

The shortest interval meeting our requirements is approximatly [18.65, 23.82] which contains about 95% of womens knees and 95% of mens knees. This was found by brute force by computing the probability of a range of intervals for men and women plotting the 95% contours and keeping the point at which the contours cross!

CB

4. ## Population Knee Heights

You are correct, Captain, I need to find common max and min that will contain 95% of all men and all women combined (as if I'm designing a carseat).

When you say you found a smaller interval using "brute force by computing the probability of a range of intervals", can you give an example on how to even pick an interval to try and how you know when contours cross?

I was aware that some guessing is involved, but I'm not sure if it is lack of sleep or lost brain cells, but I'm not sure where to begin guessing outside of givens.

Thanks.

5. Originally Posted by stats08
You are correct, Captain, I need to find common max and min that will contain 95% of all men and all women combined (as if I'm designing a carseat).

When you say you found a smaller interval using "brute force by computing the probability of a range of intervals", can you give an example on how to even pick an interval to try and how you know when contours cross?

I was aware that some guessing is involved, but I'm not sure if it is lack of sleep or lost brain cells, but I'm not sure where to begin guessing outside of givens.

Thanks.
Depend on what software you have available. This can be done approximatly in Excel (not that I used Excel).

Take a rectangular region with the interval half width down one edge, and the centre along the bottom edge. In each cell compute the proportion of mens knee heights that fall in the interval defined by the figures in the two edges with the interval parameters.

Repeat in another rectangular region but this time for women's knee heights.

In a third rectangular region record if the corresponding cells in the first two regions are greater than 0.95, select the cell with the with both greater than 0.95.

CB

6. Nice! Thank you. I'll try it out.

7. Originally Posted by CaptainBlack
Is there some confusion here?

Are we looking for two knee height intervals one containing at least 95% of men and the other 95% of women, or one interval which contains at least 95% of men and 95% of women simultaneously?

Working this numerically I find that the smallest interval symmetric about the mean knee height of men and women combined is approximately [18.48, 23.82] inches. This is the larger of the two symmetric intervals that contain 95% of womens knees centred at the grand mean and that contain 95% of mens knees centred at the grand mean.

The interval given contains about 95% of mens knees and 96.5% of womens knees.

(of course if we had chossen to centre the interval at another point we would have found a different result, now the question is what is the smallest interval that meets our requirements? I'm pretty sure that the interval found here is not the smallest that will do the job)

The shortest interval meeting our requirements is approximatly [18.65, 23.82] which contains about 95% of womens knees and 95% of mens knees. This was found by brute force by computing the probability of a range of intervals for men and women plotting the 95% contours and keeping the point at which the contours cross!

CB
My mistake. Yes there was confusion.

However, I'll note that if you calculate seperate 95% intervals for men and for women you get [19.84, 24.16] and [18.34, 22.16] respectively.

Then an interval (not the shortest) that contains at least 95% of both men and women is clearly [18.34, 24.16] which is not too much longer than [18.65, 23.82]. It would only add a small number of extra dollars to the total production cost of the car seat.

And at the current rate of obesity, the interval [18.65, 23.82] will probably be obselete (or should I say obese-lete) before the first car seats come off the production line.

8. Originally Posted by mr fantastic
My mistake. Yes there was confusion.

However, I'll note that if you calculate seperate 95% intervals for men and for women you get [19.84, 24.16] and [18.34, 22.16] respectively.

Then an interval (not the shortest) that contains at least 95% of both men and women is clearly [18.34, 24.16] which is not too much longer than [18.65, 23.82]. It would only add a small number of extra dollars to the total production cost of the car seat.

And at the current rate of obesity, the interval [18.65, 23.82] will probably be obselete (or should I say obese-lete) before the first car seats come off the production line.
It would depend on the marginal costs of providing a greater range (including consideration of the profit margin on the product). It is often supprising how large the impact of a 12.5% increase in some parameter in a product design can sometimes be (often because of their impact on other aspects of design).

I know of at least one product with some design parameters that could not be increased by even 5% without an additional development cost in the order of 10's if not 100's of millions of pounds!

CB

9. Originally Posted by mr fantastic
And at the current rate of obesity, the interval [18.65, 23.82] will probably be obselete (or should I say obese-lete) before the first car seats come off the production line.
I always knew things were different in Oz. In the Northern Hemisphere the obese become wider not taller!

CB


# Can you specify an explicit curve by specifying its x,y points ?

I was wondering if I can specify a curve by giving its x,y points:

(0,0),(0.5, 0.25),(0.6, 0.36),(2.0, 4.0)

and then do various other manipulation of this curve like, interpolation, derivative at some point, aread under the curve between two points (integral) etc?

edit retag close merge delete

There are essentially an infinite number of curves going through 5 points. But if you adjust say by mean squarre a polynomial of 5th degree it will run through this 5 points. Then you will have a function and you will be able to manipulate it.

