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matekon · 5 years

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The “best” definition for the sinus and cosinus functions

I use the following personal conventions:

● - Definitions - Propositions I assume are true

○ - Theorems – Propositions I deduce from the definitions

I also prefer $\tau$ which equals $2\pi$ as the circle ratio

_____

In mathematics, there is a common phenomenon: there can be multiple ways of defining the same mathematical object.

For example, here are 2 definitions for an isosceles triangle:

● 1) A triangle is isosceles if it has 2 sides of the same length.

● 2) A triangle is isosceles if it has 2 angles that are equal in measure

These 2 definitions are “equivalent” in the sense that a triangle would be isosceles according to the first definition if and only if it is isosceles according to the second definition. (If you are into analytical philosophy, specifically Frege, you might say these definitions express different “senses” but have the same “reference”.)

The case of the isosceles triangle is pretty simple, but in mathematics, there can be definitions for objects which are equivalent but where it isn’t trivial is the slightest.

Even though it is totally frequent for one mathematical object to have multiple definitions available, the way modern mathematics work (by axiomatisation), we have to choose one definition as a “starting point” and then deduce its equivalence with other definitions later on.

So, with our example of the isosceles triangle, we could either choose the first proposition as our definition and then, the second proposition would follow as a theorem, or we could just as well do the reverse.

But, is there a “starting definition” that is “better” than the others? From experience, I would say that, in the point of view of a “pure mathematician”, this is totally irrelevant and doesn’t matter. But, I do think that we can say some definitions are “better” than others if we allow ourselves to use didactic criteria to evaluate them.

In this article, I will be interested with the functions sinus and cosinus, for which I encountered many different definitions in my school years. In Section 1, I will present these definitions and say what I like and don’t like about them using a didactic approach. Then, in Section 2, I will introduce a definition of these functions that I personally think is the best one and I will show that it is “equivalent” with some of the definitions of Section 1.

__________

Section 1 - The usual definitions for sin and cos

At secondary school, I learned the following definition:

Definition 1 (by the triangle):

$\bullet$ Let $\triangle\ ABC$ be a right triangle where $\angle ABC$ is the right angle

Then we define $sin(\theta)=a/c$ and $cos(\theta)=b/c $.

The pros for this definition are the simplicity of the language and the fact that it is directly applicable to problems of geometry.

Among the cons, we have that this definition only makes senses for $\theta \in ]0,\tau/4[$ (let’s immediately work in radians). Also, I don’t think that this way of presenting sin and cos makes it obvious how to visualize the graphs of these functions. (It is possible to see the graphs but it requires us to be kind of clever.)

You might also say that this definition doesn’t immediately allows us to evaluate the functions for a given input. But I don’t think this is such a big problem and I will explain it soon later.

In CEGEP, I learned the definition involving power series:

Definition 2 (by the power series):

$\bullet \: sin(t) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}t^{2n+1} = t-\frac{t^3}{3!}+\frac{t^5}{5!}-...$

$ \bullet \:cos(t) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}t^{2n} = 1-\frac{t^2}{2!}+\frac{t^4}{4!}-...$

This definition has the main advantage of allowing us to directly calculate the values of the functions. But the inconvenience that immediately comes with these types of definitions is that they make us say: “Where is this coming from? What is its utility?”.

The reality is that I think the human mind prefers to start of with definitions that make us directly see why the object in question is interesting and relevant. And then, for a function, we should find a way to “evaluate” it later on.

Let’s also note that this definition doesn’t make the shape of the graphs any more obvious.

In university, I got introduced to the following definition:

Definition 3 (by the differential equation):

$ \bullet \: sin (t) $ and $cos(t) $ are solutions of the differential equation $ f^{\prime\prime}(t) = -f(t) $

We first note that, because this equation has infinitely many solutions, we need further specifications to precisely define what $sin$ and $cos$ are.

This definition for me just mainly shows us a new interest of the sin and cos functions: there are extremely useful tools for solving differential equations. (This is one of the main motivations behind Fourier Analysis.)

This makes us do an important realization: maybe for didactic reasons, the definition we want to use for different mathematical objects depends on the context: For doing regular problems of geometry, Definition 1 for sin and cos is this best one to use. But for the theory of differential equations, then Definition 3 is more relevant.

