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# Second order nonhomogenous ODE with Dirichlet boundary condition

I would like to know a hint for this problem. Brezis describes it as it can be "solved explicitly by a very simple calculation". The problem is:

Given $f\in C^1([a,b])$, find a function $u$ satisfying:

$$\tag{1} -u''+u=f, \text{ on }[a,b],$$ and $$u(a)=u(b)=0.\tag{2}$$

Thanks Umberto P. I found the solution for (1) is:

$$u(t)=c_1(t)e^t+c_2(t)e^{-t}, \text{ where}$$ $$c_1'(t)=-\frac{f(t)}{2e^{t}}, \text { and }$$ $$c_2'(t)=\frac{f(t)}{2e^{-t}} .$$

So functions $c_1$ and $c_2$ are determined modulo one constant. However, taking $a=0$ and $b=1$ and writing the boundary condition (2) results in a system that I can't solve.


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## TomLikesPhysics 2 years ago I have a parabola slide and I am looking for a function of the downhill force which only depends on the x and/or y coordinate. I came up with a function but I am not sure if this is correct. You can see everything on the attached picture.

1. TomLikesPhysics

Here is what I did so far.

2. henpen

I assume that it is safe to assume that the parabola is y=x^2, with the tip at (0,0) dy/dx=2x |dw:1351340058220:dw|$\tan^{-1}(\frac{dy}{dx})=\theta$ $\theta=\tan^{-1}(2x)$ $F=-mgsin(\theta)$

3. henpen

So you seem in the right area

4. TomLikesPhysics

Hmmm... actually I do not know that. It is not specified so it might be better if I write: y=ax^2.

5. henpen

Then $\theta=\arctan(a2x)$

6. TomLikesPhysics

Yes, it just looks kind of ugly if we plug that into the equation for the force. F=-mgsin(arctan(2ax)

7. TomLikesPhysics

Is this really the finaly answer? Can´t we do something to make it simpler or look better? Is there perhaps a different way to determine the force in terms of x and/or y?

8. henpen

|dw:1351340440158:dw|No, its not

9. henpen

|dw:1351340462854:dw|

10. henpen

F=-mgsqrt(1+(2ax)^2)

11. TomLikesPhysics

Ah... Pythagoras... I see...Now it looks better. So this is the best we can do?

12. henpen

Simplify sqrt(1+(2ax)^2) maybe You could try to x/y decompose the force, but that's probably uglier

13. henpen

Although I think it's unsimplifiable

14. TomLikesPhysics

Ok. A big thank you to you, henpen. I really appreciate the help. Thx a lot.

15. henpen

16. TomLikesPhysics

The Force stands always normal on the slide, right? If so, it does not point at some special point all the time?

17. henpen

|dw:1351349861448:dw| The normal (restrictive) force is always at a normal to the curve

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# First-order linear PDEs#

Mathematics for Scientists and Engineers 2

In this section we will consider first-order linear PDEs for an unknown function $$u$$ of two variables:

$a(x,y)u_x + b(x,y)u_y = f(x, y, u)$

The method which we will use to find the solution of such PDEs is called the method of characteristics. This is a very useful method with an intuitive geometric interpretation and it can be applied to even more complicated PDEs, which we will not consider here.

## Method of characteristics#

The main idea behind this method is a popular one: we want to reduce the PDE to an ODE, which are in general much easier to solve. We do that by analysing the PDE along specially chosen curves in $$xy$$ plane, which we call characteristic curves or just characteristics. We parameterise them by the parameter $$s$$, such that $$x = x(s), y = y(s)$$ and we choose them such that the following system of ODEs is satisfied:

$\frac{dx}{ds} = a, \quad \frac{dy}{ds} = b, \quad \frac{du}{ds} = f$

Then after using the chain rule and substituting this, we have

$\frac{du}{ds} = \frac{\partial u}{\partial x} \frac{dx}{ds} + \frac{\partial u}{\partial y} \frac{dy}{ds} = f(x(s), y(s), u(s))$

This simplifies our problem quite a bit, since we have reduced the PDE to an ODE. Let us now look at a few examples.

### Example: constant coefficients, homogeneous#

Let us start with the simplest example, where $$a$$ and $$b$$ are constants and $$f \equiv 0$$:

$2u_x + 3u_y = 0, \qquad u(x, 0) = \cos x$

Setting

$\frac{dx}{ds} = 2, \quad \frac{dy}{ds} = 3$

and using the chain rule we get

$\frac{du}{ds} = 2u_x + 3u_y = 0$

This means that the PDE along the characteristic line $$(x(s), y(s))$$ is a constant. We can write the parametrization invariant form of the above equations as

$ds = \frac{dx}{2} = \frac{dy}{3}$

from which

$\frac{dx}{dy} = \frac{2}{3}$

So the characteristics are given by $$2y = 3x + c$$, where $$c$$ is an integration constant which will determine which characteristic line we are on. Let us plot a couple of them.

import numpy as np
import matplotlib.pyplot as plt

x = np.linspace(-5, 5, 51)
c = np.linspace(-10, 10, 9)

fig, ax = plt.subplots(1, 1, figsize=(6, 6))

for cc in c:
ax.plot(x, 3/2 * x + cc, 'k-', alpha=0.3)

if cc > -5:
ax.text(2 - cc/1.5, 3, cc, c='darkblue', weight='bold')

ax.text(-0.1, 5.4, 'y', weight='bold')
ax.text(5.7, -0.1, 'x', weight='bold')

ax.set_aspect('equal')
ax.spines['left'].set_position('zero')
ax.spines['bottom'].set_position('zero')
ax.spines['right'].set_visible(False)
ax.spines['top'].set_visible(False)
ax.set_ylim(-5, 5)

plt.show()

Here we plotted characteristics for different values of $$c$$, which are written in bold along characteristics.

All characteristics are parallel to each other, of course, and since $$u$$ is constant along each characteristic line, $$u$$ will only vary if we jump from one characteristic to another. Since $$c$$ marks which characteristic we are sitting on, $$u$$ will therefore be a function of $$c$$.

$u(x(y), y) = f(c) = f(-3x + 2y)$

where $$f$$ is an arbitrary function of one variable.

Finally we need to use the auxiliary condition $$u(x, 0) = \cos x$$ to find the particular solution. For $$y=0$$:

$u(x, 0) = f(-3x) = \cos (x)$

Let $$w = -3x$$. Then:

\begin{align}\begin{aligned} f(w) = \cos (-w/3) .\\Hence the solution is\\ u(x, t) = f(-3x + 2y) = \cos(x - \frac{2}{3}y) \end{aligned}\end{align}
from mpl_toolkits.mplot3d import Axes3D

x = np.linspace(-5, 5, 201)
y = np.linspace(-5, 5, 201)
X, Y = np.meshgrid(x, y)

f = np.cos(X - 2/3 * Y)

fig = plt.figure(figsize=(10, 8))

ax1.plot_surface(X, Y, f, rstride=1, cstride=1, cmap='inferno', antialiased=False)
ax2.contourf(f, cmap='inferno', extent=[-5, 5, -5, 5])
ax2.contour(f, colors='k', extent=[-5, 5, -5, 5], linewidths=1.)

ax1.view_init(40, 70)
ax2.set_aspect('equal')

for ax in (ax1, ax2):
ax.set_xlabel('x')
ax.set_ylabel('y')
ax1.set_zlabel('u')

fig.suptitle(r'$u(x,y) = \cos (x - \frac{2}{3}y)$')

plt.tight_layout()
plt.show()

Notice that each of the contours is a characteristic, since $$u$$ is constant along a characteristic.

Important

A first-order linear PDE of the form

$au_x + bu_y = 0$

has a general solution of the form

$u(x, y) = f(bx - ay)$

where $$f$$ is an arbitrary function of one variable.

## Example: general case#

Let us now consider an inhomogeneous linear problem with variable coefficients.

$(x + y)u_x + y u_y = xyu$

with the boundary condition $$u(x,y) = e^{x^2/4}$$ for $$y = x$$.