( 2020-04-30 08:27:28 +0200 )edit

Thanks for the comment. That obviously can be done and its interesting. I take it that the answer to the original question is no then?

( 2020-04-30 15:20:56 +0200 )edit

I am trying to make the program but I have a problem for the optimization. I have ask a question and I wait for the answer

( 2020-04-30 17:17:43 +0200 )edit

Thank you very much for the answer! so the work is in progress. That's an interesting approach.

( 2020-04-30 17:31:25 +0200 )edit

Sort by » oldest newest most voted

Hello, @Mo! As @Cyrille mentioned, given $n$ points, there exists an infinite number of curves that pass through them. However, there is a number of things you could do in order to obtain one of those curves. Let me explain three of them:

1. You can take advantage of the particular characteristics of your set of points. In this case, a simple visual inspection will show that the points are of the form $(x, x^2)$. So, a good curve for this set is quite simply $y=x^2$.
2. Of course, visual inspection is not always as easy as in this particular case. When you can't figure out the structure of function, a classical approach is to use interpolation polynomials. (This is also mentioned by @Cyrille.) Now, Sage has many interpolation facilities; in particular, there is the spline command. You can write something like this:

points = [(0, 0), (0.5, 0.25), (0.6, 0.36), (2, 4)]
S = spline(points)

Now, S is an interpolation spline, which you can evaluate en $x=0.2$, for example, with S(0.2); you can plot S with plot(S, xmin=0, xmax=2); you can differentiate it in $x=1.3$ with S.derivative(1.3); you can integrate it in its whole domain with S.definite_integral(0, 2); you could even plot the derivative and the antiderivative with a little bit of clever programming.

It is my understanding that splines are excellent approximating real-life functions, because natural phenomena tend to have smooth behavior. If these points were the result of measurements of a natural phenomenon in the lab, I would consider using splines for interpolation.

The disadvantage of the spline command is that it doesn't give you an explicit formula for the function.

3. There is another interpolation command which is a little more complicated, but more useful, in my humble opinion (in particular, it gives you a explicit formula). It's called lagrange_polynomial. In its simplest form, it uses divided differences to obtain the polynomial coefficients (you can read about that in any Numerical Methods book). You have to do something like this:

PR = PolynomialRing(RR, 'x')
points = [(0, 0), (0.5, 0.25), (0.6, 0.36), (2, 4)]
poly = PR.lagrange_polynomial(points)
print(poly)

The first line creates a polynomial ring, which I called (creatively enough) PR, with real coefficients (thus the RR) and independent variable x (thus the 'x'). The second line is your set of points. The third line asks Sage to create the Lagrange Interpolation Polynomial in this polynomial ring. You should get something like this:

 -1.11022302462516e-16*x^3 + x^2 - 1.11022302462516e-16*x

You can use this polynomial as with any other polynomial. You can make plot(poly, (x, 0, 2)), poly.derivative(x), poly.derivative(x)(1.3), poly.integrate(x), etc.

However, as you can see, the coefficients for $x^3$ and $x$ are negligible, so you could deduce that poly is actually x^2. You can verify that using the following:

PR = PolynomialRing(QQ, 'x')
points = [(0, 0), (0.5, 0.25), (0.6, 0.36), (2, 4)]
poly = PR.lagrange_polynomial(points)
print(poly)

This is exactly the same as before, but you use rational coefficients instead of real ones (thus the QQ). Of course, the same result is obtained if you replace QQ with ZZ, since, in this case, the coefficients turn out to be integers. the result is x^2, as suspected!

There are other methods you can use to obtain curves that pass through $n$ points, but these are the simplest and more usual (and useful). Just remember the two downsides of interpolation: (1) Your are getting only one of an infinite number of possible functions that pass through your points. (2) You should only work inside the interpolation interval (in particular, S(3) will return nan or not-a-number; poly(4) will evaluate correctly to 16, but you are, in that case, extrapolating, which is a very dangerous operation, especially in the lab).

I hope this helps!

more

Although I hope this solves the question, I still would like to see @Cyrille's solution, which sounds interesting!

