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sanguinarysanguinity · 5 years

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Navigatio Britannica: Planar Trigonometry

Chapter 4: Of Trigonometry, Sections I - V

The first half of this chapter is Planar Trigonometry, which I learned in high school and have used on-and-off ever since; the second half of this chapter is Spherical Trigonometry, which I know nothing about. Consequently, I’m dividing this chapter into two parts -- before we let John Barrow attempt to teach me spherical trig (wish me luck!), I want to do a fast recap of what he has to say about planar trig...

Section I: Definitions

Everything is defined geometrically, on the unit circle, via a diagram that I have yet to find in this scan. (Also, even if I do find the page, I don’t have much hope that it will have been scanned correctly, since Google’s scanning machine can turn pages but not unfold them.) Happily, Barrow is pretty good about describing his figures in enough detail that I can reconstruct them as I go, which is the only reason I was able to understand anything in this chapter.

Not many surprises here, although I did learn that cosine, cotangent, and cosecant are the sine, tangent, and secant of the complementary angle, and likewise that the tangent of an angle is called the tangent because its physical instantiation lives on a line tangent to the circle. Also, Barrow defines the Verse-Sine, which was new to me: geometrically, it is the part of the radius that isn’t the cosine. (Algebraically, it is 1 - cosine.) Wikipedia says the versine was important to navigation, so I assume it will come up later.

Section II: Geometrical Constructions of the Tables of Sines, Tangents, Secants, &c.

In which we are instructed to build ourselves a unit circle, mark it off in 1-degree intervals, and construct ourselves a... well, it’s gonna look a bit like a ruler, but it’s going to measure 1 to 90 degrees, on several parallel scales: chords, sines, versed-sines, tangents, etc. To make this thing, you use your compass and measure the length of a chord for a 10-degree angle, then mark it on your chords-scale, and label it “10 deg.” Repeat for the other 89 degrees, and ta-dah, you have a chords-scale! Then do it again for sines, versed-sines, tangents, and so on. When we get to actually solving trig problems, how to use this scale is one of the three standard methods that Barrow is going to teach us.

Section III: Arithmetical Constructions of the Tables of Sines, &c.

First off, Barrow reassures us this is going to be easy-peasy, no need to panic -- which is our first cue that panicking will be required before we’re done.

But true to his word, Barrow starts out easy, using similar triangles to prove all the basic trigonometric identities: tan = sin/cos, sec = 1/cos, etc. All well and good, except it’s all done in proportions and nothing is called out by name, only by referring to various line-segments in his nowhere-to-be-found unit circle diagram. All of which makes it difficult to absorb at a glance, but once you finish decoding everything this is basically just SOHCAHTOA.

Then he proves two variants on the standard trigonometric sum/difference formulas (although he expresses them as proportions and via verbal descriptions, talking about the means of equi-different angles and the differences between them):

cos x = (1/2) (sin y + sin (y + 2x)) / sin (y + x)

sin x = (1/2) (sin (y + 2x) - sin y) / cos (y + x)

You can verify those via the standard trigonometric sum/difference identities if you want. (I did.) But they’re also pretty straightforward geometrically, if you take the time to very carefully reconstruct what his diagram must have been: in the end, it’s all just similar triangles. He then proves several corollaries -- which in hindsight are simple enough (just straightforward algebraic manipulations, multiplying everything by two, or both sides by the denominator), but sadly, I lost MANY HOURS to a rash of typos in them.

Then.

Oh, then.

All hell breaks loose as he endeavours to prove that a semi-circle has an arc-length of pi. I admit to not following this bit: I haven’t seen Newton’s notation for calculus since I was seventeen, when that one weirdo physics professor used it in lectures, and I didn’t really feel like re-teaching it to myself for this. Nor did I really want to get into re-teaching myself binomial expansions. Also, the type-face on all the fractions in the expansions was super-squinchy to read, and you know what, fuck it, I think it’s well-established that a semi-circle has an arc-length of pi, let’s move on.

The point of establishing that a semi-circle has an arc-length of pi is so that we can calculate the arc-length of one minute (simply divide pi by 10,800 minutes, easy-peasy), which we will then use as an approximation of the sine of one minute. ... Which, okay, I suppose if your angle is small enough and your applications are practical enough you can get away with that? But it makes the mathematician in me cry, I’m just saying. (Even as I admit that you really can get away with it for most purposes: according to my handy-dandy TI-84 Plus, pi/10800 differs from sin(1′) in the ninth significant digit. But Dr. Roberts and Dr. Chrestenson would never have let me get away with that shit, never mind that I also have an engineering degree and thus should be okay with this kind of ruthless practicality. In my soul there is a mathematician and an engineer battling to the death over questions like these, you simply don’t know how much shit like this wounds me.)

Anyway, once I finally got over my fit of vapours...

Now that we have an approximation for the sine of one minute, we can calculate the cosine of one minute via the pythagorean trigonometric identity, and then...

And now I want to cry again, because now we get to build our table of sines (and along with it, our table of cosines), minute by freaking goddamn minute, by using the above equations like so:

2 cos (1′) sin (1′) - sin (0′) = sin (2′)

2 cos (2′) sin (2′) - sin (1′) = sin (3′)

2 cos (3′) sin (3′) - sin (2′) = sin (4′)

...

Continue until I cry blood and the seas boil dry.

(At one point Barrow admits that it’s possible to build this table in 5-minute increments and interpolate the intervening minutes when you need them. While this reduces the task to 1/5th of the original, I still want to hug and rock myself and cry.)

Happily, I don’t need to cry, because Barrow includes these tables in the book? But someone cried blood to make those tables, and John Barrow wants us all to know it.

Section IV: Actual Trigonometry Problems, At Long Last!

A ton of sample problems, all worked three ways:

Geometrically: Basically, use a compass and straight edge and your scale-thingie of chords/sines/secants that you made earlier, and draw a triangle of the correct proportions. Then just read/measure your answer right off the actual triangle in question, ta-dah! No abstract math required, just pretty pictures!

Arithmetically: What you learned in high-school, using the tables that someone cried blood over but without calculators (although you can use logs if you want to skip ahead to chapter five for them!) God, it looks miserable and grindingly awful, and I admit I don’t have the strength of character to follow any of these calculations through to the end.

Magic, I mean, Gunter’s Scale: The instructions here are amazingly low-key -- use your chart-dividers (what tumblr calls a pointy-leg-man what it likes to make walk on tippy-toes across charts) to measure off an interval on one scale, and then drop that same interval across the appropriate second scale, and voila! You have found your answer!

Of course I wanted to know what this magical tool is!

Apparently it was a slide rule without the slidey parts -- you used your dividers to accomplish what the slidey bit does on a slide rule -- but with some extra scales especially chosen for the convenience of navigators.

Apparently these things were so common among navigators that they were simply called Gunters, and I WANT ONE SO BAD. Here’s a nifty article about them, complete with pictures, and did I say? I WANT ONE SO BAD. I collect old-school mathematical tools and I WANT ONE SO BAD.

Ahem.

Anyway, Section V is more trig homework, except now we’re no longer dealing with right triangles. I admit it, I skimmed this like fuck.

And ta-dah! That’s Planar Trigonomometry, according to John Barrow in Navigatio Britannica, or, A Complete Guide to Navigation, pub. 1750!

Up next: Section VI - Spherical Trigonometry, what makes William Bush cry. Do I have the fortitude to teach myself spherical trig? PLACE YOUR BETS NOW.

#navigatio britannica #john barrow #age of sail #trigonometry #long post

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somervillesivertsen74-blog · 6 years

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testblogplzignore · 3 years

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Consider a coffee that is completely insulated at both ends

Mathematically this takes for the form

$ \frac{\partial}{\partial x} T(0,t) = \frac{\partial}{\partial x} T(h,t) =0 $

With the initial condition that the top fifth of the coffee is cooled to freezing point

$ T(x,0) = T_1, 0 \leq x \leq 0.8h$ and $ T_2, 0.8h < x \leq h$

We can apply separation of variables because we have homogeneous boundary conditions and do so

$ \frac{\partial}{\partial t}T(x,t) = \frac{\partial^2}{\partial x^2} T(x,t) $

$ T(x,t) = \beta \chi(x) \tau(t) $

$ \frac{\partial}{\partial t} \chi(x) \tau(t) = \beta \frac{\partial^2}{\partial x^2} \chi(x) \tau(t) $

$ \chi(x) \tau’(t) = \beta \chi’’(x) \tau(t) $

$ \frac{1}{ \beta } \frac{\tau’(t)}{ \tau(t)} = \frac{\chi’’(x)}{ \chi(x) } $

We see that both derivatives are equal to eachother even though they differ by differentiation variable so they must be equal to a constant such that

$ \frac{1}{ \beta } \frac{\tau’(t)}{ \tau(t)} = \frac{\chi’’(x)}{ \chi(x) } = -\lambda $

Separate into two ODEs (wow!)