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Paul's Online Notes
Home / Calculus III / Line Integrals / Line Integrals - Part II
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Section 5-3 : Line Integrals - Part II

4. Evaluate $$\displaystyle \int\limits_{C}{{1 + {x^3}\,dx}}$$ where $$C$$ is the right half of the circle of radius 2 with counter clockwise rotation followed by the line segment from $$\left( {0,2} \right)$$ to $$\left( { - 3, - 4} \right)$$. See the sketch below for the direction.

Show All Steps Hide All Steps

Start Solution

To help with the problem let’s label each of the curves as follows,

Now let’s parameterize each of these curves.

$${C_1}:\,\,\vec r\left( t \right) = \left\langle {2\cos \left( t \right),2\sin \left( t \right)} \right\rangle \hspace{0.25in} - \frac{1}{2}\pi \le t \le \frac{1}{2}\pi$$
$${C_2}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {0,2} \right\rangle + t\left\langle { - 3, - 4} \right\rangle = \left\langle { - 3t,2 - 6t} \right\rangle \hspace{0.25in}0 \le t \le 1$$

Show Step 2

Now we need to compute the line integral for each of the curves.

\begin{align*}\int\limits_{{{C_1}}}{{1 + {x^3}\,dx}} & = \int_{{ - \,\,\frac{1}{2}\pi }}^{{\frac{1}{2}\pi }}{{\left[ {1 + {{\left( {2\cos \left( t \right)} \right)}^3}} \right]\left( { - 2\sin \left( t \right)} \right)\,dt}}\\ & = \int_{{ - \,\,\frac{1}{2}\pi }}^{{\frac{1}{2}\pi }}{{ - 2\sin \left( t \right) - 16{{\cos }^3}\left( t \right)\sin \left( t \right)\,dt}}\\ & = \left. {\left( {2\cos \left( t \right) + 4{{\cos }^4}\left( t \right)} \right)} \right|_{ - \,\frac{1}{2}\pi }^{\frac{1}{2}\pi } = \underline 0 \end{align*} \begin{align*}\int\limits_{{{C_2}}}{{1 + {x^3}\,dx}} & = \int_{0}^{1}{{\left[ {1 + {{\left( { - 3t} \right)}^3}} \right]\left( { - 3} \right)\,dt}}\\ & = \int_{0}^{1}{{ - 3 + 81{t^3}\,dt}}\\ & = \left. {\left( { - 3t + \frac{{81}}{4}{t^4}} \right)} \right|_0^1 = \underline {\frac{{69}}{4}} \end{align*}

Don’t forget to correctly deal with the differentials when converting the line integral into a “standard” integral.

Show Step 3

Okay to finish this problem out all we need to do is add up the line integrals over these curves to get the full line integral.

$\displaystyle \int\limits_{C}{{1 + {x^3}\,dx}} = \left( 0 \right) + \left( {\frac{{69}}{4}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{69}}{4}}}$

Note that we put parenthesis around the result of each individual line integral simply to illustrate where it came from and they aren’t needed in general of course.


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# Does a never-halting machine always loop?

A Turing machine that returns to a previously encountered state with its read/write head on the same cell of the exact same tape will be caught in a loop. Such a machine doesn't halt.

Can someone give an example of a never-halting machine that doesn't loop?

• Just a note: the tape can also be different: a sufficient (but not necessary) condition for an endless loop when the TM enters the same cell at step $t_1$ and at step $t_2 > t_1$ in the same state, is that at step $t_2$ the portion of the tape visited by the head between step $t_1$ and step $t_2$ is equal to the corresponding portion just before entering $t_1$. – Vor Jul 13 '12 at 15:51
• If a TM had to loop in order to fail to halt, you'd be able to fairly easily solve the halting problem for TMs: remember all previous configurations, and at each step, see if you're in a configuration you've seen before, and if so, you know the thing doesn't halt (otherwise, since we assume that it must loop in order to run forever, it won't run forever, i.e., it will stop, in which case we'll eventually know about it). – Patrick87 Jul 13 '12 at 16:59
• Inspired by @Niel de Beaudrap answer: a turing machine could compute the oeis.org/A014445 sequence and stop when it gets an odd number. It could compute oeis.org/A016742 as a running sum and halt when the number is odd. It could compute x^2 where x cycles between -100 and 100 and the cycling is done with a modulo and halt when the result is negative. It could compute x%2 where x ranges from zero to positive infinity and stop when the result equals 2. In assembly language do/while/for loops all come down do having a conditional jump, but cond jump alone means little. – Leonid Jul 16 '12 at 21:15
• The question's assumption is only true for deterministic machines. – Raphael Jul 18 '12 at 10:04

Consider the TM that always moves the tape head to the right and prints a special non-blank tape symbol at each step. This means that the TM never halts, since it always moves to the right, and never repeats a state, since after k steps the tape head is over the kth cell of the machine. Consequently, each configuration of the machine is different from all others, and the machine always loops.

We could also nonconstructively show the existence of such machines. Assume for contradiction that every TM that never halts eventually loops. This means that if you start a TM $M$ on a string $w$, one of the following will eventually happen:

1. $M$ halts, or
2. $M$ repeats a configuration.

In this case, the halting problem would be decidable as follows. Given a TM $M$ and string $w$, simulate $M$ on $w$, at each point writing out the contents of the tape, current state, and current tape position. If this configuration is a duplicate, output "does not halt." Otherwise, if $M$ halts on $w$, output "halts." Since one of these is guaranteed to happen eventually, this process always terminates, so we would have an algorithm for the halting problem, which we know does not exist.

Hope this helps!

• Hah, beat you to that edit. See my comment on the question. I like this way of explaining why not all non-halting TM's must loop... it builds intuition. – Patrick87 Jul 13 '12 at 17:33
• @Patrick87- Sorry, I didn't notice the comment. I thought of the addendum on my commute and sat down to enter it as soon as I got back. – templatetypedef Jul 13 '12 at 19:30
• No problem, man... I'm glad that you added it, since I think it's a good way of explaining it. I only added it as a comment, and not as an answer, since you beat me to that. :D – Patrick87 Jul 13 '12 at 20:53
• Actually, in terms of the halting problem such that go back and change the tape ad infinitum seem to be the "real problem". Those void-walkers you can detect. – Raphael Jul 18 '12 at 10:06

A Turing machine which computes all of the decimal places of π (or any other non-terminating fraction, in any base) never halts, and can be made to write on each cell only a finite number of times. Of course, the fact that there is no transition to a halting state would be a dead giveaway, but it is at least a natural example.

A more interesting (but also ambiguous) case would be a Turing machine which iteratively computes the Collatz function on its input, $$f(n) = \begin{cases} 3n+1 , & \text{if n is odd}; \\ n/2, & \text{if n is even}, \end{cases}$$ terminating if and only if it obtains the integer 1. The famous Collatz conjecture is that for any input, this procedure eventually halts. But it is not known whether this is the case. It can fail in two different ways, in principle: either it can find a sequence of integers which loops around (corresponding to the existence of an integer n such that $f \circ f \circ \cdots f(n) = n$ for some number of compositions, where n ≠ 1); or it could be that there are chains of integers n, f(n), f(f(n)), ... which asymptotically diverge to infinity. If any sequences of the latter sort exist, this would imply that the Turing machine I've described above would be non-repeating, as the tape would be continually changed to larger and larger numbers.

• I like playing with the digits of pi. A TM could halt whenever the square of any digit of pi equals exactly 7. – Leonid Jul 16 '12 at 21:18
• @Leonid: You could certainly consider a Turing machine which accepts some input, and halts on a condition determined by that input. You could even make the specification of the conditions under which it halts part of the input. And you could provide an input, as you describe, setting a constraint which is never satisfied. – Niel de Beaudrap Jul 17 '12 at 17:15

Consider any non-halting Turing machine that never moves the read/write head to the left.

• Some of them still loop. </nitpicking> – Raphael Jul 18 '12 at 10:08

If this were true, then the halting problem would be decidable. You would just record each (tape, state) pair seen when executing the Turing machine, and the machine would either halt (in which case it obviously halts), or you see a pair that you've seen before, in which case the machine does not halt.

Since the halting problem is undecidable, this cannot be true. (See other examples for counter examples.)


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###### Question:

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TIFR 2014 Problem 6 Solution is a part of TIFR entrance preparation series offered by Cheenta. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects. Generally, the exams scheduled in December.
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## Problem

Let $$f:[0,1] \to \mathbb{R}$$ be a continuous function. Which of the following statement is always true?