( 2020-05-01 05:18:01 +0200 )edit

Here is my solution more involved than the one of dsejas, but perhaps it can help (Thanks to Sebastien)

# Polynomial interpolation
# Passing the matrix
A = Matrix([[0,0],[0.5,0.25],[0.6,0.36],[2,4]])
var('a0, a1, a2, a3')
B =[[A[i][1], a3*A[i][0]^3+ a2*A[i][0]^2+ a1*A[i][0]+ a0] for i in range(0,4)]
C=[[(B[i][0]-B[i][1])^2] for i in range(0,4)]
# sum of the square of errors
ssr=sum(C[i][0] for i in range(4))
# First order condition of optimality
ssr_a0=ssr.diff(a0)
ssr_a1=ssr.diff(a1)
ssr_a2=ssr.diff(a2)
ssr_a3=ssr.diff(a3)
# solution for coefficients
sol=solve([ssr_a0==0,ssr_a1==0,ssr_a2==0,ssr_a3==0,ssr_a4==0], a0, a1, a2, a3, solution_dict=True)
f(x)= a0 + a1*x+ a2*x^2 + a3*x^3
# g is a the desired function
g(x)=f(x).subs(sol[0])
# derivative of g
gp=g.diff(x)
pl=plot(g(x),(x,A[0][0],A[3][0]))
pl+=points((A[i] for i in range(4)),color='red',size=20)
pl+=plot(gp(x),(x,A[0][0],A[3][0]),color="green")
pl+=text("$f(x)$",(1.2,1),color="blue", fontsize='small', rotation=0)
pl+=text("$f'(x)$",(1.2,2.8),color="green", fontsize='small', rotation=0)
show(pl)

more

The line

ssr_a4=ssr.diff(a4)

should be removed as well as the ,ssr_a4==0 part or an error is raised (a4 is not defined).

( 2020-05-01 16:50:39 +0200 )edit

Done. I asked the subsidiary question : and if the datas came from an external file say an xls one.

( 2020-05-02 12:31:54 +0200 )edit

Assume that we have $n+1$ data points $(x_0,y_0),\ldots,(x_n,y_n)$, with $x_i\ne x_j$ for $i\ne j$. Then, there exists a unique polynomial $f$ of degree $\leq n$ such that $f(x_i)=y_i$ for $i=0,\ldots,n$. This is the Lagrange polynomial that interpolates the given data. In his answer, @dsejas has shown a method to compute it. In the current case, since $n=3$, the interpolating polynomial has degree $\leq3$; in fact, it is $f(x)=x^2$.

If $f(x)=a_0+a_1x+\cdots+a_nx^n$, it follows from the interpolation conditions that the coefficients are the solution of the linear system $A\mathbf{x}=\mathbf{b}$, where $A=(x_i^j)$ (the element of $A$ at row $i$ and column $j$ is $x_i^j$), $\mathbf{b}=(y_i)$ and $\mathbf{x}$ is the vector of unknowns $a_0,\ldots,a_n$. This is not the best way to get $a_0,\ldots,a_n$, since $A$ is usually an ill conditioned matrix. However, it is interesting to point out that from a theoretical viewpoint.

If one has $m$ data points and still wants to fit them using a polynomial of degree $n$, with $n\leq m$, the linear system $A\mathbf{x}=\mathbf{b}$ is overdetermined (there are more equations than unknowns). In such a case, one can perform a least squares fitting, trying to minimize the interpolation errors, that is, one seeks for the minimum of $$E(a_0,a_1,\ldots,a_n)=\sum_{i=0}^m (f(x_i)-y_i)^2.$$ The minimum is given by the solution of the linear system $$\nabla E(a_0,a_1,\ldots,a_n)=\mathbf{0},$$ which is simply $A^TA\mathbf{x}=A^T\mathbf{b}$. This is what @Cyrille, in fact, proposes in his answer... computing explicitly $E(a_0,a_1,\ldots,a_n)$ as well as its partial derivatives, which is completely unnecessary. In fact, if the system $A\mathbf{x}=\mathbf{b}$ has a unique solution, which is the case under discussion, the system $A^TA\mathbf{x}=A^T\mathbf{b}$ yields exactly the same solution. So, the function in @Cyrille's answer is just the Lagrange polynomial already proposed by @dsejas.

There are functions to compute least squares approximations in libraries like Scipy. Anyway, for such a simple case as that considered here, the following (suboptimal) code may work. Please note that I have added some points to make the example a bit more interesting:

# Data
data = [(0,0), (0.2,0.3), (0.5,0.25), (0.6,0.36),
(1,1), (1.2, 1.3), (1.5,2.2), (1.7,3.1), (2,4)]
# Computation of a fitting polynomial of degree <=3
basis = vector([1, x, x^2, x^3])
A = matrix(len(data),len(basis),
[basis.subs(x=datum[0]) for datum in data])
b = vector([datum[1] for datum in data])
coefs = (transpose(A)*A)\(transpose(A)*b)
f(x) = coefs*basis
# Plot results
abcissae = [datum[0] for datum in data]
x1, x2 = min(abcissae), max(abcissae)
pl = plot(f(x), (x, x1, x2))
pl += points(data, color="red", size=20)
show(pl)

This code yields the following plot:

Of course, in addition to polynomial interpolation or fitting, there are many other methods which can be used.

more

I completely agree. Your solution is the one of the econometrics books. I don't know why when I tried it I have an uninversible matrix $^\top{X} X$. what shoul be nice is to add a way to read the values from an external file say a xls one.