$ \frac{1}{ \beta } \frac{\tau’(t)}{ \tau(t)} = -\lambda $

$ \frac{1}{ \beta } \frac{\tau’(t)}{ \tau(t)} +\lambda =0 $

$ \frac{1}{ \beta } \tau’(t) +\lambda \tau(t) =0 $

$ \frac{\chi’’(x)}{ \chi(x) } = -\lambda $

$ \frac{\chi’’(x)}{ \chi(x) } + \lambda =0 $

$ \chi’’(x) + \lambda \chi(x) =0 $

Which now has the form of two individual ordinary differential equations

Assume the solution takes the form of a wave in exponential form $ e^{rx} $

$ \chi’’(x) + \lambda chi(x) =0 $

$ r^2e^{rx} + \lambda e^{rx} =0 $

Dividing off the exponential. If you allow the imaginary r to exist then you can see dividing this off in both forms of a wave is valid (another way to look at the $ \lambda $ equivalency)

$ r^2 + \lambda =0 $

$ r^2 = - \lambda $

$ r = \sqrt{- \lambda } $

So $ r $ depends on 3 cases which consist of

$ \lambda <0 $, $ \lambda = 0 $, and $ \lambda > 0 $

Supoose $ \lambda <0 $:

$ r = \sqrt{- -\lambda} = \sqrt{ \lambda} $

which says $ \lambda $ is positive so we assume the solution takes the form

$ \chi(x) = c_1 cosh( \sqrt{ \lambda} \ x) + c_2 sinh( \sqrt{ \lambda} x) $

Using boundary conditions to determine constants

$ \frac{\partial}{\partial x} T(0,t) = \frac{\partial}{\partial x} T(h,t) =0 $

$ chi’(x) = c_1 \sqrt{\lambda} sinh(\sqrt{\lambda} x) + c_2 \sqrt{\lambda} cosh{\sqrt{\lambda} x}) $

$ chi’(0) = c_1 \sqrt{\lambda} sinh(\sqrt{\lambda} 0) + c_2 \sqrt{\lambda} cosh(\sqrt{-lambda} 0) = 0 $

$ chi’(0) = c_2 \sqrt{\lambda} cosh(\sqrt{\lambda} 0) = 0 $

$ chi’(0) = c_2 \sqrt{\lambda} = 0 $

Since we want nontrivial eigenvalues we take $ c_2 = 0 $

$ chi’(x) = c_1 \sqrt{\lambda} sinh(\sqrt{\lambda} x) $

Second boundary condition

$ chi’(0) = c_1 \sqrt{\lambda} sinh(\sqrt{\lambda} h) = 0 $

Since $ sinh( x ) $ is only zero at the vertical axis thus we require $ c_1 = 0 $

Second case $ \lambda = 0 $:

$ \chi’’(x) + \lambda chi(x) =0 $

$ \chi’‘(x) = 0 $

Integrating twice

$ \chi(x) = c_1 x + c_2 $

$ \chi(0) = 0 $

$ \chi(x) = c_1 0 + c_2 = 0 $

$ c_2 = 0 $

$ \chi(h) = 0 $

$ \chi(h) = c_1 h = 0 $

We assume $ h \neq 0 $ thus

$ c_1 = 0 $

Again we get the trivial solution where $ chi(x) = 0 $

Case 3 $ \lambda > 0 $:

$ r = \sqrt{- \lambda } $

$ r = i\sqrt{ \lambda } $

???????????????????????????? clarify

We have an imaginary exponential as the root(s) to $ r $

$ \therefore $ the solution takes the form (can you guess with euler’s formula?)

$ \chi(x) = c_1cos \big( \sqrt{ \lambda } x \big)+c_2 sin \big( \sqrt{ \lambda } x \big) $

$ \chi’(x) = \frac{\partial}{\partial x} c_1cos \big( \sqrt{ \lambda } x \big)+ \frac{\partial}{\partial x} c_2 sin \big( \sqrt{ \lambda } x \big) $

$ \chi’(x) = \sqrt{ \lambda } \frac{\partial}{\partial x} c_2 cos \big( \sqrt{ \lambda } x \big) - \sqrt{ \lambda } c_1sin \big( \sqrt{ \lambda } x \big) $

$ \chi’(0) = \sqrt{ \lambda } c_2 cos \big( \sqrt{ \lambda } (0) \big) - \sqrt{\lambda } c_1 sin \big( \sqrt{ \lambda} (0) \big) = 0 $

$ \chi’(0) = \sqrt{ \lambda } c_2 = 0 $

Non trivial eigenvalues

$ c_2 = 0 $

$ \chi’(h) = - \sqrt{\lambda } c_1 sin \big( \sqrt{ \lambda} (h) \big) = 0 $

If we set $ \sqrt{\lambda} = \frac{ n \pi}{h} $ then we get zeroes regardless of $ c_1 $ $ \therefore $ we have the non trivial solution to $ \chi (x) $

$ \chi(x) = c_n cos \big( \frac{ n \pi}{h} x \big) $

for $ 0 \leq x \leq h $

Now for the time ODE

$ \frac{1}{ \beta } \tau’(t) +\lambda \tau(t) =0 $

we know $ \lambda > 0 $ thus

$ \frac{1}{ \beta } \tau’(t) = - \lambda \tau(t) $

$ \tau’(t) = - \beta \lambda \tau(t) $

We notice this takes the form $ \frac{ d}{dx} e^{-\beta \lambda x} = -\beta \lambda e^{-\beta \lambda x} $ Therefore

$ \tau(t) = C e^{-\beta \lambda t} $

Plugging in our value for $ \lambda $

$ \tau(t) = C_n e^{-\beta \big( \frac{n \pi}{h} \big)^2 t} $

Combining our two seperate ODEs into one PDE

$ T(x,t) = \beta \chi(x) \tau(t) $

$ T(x,t) = c_n cos \big( \frac{ n \pi}{h} x \big) C_n e^{-\beta \big( \frac{n \pi}{h} \big)^2 t} $

Combining constants

$ T(x,t) = c_n cos \big( \frac{ n \pi}{h} x \big) e^{-\beta \big( \frac{n \pi}{h} \big)^2 t} $

We use the initial condition

$ T(x,0) = f(x) $

$ T(x,0) = c_n cos \big( \frac{ n \pi}{h} x \big) e^{-\beta \big( \frac{n \pi}{h} \big)^2 (0)} $

we can see the exponential term goes to zero therefore we’re left with

$ T(x,0) = c_n cos \big( \frac{ n \pi}{h} x \big) $

to determine what $ c_n $ is by integrating

$ c_n = \int_0^h f(x) cos \big( \frac{ n \pi}{h} x \big) dx $

we set $ T(x,0) = f(x) = T_1, 0 \leq x \leq 0.8h$ and $ T_2, 0.8h < x \leq h $

HOLY FUCK

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mathematicianadda · 4 years

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Geometric Similarity of Functions https://ift.tt/3jnkPC3

I am a 16 year old high school student and recently I have written a paper on a numerical approximation of distinct functions. I have shown my teachers this and they do not understand it. My questions: Is this a valid theorem to use to estimate functions with differently based functions? Has something similar already been created? Is it all useful/publishable? Any tips on how to improve? I will give an outline but you can find it here: https://www.overleaf.com/read/xjqhfgvrcrbj

Definitions

Geometric similarity refers to the dilation of a particular shape in all its dimensions. Proofs of geometric similarity are included in congruence proofs of triangles with AAA (Angle-Angle-Angle) proofs. Knowing the sizes of all sides of both triangles: $\triangle{ABC}$ and $\triangle{A'B'C'}$, to find the dilation factor and prove geometric similarity the following must be true: $\frac{\mid A' \mid}{\mid A \mid} =\frac{\mid B' \mid}{\mid B \mid}=\frac{\mid C' \mid}{\mid C \mid}$.

Interpreting functions as shapes on the Cartesian plane and using geometry, geometrically similar functions can be calculated. Analytically this would imply for a function $y=f(x)\; \{x_0\leq x \leq x_1\}$ a geometrically similar function would be of the form $ny=f(nx)\;\{\frac{x_0}{n}\leq x \leq \frac{x_1}{n}\}$ where $n\in {\rm I\!R}$. This is because the function is scaled by the same factor in the $x$ and $y$ direction thus would be geometrically similar.

$y_1=\sin(x)\;\{0\leq x \leq 2\pi\}$ and $y_2=\frac{1}{2}\sin(2x)\; \{0 \leq x\leq \pi\}$" />

However to compare two functions which are distinct, multiplying $x$ and $y$ by $n$ will not suffice for proving similarity. The formula to find the dilation factor can be used to prove similarity between two functions. By describing a function geometrically it has three superficial 'edges' which can be represented as sets. Two of the edges are the two axis $x$ and $y$. The length of the side '$y$' is the $\max \{ f(x) : x = 1 .. n \}-\min \{ f(x) : x = 1 .. n \}$ and the length of the side $x$ is $b_1$-$a_1$ where $b_1$ is the upper bound and $a_1$ is the lower bound. Finally the third side of the function will be the arc length over the interval $\{a_1\leq x\leq b_1\}$. Another characteristic for two shapes to be geometrically similar is the area is increased by the dilation factor squared.Thus from the formula for the dilation factor for two similar triangles the following theorem can be derived:

Theorem Let $y_1\;\{a_1\leq x \leq b_1\}$ and $y_2\;\{a_2\leq x \leq b_2\}$ be functions whose derivative exists in every point. If both functions are geometrically similar then the following system holds: \begin{equation} \frac{1}{\big(b_1-a_1\big)}\int_{a_1}^{b_1} \sqrt{1+\bigg( \frac{dy_1}{dx} \bigg) ^{2} } dx= \frac{ 1 }{ \big(b_2-a_2\big) } \int_{a_2}^{b_2} \sqrt{1+\bigg( \frac{dy_2}{dx} \bigg) ^{2} } dx \end{equation} \begin{equation} \frac{1}{\big(b_1-a_1\big)^2} \int_{a_1}^{b_1} y_1 dx= \frac{1}{\big(b_2-a_2\big)^2}\int_{a_2}^{b_2} y_2dx \end{equation}

Similarity Between Distinct Functions

When describing a function as distinct it denotes that the functions have different bases, i.e. sinusoidal and exponential. As mentioned above, for geometric similarity to exist of a function $y=f(x)$ the resultant function will become $ny=f(nx)$. However if comparing functions of different bases, equations (1) and (2) are necessary to find the bounds of similarity. For example, the problem:

Find the bounds $b$ and $a$ where $e^x\;\{0\leq x\leq 1\}$ is similar to $x^2 $.