A. $$\int_{0}^{1} f^2(x)dx = (\int_{0}^{1}f(x)dx)^2$$

B.  $$\int_{0}^{1} f^2(x)dx \le (\int_{0}^{1}|f(x)|dx)^2$$

C.  $$\int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2$$

D.  $$\int_{0}^{1} f^2(x)dx < (\int_{0}^{1}f(x)dx)^2$$

## Discussion:

We first recall the Cauchy-Schwartz inequality for an inner product space $$V$$ and two vectors $$a,b\in V$$

$$<a,b> \le ||a||||b||$$.

Here, we also remember the fact that $$C[0,1]$$ (the set of all continuous real (/complex) valued functions on [0,1] ) forms an inner product space with respect to the inner product

$$<f,g>= \int_{0}^{1} f(x)g(x)dx$$ (We are only taking real valued functions so we neglect the conjugation…)

We want to apply this inequality to suitable functions so that we get some inequality from the options above.

Let’s try $$|f|$$ and $$1$$ (the constant function).

We get: $$\int_{0}^{1} |f(x)|1dx \le (\int_{0}^{1} |f(x)|^2dx)^{1/2} (\int_{0}^{1} 1^2dx)^{1/2}$$

Squaring, we get

$$\int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2$$.

So option C $$\int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2$$ is correct.

## Helpdesk

• What is this topic: Real Analysis
• What are some of the associated concept: Inner Product Space, Cauchy – Schwartz inequality,continuous real (/complex) valued functions
• Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert


text:
# Excitation of electron in Niels Bohr's atomic model

Professor taught us that an electron gains or looses only those energies which are equal to difference in two energy levels. That is $$E_1 + \Delta E = E_2$$ or $$E_1 + \Delta E = E_3.$$

What if we give, say, $$\pu{11.3 eV}$$ and the remainder is $$\pu{-2.2 eV},$$ so the electron jumps to $$n = 2$$ where the energy is $$\pu{-3.4 eV}$$ and the remainder $$\pu{-1.2 eV}$$ is released?

I'm really confused with the working here.

• Have you thought about why atomic absorption spectra contain just narrow lines and not wide bands ? Apr 21 at 8:01
• Yes, that's because we have specific 'energy levels' and not continuous i.e. only for n=1,2,3.... and not n=1,1.1,1.2..... and because the energy levels aren't continuous the spectra isn't continuous but discrete? Am I right? Apr 21 at 9:17
• Yes, and also atoms absorb photons of just specific energy levels, equal to differences between atom discrete allowed energy levels. Apr 21 at 9:20
• It can't reject the excess energy because it can't absorb it in the first place. You need to understand that photons can provide all energy they have or none. There's no absorb half reject other half. Apr 21 at 14:35
• @NisargBhavsar So let me get this straight. So it's just like how Bohr explained it. We have a ladder that the electron can climb up to. It can either make its way up by stepping on the next or can go down by stepping down, but only on the steps and not in between. So either give me the required amount of energy or don't. Right?! Apr 24 at 6:24


text:
# How can I draw a curved surface in a polar coordinate system?

I'm fairly new to using TikZ and so far I've got a spherical coordinate system, and labelled the point P at which I want to draw a curved surface. A full sphere of unit radius, for example, would be a 'curved surface' of radius 1, with a polar angular width, del_theta, of 180 degrees and an azimuthal angular width, del_phi, of 360 degrees.

The curved surface is to be drawn at point P, (xp, yp, zp), with del_theta of say 10 degrees and del_phi of say 20 degrees. In summary, I want to express where this curved section is in Cartesian coordinates, and how big it is in terms of polar coordinates.

If not del_theta, and del_phi, I could instead provide theta_1, theta_2, phi_1, and phi_2 that would define the boundaries of the curved surface [del_theta being theta_2 - theta_1 and so on].

The spherical grid shown at http://www.texample.net/tikz/examples/spherical-and-cartesian-grids/, and copied below, is sort of what I want to create, but that example doesn't use the tikz-3dplot package. I'm not sure how to create this surface. Please help!

I want a shape that looks similar to the red spherical grid

My image so far with coordinate system and showing the location of P where I want the spherical shape

\documentclass{article}
\usepackage{verbatim}

\usepackage{tikz}
\usepackage{tikz-3dplot}
\usepackage[active,tightpage]{preview}  % generates a tightly fitting border around the work
\PreviewEnvironment{tikzpicture}
\setlength\PreviewBorder{2mm}

\begin{document}

%Angle Definitions
%-----------------

%set the plot display orientation
%syntax: \tdplotsetdisplay{\theta_d}{\phi_d}
\tdplotsetmaincoords{60}{25}

%define polar coordinates for some vector
%TODO: look into using 3d spherical coordinate system
\pgfmathsetmacro{\rvec}{1}
\pgfmathsetmacro{\thetavec}{50}
\pgfmathsetmacro{\phivec}{15}

%start tikz picture, and use the tdplot_main_coords style to implement the display
\begin{tikzpicture}[scale=5,tdplot_main_coords]

%set up some coordinates
%-----------------------
\coordinate (O) at (0,0,0);
\fill (O) circle[radius=0.4pt] node[below left] {O};

%determine a coordinate (P) using (r,\theta,\phi) coordinates.
%syntax: \tdplotsetcoord{Coordinate name without parentheses}{r}{\theta}{\phi}
\tdplotsetcoord{P}{\rvec}{\thetavec}{\phivec}
\fill (P) circle[radius=0.4pt] node[anchor=south west] {P};

%draw figure contents
%--------------------

%draw the main coordinate system axes
\draw[thick,->] (0,0,0) -- (1,0,0) node[anchor=north east]{$x$};
\draw[thick,->] (0,0,0) -- (0,1,0) node[anchor=north west]{$y$};
\draw[thick,->] (0,0,0) -- (0,0,1) node[anchor=south]{$z$};

%draw a vector from origin to point (P), projection on xy plane, and a connecting line
\draw[-stealth,color=red] (O) -- (P);
\draw[dashed, color=red] (O) -- (Pxy);
\draw[dashed, color=red] (P) -- (Pxy);

%draw the angle \phi, and label it
%syntax: \tdplotdrawarc[coordinate frame, draw options]{center point}{r}{angle start}{angle end}{label options}{label}
\tdplotdrawarc{(O)}{0.2}{0}{\phivec}{anchor=north}{$\phi$}

%set the rotated coord system so the x'y' plane coincides with the"theta plane" of the main coord system
%syntax: \tdplotsetthetaplanecoords{\phi}
\tdplotsetthetaplanecoords{\phivec}

%draw theta arc and label, using rotated coordinate system
\tdplotdrawarc[tdplot_rotated_coords]{(0,0,0)}{0.2}{0}{\thetavec}{anchor=south west}{$\theta$}

%draw some dashed arcs, demonstrating direct arc drawing
\draw[dashed,tdplot_rotated_coords] (\rvec,0,0) arc (0:90:\rvec);
\draw[dashed] (\rvec,0,0) arc (0:90:\rvec);

\end{tikzpicture}
\end{document}

• The are many example here for drawing spheres. See tex.stackexchange.com/questions/196902/… for example. – John Kormylo May 20 '17 at 19:12
• @JohnKormylo I know how to draw arcs using 3d polar coordinates (I've done that for labeling theta for example). I just don't know how I would draw 4 arcs and then colour the curved shape so it looks 3 dimensional. I would also have to define rotations before drawing each of the four arcs. 'Scope' is new to me, I will check it out. – bluprince13 May 21 '17 at 17:06
• You can jump out of the tikz-3dplot coordinate system temporarily to draw the sphere. Wouldn't that do what you want? Have you considered using software which knows about 3D? – cfr May 21 '17 at 20:25
• @cfr well, I don't want to draw a sphere, only a slice of it. What software are you referring to? CAD? – bluprince13 May 21 '17 at 21:58
• Things like asymptote know about 3D, whereas TikZ/PGF, as we know, does not. – cfr May 21 '17 at 22:44

So all you really need is a couple of \foreach loops. Note that lines of latitude do not match rotated arcs through the origin (great circle routes).