( 2020-05-02 12:27:44 +0200 )edit


Basic Examples

Proofs for the problems below follow a common structure, as described on the prior page.
We strongly encourage readers attempt each problem by first drawing a picture, and then attempting a formal write-up before reading the example write-up provided below.
Example 1.
Define the sequence
$\{a_n\}$
by
$a_n \triangleq \dfrac{n}{n^2+9}$
for all
$n\in\mathbb{N}.$
Prove
$\{a_n\}$
converges to zero.
Ex 1: Comment
Hint
Visual
Formal Write-Up
Follow-Up
Although the result may be clear from calculus, the point here is to provide formal arguments.
Note
$0 < a_n < \dfrac{1}{n}.$
Plot of and . We see , and so as . Also note for large .
We claim
$\{a_n\}$
converges to zero and prove this as follows. Let
$\varepsilon > 0$
be given. It suffices to show there exists
$N\in\mathbb{N}$
such that
$|a_n - 0 | \leq\varepsilon$
for all
$n\geq N.$
For each
$n\in \mathbb{N}$
, note
$n > 0$
and so
$1/n > 0$
, which implies
 $n < n + \dfrac{9}{n} \ \ \implies \ \ 1 < \dfrac{n + \frac{9}{n}}{n} \ \ \implies \ \ \dfrac{1}{n + \frac{9}{n}} < \dfrac{1}{n}.$
Consequently,

where the second equality holds since
$n > 0$
and
$n^2 + 9 > 0$
. By the Archimedean Property of
$\mathbb{R}$
, there exists
$\tilde{N}\in\mathbb{N}$
such that
$1 < \frac{\tilde{N}}{\varepsilon}$
, which implies
$\frac{1}{\tilde{N}}< \varepsilon$
. Thus, combining the above results,

which verifies the desired inequality, taking
$N=\tilde{N}.$
 $\blacksquare$
• We emphasize each statement is a complete sentence (creativity is not required).
• It is completely acceptable to mirror this structure, but change the inequalities/terms for another problem.
• Readability is greatly aided by concluding with a statement that explicitly states how an initial task (which we set out to do) was completed.
Example 2.
Define the sequence
$\{a_n\}$
 by
$\dfrac{4n^3+n}{n^3+6}$
 for all
$n\in\mathbb{N}.$
 Prove
$\{a_n\}$
converges.
Ex 2: Comment
Hint
Visual
Formal Write-Up
We can show
$\{a_n\}$
 converges by verifying it converges to a limit (that you should find).
As
$n$
 gets big, the dominant term in the numerator is
$4n^3$
 and in the denominator is
$n^3$
, which should give insight about the limit of
$\{a_n\}.$
Animation of the sequence converging.
We claim
$\{a_n\}$
 converges to four and prove this as follows. Let
$\varepsilon>0$
 be given. If suffices to show there is
$N\in\mathbb{N}$
 such that
$|a_n-0|\leq\varepsilon$
 for all
$n\geq N.$
 For each
$n\in\mathbb{N}$
, observe
 $|a_n-4|=\left|\dfrac{4n^3+n}{n^3+6}-4\right|=\left| \dfrac{(4n^3+n)-(n^3+6)4}{n^3+6}\right|=\left|\dfrac{n-24}{n^3+6}\right|.$
Building on this inequality,
$n\geq 24$
 implies
 $|a_n - 4| = \dfrac{n-24}{n^3+6} \leq \dfrac{n}{n^3+6} \leq \dfrac{n}{n^3} \leq \dfrac{n}{n^2}=\dfrac{1}{n},$
where we note
$n^2=n\cdot n \geq n\cdot 1=n$
 implies
$1/n^2\leq 1/n.$
 By the Archimedean property of
$\mathbb{R}$
, there exists a natural number
$N_1\in\mathbb{N}$
 such that
$1 \leq N_1 \varepsilon$
, and so
$1/N_1 \leq \varepsilon$
. Setting
$N_2 = \max(24,\ N_1)$
yields, by the above results,

This verifies the desired inequality, taking
$N=N_2.$
 $\blacksquare$
Example 3.
Suppose
$\{c_n\}$
 is a sequence in
$[-3,7)$
, i.e.
$-1\leq c_n<7$
for all
$n\in\mathbb{N}.$
 Define the sequence
$\{a_n\}$
 by
$a_n\triangleq c_n/n$
 for all
$n\in\mathbb{N}.$
 Prove
$\lim_{n\rightarrow\infty} a_n = 0.$
Ex 3: Comment
Hint
Visual
Formal Write-Up
Follow-Up
Be careful to formally state what must be shown (to ensure no confusion between the use of each of the listed sequences).
Here is it helpful to bound
$a_n$
 in terms of a bound for
$c_n$
.
Example sequence convergence within bounds for 7/n and -7/n.
Let
$\varepsilon > 0$
 be given. It suffices to show there is
$N\in\mathbb{N}$
 such that
$|a_n - 0|\leq \varepsilon$
 for all
$n\geq N.$
 By our hypothesis on
$\{c_n\}$
,

By the Archimedean property of
$\mathbb{R}$
, there is
$N_1\in\mathbb{N}$
 such that
$7 < N_1 \varepsilon$
, which implies
$7/N_1 < \varepsilon.$
 Hence

This shows the desired result, taking
$N=N_1$
.
 $\blacksquare$
A trend in the previous three problems is to use the Archimedean property of
$\mathbb{R}$
. This is not a coincidence! Be sure to know how to use this property so you may obtain the existence of the natural number needed in the definition of limit convergence.