To see examples go to the above link. Any help would be much appreciated and apologies if this is crude mathematics.

from Hot Weekly Questions - Mathematics Stack Exchange hwood87 from Blogger https://ift.tt/2FYTbOm

#Blogger

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juniperpublishers-robotics-blog · 5 years

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Research on Electromagnetic Driving Robot Fish-Juniper Publishers

Abstract

Based on the actual physical appearance of the tuna and the wave equation derived from observing its swimming locomotion, this paper proposes a new idea on the simulation and modeling of the robot fish and a novel way of thinking on kinetic analysis. The motion law for each joint of the robot fish was first obtained through the discrete fitting method and the motion law was subsequently used as the basis for the robot fish electromagnetic drive control signal. The fluent flow field analysis software was utilized. The meshing of the fluid around the fish body was performed by using the dynamic grid technique. The surface pressure values of the fish body during the steady state forward movement were analyzed, and the values of the driving force of each part of the fish were obtained. It was then determined that the electromagnetic drive caudal fin robot fish was the optimal design. Based on the idea of digital-analog conversion, the drive control signal waveform was digitally discretized, then the C51 single-chip microcomputer and DAC digital-to-analog converter was used for conversion. The OPA544 op amp chip was then used to simulate the amplified output in accordance with the control signal waveform. Electromagnetic coil drive signal with variable frequency and voltage can thus be achieved.

Keywords: Robot fish; Electromagnetic drive; Fluent flow field analysis; Kinetic analysis; Control signal; Swimming experiment

Construction of the Robot Fish Model

Determination ofthe external physical characteristics of the fish body and the number of joints

A real tuna fish was selected, the total length was measured and the external curve of the fish body was fitted; the following data can be obtained:

Total length l = 260mm

Where R(x) is the function of the longitudinal fish body curve, r(x) is the function of the transverse fish body curve

In the selection of the number of joints of the robot fish, the higher the number of joints, the higher the degree of fit between the swimming curve and the known fish body wave equation, the closer it is to the real fish-swimming situation. On the other hand, the cumulative error of the series structure and the complexity of the structure should also be considered. Usually the range for the number of robot joints N is 2-10. The number of robot fish joints designed in this paper is N = 6.

In the design of the length of each joint, considering that the movement of the fish body is mainly concentrated in the rear half of the fish body, therefore the first half 130 mm of the fish can be viewed as not swinging. Also, according to the actual measurement, it is known that the fish tail length is 47 mm. Based on the joint size parameter optimization design method proposed by Chai Zhikun [1], except for the tail joint, the proportion of the remaining five joints should be 83: 67: 59: 53: 48. The length of each joint after calculation and rounding is: joint 1 is 22mm, joint 2 is 18mm, joint 3 is 16mm, joint 4 is 14mm, joint 5 is 13mm, and the joint 6 (caudal fin) is 47mm long.

Calculation of the mass and moment of inertia of each part

Based on the above geometric parameters of the fish, the mass of the various parts of the fish can be calculated:

Where, ρ is the fish density (kg/m3), which can be approximated as the density of water, a, b are the abscissa values of the starting point and the end of each joint.

During the swimming process of the fish, due to interaction with the surrounding fluid, the fish has a virtual mass. According to Lighthill's "Large-amplitude elongated-body theory of fish locomotion,” the equation for calculation of the virtual mass of the fish body cross-section is [2]:

Where ρf is the fluid density, β is the virtual mass coefficient; when there is no fish fins on the fish cross-section, β = 1.

As the robot fish cross-section can be regarded as an oval with the long axis of 2R (x), the short axis of 2r (x), the inertia of each part around the z-axis can be calculated according to the following formula:

Where m is the sum of the mass of each joint and the virtual mass, and do is the length of each joint. As this robot fish has six joints, therefore there is seven parts in total. The following is the calculated mass of each joint, virtual mass, and moment of inertia:

Kinematical Analysis of the Robot Fish

According to the observations of the tuna fish by Donley and Sepulveda and others [3], the fish body wave equation corresponding to the swimming locomotion of this type of fish is: h (x, t )= H (x) sin (ωt - kx) , where, h (x, t) is the lateral displacement of the fish. H (x) is the envelope equation of the fish’s swimming locomotion. ω is the swing frequency of each fish joint, k is the wave number of the fish body. k = 5.7/1 . H (x) = α1 + α2 + α3x2 , Where α1 = 0.02/ α2 =-0.12 α3 = 0.3/1.

When the frequency of each joint is 2Hz, the swing period is 0.5s. The swing cycle is divided into 10 moments. With the starting point of each joint as the center, the joint length is the intersection of the radius and the fish wave equation at that moment. The coordinate position of the six joint starting points at each moment was obtained (In this coordinate system, the front end of the fish head is the coordinate origin, the forward direction of the swimming direction is the negative direction of the x axis, and the direction of the lateral displacement is the y direction, dynamic body coordinate system). The following Table 1 is the calculated coordinates of each joint starting point at each moment.

The swing movement law of each joint can be obtained by using curve fitting toolbox cftool of Matlab, As shown in Figure 1.

Fluent Flow Field Simulation and Kinetic Analysis of Robot Fish

The Cartesian coordinate system was established: the negative direction of the x-axis was the advancing direction of the fish, the z-axis was the transverse displacement direction of the fish, and the y-axis was the longitudinal displacement direction of the fish. Ignoring the force in the y-axis direction of the fish (gravity, buoyancy, lift due to pectoral fins, and others), and considering only the x-z plane force on the body during the advancing process, the fish is only subjected to two forces: the resulting thrust due to swing water and the resistance in the forward movement.

According to the "large-amplitude elongated-body theory of fish locomotion,” the fish's thrust is the component of the hydrodynamic force generated by the fish's swing in the forward direction [4]. However, due to the complexity of the flow field, there is no standard and uniform thrust formula. It is difficult to obtain relatively accurate simulation results. Therefore, the fluent flow field analysis method was used. The dynamic grid technology was used for the meshing of the fluid around a certain area of the fish, and turbulence model was used to obtain the solution. Hence, the forces exerted by the surrounding fluid on the fish body during the steady state forward movement can be obtained by simulation (surface pressure and viscous force). Further, the thrust formula can be obtained.

As shown in Figure 2a is a three-dimensional model of the extracted robot fish, Figure 2b is a plane view intercepted at y = 0 after the static grid division is completed. In the choice of computing domain, on the one hand, the length of the calculation area must be large enough to meet the needs of the continuous advance of the robot fish; on the other hand, the smaller calculation area can improve the calculation efficiency. After several trials, the calculation domain was selected as rectangular, the length, width, and height were selected to be 10 times of the corresponding maximum size of the biomimetic robot fish.

In the above model, the triangular/tetrahedral mesh is used to divide the area/body domain of the region. Since the grid boundary changes with time according to the variation trend of the fish wave equation, it is necessary to define the grid as the dynamic grid in the computing domain. Also, the spring smooth method and the local mesh reconstruction method were used to ensure the quality of dynamic grid. The following is the dynamic grid parameters: the spring coefficient is 0.2, the boundary node relaxation factor is 0.4, the maximum length of the local reconstruction is 0.021, the minimum length is 0.0004, the maximum cell skew rate is 0.76, and the size function is activated, and the remaining values are default.

On the boundary condition, the boundary conditions of the six surfaces of the rectangular parallelepiped and the surfaces of the biomimetic robot fish are all non-slip condition on the wall surface, but the dynamic region is defined by the UDF (User Defined Functions) function. For the robot fish model, DEFINE-GRID-MOTION Macro was used to define the movement of fish. Take the movement at the end of the joint as an example; the custom program of this dynamic region can be defined as follows:

#include "udf.h”

#define PI 3.1415926

DEFINE_GRID_MOTION (guanjie6,domain,dt,time,dtime)

{

Thread *tf = DT_THREAD(dt);

face_t f;

Node *v;

real NV_VEC(omega), NV_VEC(axis), NV_VEC(dx);

real NV_VEC(origin), NVVEC(rvec);

int n;

SET_DEFORMING_THREAD_FLAG(THREAD_T0(tf));

begin_f_loop(f,tf)

{

f_node_loop(f,tf,n)

{

v = F_NODE(f,tf,n);

if (NODE_POS_NEED_UPDATE (v))

{

NODE_POS_UPDATED(v);

NV_S(omega, =, 0.0);

NV_D(axis, =, 0.0, 1.0, 0.0);

origin[0] = 0.213;

origin[1] = NODE_Y(v);

origin[2] = 0;

omega[1] = 0.2587*PI*PI*sin(4*PI*time-0.3848+PI/2);

NV_VV(rvec, =, NODE_COORD(v), -, origin);

NV_CROSS(dx, omega, rvec);

NV_S(dx, *=, dtime);

NV_V(NODE_COORD(v), +=, dx);

}

end_f_loop(f,t)

}

In the choice of the calculation model, because the robot fish movement belongs to the large Reynolds number movement mode, therefore the transient three-dimensional turbulence model was used for solution. Due to the wide range of applications (especially the situation of high Reynolds number), economic, reasonable accuracy, and other characteristics, k-epsilon turbulence model was used.