\documentclass{article}
\usepackage{verbatim}

\usepackage{tikz}
\usepackage{tikz-3dplot}
\usepackage[active,tightpage]{preview}  % generates a tightly fitting border around the work
\PreviewEnvironment{tikzpicture}
\setlength\PreviewBorder{2mm}

\begin{document}

%Angle Definitions
%-----------------

%set the plot display orientation
%syntax: \tdplotsetdisplay{\theta_d}{\phi_d}
\tdplotsetmaincoords{60}{25}

%define polar coordinates for some vector
%TODO: look into using 3d spherical coordinate system
\pgfmathsetmacro{\rvec}{1}
\pgfmathsetmacro{\thetavec}{50}
\pgfmathsetmacro{\phivec}{15}

%start tikz picture, and use the tdplot_main_coords style to implement the display
\begin{tikzpicture}[scale=5,tdplot_main_coords]

%set up some coordinates
%-----------------------
\coordinate (O) at (0,0,0);
\fill (O) circle[radius=0.4pt] node[below left] {O};

%determine a coordinate (P) using (r,\theta,\phi) coordinates.
%syntax: \tdplotsetcoord{Coordinate name without parentheses}{r}{\theta}{\phi}
\tdplotsetcoord{P}{\rvec}{\thetavec}{\phivec}
\fill (P) circle[radius=0.4pt] node[anchor=south west] {P};

%draw figure contents
%--------------------

%draw the main coordinate system axes
\draw[thick,->] (0,0,0) -- (1,0,0) node[anchor=north east]{$x$};
\draw[thick,->] (0,0,0) -- (0,1,0) node[anchor=north west]{$y$};
\draw[thick,->] (0,0,0) -- (0,0,1) node[anchor=south]{$z$};

%draw a vector from origin to point (P), projection on xy plane, and a connecting line
\draw[-stealth,color=red] (O) -- (P);
\draw[dashed, color=red] (O) -- (Pxy);
\draw[dashed, color=red] (P) -- (Pxy);

%draw the angle \phi, and label it
%syntax: \tdplotdrawarc[coordinate frame, draw options]{center point}{r}{angle start}{angle end}{label options}{label}
\tdplotdrawarc{(O)}{0.2}{0}{\phivec}{anchor=north}{$\phi$}

%set the rotated coord system so the x'y' plane coincides with the"theta plane" of the main coord system
%syntax: \tdplotsetthetaplanecoords{\phi}
\tdplotsetthetaplanecoords{\phivec}

%draw theta arc and label, using rotated coordinate system
\tdplotdrawarc[tdplot_rotated_coords]{(0,0,0)}{0.2}{0}{\thetavec}{anchor=south west}{$\theta$}

%draw some dashed arcs, demonstrating direct arc drawing
\draw[dashed,tdplot_rotated_coords] (\rvec,0,0) arc (0:90:\rvec);
\draw[dashed] (\rvec,0,0) arc (0:90:\rvec);

\foreach \theta in {80,75,...,\thetavec} {%
\pgfmathsetmacro{\z}{\rvec*cos(\theta)}
\pgfmathsetmacro{\r}{\rvec*sin(\theta)}
\tdplotdrawarc[blue]{(0,0,{\z})}{\r}{15}{30}{}{}
}

\foreach \phi in {15,20,25,30} {\tdplotsetthetaplanecoords{\phi}%
\tdplotdrawarc[tdplot_rotated_coords,red]{(0,0,0)}{\rvec}{\thetavec}{80}{}{}
}

\end{tikzpicture}
\end{document}

To fill the area, I stored points along the border. I also used the plot[smooth cycle] option to make the joins less noticeable. Note that the corner points are entered twice; e.g. (A50) and (B15) are at the same location.

\documentclass{article}
\usepackage{verbatim}

\usepackage{tikz}
\usepackage{tikz-3dplot}
\usepackage[active,tightpage]{preview}  % generates a tightly fitting border around the work
\PreviewEnvironment{tikzpicture}
\setlength\PreviewBorder{2mm}

\begin{document}

%Angle Definitions
%-----------------

%set the plot display orientation
%syntax: \tdplotsetdisplay{\theta_d}{\phi_d}
\tdplotsetmaincoords{60}{25}

%define polar coordinates for some vector
%TODO: look into using 3d spherical coordinate system
\pgfmathsetmacro{\rvec}{1}
\pgfmathsetmacro{\thetavec}{50}
\pgfmathsetmacro{\phivec}{15}

%start tikz picture, and use the tdplot_main_coords style to implement the display
\begin{tikzpicture}[scale=5,tdplot_main_coords]

%set up some coordinates
%-----------------------
\coordinate (O) at (0,0,0);
\fill (O) circle[radius=0.4pt] node[below left] {O};

%determine a coordinate (P) using (r,\theta,\phi) coordinates.
%syntax: \tdplotsetcoord{Coordinate name without parentheses}{r}{\theta}{\phi}
\tdplotsetcoord{P}{\rvec}{\thetavec}{\phivec}
\fill (P) circle[radius=0.4pt] node[anchor=south west] {P};

%draw figure contents
%--------------------

%draw the main coordinate system axes
\draw[thick,->] (0,0,0) -- (1,0,0) node[anchor=north east]{$x$};
\draw[thick,->] (0,0,0) -- (0,1,0) node[anchor=north west]{$y$};
\draw[thick,->] (0,0,0) -- (0,0,1) node[anchor=south]{$z$};

%draw a vector from origin to point (P), projection on xy plane, and a connecting line
\draw[-stealth,color=red] (O) -- (P);
\draw[dashed, color=red] (O) -- (Pxy);
\draw[dashed, color=red] (P) -- (Pxy);

%draw the angle \phi, and label it
%syntax: \tdplotdrawarc[coordinate frame, draw options]{center point}{r}{angle start}{angle end}{label options}{label}
\tdplotdrawarc{(O)}{0.2}{0}{\phivec}{anchor=north}{$\phi$}

%set the rotated coord system so the x'y' plane coincides with the"theta plane" of the main coord system
%syntax: \tdplotsetthetaplanecoords{\phi}
\tdplotsetthetaplanecoords{\phivec}

%draw theta arc and label, using rotated coordinate system
\tdplotdrawarc[tdplot_rotated_coords]{(0,0,0)}{0.2}{0}{\thetavec}{anchor=south west}{$\theta$}

%draw some dashed arcs, demonstrating direct arc drawing
\draw[dashed,tdplot_rotated_coords] (\rvec,0,0) arc (0:90:\rvec);
\draw[dashed] (\rvec,0,0) arc (0:90:\rvec);

\foreach \theta in {80,75,70,65,60,55,50} {% left and right borders
\pgfmathsetmacro{\z}{\rvec*cos(\theta)}
\pgfmathsetmacro{\r}{\rvec*sin(\theta)}
\tdplotsetcoord{temp}{\r}{90}{\phivec}
\path (0,0,{\z}) ++(\phivec:{\r}) coordinate(A\theta);
\path (0,0,{\z}) ++(30:{\r}) coordinate(C\theta);
}

\foreach \phi in {15,18,21,24,27,30} {% top and bottom borders
\tdplotsetcoord{B\phi}{\rvec}{\thetavec}{\phi}%
\tdplotsetcoord{D\phi}{\rvec}{80}{\phi}%
}

\fill[green,opacity=.5] plot[smooth cycle] coordinates{(A80) (A75) (A70) (A65) (A60) (A55) (A50)
(B15) (B18) (B21) (B24) (B27) (B30)
(C50) (C55) (C60) (C65) (C70) (C75) (C80)
(D30) (D27) (D24) (D21) (D18) (D15)};

\end{tikzpicture}
\end{document}

• Thanks John, that helps! But, what I was after is a filled in shape than a grid. Is it perhaps a minor tweak to achieve that? – bluprince13 May 22 '17 at 14:56
• I suggested an edit with a nicer screenshot. Why does yours have such ungly alising? Just curious, I see these around here sometimes. – lblb May 22 '17 at 16:26
• @lblb - It's not a screen shot. I used Acrobat to save the PDF as PNG. I suppose I could get a higher resolution, but I do this a lot and this is faster. – John Kormylo May 22 '17 at 22:21
• @bluprince13 - Tricky! To fill things you need to set up a path around the perimeter, but in this case the perimeter involves changing the coordinates system several times. – John Kormylo May 22 '17 at 22:30
• Ah! I was afraid of that. Thanks though. I guess I should I use asymptote rather than TikZ for this. – bluprince13 May 22 '17 at 23:01


text:
# one one and onto

• March 19th 2011, 08:35 AM
one one and onto
is monotonicity is sufficient condition for proving one one .do every function must satisfy this condition.
also what is best method to prove onto
• March 19th 2011, 08:49 AM
TheEmptySet
Quote:

is monotonicity is sufficient condition for proving one one .do every function must satisfy this condition.
also what is best method to prove onto

If by monotone increasing you mean (the opposite for decreasing) the important part is the strict inequality.