Exercises.

Exercise 1. Prove the sequence
$\{a_n\}$
defined by
$a_n\triangleq \dfrac{2n^2+7}{3+n^2}$
converges to two.
• State the definition of what must be shown.
• Rewrite the inequality with
$a_n - 2.$
• Apply the Archimedean property of
$\mathbb{R}$
.
• Combine relations to verify the inequality that must be shown.


# Series of characteristic polynomials

Consider the sequence of symmetric matrices with diagonal 2 and second-diagonal s $$-1$$, e.g. $$M_4= \begin{pmatrix} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0\\ 0 & -1 & 2 & -1\\ 0 & 0 & -1 & 2\\ \end{pmatrix}$$

I've found out that the characteristic polynomials are $$\begin{cases} P_1(x)=2-x\\ P_2(x)=(2-x)^2-1\\ P_n(x) = (2-x)P_{n-1}(x)-P_{n-2}(x) \end{cases}$$

Or with a variable change $$\begin{cases} Q_1(y)=y\\ Q_2(y)=y^2-1\\ Q_n(y) = y Q_{n-1}(y)-Q_{n-2}(y) \end{cases}$$

Looking at the first 8 $$P_n$$

I see that all eigenvalues are real (as for any symmetric matrix), they are between 0 and 4.

1. How can I prove that all eigenvalues are between 0 and 4?
2. Are these polynomials known (have a name)?
3. How can I prove that the polynomial are sandwitched between $$\frac{1}{x}+\frac{1}{4-x}\quad\text{and}\quad -\frac{1}{x}-\frac{1}{4-x}$$

• I’m not with my laptop right now, so i’ll just try to comment a little bit on this beautifully written question. So, your question comes down to a question of considering a real sequence with a linear recursive relation, which is not hard to control – Paresseux Nguyen Dec 4 '20 at 16:34
• for example, for your first question, you can see that if $x_0\le 0$ you can prove easily that $P_n(x_0)>= P_{n-1}(x_0)$ etc – Paresseux Nguyen Dec 4 '20 at 16:38
• you can even deduce a compact expression for your Pn(x) which would be $a(x)u(x)^n+b(x)v(c)^n$ something except some critical cases – Paresseux Nguyen Dec 4 '20 at 16:43
• @ParesseuxNguyen thank you. I proved $P_n(x_0)\ge P_{n-1}(x_0)$ for $x_0\le 0$. Then using the symmetry of the polynomial is easy to show that $|P_n(x_0)|\ge |P_{n-1}(x_0)|$ for $x_0\ge 4$. Thus the first point is proven. Now I look for your last suggestion. – Agyla Dec 4 '20 at 17:04
• Best question ever from a <1k user? – Randall Dec 5 '20 at 1:08

The Chebyshev polynomials of the second kind satisfy the recurrence relation $$\begin{cases} U_0(y) = 1 \\ U_1(y)=x\\ U_n(y) = 2y U_{n-1}(y)-U_{n-2}(y) \end{cases}$$ so that $$Q_n(y) = U_n(y/2)$$ and $$P_n(x) = U_n(1-x/2)$$.

The zeros of $$U_n$$ are $$y_k = \cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n$$ in the range $$(-1, 1)$$, so that $$x_k = 2 - 2\cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n$$ are the zeros of $$P_n$$ in the range $$(0, 4)$$.

Also for $$|x| < 1$$ $$U_n(x) = \frac{\sin((n+1)\arccos(x))}{\sqrt{1-x^2}}$$ which implies $$|U_n(x)| \le \frac{1}{\sqrt{1-x^2}}$$ and therefore $$| P_n(x)| \le \frac{2}{\sqrt{x(4-x)}} \le \frac 1x + \frac{1}{4-x}$$ for $$0 < x < 4$$, the last estimate follows from the inequality between harmonic and geometric mean.