I do think this argument is very important. But I also have a weak spot for definitions that are kind of more intuitive, more visual and just more “neutral” and “universal” I would say. I think all these criteria apply to the definition I will introduce in Section 2, which I will call “Definition 4 (by the circle)”. In fact, this definition is quite common, but I have never seen it being formalized the way I am about to do.

____________

Section 2 - The “best” definition for sin and cos

Let’s start with a simple question: “How do I describe a circle?”.

Algebraically, the simplest circle is the one of radius 1 centered at the origin. We define it this way:

$ \bullet \: S^1 = \{ (x,y) \in \mathbb{R}^2 | \: ||(x,y)||=1\} $

We see this way of defining a circle works the following way: We take as points $ (x,y) $ on the circle all the solutions to the equation $ x^2+y^2=1 $.

But what I would like to do instead is to describe the circle with a function, not an equation.

I want a function $f $ that outputs a point on the unit circle given a real number input.

$\bullet\:f:\mathbb{R}\to\mathbb{R}^2$ where $Im(f)=S^1$

We immediately see that $f$ is a vector function (has vectors as outputs). Because of that, it can be separated into 2 scalar functions (have real numbers as outputs).

$ f(t) = (x(t),y(t)) $ where $ x: \mathbb{R} \rightarrow \mathbb{R} $ and $ y: \mathbb{R} \rightarrow \mathbb{R} $

In case you didn’t guessed it, $x(t)$ will become $cos(t)$ and $y(t)$ will become $sin(t)$. I will continue to write them as $x(t)$ and $y(t)$ mainly because this notation makes their role clearer and because they aren’t fully defined yet.

Now, what I want to do is to fully define the functions $x(t)$ and $y(t)$. To do that, I will enumerate a list of properties that I want these 2 functions to have.

Because I said I wanted $f$ to output a point on the unit circle, that implies:

$ f(t) \in S^1 \iff ||f(t)||=1 \iff ||(x(t),y(t))||=1$

$\iff \sqrt{x^2(t)+y^2(t)}=1 \iff x^2(t)+y^2(t)=1 $

With this, I will state the first property to these functions, which is the first part of their Definition 4:

$ \bullet \: (1) \: x^2(t)+y^2(t)=1 $

This property isn’t enough. To illustrate this, let’s remark that the following function $f^{*} $ does obey the property $ (1) $ but isn’t the “nicest” function we could think of:

Basically, we see that $f^{*} $ isn’t “continuous”, because it occasionally “jumps”. But, let’s say I want $f$ to be a function that goes continuously around the circle.

In fact, I want $f $ to be something more specific: “parameterized by arc length”.

This means the following:

● Let $ g(t) $ be a curve in space (2d in this case). Then $ g(t) $ is parameterized by arc length if the length of the arc between $ g(t_0) $ and $g(t_1) $ (where $t_1 > t_0 $ ) is precisely $ t_1 – t_0$.

(In the case of the unit circle, how is this idea related to “radians”?).

I won’t go behind all the theory behind it. It is easy to google anyway. For our purposes, I need to know the theorem that says:

$ \circ \: f $ is “parametrized by arc length” $ \iff ||f’(t)||=1 $

From that I deduce:

$ ||f’(t)||=1 \iff ||(x’(t),y’(t))||=1 $

$\iff (x’(t))^2+(y’(t))^2=1 $

From this, we get the second part of Definition 4:

$ \bullet \: (2) \: (x’(t))^2+(y’(t))^2=1 $

$ f $ being “parameterized by arc length” implies that $ f $ is countinous, as we wanted.

The properties (1) and (2) largely define $ x(t)$ and $y(t)$. In fact, with just these 2 properties, we can show that $ x(t)$ and $y(t)$ must obey Definition 3 (by the differential equation). To prove this is a very fun mathematical exercise. Anyway, here’s my demonstration:

We know:

$ \bullet \: (1) \: x^2+y^2=1 $ $ \bullet \: (2) \: x’^2+y’^2=1 $

We want to show that $ y ^{\prime\prime} =-y $ (The proof for $ x ^{\prime\prime} =-x $ is analogous)