$x_1 < x_2 \implies f(x_1) < f(x_2)$ then this is sufficient to prove 1-1.

Suppose $f$ is monotone. Now suppose by way of contradiction that $f$ is not 1-1 then there exists $x_1 \ne x_2$ such that $f(x_1)=f(x_2)$ without loss of generality assume that $x_1 < x_2$ then by the monotone property we have that

$f(x_2) = f(x_1) < f(x_2)=f(x_1)$ but this is a contradiction!

Onto is false consider any piecewise increasing function with jump discontinuities.
• March 19th 2011, 09:09 AM
I think there is a slight bit of confusion going on here. It's true, for example, that for continuous $f:\mathbb{R}\to\mathbb{R}$ one has that strict monotonicity is equivalent to injectiveness.


text:
# zbMATH — the first resource for mathematics

Families of finite sets in which no set is covered by the union of two others. (English) Zbl 0489.05003
From the summary: “Let $$f^*(n)$$ denote the maximum of $$k$$-subsets of an $$n$$-set satisfying the condition in the title. It is proved that $$f^{2t-1}(n)\leq f^{2t}(n+1)\leq\binom{n}{t}/\binom{2t-1}{t}$$ with equalities holding iff there exists a Steiner system $$\mathcal{S}(t,2t-1,n)$$. The bounds are approximately best possible for $$k\leq 6$$ and of correct order of magnitude for $$k\geq 7$$, as well, even if the corresponding Steiner systems do not exist.”
Reviewer: J.Libicher

##### MSC:
 05A05 Permutations, words, matrices 05B07 Triple systems 05C65 Hypergraphs
subsets
Full Text:
##### References:
 [1] Erdös, P; Hanani, H, On a limit theorem in combinatorial analysis, Publ. math. debrecen, 10, 10-13, (1963) · Zbl 0122.24802 [2] {\scP. Frankl and Z. Füredi}, A new generalization of the Erdös-Ko-Rado theorem, Proc. J. Bull. London Math. Soc., submitted. · Zbl 0529.05001 [3] Hanani, H, On some tactical configurations, Canad. J. math., 15, 702-722, (1963) · Zbl 0196.29102 [4] Katona, G.O.H; Katona, G.O.H, A theorem of finite sets, (), 187-207 · Zbl 0878.05079 [5] Kirkman, T.P, On a problem in combinations, Cambridge and Dublin math. J., 2, 191-204, (1847) [6] Kleitman, D.J; Shearer, J; Sturtevant, D, Intersections of k-element sets, Combinatorica, 1, 381-384, (1981) · Zbl 0489.05005 [7] Kruskal, J.B, The number of simplices in a complex, (), 251-278 [8] Erdös, P; Füredi, Z, The greatest angle among N points in the d-dimensional Euclidean space, (), in press · Zbl 0534.52007
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
# Calculating the break-even point between traveling around the inner / outer circle

Most days when I go to work, I ride a matatu (this is a Kenyan word. I am not in Kenya but the concept is the same). When I get off the matatu, I need to walk around the circle to get to the street that my office is on. And on occasion while I am making this trip, I think about whether it is shorter to get off before the light and walk around the smaller circle, or to get off after the light; in reality, it is probably not a perfect circle, but for my purposes I'm simplifying. In the image below, I looking to find the breakeven angle in terms of traveling to the blue point where the distance around the inner circle plus the distance between the circles (crossing the street) is equal to the distance walking around the outer circle. (The green point is my office)

Today I actually took to calculating the answer to this.

The way I am setting up the equation:

r1 * x * PI + (r2 - r1) = r2 * x * PI

The inner radius times x radians times PI plus the difference between the radius (to cross the street) equals the outer radius times x radians time PI.

In which case the answer is 1/PI radians, right?

I just wanted to confirm that my math is still working and that this is the right answer

• Could you please specify the start and end? Before we find the shortest path from A to B we should know what A and B are! – Unit Jun 17 '19 at 13:55
• @Unit there are actually 2 point A's: the intersection between line (street) going straight up on the inner circle and the outer circle. Point B is at the intersection between the outer circle and the line that's offset. I am interested in finding the angle at which the distance from both point A's to point B is the same... The break even angle – ControlAltDel Jun 17 '19 at 14:28
• Where is your office? On one of those radii, I guess but on the inner circle, the outer circle, or in between? You have a choice of two places to get out, I guess that they are on the other radius. Are the choices, the two intersections with the circles? If you could add labels to the diagram, it would help a lot. – badjohn Jun 17 '19 at 14:44
• Also, if $x$ is in radians then you probably don't want to multiply it by $\pi$. $x r$ is already the distance for $x$ radians around a circle of radius $r$. – badjohn Jun 17 '19 at 14:47
• @badjohn so that would make the answer 1 radian, or? – ControlAltDel Jun 17 '19 at 15:02

## 1 Answer

I still need to make an assumption: You may get off at either of the red circles and need to get to the blue one.

I will use $$\theta$$ for the angle between the two radii as that is a very common choice for angles. I will assume that it is in radians. So, travelling around a circle of radius $$r$$ for $$\theta$$ radians is a distance of $$r \theta$$.

The inner circle has radius $$r_1$$ and the outer $$r_2$$.

If you get off at the inner blue circle then you will walk $$r_1 \theta + r_2 - r_1$$. If you get off at the outer blue circle then you will walk $$r_2 \theta$$. The break even will be when $$r_1 \theta + r_2 - r_1 = r_2 \theta$$

So, indeed $$\theta = 1$$. Surprisingly neat and simple. A nice problem.

• Something else I think is neat about this: since 1 radian is irrational, in real life one of the two options must always be better (never mind that we don't walking in infinitely thin lines, nor that in real life this is an oval not a circle) – ControlAltDel Jun 17 '19 at 17:00
• $1$ radian is irrational in terms of degrees but rational in radians (it is $1$). Whether it is irrational or not depends on your ruler. Think of ordinary straight lines. You make a ruler and I make one. Mine happens to match the diagonal of a $1 \times 1$ square drawn with yours. Lines that I think have a rational length will seem irrational to you. – badjohn Jun 17 '19 at 18:11
• Except that you can define 1 inch or 1 centimeter as X number of wavelengths of a certain light coming from X atom. Whereas root 2 (or 1 radian) is irrational (cannot be defined with a certain number of wavelengths of anything) when you've already defined 1 inch/centimeter using wavelengths – ControlAltDel Jun 17 '19 at 19:43
• What if we use different isotopes or transitions for our length standard? Would they necessarily have a rational relationship? We could never know since we could never perfectly compare them. – badjohn Jun 17 '19 at 20:24