Since the surface of the fish is divided into seven parts according to the joint position when drawing the three-dimensional map, the force component of the wall area of the fish body in the specified direction (the advancing direction of the robot fish) can be obtained in the processing of the results.

In Table 2, x(p) represents the fish surface pressure value, and x (v) represents the viscous force value. It can be seen from Figure 3 and Table 2 that the positive and negative surface pressure of the fish in the z-direction is cancelled in one cycle, and the viscous force is usually in the resistance state, and its value is small relative to the surface pressure, the difference is 2-3 orders of magnitude. Consequently, it can be neglected when fitting the thrust formula. The thrust value of each part of the fish body can be obtained as follows (N):

Fish front end: T0 = 0.1094sin(4nt + 5.098)-0.0410 ;

Joint 1: T1 = 0.0875sin (8πt - 0.3221)-0.0477 ;

Joint 2: T2 = 0.1639sin(4πt + 0.9005)-0.0835 ;

Joint 3: T3 = 0.2328sin(4πt + 0.2085)-0.1045 ;

Joint 4: T4 = 0.2410sin(4πt-0.3612)-0.1170 ;

Joint 5: T5 = 0.2247sin(4πt + 5.575)-0.1360 ;

Joint 6: T6 = 1.638sin (8πt +1.881)-1.571 .

The influence of viscous resistance (frictional resistance) is small and negligible in the case of fish swimming locomotion that has a large Reynolds number (As demonstrated in Table 2). The pressure resistance, which played a major role, can be calculated by the standard Newton equation [5]:

Dp = 0.5CpSpU 2ρ

WhereCp is the resistance coefficient, usually 1.2; Sp is the effective area, in the practical application ( is the displacement) can be used as the effective area of action as the fish moves in the water, because the density of fish and water are basically the same, Sp =(m/ρ)2/3 [6].U is the average speed of swimming locomotion; ρ is the density of water.

It can be concluded that the rear joint (caudal fin) provides the maximum thrust, accounting for more than 80% of the total thrust, so the caudal fin drive mode is selected; electromagnetic drive has advantages of compact structure, pollution-free, high reliability, and large driving force, and accurate process control can be achieved. The cyclic oscillation-driving mode is selected to be the electromagnetic drive.

Research on Control Signal of Electromagnetic Drive Robot Fish

Signal transformation

The standard electromagnetic coil current waveform was digitized, and then through the MCU processing the signal was transferred to the DAC module for digital-analog transformation to reproduce the analog control signal. The realization of the control signal was the acquisition, storage, and reproduction of the standard control waveform, which was used to drive the electromagnetic coil that further drove the caudal fin movement. The basic flow of the control drive circuit is shown in Figure 4. The components used in the Multisim simulation are described in Table 3.

Simulation circuit has been completed. A waveform generator and an oscilloscope were used as waveform generation and acquisition circuit (sinusoidal wave as an example) in Multisim. The amplitude was set to 255V, the cycle was 1s, as shown in Figure 4 & 5.

After the standard sine wave was obtained, the collection and digital discretization was required. As shown in Figure 6, the oscilloscope in Multisim provided the function of waveform sampling. Click "Save” on the oscilloscope in the figure, the settings of the sampling time interval and number of sampling points of the oscilloscope will be displayed. To ensure signal accuracy, 120 points were sampled in a cycle. The sampled data was set as a binary output text file.

After the sampling data was obtained, start the singlechip microcomputer-programming interface of the completed circuit simulation platform. The collected discrete value's output was through the single-chip microcomputer one by one cyclically. The programming interface and the program are shown in Figure 7.

After programming, the program is successfully downloaded to the single-chip microcomputer, click on the "start simulation,” back to Figure 5 interface and open the two oscilloscopes in the figure, waveform as shown in Figure 8 can be observed (Figure 9).

Oscilloscope XSC2 displays the 8-bit DAC direct output of the control signal; the oscilloscope XSC1 displays the output control signal of the DAC output after passing through the polarity converter. For the frequency of the waveform, delay value in the parentheses of the delay statement in the program can be adjusted. When calling the delay program, the singlechip microcomputer will adjust the output hold time of each digit according to the delay value in parentheses, and thus the frequency of the whole waveform can be adjusted. The amplitude needs to be matched based on the characteristics of the power amplifier module in the actual circuit and the DAC module in the actual circuit has a reference voltage setting [7-10].

The simulation program as shown in Figure 3 & 4 was then transplanted to the Keil single-chip microcomputer programming software. After the program was successfully compiled, the crystal resonator frequency was set to be the crystal resonating frequency of the single-chip microcomputer, which was 12Mhz. The Keil debugging interface was started, break points out of the cycles were set, and the delay parameter in the brackets of the "delay ()” statement was adjusted continuously until the output cycle reached 1s, namely, control signal so that the caudal fin swing cycle was 1Hz. The debugging interface of Keil as shown in Figure 4 & 5 was entered. In every click of "run”, the program executed a cycle, and "sec” increased accordingly, and the added value was the time required to output a complete waveform [11-14].

After debugging, the program was downloaded to the single-chip microcomputer. The DAC output control waveform can be observed by connecting the DAC module output with the oscilloscope. It can be seen that the actual circuit output control signal and the simulation circuit output signal were exactly the same.

Signal adjustment

The DAC module was connected with the STC89C52 development board with the downloaded test program. The output of the DAC module was connected to the input of the power amplifier module. The DAC module and the power amplifier module shared the DC ± 15V power supply. The overall circuit of the control platform is shown in Figure 10. The "delay ()” delay parameter was modified in the Keil programming debugging interface, the signal frequency was the electromagnetic coil operating frequency.

Experimental Study

Development of experimental prototype of robot fish

The external fish body shape curve was imported into the Pro/E three-dimensional modelling software; closed fish surface shape can be obtained by the use of "variable crosssection scanning” tool. Then, "thickening” operation was performed on the surface, with the set thickness of 1.8 mm. The closed fish body shell was thus obtained. Considering the fish body assembly, weight allocation, fish body was designed into a removable form, as shown in Figure 11.

After modeling, the fish body model was obtained with the use of precision 3D printing, the printing material was resin. The polypropylene caudal fins were tailored to thin caudal fins according to the caudal fins of the actual tuna fish, and the flexible caudal fins were then fixed on the connecting parts as shown in Figure 12. Underwater experiments are shown in Figure 13.

Electromagnetic drive robot fish underwater experiment

Fish body linear curve swimming experiment: In the experiment, the swing frequency of the fish when swimming along straight lines is set within 1-8Hz. The amplitude of the drive voltage is set between 2-6V. After adjustment of the fish weight, the control signal waveform acquisition was used to generate 1-8Hz sinusoidal signal, the signal waveform are shown in Figure 14. By adjusting the frequency of the control signal through adjusting the delay time of the sinusoidal output program, the obtained experimental data is recorded in Table 4, the distribution diagram of velocity - frequency under different voltage amplitude is shown in Figure 15.

The above data is shown in Figure 15. The greater the amplitude of the driving voltage signal, the faster the fish swims. At different drive voltage amplitude, the best swing frequency was basically the same; at about 2Hz, the fastest swimming speed was achieved. As the high drive voltage heats up the coil, it is desirable to operate at low voltage, hence at 2 V voltage, 2Hz, the swimming speed of 81.74mm/s was selected as the best electromagnetic drive control signal [15-18].

Conclusion

Based on the actual physical appearance of the tuna and the wave equation derived from observing its swimming locomotion, the motion law of each joint of the robot fish is obtained through the discrete fitting method, and the motion law is utilized as the basis of the control signal of the electromagnetically driven robot fish. With the utilization of the fluent flow field analysis software, a dynamic grid technique was used to mesh the fluid around the fish body. By analyzing the surface pressure value of the simulated fish body in the steady state of forward movement, the thrust values of each part of the fish body were obtained. Hence, the electromagnetic-driven caudal fin robot fish was determined as the optimal design. Based on the idea of digital-analog conversion, the drive control signal waveform was first digitally discretized. Then, it was converted through the C51 single-chip microcomputer and DAC digital-to-analog converter. Finally, the simulated amplified analog output was obtained through the OPA544 op amp chip in accordance with the waveform of the control signal. Electromagnetic coil drive signal with variable frequency and voltage was thus achieved. An electromagnetic driven caudal fin robot fish prototype was successfully produced, a number of robot fish underwater experiments were designed, and a series of experimental data was obtained. Based on the linear swimming experiment of the robot fish prototype, it was obtained that the optimal value of the electromagnetic drive signal is 2V, 2HZ when the electromagnetic driven caudal fin robot fish was at a relatively high speed. Based on this, the next steps of the study lays on the following two aspects:

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craigbrownphd-blog-blog · 6 years

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Some Irresistible Integrals, Computed Using Statistical Concepts