From a very attenuated literature search, I see that the Lucas polynomials of the second kind obey the recursion $$L_{k+1}(x) = x \ L_k(x) - L_{k-1}(x), \quad L_1(x) = 1 \ , L_2(x)=x$$ This matches the recursion for $$Q_n,$$ with an index shift of 1. The one reference I've seen (I don't have access to journal papers from home) states that this polynomial set obeys the relation

$$L_k(2 \cos(\theta) ) = 2 \cos( k \ \theta)$$

This functional relationship would explain the location of the zeros, and likely the envelope property as well. I suggest that the proposer start his research with 'Lucas polynomials of the second kind.'

• I don't know the Lukas polynomials, but apparently $Q_n(y) = U_n(y/2)$ where $U_n$ are the Chebyshev polynomials of the second kind, and those have all zeros in the interval $(-1, 1)$. – Martin R Dec 4 '20 at 17:56
• @MartinR Well, that will solve the problem completely. I thought with the $\cos(k \theta)$ in my second formula the Chebyshev polynomials would appear. – skbmoore Dec 4 '20 at 17:59


# How do I compute this integral with a Dirac's delta?

While studying probability I encountered this integral $$I=\int_{\mathbb{R}^2}\exp\left({-\frac{x_1^2+x_2^2}{2}}\right)\delta\left(r-\sqrt{x_1^2+x_2^2}\right)\,dx_1\,dx_2$$ If I compute this in polar coordinates i get $$I=\int_0^{2\pi}\,d\theta \int_0^{+\infty}\exp\left(-\dfrac{\rho^2}{2} \right)\rho\delta(r-\rho)\,d\rho=2\pi r\exp\left(-\dfrac{r^2}{2}\right)$$ but in cartesian coordinates I only get $$I=\exp\left(-\frac{r^2}{2}\right)$$ I don't understand why. I just thougth that I was using the Dirac's delta's properties in both cases. I think the first result is the correct one and there is something I don't know about Dirac's delta with more than one variable.

Which result is correct and why?

You may also make direct calculations in Cartesian coordinates - integrating, for instance, over $$x_1$$ first and then over $$x_2$$. We can multiply and divide the argument of $$\delta$$-function ($$r-\sqrt{x_1^2+x_2^2}$$) by ($$r+\sqrt{x_1^2+x_2^2}$$) (because ($$r+\sqrt{x_1^2+x_2^2}$$) is always positive). We can also replace the power of exponent by $$-\frac{r^2}{2}$$ - due to the condition imposed by $$\delta$$-function

$$I(r)=\int_{\mathbb{R}^2}\exp\left({-\frac{x_1^2+x_2^2}{2}}\right)\delta\left(r-\sqrt{x_1^2+x_2^2}\right)\,dx_1\,dx_2=\int_{\mathbb{R}^2}\exp\left({-\frac{r^2}{2}}\right)\delta\left(\frac{r^2-(x_1^2+x_2^2)}{r+\sqrt{x_1^2+x_2^2}}\right)\,dx_1\,dx_2$$

We see that $$x_2$$ contributes if only $$x_2\in[-r,r]$$, otherwise $$\delta()=0$$

$$I(r)=\int_{-r}^rdx_2\int_{-\infty}^{\infty}dx_1\exp\left({-\frac{r^2}{2}}\right)\delta\left(\frac{(\sqrt{r^2-x_2^2}-x_1)(\sqrt{r^2-x_2^2}+x_1)}{r+\sqrt{x_1^2+x_2^2}}\right)$$

But $$\delta(\frac{(a-x_1)(x_1+b)}{A})=|\frac{A}{a-x_1}|\delta(x_1+b)+|\frac{A}{x_1+b}|\delta(a-x_1)=|\frac{A}{a-x_1}|\delta(x_1+b)+|\frac{A}{x_1+b}|\delta(x_1-a)$$

We get

$$I(r)=\int_{-r}^rdx_2\int_{-\infty}^{\infty}dx_1\exp\left({-\frac{r^2}{2}}\right)\left(r+\sqrt{x_1^2+x_2^2}\right)\left(\frac{1}{\sqrt{r^2-x_2^2}-x_1}\delta(\sqrt{r^2-x_2^2}+x_1)+\frac{1}{\sqrt{r^2-x_2^2}+x_1}\delta(\sqrt{r^2-x_2^2}-x_1)\right)=$$ $$\int_{-r}^rdx_2\int_{-\infty}^{\infty}dx_1\exp\left({-\frac{r^2}{2}}\right)2r\left(\frac{1}{2\sqrt{r^2-x_2^2}}\delta(\sqrt{r^2-x_2^2}+x_1)+\frac{1}{2\sqrt{r^2-x_2^2}}\delta(x_1-\sqrt{r^2-x_2^2})\right)=\int_{-r}^r\exp\left({-\frac{r^2}{2}}\right)\frac{2r}{\sqrt{r^2-x_2^2}}dx_2$$ $$I(r)=\int_{-1}^1\exp\left({-\frac{r^2}{2}}\right)\frac{2r}{\sqrt{1-t^2}}dt=2\pi{r}e^{-\frac{r^2}{2}}$$