text:
Pro
19

However, they are as numerous as photons in our universe and play an important role in the formation and evolution of the universe. a, Kam-Biu Luk. Received 18 February 2016; received in revised form 21 April 2016; accepted 21 April 2016. Coordinates. Jen-Chieh Peng. The Daya Bay reactor neutrino experiment was designed to precisely measure the neutrino oscillation parameter θ 13 via the relative comparison of neutrino rates and spectra at different baselines. ABSTRACT: The Daya Bay Reactor Neutrino Experiment has measured the last unknown neutrino mixing angle, q 13, to be non-zero at the 7:7s level. Daya Bay Neutrino Experiment. Eight identically designed detectors were deployed in two near experimental halls and a … The Daya Bay Reactor Neutrino Experiment discovered an unexpectedly large neutrino oscillation related to the mixing angle θ13 in 2012. The experiment will look for electron antineutrinos from the reactors via the inverse beta-decay reaction in a gadolinium-doped liquid scintillator target (figure 1). b. a. This nding paved the way to the next generation of neutrino oscillation experiments. Institute of High Energy Physics, Beijing, China. August 15, 2011. enlarge. The Daya Bay Reactor Neutrino Experiment, located in Shenzhen, China, is going to probe the last unknown neutrino mixing angle q 13 with a sensitivity of 0.01 or better in sin2 2q 13 at 90% C.L. The Daya Bay Reactor Neutrino Experiment began data taking in 2011 and soon after dis-covered a non-zero value of lepton mixing angle q 13 [2]. The Daya Bay Neutrino Experiment is a neutrino-oscillation experiment designed to measure the mixing angle q 13 using anti-neutrinos produced by the reactors of the Daya Bay Nuclear Power Plant (NPP) and the Ling Ao NPP. Location: Daya Bay Nuclear Power Plant complex, Shenzen, China. This Letter presents the combined results of these searches, along with exclusion results from the Bugey-3 reactor experiment, framed in a minimally extended four-neutrino scenario. The reactor complex near Daya Bay, China, about 70 km northeast of the Hong Kong airport, is an excellent site for the experiment, and one of the few appropriate sites worldwide. Daya Bay Reactor Neutrino Experiment, China. A first-of-its-kind equal partnership agreement between the U.S. and China launched the Daya Bay Reactor Neutrino Experiment northeast of Hong Kong, and it didn’t take long for the experiment to experience its first success. The observation of neutrino oscillation in experiments involving solar, atmospheric, accelerator, and reactor neutrinos has demonstrated that the masses of neutrinos are nonzero, and neutrinos with different flavors can mix with each other to form the mass eigenstates. Daya Bay Reactor Neutrino Experiment. Antineutrino detectors in Daya Bay Hall #3. Background The Daya Bay experiment conducts precision measurements of reactor neutrinos to find new physics beyond the Standard Model of Particle Physics. Multiple antineutrino detectors are deployed in three underground water pools at different distances from the reactor cores to search for deviations in the antineutrino rate and energy spectrum due to neutrino mixing. After 4 years of gestation, 4 years of construction and 9 years of operation since 2003, it was officially retired on the 12th after achi About the Daya Bay Experiment. Jon Link's Experimental Neutrino Physics Group News About Daya Bay VT Group Internal Pictures Presentations Daya Bay Reactor Neutrino Experiment. The Daya Bay Reactor Neutrino Experiment was designed to study the oscillation of neutrinos coming from nuclear reactors and in 2012 was the first experiment to conclusively determine that $$\theta_{13}$$ is non-zero. This finding paved the way to the next generation of neutrino oscillation experiments. The Daya Bay Reactor Neutrino Experiment is a China-based multinational particle physics project studying neutrinos.The multinational collaboration includes researchers from China, Chile, the United States, Taiwan (Republic of China), Russia, and the Czech Republic. It has provided the most precise measurement of $$\theta_{13}$$ to date. This is the most precise measurement to q 13 to date [1,2]. b. Download Citation | The Daya Bay Reactor Neutrino Experiment | The 3x3 PMNS leptonic mixing matrix relates the mass and flavor eigenstates of the 3 known neutrinos. On December 12, Shenzhen, Guangdong, the Daya Bay Reactor Neutrino Experiment (Daya Bay Experiment) device that discovered a new type of neutrino oscillation and accurately measured the energy spectrum of the reactor neutrino was obtained. In this article, we review the history, featured design, and scientific results of Daya Bay. 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text: How to find formulae for Fibonacci numbers. Calculus 1 — A Must Know Concept For Every Professional. Prove that for all positive integers $n$, $$\sum_{i=0}^n f_{i}^2 = f_n f_{n+1}.$$. Program to print first n Fibonacci Numbers | Set 1; Program to implement Inverse Interpolation using Lagrange Formula; Roots of the quadratic equation when a + b + c = 0 without using Shridharacharya formula; Check if a M-th fibonacci number divides N-th fibonacci number; Check if sum of Fibonacci elements in an Array is a Fibonacci number or not We study the Fibonacci numbers and show how to use mathematical induction to prove several Fibonacci identities. You will have one formula for each unique type of recursive sequence. From above, we have $$\phi^n = \phi \, f_n + f_{n-1}\quad \text{ and } \quad \tau^n = \tau \, f_n + f_{n-1}.$$ It follows that $f_n=\frac{\phi^n-\tau^n}{\phi-\tau}$. What Did Newton Do with his Time During Quarantine? Now observe that the Euler-Binet Formula follows since $\phi-\tau=\sqrt{5}$. The denominator is a quadratic equation whose roots can easily be obtained to be, (For an easy graphical method of finding roots, check out this article), Using these roots, it is possible to write the denominator as, We can split the denominator and write this as, Before we proceed, we need to understand a useful fact about geometric series. \end{align*} By mathematical induction, the equation holds for all positive integers $n$. The iterative approach depends on a while loop to calculate the next numbers in the sequence. In reality, rabbits do not breed this… It is easy to check that this modification still produces the same sequence of numbers, starting from n=1, instead of n=0. In order to make use of this function, first we have to rearrange the original formula. Assume, for some positive integer $k$, that the equation holds. The first 21 Fibonacci numbers are \begin{equation*}\begin{matrix} 1 & 1 & 2 & 3 & 5 & 8 & 13 \\ 21 & 34 & 55 & 89 & 144 & 233 & 377 \\ 610 & 987 &1597 &2584 &4181 & 6765 & 10946 \end{matrix} \end{equation*}. Find the first 50 Lucas numbers. How many digits does Fib(100) have? All rights reserved. Exercise. Show that $f_{n-2}+f_{n+2}=3f_n$ whenever $n$ is a positive integer with $n\geq 2.$, Exercise. Then prove that $b_n=(3^{n}+1)/2$ for all positive integers. \end{align*} By mathematical induction, the equation holds for all positive integers $n$. Example. Then we find \begin{align*} \sum _{j=1}^{k+1} f_j & =\sum _{j=1}^k f_j+f_{k+1} \\ & =f_{k+2}-1+f_{k+1} \\ & =f_{k+1}+f_{k+2}-1 \\ & =f_{k+3}-1. This formula is a simplified formula derived from Binet’s Fibonacci number formula. This field is for validation purposes and should be left unchanged. Exercise. ½ × 10 × (10 + 1) = ½ × 10 × 11 = 55. Fibonacci sequence. We also develop the Euler-Binet Formula involving the golden-ratio. Prove that for all positive integers $n,$ $$\sum _{i=1}^n f_i = f_{n+2}-1. That is tha… Prove that L_n=\phi^n+\tau^n where \phi is the golden ratio and \tau its conjugate. In this book, Fibonacci post and solve a … He used the number sequence in his book called Liber Abaci (Book of Calculation). The formula for nth triangular number is: ½n(n + 1) For example, to get the 10th triangular number use n = 10. We now study the Fibonacci Numbers and the Euler-Binet Formula. The golden ratio is the positive root of the quadratic equation x^2-x-1=0, that is,$$\phi = \frac{1+\sqrt{5}}{2}.