$ (1) \Rightarrow \frac{d}{dt}(x^2+y^2)= \frac{d}{dt}(1) $

$\Rightarrow 2xx’+2yy’=0 $

$\Rightarrow xx’=-yy’$

So, we have: $ \circ \: (A)\: xx’=-yy’\\ $

$ (2) \Rightarrow x^2( x’^2+y’^2)=x^2(1)$

$ \Rightarrow (xx’)^2+(xy’)^2=x^2 $

$\Rightarrow^{(A)} (-yy’)^2+(xy’)^2=x^2 $

$\Rightarrow y’^2(y^2+x^2)=x^2 $

$\Rightarrow^{(1)} y’^2(1)=x^2 \Rightarrow y’^2=x^2 $

Consequently, $ \circ \: (B)\: y’^2=x^2\\ $

$ (B)\Rightarrow \frac{d}{dt}(y’^2)= \frac{d}{dt}(x^2) $

$\Rightarrow 2y’y ^{\prime\prime} =2xx’$

$ \Rightarrow^{(A)} y’y ^{\prime\prime} =-yy’ $

$ \Rightarrow_{*} y ^{\prime\prime} =-y $

(I will call this final equation $ (E_y) $ and use it later)

$QED $

(The last implication with an asterisk below really needs a bit of justification. Because $ y’ $ can equal 0 for some inputs, we can’t just divide by it. But, there is a way to clean up the mess and make the deduction valid.)

As I said before, Definition 3 (by the differential equation) is incomplete in its formulation. So, it’s not because I was able to deduce it from $(1)$ and $(2)$ that these 2 properties are enough to define $ x(t) $ and $ y(t) $.

What I will do now is that I will try do deduce Definition 2 (by the power series). Trying to do so, it will show me what I have to add to $(1)$ and $(2)$ to make the Definition 4 (by the circle) complete.

What I know:

$ (1) \: x^2+y^2=1 $ $ (2) \: x’^2+y’^2=1 $

$ (A)\: xx’=-yy’\\ $ $ (B)\: y’^2=x^2\\ $

$ (E_x) \: x ^{\prime\prime} =-x $ $ (E_y) \: y ^{\prime\prime} =-y $

To deduce Definition 2, I would like to find the Maclaurin Series for $ x(t) $ and $ y(t) $:

$ x(t) = \sum_{n=0}^{\infty} \frac{x^{(n)}(0)}{n!}t^n $

$ y(t) = \sum_{n=0}^{\infty} \frac{y^{(n)}(0)}{n!}t^n $

So, I need to find $ x(0)$, $x’(0)$, $x ^{\prime\prime} (0)$, ... and $y(0)$, $y’(0)$, $y ^{\prime\prime} (0)$,...

$(1)$ and $(2)$ aren’t enough to find the Maclaurin Series. So, I will add the following defining property for $sin$ and $cos$ that is honestly only justifiable as a convention:

$ \bullet \: (3) \: f(0) = (x(0), y(0)) = (1,0) $

So we’ve essentially just chosen the starting point for our curve $ f$. It could just as easily have been $ (0,1)$ or $ (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} ) $, which also are on the unit circle. This choice for $ f(0) $ is, I think, mainly justifiable as a way to make Definition 4 (by the circle) equivalent to Definition 1 (by the triangle).

Now, let’s use this equation in the quest of finding the Maclaurin Series of $x(t) $ and $ y(t)$ :

From $(3) \: x(0)=1 $ and $ (E_x) \: x ^{\prime\prime} =-x $, I deduce:

$ x^{(4k)}(0)=1 $ and $ x^{(4k+2)}(0)=-1 $ where $k \in \mathbb{N}$

From $ (3) \: y(0)=0 $ and $ (E_y) \: y ^{\prime\prime} =-y $, I deduce:

$ y^{(2k)}(0)=0 $ where $k \in \mathbb{N}$

We are halfway done. But, for the next step, we need to be kind of clever and use an old equation we proved earlier:

$ (B) \: (y’(t))^2= (x(t))^2$

$ \Rightarrow (y’(0)^2=(x(0))^2$

$ \Rightarrow^{(3)} (y’(0))^2=(1)^2 $

$ \Rightarrow y’(0) = \pm 1 $

Again, there is a choice to be made, and again, it is a matter of convention.

It can be shown that choosing $ y’(0)= 1$ will make the vector function $f$ go counterclockwise around the unit circle and that choosing $y’(0)= -1$ will make it go clockwise instead. Yes, we will choose the first option because of the convention of how we measure angles.