#ICYDK: Below are a few integrals that you won't find in textbooks. Solving them is a good exercise for college students with some advanced calculus training. We provide the solution, as well as a general framework to compute many similar integrals. Maybe this material should be part of the standard math curriculum. Here, p, q, r are positive real numbers, with q larger than p. The Gamma symbol represents the gamma function. It is possible that these results are published here for the first time. These are known as Frullani integrals, although the ones mentioned here are not covered by Frullani's theorem, nor any recent generalization that I am aware of. Indeed, AI-based automated integration platforms such as WolframAlpha can not find the exact value (only an approximation) while they are able to compute standard Frullani integrals exactly. My approach to derive the exact values is different from the classical approaches, as it relies on the statistical concept of expectation, possibly leading to interesting areas of research. How to compute such integrals? These integrals are a particular case of the following main result, proved in the next section: where g(x) / x tends to 1 as x tends to infinity, and f is a bounded function with a finite expectation. Some additional conditions may be required, for instance the fact that there is no singularity point in the above quotient, and that g(x) has a lower bound that is strictly positive. The expectation of f, also called average value, is defined as For instance, if f(x) = sin(SQRT(x)), then the expectation exists, and it is equal to E(f) = 2 / Pi. (Prove it!) The main result introduced at the beginning of this section, is rather intuitive but needs great care to prove it rigorously, including correctly stating the required assumptions on f and g to make it valid. Some cases might require working with non-Riemann integrals. Here we only provide the intuitive explanation. Proof of the main result (sketch) Here p, q and n are integers, with q greater than p. We are interested in the case where n tends to infinity. We approximate integrals using the Euler-Maclaurin summation formula. The approximations below become equalities as n tends to infinity. We used the classic approximation of the harmonic series to make the logarithm terms appear. Note that for large values of k, g(k) is asymptotically equal to k. This was one of the requirements for the formula to be valid. We also have: Using the change of variable y = x / q in the first integral, and y = x / p in the second integral, we obtain: Let us remark that: * q / g(qy) is asymptotically equivalent to 1 / y (for large values of y) * p / g(py) is asymptotically equivalent to 1 / y * both integrals diverge, so the impact of small values of y eventually vanishes in each integral separately * the difference between the two integrals converges In view of this, we have: This concludes the proof. Related Problems * Four Interesting Math Problems * Curious Mathematical Problem * Two Beautiful Mathematical Results - Part 2 * Two Beautiful Mathematical Results * Number Theory: Nice Generalization of the Waring Conjecture * Yet Another Interesting Math Problem - The Collatz Conjecture To not miss this type of content in the future, subscribe to our newsletter. For related articles from the same author, click here or visit www.VincentGranville.com. Follow me on on LinkedIn, or visit my old web page here. DSC Resources * Book and Resources for DSC Members * Comprehensive Repository of Data Science and ML Resources * Advanced Machine Learning with Basic Excel * Difference between ML, Data Science, AI, Deep Learning, and Statistics * Selected Business Analytics, Data Science and ML articles * Hire a Data Scientist | Search DSC | Find a Job * Post a Blog | Forum Questions https://goo.gl/RXZFDe

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devphilamaths · 3 years

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Solve sec(pi/2+theta) | sec(pi/2 +x) | sec pi/2 + x formula, Find Exact value sec pi by 2 + x

#Solve sec(pi/2+theta)#sec(pi/2 +x)#sec pi/2 + x formula #Find Exact value sec pi by 2 + x #sec(pi/2+theta)#se #si #sec #sin #cos #ten #cot #math #nda #cds #pi/2 #180 #90 #1200 #classy #gdlmx #gd

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top1course · 5 years

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How To Sell Millions Without Ever Opening Your Mouth! Copywriting Secrets Simple 7-Step Formula Pt.1

youtube

Is it possible to sell millions of dollars without opening your mouth, well is he by the end of this, video I’m going to prove to you, yes is very, possible today we are going to talk about cooperating, secrets I have gave joining me, am I, cole director of social media, and gave manages all might, all my social media especially YouTube, so if you see that how I’m blowing up on YouTube how we went from, in the beginning of this year, with 100000 subscribers now almost the end, office year we’re going to hit a million Subs, subscribers, 10 x growth, is because, the gates help, okay so, he is going to act as almost as an audience or entering me on Sundays covered in question, i’ve been doing this for a long long time so sometimes the concepts, the preceptor I know that it’s, soul, so be with me for so long as I can conscious competent but maybe gave for him it’s a, Copper dick is a new thing for you right, is a new world, he will almost asked questions on your, behalf, now before I go into the, 7 step, on how to create, gray coffee, let’s just defy what cop rating, because I believe most people they could confuse what comprehend, you think copywriting, he’s about, writing, right and we hear I hear this a lot, odana I’m not a writer, my English is not, or I am not very, creative, what is not a, copywriting, what about closing, the way I Define carpeting is, closing in, print, okay closing in print or, closing the introvert right you can close using instead of using your, spoken words, you’re using your written words, imprint, to persuade convince and employing someone, want to make a Vine perch, do do a move to to do a make a decision whatever it might be what a goal that you might have, That’s all that all cooperating is, so if it’s closing you means it is about communication, it’s not about writing, right it’s not about, so don’t be intimidated by the whole concept of, writing cuz I flunked English twice when I was, in high school, but it doesn’t stop me from creating coffee, does May, millions and millions of dollars, deep Eddy questions, so I would say I see a lot in the YouTube comment where people they don’t really, noah copywriting is and how to, pi, and how, actually the origins of copywriting how, for any design or anything part of the business at all goes down to, word of mouth and being spoken and being read, correct, so how would someone I would, say how would they, Shift their thinking to start from a.

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I’m just doing some research about my company, i want to do some marketing can pay you your my best and most loyal customer, can I have, 30 minutes, and if they do you can get them some morphe, certificate, starbucks whatever it is doesn’t matter what it is, and you talk, and you ask him questions like hey, you look before you, you do business with, company, what was your problem, right why did you choose us, and why do you stay with, they would tell you all these things, and which one of you want to record it ask for the commission you record it, and this is so easy, you take what they say, and you turn it into, coffee, so let’s say, give me example give me an example of something that people, alternate so let’s say, Yeah like, i can’t find any leads to my clients, and I’m really struggling finding leads and customers I go to these conference, and I talk to other people but none of them that really follow, oh okay so, leads all that so exactly that’s a pain that they have less a yellow product to solve their problem, easley, he’s the light. Could be a potential hotline, it could you want to call I get one of those, benefits team in a bullet point, how to quickly and easily generate more leads for your business without going to these, networking, boom right there you have, how to cook a medium talking about exactly what they say and it turned it into your copy, you do that for a number of customers song you are good, to go, see how easy this is not as complex as you think, right they would tell you your, Your customers they want to buy from you, but what do you want to know is the assurance that you know what, are going through, they want the shoes that, you’ll probably service can help himself., number one, look at 3 I want to teach you, 7 steps that you can take, every single time you buy coffee, you go through these steps, i Promise You by the end of the seven steps, you’ll coffee would be so much more powerful so much more compelling, . one and that is, identify your ideal, this is the most important part, because most people when did Duke operating I used to make this mistake, you so excited, you get you could tell it on the computer and you stopped typing, no you do not do that, you need to First understand truly understand, who your ideal customers, now that the way that is fine I do customers, Is three things, number one your ideal customer, they have a need, volume, so if I am a vegan, if you try to sell me steaks, we got a problem here right, you got to sell stick to a stick lighter so I should have a meeting at 1 for your part at this number one, number 2 is IHOP the ability, biopod, so I want it but also I could pay for, and number three if I have the authority, dubai Abaya service, so if, i am selling something to this a husband and wife, and I’m only talking to husband but we leased a wife that makes it, decision, that is not good, so it’s, they have their Nita want they have their building at the store if that’s the ideal customer, now once you know this, who do custom is even Facebook, This so much.

Targeting that we can do nowadays, with, he could be your interest it could be, gucci belong to it could be your age, ditto, hundreds of variations hundreds of, reference point hundred things you can do to Target, what you know exactly who you talkin bout that is, now you go, talk to them in your copy, one-on-one like this, and that’s what you do, i believe, good copy, 80% of research, and 20% lighting, so if Andrea Spinelli a month, sometimes sometimes 3 weeks a month, working on, coffee walk away campaign, you should spend like 2-3 weeks just in, before you write a single, you should have a very clear idea, exactly what the offer is, all these things, before you, i think another helpful tip, tipsy food that you gave the other day was, you really want to think about, who that ideal customer is in that, Customer profile and really Envision you talking to them one-on-one, like if you were just sitting down talking to them casual, don’t try to use very big mumbo jumbo, words just casual conversation as if, if you’re really trying, talk with them you want to talk to him on a one-on-one basis right you don’t want to be like right now I’m communicating with, with you, at any given time, your reader, audience, oil plus, they are, reading your coffee, the off, watching a video, by themselves, usual Toyota 101 connection, so I’m talking to you, i’m not talking like, speaking to a thousand a million people I’m not talking to groups, that public speaking is, completely different, what way comes to coffee, you want to be very personal I want to be like, one on one, assuming now you know your customer and I could tell you the amount of money you make is indirect, Proportion to how well you understand your car, i’ll say that one more time, the amount of money you make, it indirect proportion, 2 how well you understand, you know this business to fail, that, be entrepreneurs the business owner if they have this idea, on this widget I have this adventure time I am so excited about it, they go they take it to the market place nobody gets, because they don’t understand their, ideal, costume, it’s not enough to say hey my mom likes it my wife likes, did you say, when people typically say while everybody is my ideal cussed, i want to sell to every the white there is a bigamist, he kanaka even company ice Vegas Apple, even a couple days because she likes a Walmart that serves, everyone is my costume, no, yep Walmart, right you have so many your target, They all have different demographics so you know one could say that everyone is, especially for like small business are you mija medium-sized business you got to think you got to narrow it down, very much narrow down because the more you can narrow down the more personal, you can make it message, in Dominion, one of the best reactions you could get and I love this one man to Allen, he said let’s say you have an idea you can offer you want to test it out is that it is, you hang out with your target market, can you bring your offer and you show it to them, do you know what checkout is at 4, right, and if they say if they read the ad and they say to you, this is just as good as pretty good, your ex sucks, images no good, okay, i think if you have this out, No that’s no good.