$$The conjugate of \phi is denoted by$$\tau = \frac{1-\sqrt{5}}{2}.It is easily verified that \phi\tau=-1 and \phi-\tau=\sqrt{5}. Assume, for some positive integer k, that the equation holds. We can see from the following table, that by plugging the values of n, we can directly find all Fibonacci numbers! "Fibonacci" was his nickname, which roughly means "Son of Bonacci". Then, the idea is to provide several interesting examples of how mathematical induction is applied to prove Fibonacci Identities. Show that f_{2n}=f_{n}^2+2f_{n-1}f_n whenever n is a positive integer with n\geq 2.. Secondly, many of the results obtained are quite surprising. 6. Solution. Write out the first 50 Fibonacci numbers. Prove that f_{m+n}=f_m f_{n+1}+f_n f_{m-1} whenever n is a positive integer. Then we have \begin{align*} f_k = \ & f_{k-1}+f_{k-2} \\ & > \left(\frac{3}{2}\right)^{k-2}+\left(\frac{3}{2}\right)^{k-3} \\ & =\left(\frac{3}{2}\right)^{k-3}\left(\frac{3}{2}+1\right)\\ & = \left(\frac{3}{2}\right)^{k-3}\left(\frac{5}{2}\right) \\ & >\left(\frac{3}{2}\right)^{k-3}\left(\frac{9}{4}\right) \\ & =\left(\frac{3}{2}\right)^{k-3}\left(\frac{3}{2}\right)^2 \\ & =\left(\frac{3}{2}\right)^{k-1} \end{align*} as desired. For n=1, f_2 f_0 -\left(f_1\right)^2=(-1)^1 so the base case holds. Next, we multiply the last equation by xⁿ to get, Let us first consider the left hand side of this equation —, Now, we try to represent this expansion in terms of F(x), by doing the following simple manipulations -, Using the definition of F(x), this expression can now be written as, Therefore, using the fact that F₁=1, we can write the entire left hand side as, Let us now consider the right hand side —, By taking out a factor of x from the second expansion, we get, Using the definition of F(x), this can finally be written as. Using the LOG button on your calculator to answer this. The Fibonacci calculator uses the following generalized formula for determining the n-th term: Fₙ = aφⁿ + bψⁿ. The recursive approach involves defining a function which calls itself to calculate the next number in the sequence. Exercise. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Our job is to find an explicit form of the function, F(x), such that the coefficients, Fₙ are the Fibonacci numbers. \end{align*} By mathematical induction, the equation holds for all positive integers n. . @Calvin Lin I learned this method from my math teacher, but is there a much easier way to derive the explicit formula for the Fibonacci Sequence? Prove that L_{m+n}=f_{m+1}L_n+f_mL_{n-1} whenever m and n are positive integers with n >1. Fibonacci series in Java. Example. Example. Male Female Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school student In fibonacci series, next number is the sum of previous two numbers for example 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 etc. In addition, here is a large (29 digit) Fibonacci number:19,134,702,400,093,278,081,449,423,917.$$While it remains whether Fibonacci primes are infinite, the conjecture is that there are infinitely many Fibonacci prime numbers. The proof is by induction. It has long been noticed that the Fibonacci numbers arise in many places throughout the natural world. Exercise. A Formula For Fibonacci Sequence. With this formula, if you are given a Fibonacci number F, you can determine its position in the sequence with this formula: n = log_((1+√5)/2)((F√5 + √(5F^2 ± 4)) / 2) Whether you use +4 or −4 is determined by whether the result is a perfect square, or more accurately whether the Fibonacci number has an even or odd position in the sequence. Recursion. where n is a positive integer greater than 1, Fₙ is the n−th Fibonacci number with F₀=0 and F₁=1. The first few terms of the sequence are 1,3,7,15. His real name was Leonardo Pisano Bogollo, and he lived between 1170 and 1250 in Italy. Interpretations, questions, and a few speculations from “Deep Learning with Python” by François…. Surprisingly enough, there is also a simple non-recursive formula for the nth Fibonacci number (the ^ operator is exponentiation): fib(n) = (Phi^n - (-phi)^n) / sqrt(5). Therefore, by equating the left and the right hand sides, the original formula can be re-written in terms of F(x) as, Let us now simplify this expression a bit more. Find the first five terms of the sequence. [4] It goes by the name of golden ratio, which deserves its own separate article.). The golden ratio and the fibonacci sequence are inextricably linked by the Euler-Binet Formula. Simply open the advanced mode and set two numbers for the first and second term of the sequence. (1) \quad L_n-5 f_n^2=4(-1)^n$$(2) \quad L_{2n}=L_n^2-2(-1)^n$$(3) \quad f_{n+1}+f_{n-1}=L_n$$(4) \quad L_{n+1}+L_{n-1}=5 f_n$$(5) \quad L_{2n+1}=L_n L_{n+1}-(-1^n)$$(6) \quad L_{n-1}L_{n+1}-L_n^2=5(-1)^{n-1}$,$n\geq 2. They hold a special place in almost every mathematician’s heart. He used the number sequence in his book called Liber Abaci (Book of Calculation). David Smith (Dave) has a B.S. The Fibonacci sequence of numbers “F n ” is defined using the recursive relation with the seed values F 0 =0 and F 1 =1:. Then we find \begin{align*} \sum _{j=1}^{k+1} f_{2j} & =\sum _{j=1}^k f_{2j}+f_{2k+2} \\ & =f_{2k+1}-1+f_{2k+2} \\ & =f_{2k+3}-1. Exercise. In the comments, the OP said he means some explicit formula involving the index $n$ (rather than, say, a recursion). It is not hard to imagine that if we need a number that is far ahead into the sequence, we will have to do a lot of “back” calculations, which might be tedious. The Fibonacci formula is used to generate Fibonacci in a recursive sequence. Prove thatf_n > \left(\frac{5}{3}\right)^{n-1}$, for all$n\geq 2$. There are two obvious reasons for this. Exercise. Leonardo Pisano Bigollo (1170 — 1250) was also known simply as Fibonacci. Fibonacci Sequence Formula. In this article, we are going to discuss another formula to obtain any Fibonacci number in the sequence, which is (arguably) easier to work with. The idea is to provide several interesting examples on how mathematical induction can be applied to give rigorous arguments on (perhaps) experimentally found identities. (Euler-Binet Formula) For all positive integers$n$, $$f_n=\frac{1}{\sqrt{5}}\left(\phi^n-\tau^n\right).$$. Show that$f_{n+3}-f_n=2f_{n+1}$whenever$n$is a positive integer. Fibonacci Numbers and the Euler-Binet Formula. Copyright © 2020 Dave4Math LLC. Firstly, we notice that the Euler-Binet formula gives us a closed form for calculating the$n$-th Fibonacci number. This fact is easily provable by mathematical induction. You can also use the Fibonacci sequence calculator to find an arbitrary term of a sequence with different starters. We verify that $$f_3=3>\phi^1\approx 1.618\qquad \text{and}\qquad f_4=5>\phi^2\approx 2.618.$$ Now let$k>4$, and so assume the inductive hypothesis that$f_i >(3/2)^{i-1}$for$i=k-1$and$i=k-2$. Prove that for all positive integers$n$, $$\sum _{j=1}^n f_{2j}=f_{2n+1}-1.$$. Exercise. Here we have omitted F₀, because F₀=0, by definition. The Fibonacci Prime Conjecture and the growth of the Fibonacci sequence is also discussed. For$n=1,f_2=1= f_3-1so the base case holds. $$. A natural derivation of the Binet's Formula, the explicit equation for the Fibonacci Sequence. Example. Let us define a function F(x), such that it can be expanded in a power series like this. Fibonacci numbers are defined as a recursive sequence by starting with 0 and 1, and then adding the previous two integers together. Commentdocument.getElementById("comment").setAttribute( "id", "a7d7ec7135165028f8ebdd5fd4eb63b4" );document.getElementById("bee3c713a7").setAttribute( "id", "comment" ); With Daveâs Math Help Service, you send in your problems, and heâll solve them for you. After that, I explore the Euler-Binet Formula and discuss the Fibonacci Prime Conjecture and the growth of the Fibonacci Sequence. For instance, the first seven Fibonacci primes are 2, 3, 5, 13, 89, 233, 1597. To improve this 'Fibonacci sequence Calculator', please fill in questionnaire. Here are more exercises to help you practice with Fibonacci numbers. This is the simplest nontrivial example of a linear recursion with constant coefficients. . In this exciting article, I introduce the Fibonacci numbers. Thanks to todayâs technological advances, getting math help online is the easiest it has ever been. Let a_0=1 and, for n>0, let a_n=2 a_{n-1}+1. The Fibonacci sequence can be written recursively as and for . Is there an easier way? Fibonacci omitted the first term (1) in Liber Abaci. n = 6. p˚6 5 = , so F6 = n = 13. Exercise. Observe the following Fibonacci series: Show that for n\geq 2,$$f_n=\frac{f_{n-1}+\sqrt{5 f_{n-1}^2+4(-1)^n}}{2}.Notice that