$ \bullet \: (4) \: y’(0) = 1 $

This will be the last property we add to Definition 4. Let’s see what we can do with it:

From $ (4) \: y’(0)=1 $ and $ (E_y) \: y ^{\prime\prime} =-y $, I deduce:

$ y^{(4k+1)}(0)=1 $ and $ y^{(4k+3)}(0)=-1 $ where $k \in \mathbb{N}$

From $ (4) \: y’(0)=1 $ and $ (2) \: x’^2+y’^2=1 $, I deduce:

$ (*) \: x’(0) = 0 $

Finally, from $ (*) \: x’(0)=0 $ and $ (E_x) \: x ^{\prime\prime} =-x $, I deduce:

$ x^{(2k+1)}(0)=0 $ where $k \in \mathbb{N}$

Putting all of this together, we finally get the MacLaurin Series:

$ x(t) = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k)!}t^{2k} $

$ y(t) = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)!}t^{2k+1} $

And so we’ve just proven that Definition 4 (we’ve just completed) is equivalent to Definition 2!

For clarity, let’s put all the parts of Definition 4 together and we will posture $x(t) = cos(t) $ and $y(t) = sin(t)$.

Definition 4 (by the circle):

$ \bullet \: sin: \mathbb{R} \rightarrow \mathbb{R} $ and $ cos: \mathbb{R} \rightarrow \mathbb{R} $, where

$ (1)\: cos^2(t)+sin^2(t)=1 $

$ (2)\: (\frac{d}{dt}cos(t))^2+ (\frac{d}{dt}sin(t))^2 =1 $

$ (3)\: sin(0) = 0 $ (which implies $ cos(0) =1) $

$ (4) \: ( \frac{d}{dt}sin(t))|_{t=0} = 1 $

We can also put it on words like that:

Definition 4 (by the circle):

$ \bullet \: f(t) = (cos(t),sin(t)) $ is a function from $\mathbb{R}$ to $\mathbb{R}^2$ where $ Im(f) = S^1$ (unit circle). Also, $ f$ is parameterized by arc length, starts on $ (1,0) $ and goes counterclockwise.

The language can really be seeing as harsh, but once this definition is really understood, it allows us to directly visualize what $sin(t)$ and $cos(t)$ mean. It is also not hard to see the shape of their graphs, especially with the help to the following gif: http://i.imgur.com/jvzRYnC.gif

This is the reason why I think this is the best definition in a didactic point of view.

If we want a more accessible language for people who aren’t specialized in math, this formulation would also be valid:

Definition 4: ● If I start on the unit circle at $ (1,0)$ and I walk t units of distance counterclockwise while staying on the unit circle, my x position will be $cos(t) $ and my y position will be $ sin(t)$

I let to you the proof that Definition 4 is equivalent to Definition 1 (for $ t \in ]0,\tau/4[) $. It is definitely the most straightforward proof on the bunch.

#mathematics #trigonometry #differential equations #calculus

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testblogplzignore · 3 years

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I used that necessary boundary condition

http://jwilson.coe.uga.edu/EMT668/EMAT6680.2003.Su/Daly/write_ups/WR_12/Asmt_12.htm

The assumptions for this are spicy and quite amusing so if I can get it to work I can show direct numbers/functions. Legit making me laugh thinking about it and what you’d physically have to do. If it doesn’t work I’ll tell you in this post

$ f(x) = H(x) $

where

$ H(x) = 0$ when $ x=h=0$ and $ H(x) = T_1 $ when $ 0<x<h $

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

$ w(x,t) = H(x)-\Big[ \frac{T_2 - T_1}{h} x + T_1 \Big] $

$ w(0,t) = w(h,t) = 0 $ indeed (how did I not notice this egregious mistake)

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

$ w(x,0) = H(x) + \frac{T_1 - T_2}{h} x - T_1 $

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

$ c_n = \frac{2}{h} \int_{0}^{h} \Big[ H(x) + \frac{T_1 - T_2}{h} x - T_1 \Big] sin \big( \frac{n \pi x}{h} \big)dx $

$ c_n = \frac{2}{h}\Big[ \int_{0}^{h} H(x) sin \big( \frac{n \pi x}{h} \big) dx + \frac{T_1 - T_2}{h} \int_{0}^{h} x sin \big( \frac{n \pi x}{h} \big)dx - T_1 \int_{0}^{h} sin \big( \frac{n \pi x}{h} \big) dx \Big] $

first integral

$ \int_{0}^{h} H(x) sin \big( \frac{n \pi x}{h} \big) dx = 0 $ by space

Just goes to show brute forcing is legit retarded. I missed it in my notes so I got punished. Also we have