Do you actually want is they read at, how can I buy some, is my credit card I want to buy some that’s when you know you got something, okay you need the real reaction because people vote with their wallet, don’t listen to Just what they say watch what they do they vote with a wallet, when did minute they put some money online okay I’ve got something, before that is all just like lips, so that’s number one research, knowing identifying do I do, step number, q and that is create an accessible offer, a compelling offer, this is probably the one of the most important steps Indian, higher formula, because you look at most businesses, the biggest challenges, they sound the same as everybody, the office not very compelling, oh you know what, this is my glock my competitors, And didn’t cuddle with you as well, maybe I’m a little bit better and maybe a little bit cheaper, what day is not, huge differentiator, so with what you do not think about how could you make, your offer as compelling as pause, i was talking with, one of a copywriter, and he was asking me well then do you believe, what is Montour, the audience, all the message, i say without a doubt, the market, i could have the greatest message, what I said the wrong people, what is distribution, we will just like yesterday last night we having dinner right, love’s like Rhett meet and all that doesn’t like Steve has so much, so I could have the great like lobster and crab and a great message this is awesome man, i don’t know it’s the right message, but wrong Market, play doesn’t work, so when it comes to offer you have to think about, What is it that your audience truly wants, right what is it that they did cleaning, what more what pain, what frustrations that they have they they just they’re sick and tired of that, so when you can come up with an offer this so irresistible, is easy to sell, in fact a great, offer, you can see, less, anyone still sell, you’ll need to use as many words, let me give you perfect, let’s say it’s a gate he has $100 bill, okay, can you give me $10 in exchange of $100, give me ten bucks I’ll give you $100 right now, that isn’t resistible, right there, what am I doing there is called selling money at a disco, so if you’re selling any kind of, business Improvement, offer or any offer that helps people to make money save money on investment, the concept, money saving money at a discount, Spandex with me, i’m going to teach you how to make, why, right, essay example people who teaches real estate investing, hey spend $5,000 on Discord, i’m going to teach you how to generate you know, 10000 a month in rental income, oh wow that’s a no-brainer, that’s only money at a discount, that’s an irresistible offer, maybe you’re not in the education space maybe you have, do some physical product how do you do this, in so many ways, maybe your irresistible offer is, a free 30-day trial, okay if you have seen those infomercials, where they have those are, skin care, the trifle 30 they see how you like it, example, or a strong guarantee, irresistible, add certain bonuses, jamaican resistant material quick story, the one time many years ago I was late night watching TV, that’s all this, eva Marcille just popped up my loved ones, And this this guy is demonstrating he has a pair of shoes on the desk and his.

Cutting it with a knife, and I’m like, this is cool, what is this is a was talking about house shopping Ibis, is it yeah that’s cool, okay I got the knife and then he cut actually a can, with with a knife, holy s*** this is cool, not going to use it but it’s cool right, so I was watching a commercial and then people like and then he has a lot of, people coming onto the TV talk with a testimonial, using this naive housewife, like people do any kind all kinds of, background that uses knife, and he said well you know what, can you buy today, you don’t just get one night, you get to knife, homelite holyfuck, this is awesome and then he goes on, this is the big knife, But what about All the Small Things, and then you get this knife, and then you get this knife, and then you get this knife, and then before you know what he’s talking about like, twenty f****** divinized, i don’t like this is crazy, this is awesome before the infomercial I don’t need a knife, i don’t need a knife, but offered it is so dramatic and so cool, and then he just dropped the mic and see if you order within the next 5 minutes, you don’t get, these 29th, we going to send you two sets of knives, that’s it I’m done right call number call immediately and by that is, list of offer it when did it the office so irresistible, logic goes out the window, outside bought to set up nice actually the lady cuz I’m like, what the f*** I need like 40 pairs of nice voices stupid I just dumb, It was at that moment it was so compelling, so think about how you could use that, maybe sometimes it’s, faster shipping Amazon that’s good Amazon, proserpine you get a fast, i don’t want to wait when I could also speak about how, you were saying how Amazon offers that fit did you pay them Accenture, that’s right cuz Amazon if you have Prime Membership Eva Prime member, hello, NC, that, with fine what they’re doing is essentially they’re charging you for membership fee, so they can buy more, applicants applying to get your product faster, so think about what Amazon does, and how they encourage you, a simple thing in your system, and it’s not like a hundred bucks or so right they raise the price people, put up a hole stink but everybody’s, bill pay the higher, i have seen people like us now, Amazon, hundred, housing people that you would think Prime is like, it’s a, luxury, everything nice and membership it’s not like a, necessity you’re paying, a hundred bucks you get some benefit in Amazon and, ended a video and entertainment, william painter sponsorship, so you’re paying them so I can buy more stuff, you got understand, i don’t know people, who are like, literally you would think like lower-income, the living in like, trailer park and stuff like that, neither would not think they would spend money on Amazon and yet their Prime members, it’s the last thing you think about, but to them, destify member, and I like you spend $100 a year, but to be a Prime member, is Solaris, it’s like a no-brainer, when you are a Prime member you thinking while you know, i’m going to pay for the shipping anyway, but I’m going to get my stuff faster, Plus you’re not getting all these other benefits I get Amazon need a video I can.

Music, and you may watch email watch it but you like the fact that, you have access, right, that’s irresistible,

from WordPress https://top1course.com/how-to-sell-millions-without-ever-opening-your-mouth-copywriting-secrets-simple-7-step-formula-pt-1/

#People & Blogs #become a copywriter #copywriter #copywriting #copywriting course #copywriting exampl

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arrayschism · 5 years

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Testing LaTeX

Step 2: Use LaTeX as normal: MathJax has two different ways to begin a TeX section, $ for in-line TeX and \[ for paragraph formatting. These sections would be closed using $ and \] respectively. The equations show above, are produced by writing this in the text of the post: #Title: Embedding $\LaTeX$ in Tumblr The classic Pythagorean identity is: $sin^2(\theta) + cos^2(\theta) =1 $ The binomial formula which calculates the probability of obtaining k tails when flipping a coin n times, with a assumed probability p for each trial is: $ P(E) = {n \choose k} p^k (1-p)^{ n-k} $ Finally, the normal (Gaussian) distribution defined by mean $\mu$ and variance $\sigma^2$ is: \[f(x; \mu, \sigma^2) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{1}{2} (\frac{x-\mu}{\sigma})^2} \]

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physicscatalyst · 6 years

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Fourier Series formula sheet

Fourier series is an expansion of a periodic function of period $2\pi$ which is representation of a function in a series of sine or cosine such as

$f(x)=a_{0}+\sum_{n=1}^{\infty }a_{n}cos(nx)+\sum_{n=1}^{\infty }b_{n}sin(nx)$

where $a_{0}$ , $a_{n}$ and $b_{n}$ are constants and are known as fourier coefficients. In applying fourier theorem for analysis of an complex periodic function , given…

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testblogplzignore · 3 years

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coffee

The heat equation in three dimensions

$ \frac{\partial u}{\partial t} = \beta \Delta u $

Where $ u $ is an arbitrary function, $\Delta $ is the Laplace operator equaling

$ \Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} $

In one dimension the heat equation simplifies to

$ \frac{\partial u}{\partial t} = \beta \frac{\partial^2 u}{\partial x^2} $

Looking at a cup of coffee with the coffee filled up to height $ h $

We assume the coffee is the same temperature everywhere in the cup before the ice cube (1), cup is the same temperature as the coffee (2), ignore the temperature lost to air hitting the cup and surface of liquid (3), the temperature at a given level of the cup is the same everywhere for that area (4), and the height from the bottom of the cup once the ice cube is added is the same height as before it’s added. The rationale for (1) is that the coffee is roughly the same temperature as heat flows fast in a liquid. The rationale for (2) is that the cup is roughly the same temperature of the coffee depending on the cup, If it’s not then the coffee cools down even faster than our derivation. The rationale for (3) is because the temperature difference of the ice cube is more than the temperature difference of the air. The rationale for (4) is that because the the temperature of a little disc of coffee is the roughly the same across the disc (a consequence pf this is that it makes the coffee’s temperature a bunch of discs stacked on eachother). The rationale for (5) is the height difference of adding an ice cube is mostly negligible when looking at the height change (volume increases as you add ice)

We say

$ u = T(x, t) $

Where $ T $ is temperature of the coffee at a certain height $x$ and time $t $

Therefore in one dimension with those assumptions the temperature of a coffee cup takes the form

$ \frac{\partial}{\partial t} T(x,t) = \beta \frac{\partial^2 }{\partial x^2} T(x,t) $

and has the boundary conditions

$ T(0,t)= T_1 $ which says at the bottom of the coffee in the cup the temperature is $ T_1 $ for $ t > 0 $

$ T(h,t)=T_2 $ which says at the top of the coffee in the cup the temperature is $ T_2 $ for $ t>0 $

$ T(x,0) = f(x) $ which says that the temperature at time zero is a function of height $ f(x) $ for $ 0<x<h$

We assume that the solution takes the form $ T(x,t) = u(x)+w(x,t) $

where $ u(x) $ is called the steady state term or the term that describes the default heat exchange assuming no heat differences are applied to the system

$ w(x,t) $ is called the transient term which applies to when a heat difference is applied to the system and goes away as the system’s temperature equalizes

Plugging into our heat equation in one dimension

$ \frac{\partial}{\partial t} T(x,t) = \beta \frac{\partial^2 }{\partial x^2} T(x,t) $