this formula gives f_n in terms of one predecessor rather than two predecessors. Show that f_{2n}=f_n L_n whenever n is a positive integer. Exercise. Discussion and conclusion. Proof. The Fibonacci Sequence can be generated using either an iterative or recursive approach. Prove that a_n=2^{n+1}-1 for all positive integers. Solution. The Fibonacci Prime Conjecture. You can also solve this problem using recursion: Python program to print the Fibonacci sequence using recursion. The recurrence formula for these numbers is: F(0) = 0 F(1) = 1 F(n) = F(n − 1) + F(n − 2) n > 1 . Then we find\begin{align*} \sum_{i=1}^{k+1} f_{2i-1} & =\sum_{i=1}^k f_{2i-1} +f_{2(k+1)-1} \\ & =f_{2k}+f_{2k+1} \\ & =f_{2k+2} \\ & =f_{2(k+1)}. Exercise. We now give the closed formula to compute the n-th Fibonacci number. Solution. Solution. If we have an infinite series, This means, if the sum of an infinite geometric series is finite, we can always have the following equality -, Using this idea, we can write the expression of F(x) as. The Fibonacci Sequence is one of the cornerstones of the math world. It is: a n = [Phi n – (phi) n] / Sqrt[5]. Theorem. Male or Female ? [ The 11 Most Beautiful Mathematical Equations ] Exercise. Complete with step-by-step solutions with a video option available. The golden ratio \phi is approximately, 1.61803398875. However, if I wanted the 100th term of this sequence, it would take lots of intermediate calculations with the recursive formula to get a result. Nick Lee - … Finding the Moment of Inertia from a Point to a Ring to a Disk to a Sphere. If we make the replacement. Exercise. Fibonacci initially came up with the sequence in order to model the population of rabbits. Formula for the Fibonacci numbers: But the Greeks had a more visual point of view about the golden mean. Exercise. The Fibonacci sequence can be written recursively as and for .This is the simplest nontrivial example of a linear recursion with constant coefficients. Exercise. Now observe that the Euler-Binet Formula follows since \phi-\tau=\sqrt{5}. Establish each of the following for all n\in \mathbb{N}. There is one thing that recursive formulas will have in common, though. ˚p13 5 = , so F13 = In fact, the exact formula is, Fn = 1 p 5 ˚n 1 p 5 1 ˚n; (+ for odd n, for even n) 6/24 Lower case asub 2 is the second number in the sequence and so on. Let F=\begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix}. To recall, the series which is generated by adding the previous two terms is called a Fibonacci series. Now we assume k\in P. Recursive sequences do not have one common formula. Find and prove a formula for the sum of the first n Lucas numbers when n is a positive integer. Show that f_{2n+1}=f_{n+1}^2+f_{n}^2 whenever n is a positive integer. Definition. Proof. Since f_1=1=f_2 we see the base case holds. Let P be the set of positive integers for which the equation is true. The Fibonacci sequence first appears in ancient Sanskrit texts as early as 200 BC, but the sequence wasn't widely known to the western world until 1202 when Italian mathematician Leonardo Pisano Bogollo published it in his book of calculations called Liber Abaci.Leonardo also went by the moniker Leonardo of Pisa, but it wasn't until 1838 that historians gave him the nickname Fibonacci … Lower case a sub 1 is the first number in the sequence. Receive free updates from Dave with the latest news! The term “regular formula” doesn't have any common meaning. Exercise. Firstly, many of the results exercises are relatively straight-forward. Exercise. Exercise. Solution. How can we compute Fib(100) without computing all the earlier Fibonacci numbers? \begingroup Possible duplicate of Prove this formula for the Fibonacci Sequence \endgroup – Watson Jan 12 '17 at 14:27 add a comment | 4 Answers 4 We then interchange the variables (update it) and continue on with the process. \end{align*} By mathematical induction, the equation holds for all positive integers n. where the GoldenRatios Phi and phi are defined as It follows that 0\in P, since f_0^2 =0=f_0 f_1. This change in indexing does not affect the actual numbers in the sequence, but it does change which member of the sequence is referred to by the symbol and so also changes the appearance of certain identitiesinvolvin… After having studied mathematical induction, the Fibonacci numbers are a good subject to test ones abilities to perform induction. where: It's easy to create all sorts of sequences in Excel.For example, the Fibonacci sequence.. 1. Exercise. The first and second term of the Fibonacci series is set as 0 and 1 and it continues till infinity. You may also attempt to solve the following exercises. David is the founder and CEO of Dave4Math. Leonardo Pisano Bigollo (1170 — 1250) was also known simply as Fibonacci. The Fibonacci sequence exhibits a certain numerical pattern which originated as the answer to an exercise in the first ever high school algebra text. How many numbers are required to define a sequence without stating a rule/function for generating the next term in the sequence? Example. (Issues regarding the convergence and uniqueness of the series are beyond the scope of the article). Fibonacci primes with thousands of digits have been found. Formula for the n-th Fibonacci Number Rule: The n-th Fibonacci Number Fn is the nearest whole number to ˚ n p 5. Exercise. We verify thatf_6=8>\left(\frac{3}{2}\right)^5\approx 7.59375.$$We also must verify$$f_7=13>\left(\frac{3}{2}\right)^6\approx 11.390625.$$Now let k>7, and assume the inductive hypothesis that f_i >(3/2)^{i-1} for i=k-1 and i=k-2. Exercise. Fibonacci was not the first to know about the sequence, it was known in India hundreds of years before! Exercise. Fibonacci Sequence is popularized in Europe by Leonardo of Pisa, famously known as "Leonardo Fibonacci".Leonardo Fibonacci was one of the most influential mathematician of the middle ages because Hindu Arabic Numeral System which we still used today was popularized in the Western world through his book Liber Abaci or book of calculations. In this article, we are going to discuss another formula to obtain any Fibonacci number in the sequence, which is (arguably) easier to work with. Show that$$F^n=\begin{bmatrix}f_{n+1} & f_n \\ f_n & f_{n-1}\end{bmatrix}$$for all n\geq 1. Show that f_1 f_2+f_2f_3+\cdots +f_{2n-1}f_{2n}=f_{2n}^2 whenever n is a positive integer. Fibonacci numbers are one of the most captivating things in mathematics. Example. Readers should be wary: some authors give the Fibonacci sequence with the initial conditions (or equivalently ). Show that f_{n+2}^2-f_{n+1}^2=f_n f_{n+3}, whenever n is a positive integer greater than 1. and M.S. F n = F n-1 +F n-2. Clearly, x^1= x \, f_1 + f_{0} = x (1) + 0 = x so the base case holds. Example. What are the next fews terms? Fibonacci numbers have many special mathematical properties. Let b_0=1 and, for n>0, let b_n=3 b_{n-1}-1. The Fibonacci sequence is attributed originally to Indian mathematics. Note that we have also introduced Lucas numbers in this section, so feel free to look it up. Prove that for all positive integers n,$$\sum_{i=1}^n f_{2i-1} = f_{2n}.$$. Call us at 817-241-0575 or Order now! in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. Show that f_n > \left(\frac{3}{2}\right)^{n-1} whenever n is a positive integer greater than 6. Sum of Fibonacci numbers is : 7 This article is contributed by Chirag Agarwal . Assume now that, for some positive integer k, that$$ x^k=x \, f_k+f_{k-1} We wish to prove that x^{k+1}=x\, f_{k+1}+f_{k}. To this end, multiply the identity by x to obtain \begin{align*} x^{k+1} = x^2 \, f_k+ x, f_{k-1} & = (x+1) \, f_k + x \, f_{k-1} \\ & = x\, (f_k + f_{k-1}) + f_k \\ & =x \, f_{k+1} + f_k \end{align*} as needed. It then follows k+1\in P by \begin{align*} \sum_{i=0}^{k+1} f_i^2 & =\sum_{i=0}^{k} f_i^2+ f_{k+1}^2 \\ & =f_k f_{k+1}+f_{k+1}^2 \\ & =f_{k+1}(f_k+f_{k+1}) =f_{k+1} f_{k+2}. There is also an explicit formula below. For any solution x of x^2-x-1=0 and any positive integer n,x^n=x \, f_n+f_{n-1}.\$. To develop a better understanding of the unique behaviors of Fibonacci numbers, here are a few more examples. X Research source The formula utilizes the golden ratio ( ϕ {\displaystyle \phi } ), because the ratio of any two successive numbers in the Fibonacci sequence are very similar to the golden ratio. Fibonacci did not discover the sequence but used it as an example in Liber Abaci.