$ \frac{\partial}{\partial t} w(x,t) = \beta \frac{\partial^2}{\partial x^2} w(x,t) $ which is true because it’s not dependent on time and its spatial derivative the spatial Heaviside derivative is zero after the first. We have this because $ u(x) $ has no time dependence so left side is zero and it’s a straight line in one spatial dimension so double derivative is zero

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

I see now why they characterize it with a function dependent on x only as the no function with dependence on t has the first derivative equal to zero which we require for this method. They could explain that or maybe they did and didn’t write it down for people who terribly take notes. Anyhoo

second integral

$ \frac{T_1 - T_2}{h} \int_{0}^{h} x sin \big( \frac{n \pi x}{h} \big)dx = \frac{T_1 - T_2}{h} \Big[ \frac{h^2}{(n \pi )^2} \big[ sin \big( \frac{n \pi x}{h} \big)- \frac{n \pi x}{h} cos \big( \frac{n \pi x}{h} \big)\big] \Big]_{0}^{h} $

Gonna show full steps in case this one works (it should by the extra condition I saw but I’ve done this integration before and it didn’t work out, can try small angle to see if it does and also can put faith in the Fourier decomposition and taking a specific n doesn’t necessarily have to cancel off in that term)

Notice at $ sin \big( \frac{n \pi x}{h} \big) \Big|_0^h $ you get $ sin \big( \frac{n \pi h}{h} \big) - sin \big( n \pi 0 \big) $ and since we’re summing over n that first term always goes to zero hence

$ sin \big( \frac{n \pi x}{h} \big) \Big|_0^h =0 $

$ \Big[ \frac{n \pi x}{h} cos \big( \frac{n \pi x}{h} \big) \Big]_0^h = - \frac{n \pi h}{h} cos \big( \frac{n \pi h}{h} \big) - - \frac{n \pi 0}{h} cos \big( \frac{n \pi 0}{h} \big) $

As you can see the right term goes to zero (I believe at this point being this in detail is retarded) and the left term is proportionate to $ cos ( n \pi ) $ which again is being summed over so it’s always equal to $ 1 $

$ \Big[ \frac{n \pi x}{h} cos \big( \frac{n \pi x}{h} \big) \Big]_0^h = - n \pi $

third integral

$ - T_1 \int_{0}^{h} sin \big( \frac{n \pi x}{h} \big) dx = 0 $ by space

so final form looks like

$ c_n = \frac{2}{h} \frac{T_1 - T_2}{h} \frac{h^2}{(n \pi )^2}(- n \pi ) $

$ c_n = \frac{2}{h} \frac{T_2 - T_1}{h} \frac{h^2}{(n \pi )^2}( n \pi ) $

$ c_n = \frac{2(T_2 - T_1)}{n \pi} $

$ 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} $

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

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

$ T(x,t) = \frac{2(T_2 - T_1)}{\pi} \sum_{n=1}^{\infty} \frac{1}{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 $

$ T(0,t) = \frac{2(T_2 - T_1)}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} sin \big( \frac{n \pi 0}{h} \big) e^{-\beta \Big( \frac{n \pi}{h} \Big)^2 t} + \frac{T_2 - T_1}{h} 0+ T_1 $

$ T(0,t) = T_1 $

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

$ T(h,t) =T_2 $

$ T(x,0) = \frac{2(T_2 - T_1)}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} sin \big( \frac{n \pi x}{h} \big) e^{-\beta \Big( \frac{n \pi}{h} \Big)^2 0} + \frac{T_2 - T_1}{h} x + T_1 $

$ T(x,0) = \frac{2(T_2 - T_1)}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} sin \big( \frac{n \pi x}{h} \big) + \frac{T_2 - T_1}{h} x + T_1 $

Let’s check $ n=1 $

$ T(x,0) \Big|_{n=1} = \frac{2(T_2 - T_1)}{\pi} \sum_{n=1}^{1} \frac{1}{n} sin \big( \frac{n \pi x}{h} \big) + \frac{T_2 - T_1}{h} x + T_1 $

$ T(x,0) \Big|_{n=1} = \frac{2(T_2 - T_1)}{\pi} \frac{1}{1} sin \big( \frac{1 \pi x}{h} \big) + \frac{T_2 - T_1}{h} x + T_1 $