In Cartesian coordinates, you have \begin{align} I=&\ \int dx\ \exp\left(-\frac{x_1^2+x_2^2}{2} \right)\delta(r-\sqrt{x_1^2+x_2^2})\\ =&\ \int_{r=\sqrt{x_1^2+x_2^2}} d\sigma\ \exp\left(-\frac{x_1^2+x_2^2}{2} \right) \end{align} since \begin{align} |\nabla (r- \sqrt{x_1^2+x_2^2})| = 1. \end{align} Hence, it follows that \begin{align} I = \int_{r=\sqrt{x_1^2+x_2^2}} d\sigma\ \exp\left(-\frac{r^2}{2} \right) = 2\pi r\exp\left(-\frac{r^2}{2} \right). \end{align}

• Is it like a surface integral in one dimension instead of two? Why did you compute the gradient? Feb 15 at 21:53
• @Rhino Please consult en.wikipedia.org/wiki/…. Feb 15 at 22:00


graph to the left by three. With this mixed transformation, we need to perform the inner absolute value first: For any original negative x ’s, replace the y value with the y value corresponding to the positive value (absolute value) of the negative x ’s. '&https=1' : ''); If you're shifting in And we should expect to need to plot negative x-values, too. that I'm shifting to the right but I encourage you to try numbers and think about what's happening here. In particular, they don't include any "minus" inputs, so it's easy to forget that those absolute-value bars mean something. x plus three, plus two. Because of this, absolute-value functions have graphs which make sharp turns where the graph would otherwise have crossed the x-axis. value of something and so you say, okay, if x is three, how do I make that equal to zero? y = ∣ x ∣. Absolute values are never negative. It's gonna look something like this. for the black function, I'm gonna have to get two more than that. $\begin{cases}f\left(x\right)=2\left|x - 3\right|-2,\hfill & \text{treating the stretch as a vertical stretch, or}\hfill \\ f\left(x\right)=\left|2\left(x - 3\right)\right|-2,\hfill & \text{treating the stretch as a horizontal compression}.\hfill \end{cases}$, $f\left(x\right)=a|x - 3|-2$, $\begin{cases}2=a|1 - 3|-2\hfill \\ 4=2a\hfill \\ a=2\hfill \end{cases}$, http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175. All right reserved. Web Design by. of this graph if we shift, if we shift three to the right and then think about how that will change if not only do we shift three to the right but we also shift four up, shift four up, and so once again pause this At this vertex right over here, whatever was in the absolute So in particular, we're And then instead of having going up here. of x plus three, plus two. You remember that absolute-value graphs involve absolute values, and that absolute values affect "minus" inputs. This is the graph of y If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for $x$ and $f\left(x\right)$. The general form of the absolute value function is: f (x) = a|x-h|+k. 'https:' : 'http:') + '//contextual.media.net/nmedianet.js?cid=8CU2W7CG1' + (isSSL ? You could have shifted up two first, then you could have Yes. But mostly we need to take the time to plot quite a few points, so that we can "see" the shape before we start sketching it in. Well, one way to think about it is, well, something interesting is happening right over here at x equals three. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. value of x plus three. thing was happening at x equals zero. Instead, the width is equal to 1 times the vertical distance. var mnSrc = (isSSL ? medianet_versionId = "111299"; And we're gonna do that right now and then we're gonna just gonna If k<0, it's also reflected (or "flipped") across the x-axis. For instance, suppose we are given the equation y = | … And then, so here instead There are any number of things we can do to help ourselves graph this correctly. Now, it's happening at x equals three. Graph an absolute value function. signs to positive signs. The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. to absolute value of x which you might be familiar with. the right and four up? Alright, now let's do this together. Now let's see if we can graph y is equal to two times the So there's multiple, there's three transformations You could view this as the same thing as y is equal to the absolute

.

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# “Differs in only finitely many terms” an equivalence relation on sequences?

Consider sequences $a : \mathbb{Z}^+ \rightarrow A$ on a set $A$. Define the relation $\sim$ over sequences by $a \sim b$ iff there are only finitely many indices $i$ at which $a_i \neq b_i$. Clearly $\sim$ is reflexive and symmetric.

It appears to me that it is also transitive! Suppose $a \sim b$, and $b \sim c$. Let $I$ be the set of indices $i$ such that $a_i \neq b_i$. Let $J$ be the set of indices at which $b_i \neq c_i$. Both $I$ and $J$ are finite, so $K = I \cup J$ is finite. Let $i \notin K$. Then $a_i = b_i$, and $b_i = c_i$, so $a_i = c_i$. So there are only finitely many points at which $a$ and $c$ differ (all of them are in $K$), so $a \sim c$.

This is extremely surprising to me, because it implies that $\sim$ is an equivalence relation. My difficulty is in understanding what equivalence classes this relation could possibly define.