$ \frac{\partial}{\partial t} T(x,t) = \beta \Big[ \frac{\partial^2 }{\partial x^2}w(x,t) + \frac{\partial^2 }{\partial x^2} u(x) \Big] $

With these new terms the boundary values become

$ u(0, t) = u(0)+w(0,t) = T_1 $

$ u(h, t) = u(h)+w(h,t) = T_2 $

$ u(x, 0) = u(x)+w(x,0) = f(x) $

Work in a frame far in the future

Take limit of $ T(x,t ) $ as $ t \to \infty $

$ \lim_{t \to \infty} \frac{\partial}{\partial t} T(x,t) = \lim_{t \to \infty} \Big[ \frac{\partial^2 }{\partial x^2}w(x,t) + \frac{\partial^2 }{\partial x^2} u(x) \Big] $

By definition $ w(x,t) \to 0 $ as $ t \to \infty $ because $ w(x,t) $ is the transient term that eventually disappears as time goes on (has a weaker and weaker contribution to the temperature)

so in the frame of the future

$ T(x,t) = u(x) $

In this setting $ u(x) $ doesn’t change in time thus

$ \frac{\partial}{\partial t} u(x) = \beta \frac{\partial^2}{\partial x^2} u(x) = 0 $

Using this and the boundary conditions far in the future

$ u(0)=T_1 $ and $ u(h) = T_2 $

Integrating both sides twice

$ \int \frac{\partial^2}{\partial x^2} u(x)dx = 0 $

$ \frac{\partial}{\partial x} u(x) = m $

$ \int \frac{\partial}{\partial x} u(x)dx = \int m dx $

$ u(x) = mx+ b $

$ u(0)=m(0)+b = T_1 $

$ u(h)=m(h)+T_1 = T_2 $

$ m = \frac{T_2-T_1}{h} $

Therefore

$ u(x) = \frac{T_2-T_1}{h} x + T_1 $

Using the fact that $ \frac{\partial^2}{\partial x^2} u(x) = 0 $

$ \frac{\partial}{\partial t} T(x,t) = \beta \Big[ \frac{\partial^2 }{\partial x^2}w(x,t) + 0 \Big] = \beta \frac{\partial^2 }{\partial x^2}w(x,t) $

$ T(0,t) = w(0,t) + u(0,t) = w(0,t) + T_1 = T_1 \implies w(0,t)=0 $

$ T(h,t) = w(h,t) + u(h,t) = w(0,t) + \frac{T_2-T_1}{h}h + T_1 = T_2 \implies w(h,t)=0 $

$ w(x, 0) = f(x)-u(x) $

because $ w(x,t) + u(x) = f(x) $

$ w(x,0) = f(x) - \Big[ \big( \frac{T_2 - T_1}{h} \big) x+T_1 \Big] $

$ w(x,t) = \chi (x)\tau (t) $

Separation of variables

(this step is tedious and I always skipped writing it down for my assignments)

Copying from my notes as I forgot how to solve Partial Differential Equations besides the fact you assume a solution (in all cases for PDE’s you assume the solution is a fourier series or a series of exponentials). It is legitimately a ton of work to solve a PDE in one dimension let alone two dimensions. In 3 dimensions it’s quite literally left to a computer and you don‘t try and solve them but you do learn the pattern (separation of variables) from solving one dimensional to two dimensional and this continues to 3 dimensions. Suffice to say this specific case has the solution for the transient $ w(x,t) $

$ w(x,t) = \sum_{n=1}^{\infty} c_n sin \big( \frac{n \pi x}{h} \big) e^{\beta \Big( \frac{n \pi}{h} \Big)^2 t} $

Using initial conditions

$ w(x, 0) = f(x)- \Big[ \big( \frac{T_2 - T_1}{h} \big) x+T_1 \Big] = \sum_{n=1}^{\infty} c_n sin \big( \frac{n \pi x}{h} \big) $

Call

$ f(x)- \Big[ \big( \frac{T_1 - T_2}{h} \big) x+T_1 \Big] = r(x) $

$ c_n = \frac{1}{h} \int_{-h}^{h} f_0 (x) sin \big( \frac{n \pi x}{h} \big)dx $

$ c_n = \frac{2}{h} \int_{0}^{h} r (x) sin \big( \frac{n \pi x}{h} \big)dx $

Combining the transient and steady state solutions to our formula for temperature

$ T(x,t) = w(x,t) + u(x) $

$ T(x,t) = \sum_{n=1}^{\infty} c_n sin \big( \frac{n \pi x}{h} \big) e^{\beta \Big( \frac{n \pi}{h} \Big)^2 t} + \frac{T_2-T_1}{h} x + T_1 $

with $ c_n = \frac{2}{h} \int_{0}^{h} r (x) sin \big( \frac{n \pi x}{h} \big)dx $

and $ r(x) = f(x)- \Big[ \big( \frac{T_2 - T_1}{h} \big) x+T_1 \Big] $

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allalfavincerho · 6 years

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opticks - prima puntata

For understanding therefore how the Bow is made, let a Drop of Rain, or any other spherical transparent Body be represented by the sphere BNFG, described with the Center C, and Semi-diameter CN. And let AN be one of the Sun’s rays incident upon it at N, and thence refracted to F, where let it either go out of the sphere by Refraction towards V, or be reflected to G [...] Parallel to the incident Ray AN draw the Diameter BQ, and let the Sine of Incidence out of Air into Water be to Sine of Refraction as I to R. Now, if you suppose the Point of Incidence R to move from the Point B, continually till it come [on the vertical over C], the Arch QF will first increase and then decrease, and so will the Angle AXR which the Rays AN and GR contain; and the Arch QR and Angle AXR will be biggest when [(calling x the angle of incidence), we have `\cos x = \frac{ \sqrt{I^2-R^2} }{ \sqrt{3R^2} }`], in which case [(calling r the refraction angle) one has ` \frac{ \cos r }{ \cos x } = \frac{2R}{I} `]

(da I. Newton, Opticks, Book I Part II Prop. IX)

Newton comincia così la sua celebre trattazione dell’arcobaleno, nella sua Opticks. Ovviamente si guarda bene dal fornire giustificazioni o dettagli più o meno tecnici sul modo di ottenere i risultati che va disvelando al lettore. Anzi, a un certo punto se ne esce perfino con una frase come “The Truth of all this Mathematicians will easily examine”, e se ne lava le mani.

Qui vorrei proporre una illustrazione in chiave moderna dei passaggi lasciati sottintesi da Ser Isacco.

Egli sembra suggerire di studiare il variare dell’arco QF, e contestualmente dell’“angolo d’uscita” AXR che si forma tra il raggio incidente e quello uscente dopo due riflessioni interne alla sfera, in funzione della posizione del punto di incidenza N, facendo muovere quest’ultimo lungo un quadrante della circonferenza, a partire da B in senso orario.

Quel che mi ripropongo di fare è scrivere l’angolo `AXR = Y_1 ` in funzione dell’angolo di incidenza x, che, quando il punto B varia come detto, varia da 0 a `\pi/2`. L’obiettivo è poi verificare che tale funzione ha un massimo nel punto detto da Newton.

In via preliminare, però, trovo la dipendenza da x anche dell’arco QF, che secondo Newton ha un massimo nello stesso punto di AXR. In base al valore di x, l’estremo F può trovarsi sopra (come in figura) o sotto Q: poiché l’angolo BCN = x (angoli corrispondenti) e BCQ è piatto, si sarà nel primo caso (F sopra Q) se è `x - 2r < 0`, e allora

`QF = \pi - (x + \pi - 2r) = 2r - x`;

si sarà nel secondo caso (F sotto Q) se è `x - 2r > 0`, e allora

`QF = x - 2r`.

In breve, si può scrivere che `QF = | x - 2r |`.

Passo ad `Y_1 = AXR` (più oltre sarà chiara la relazione con QF). Nel triangolo isoscele NXG di base NG, l'angolo `Y_1` ne forma uno piatto insieme coi due angoli alla base (congruenti). Ciascuno di questi è la somma di x con un angolo `\beta` che si può calcolare osservando che, nel triangolo isoscele NCG, l'angolo al vertice C vale 4r, pertanto `2\beta = \pi - 4r `. Si conclude allora che

`Y_1 = \pi - 2x - 2\beta = 4r - 2x.`

(Si vede allora che, almeno nel caso illustrato in figura, con `x - 2r < 0`, vale la relazione AXR = 2 QF. Nell'altro caso si ha la medesima conclusione procedendo in maniera analoga. Angolo ed arco essendo proporzionali, basta studiare l’andamento di uno solo dei due per conoscerli entrambi.)

Newton assume con noto quel che in termini moderni diremmo l’indice di rifrazione tra i mezzi aria ed acqua, esprimendo ciò nella formula `\frac{\sin x }{ \sin r } = \frac{ I }{ R } > 1` (dove la disuguaglianza finale non è espressa nel testo di Newton, ma è risultato noto dalla fisica, che può essere ridetto così: quando passa dall'aria all'acqua, un raggio di luce si avvicina alla normale alla superficie di separazione tra i due mezzi). I ed R sono da intendersi parametri noti.

Da questa relazione si ricava: `r = arcsin (\frac{ R }{ I } \sin x)`. Da cui allora

`Y_1 = 4 arcsin (\frac{ R }{ I } \sin x) - 2x, x \in [0; \pi/2]. `

La derivata positiva per questa funzione di x si ha se (ometto alcuni passaggi di calcolo) `2R \cos x \ge \sqrt{ I^2 - (R \sin x)^2 } `, che - con qualche altro passaggio algebrico - porta a

`\cos x \ge \sqrt{\frac{ I^2 - R^2 }{3R^2} }.`

Quest’ultima disequazione è risolta da valori di x contenuti in un intervallo `(0; \theta_1)`, in cui l'arco `Y_1` cresce, per raggiungere il valore massimo quando è soddisfatta l'uguaglianza `\cos x = \sqrt{\frac{ I^2 - R^2 }{ 3R^2 } }`, così come detto da Newton.