Question 1: Is it really true that $\sim$ is an equivalence relation?

Question 2: Can anybody give me some kind of concrete description of what equivalence classes $\sim$ defines?

Thank you!

-

Yes, it’s an equivalence relation. Another way to describe is to say that $a\sim b$ iff there is an $n\in\Bbb N$ such that $a_k=b_k$ for all $k\ge n$. About the only way that I’ve ever found to visualize an equivalence class is to pick one member of it; if that member is $a$, then the class consists precisely of all sequences that agree with $a$ from some point on.

It turns out to be very important in studying the box topology on products of infinitely many topological spaces.

-
That alternative definition helps a lot, Brian! Thanks for the answer! –  Nick Thomas Feb 12 '13 at 7:05
@Nick: You’re welcome! –  Brian M. Scott Feb 12 '13 at 7:06
Do you think the weak direct sum could visualize the OP this relation? thanks. –  Babak S. Feb 12 '13 at 7:12
@Babak: Maybe: its elements are exactly one equivalence class in the full product. –  Brian M. Scott Feb 12 '13 at 7:14

This is a very small hint in the light of Brian's complete one. I think the weak direct sum $\sum A_k$ when, for example, $\{A_k\}$ is a family of groups indexed by a set $K$, can help you to find out what is happening in the relation. Try to Google it for details.

-
Nice hint...+ 1 ;-) –  amWhy Feb 12 '13 at 15:55

And as to what that equivalence class might be/look like?

Well, that depends on the nature of set A.

If A={0,1} then your function can be interpreted as producing a binary Real number corresponding to each function a, such as .01110011.. or .1101110... etc. In this case, the equivalence classes are Real numbers which vary by a finite sum of negative powers of two. So x and x + 1/4 + 1/32 would be in the same equivalence class. A different set A and a different interpretation of the sequence as a Real might produce (for example) equivalence classes of Reals which differ by a rational, such that x, x + 1/7, x + 23/788 etc.

-

It is an equivalence relation. There's nice Brian answer, but it is short on examples, so here is something to gain more intuition:

• Try to imagine what is the class of abstraction of sequence of zeros $(\ldots, 0, 0, 0, \ldots)$.

• Another example would be a convergent sum $\sum_{i \in \mathbb{Z}} a_i < \infty$, what can you tell about other sequences in its class of abstraction?

-


Problem An airplane is flying at a constant speed at a constant altitude of $10$ km in a straight line directly over an observer. At a given moment the observer notes that the angle of elevation $\theta$ to the plane is $54^\circ$ and is increasing at $1^\circ$ per second. find the speed, in kilometres per hour, at which the airplane is moving towards the observer.

I'm working on a related rates problem, and the equation that i'm using to relate the two variables is $$\tan \theta= \frac{10}{x}.$$ so I can differentiate this, and after simplifying, I end up with $$\frac{dx}{dt}=-\frac{1}{10} \cdot x^2\sec^2 \theta\frac{d\theta}{dt}.$$ But if I rearrange the original equation so that it's $x=\frac{10}{\tan \theta}$, I get $$\frac{dx}{dt}=(-10\csc^2 \theta) \frac{d\theta}{dt},$$ which is different from the other derivative. Is there something quirky about rearranging the equation, or am I just blindly messing something up?

• Where does the variable $t$ come into play? Is $x$ a function of $t$? Please post the problem statement (the exercise you are working on). You're missing something, or your differentiation makes no sense. Jul 2 '18 at 22:42
• Welcome to MSE! Could you provide a little more context, like the original statement of the problem? Also, your question will be much easier to read (and therefore much more likely to get answered!) if you use MathJax formatting to format the math in the question: math.meta.stackexchange.com/questions/5020/…. (Kind of a long tutorial, but the basics are close to the top, and the whole thing is well worth reading if you have the time.) Jul 2 '18 at 22:44
• ok yeah sorry my bad, the problem goes like this: An airplane is flying at a constant speed at a constant altitude of 10 km in a straight line directly over an observer. At a given moment the observer notes that the angle of elevation θ to the plane is 54° and is increasing at 1° per second. find the speed, in kilometres per hour, at which the airplane is moving towards the observer. Jul 2 '18 at 22:53
• Hint: if you substitute $x = \frac{10}{\tan \theta}$ into the first derivative, what do you get? Jul 2 '18 at 23:07
• omgggg i can't believe i didn't see that before, thank u so much! Jul 2 '18 at 23:21

Though the two derivatives may seem different from one another, they are actually the same! This is because $$x^2=\left(\frac{10}{\tan\theta}\right)^2=\frac{100}{\tan^2\theta}=100\frac{\cos^2\theta}{\sin^2\theta}=100\frac{\csc^2\theta}{\sec^2\theta}.$$
• By the way, welcome to Math.SE! After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. Jul 2 '18 at 23:55