Per concludere, resta da calcolare il rapporto tra i coseni degli angoli di rifrazione e di incidenza nel punto di massimo appena trovato. Dalla formula dell'indice di rifrazione si ottiene con qualche passaggio che

`\cos^2 r = 1 - \sin^2 r = ... = \frac{ 4 }{ 3 } \frac{ I^2 - R^2 }{ I^2 }.`

E allora

`\frac{ \cos^2 r }{ \cos^2 x } = \frac{ 4 }{ 3 }\frac{ I^2 - R^2 }{ I^2 } \frac{ 3R^2}{ I^2 - R^2} = \frac{ 4R^2 }{ I^2 }`

che è il valore indicato da Newton.

E bravo Newton. Sintetico, direi.

(fine prima puntata. la seconda qui)

#newton #opticks #ottica #ottica geometrica #arcobaleno #rifrazione #derivata #citazioni #esercizi #fisica #geometria

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local-to-global-blog · 7 years

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On Approximating Pi

So, you want to approximate $\pi$ from first principles (ie, no fancy formulas that you don’t know how to prove). One of the most straightforward ways to do this is through the method that Archimedes used thousands of years ago. He did this by looking at the perimeter of regular polygons, which he could compute. The more sides that the polygon had, the closer the perimeter was to $\pi$. In modern notation, his method amounts to the formula

$$ \lim_{x \rightarrow 0} \frac{\sin(\pi x)}{x} = \pi = \lim_{x \rightarrow 0} \frac{\tan(\pi x)}{x} $$

Since we can compute things like $\sin(\pi/2^n)$ without actually knowing what $\pi$ is, we can use this to find an approximation for $\pi$. But just knowing this formula is not good enough. For instance, how do we know how close we are to $\pi$? If we use $x=2^{-100}$ to do this, then exactly how close we are? If we want to get $\pi$ to, say, ten digits, then what power of two do we need in the denominator?

These questions can be answered by going through the proofs and keeping track of how these limits actually work. Through this, we can get a refined look at exactly how to approximate $\pi$. (And gain an appreciation for $\epsilon-\delta$ in limits.)

Our Fundamental Identity

Firstly, we need to show that for any $-\frac{\pi}{2} < x < \frac{\pi}{2}$ we have

$$ |x\cos(x)| \leq |\sin(x)| \leq |x| \leq |\tan(x)| \leq |x\sec(x)|$$

where we have equality only at $x=0$.

This is where the diagram in the picture comes into play. Note that we clearly have equality at $x=0$ and, by symmetry, we just have to check this for $x>0$. In the picture, we have a quarter-circle of radius 1, with a line from the origin to the circle that makes an angle of $\theta$ with the horizontal axis. We can make this to make a triangle as shown. The length of the horizontal leg of this triangle is $\cos(\theta)$ and the length of the vertical leg of this triangle is $\sin(\theta)$. The blue arc in the picture is clearly larger than the vertical leg of the triangle and, since we are in radians, this arc has length $\theta$. This gives us that $\sin(\theta)<\theta$. On the other hand, the pink highlighted arc has radius $\cos(\theta)$ which means that its length is $\theta\cos(\theta)$. Since this arc is less than the vertical leg of the triangle, we get the leftmost inequality. Finally, through similar reasoning with the larger triangle and arc, we get the final two inequalities.

The Limit Lemma

Next we will look at our first limit. We want to show that

$$\lim_{x\rightarrow 0}\cos(x) = 1 = \lim_{x\rightarrow 0} \sec(x) $$

This means working with $\epsilon$s and $\delta$s, and it will be important to keep track of this information, as it tells us how to actually do the approximation.

Let $\epsilon >0$. We have the basic trig identity: $1-\cos(x) = 2\sin^2(x/2)$. This means that, using our above inequality, that $|\cos(x)-1| <x^2/2$. Therefore if $0<|x|<\sqrt{2\epsilon}$, then $|\cos(x)-1|<\epsilon$. This proves the first limit.

We will use this to prove the second limit. We have $\sec(x)-1 = (1-\cos(x))/\cos(x)$. Suppose that $0<|x|<\sqrt{2\epsilon’}$. Using what we just did, we have $1-\cos(x) < \epsilon’$ and $1-\epsilon’ < \cos(x)$. Combining these gives $\sec(x)-1 < \frac{\epsilon’}{1-\epsilon’}$. Choose $\epsilon’=\epsilon/(1+\epsilon)$. Plugging this into $\frac{\epsilon’}{1-\epsilon’}$ simplifies to $\epsilon$. So, if $0<|x|< \frac{\epsilon}{1+\epsilon}$ then $|1-\sec(x)|<\epsilon$. This proves the second limit.

To summarize the important bits of what we just did, since $ \frac{\epsilon}{1+\epsilon} < \epsilon$, it follows that if $0<|x|< \sqrt{\frac{2\epsilon}{1+\epsilon}} $ then

$$ |1-\cos(x)|< \epsilon \;\;\;\;\&\;\;\;\;\; |1-\sec(x)| < \epsilon$$

The Main Limit

We’re now done with lemmas and can prove the main result, which is: For $0<|x|<\frac{\pi}{2}$ we have that

$$ \frac{\sin(x)}{x} < 1 < \frac{\tan(x)}{x}$$

Moreover, we have the limits

$$ \lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1 = \lim_{x\rightarrow 0}\frac{\tan(x)}{x}$$

Dividing our original inequality through by $x$ gives

$$ \cos(x) < \frac{\sin(x)}{x} < 1 < \frac{\tan(x)}{x} < \sec(x)$$

This holds for all $x$ in our range because we are able to use the evenness of these functions to get rid of the absolute value. We then immediately have the first statement of this, and we just have to do the limit.

For the limits, we will keep track of our $\epsilon$s and $\delta$s. Obviously, using the divided expression, the Squeeze Lemma will give our result, but we want a more refined result, with errors in place. Take the first two inequalities, and rearrange to get

$$ 0 < 1-\frac{\sin(x)}{x} < 1-\cos(x)$$

By subtracting through by $1$ and multiplying by $-1$. Similarly for the latter two inequalities we have:

$$ 0 < \frac{\tan(x)}{x} -1 < \sec(x) -1$$

If we then take $0<|x|< \sqrt{\frac{2\epsilon}{1+\epsilon}}$, we find, using our previous results, that

$$ 0 < 1- \frac{\sin(x)}{x} < \epsilon$$

$$ 0 < \frac{\tan(x)}{x} - 1 < \epsilon$$

This proves the result, and we know how the error works.

Approximating Pi

We will now use this to approximate $\pi$. In our final result, replace $x$ with $\pi x$, multiply our equations through by $\pi$ and replace $\epsilon$ with $\epsilon/\pi$. This results in the statement: If $0<|x|<\frac{1}{\pi} \sqrt{\frac{2\epsilon}{\pi+\epsilon}}$ then

$$ 0 < \pi- \frac{\sin(\pi x)}{x} < \epsilon$$

$$ 0 < \frac{\tan(\pi x)}{x} - \pi < \epsilon$$

This means that the sine inequality gives a lower bound for $\pi$ with error $\epsilon$ and the tangent inequality gives an upper bound for $\pi$ with error $\epsilon$. Explicitly: Whenever $0<|x|<\frac{1}{\pi} \sqrt{\frac{2\epsilon}{\pi+\epsilon}}$ we have

$$\pi\in \left( \frac{\sin(\pi x)}{x}, \frac{\tan(\pi x)}{x} \right) \subseteq (\pi-\epsilon , \pi +\epsilon)$$

Follow this recipe and you’ll be able to approximate $\pi$ to your heart’s content.

Example

We can compute trig functions of angles $\pi/(3\cdot 2^n)$ by hand. If we want to compute $\pi$ to within $k$ digits, then the question we need to answer is: What does $n$ have to be?

Using $\epsilon = 1/10^{k+1}$, $x=3^{-1}2^{-n}$, and the trivial bound $3<\pi$, in our main result, we find that:

$$ \frac{1}{2\cdot 3^n} < \frac{1}{3} \sqrt{\frac{2\cdot 10^{-k-1}}{3+10^{-k-1}}}$$

which simplifies to

$$n> \frac{k+1}{2}\log_2(10) + \frac{1}{2}\log_2(3/2) $$

$$n > 1.66097k +1.95345$$

This gives a linear relationship between the number of digits that we get and how many powers of two we have to take.So, if, say, we want to get $k=3$ digits, then we need $n>6.9$, which means that we need $x=\frac{1}{384}$ will do. Archimedes did it for $x=\frac{1}{96}$, which means that $n=5$, which means that we would have to have $k<1.83$, or we can be sure his result is accurate with error $\epsilon = 0.01$. This isn’t too tight of a bound, unfortunately, but it’s what we can prove using elementary trigonometry.

The record is $k=22,459,157,718,361$ digits. If we were to do this using this method, we would have to have $n=3.73\times 10^{13}$, which means $x\approx 3^{-1}\cdot 2^{-10^{13}}$, which would be really tough.

Using Calculus and Taylor Series, we can show that the error is actually much tighter. The question and challenge would then be to figure out if you can tighten this error using elementary methods.

#pi #math #maths #approximation #trig #archimedes #limits

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