$ T(x,0) \Big|_{n=1} = \frac{2(T_2 - T_1)}{\pi} sin \big( \frac{ \pi x}{h} \big) + \frac{T_2 - T_1}{h} x + T_1 $

small angle approximation says

$ sin(\theta ) \approx \theta$ so

$ T(x,0) \Big|_{n=1} \approx \frac{2(T_2 - T_1)}{\pi} \frac{ \pi x}{h} + \frac{T_2 - T_1}{h} x + T_1 $

$ T(x,0) \Big|_{n=1} \approx \frac{ 2(T_2 - T_1) }{h}x + \frac{T_2 - T_1}{h} x + T_1 $

well shit

$ T(x,0) \Big|_{n=1} \approx 3\frac{T_2 - T_1}{h} x + T_1 $

Let’s invoke some physicality or bullshit, $ f(x) $ says the temperature of our coffee is exactly $ T_1 $ everywhere in the cup except the endpoints being exactly $ 0 $. Physically we take this to mean that $ T_1 = T_2 $ at exactly $ T(x,0) $ so (again I completely missed it

$ T(x,0) \Big|_{n=1} \approx T_1 $ which is our initial condition

fuck that feels good

$ T(x,t) = \frac{2(T_2 - T_1)}{\pi} \sum_{n=1}^{\infty} \frac{1}{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 $

Technically if you were to graph this it would be flat for a millisecond then one fixed endpoint (in our case at $h$) you’d see an instantaneous drop on one side to $ T_2 $ and then the heat traveling through it from the cold side to the warm side like the opposite of this graph (going into more detail in my actual post and not shitty latex posts

The way this would be achieved is physically impossible but let’s see your coffee cup

started at 4:27 am took way too long for such a simple pde but damn does the latex take work

8:36 am turns out every single time I typed $ T(x,t) $ it was missing a negative sign in the exponential. Fixing for this post and probably the other one where I make the actual post

9:40 am

fug maple says it’s wrong

shouldn’t have tried to bullshit the obvious contradiction

9:58 am I see it’s because I did the Fourier decomposition of only the steady state solution without a term to offset that. XD

I’m at a point where I’m starting to see it as too complex for an analytical solution

10:13 considering dropping the physicality just so i can do it

It’s the mother fucking Fourier decomposition that fucks this. Perhaps I do actually need to do separation of variables but again I need homogeneous boundary conditions in at least one dimension

12:39 pm

aight might as well tell you. Sadge about thinking I solved it. Anyway it required the ice cube to melt literally instantly and required constant ice pouring at the top of the coffee to consistently keep it this temperature. Now that I think about it an insulated coffee makes more sense with an impulse for the ice. I thought it was amusing thinking of somebody constantly putting in ice to keep the temperature constant. Probably only me who finds it that amusing. Likewise it required the bottom of the coffee to be constantly heated to a certain temperature. The other assumptions are on the original post. I suppose I should do something with my time

12:27 am 21/08/04

Did I fuck up my step function. It was supposed to have $ H(0)=T_1 $ instead of $ 0 $

The fixes the integral and you have an offset term. Let me try putting it into maple

12:29 am

No because the integral still goes to zero for that step function. I wonder if it makes a difference for insulated endpoints. I don’t see why it would and it would probably still fail

12:46 am

It occurs to me that I messed up my step function and am writing it down on paper now

1:24 am

This is actually the solution I just messed up my $ H(x) $ which is actually just a step function instead of the Heaviside function. I had multiple warnings too at both the bullshit step and the initial checking that $ w(0,t) = w(h,t)=0 $

12:00 pm 21/08/04

Is very close to if not the right answer. Maple doesn’t like the step function so it bugs out. I did multiple integrals and one of them is right

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mathematicianadda · 4 years

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A circle is divided into $5$ parts as shown in the diagram and parts are colored either red or green. Find which area is bigger. https://ift.tt/2PeDy6n

In the given diagram, there are $5$ points $A, B, C, D$ and $E$ on the circumference of the circle such that $\angle ABC = \angle BCD = \angle CDE = 45^{\circ}$ and $O$ is the center of the circle.

Sectors made by $AB$ and $DE$, and area of the circle between $BC$ and $CD$ are highlighted in green. Area of the circle between $AB$ and $BC$, and between $CD$ and $DE$ are highlighted in red.

Which area is bigger, the area highlighted in red or the area highlighted in green?

This was sent to me by someone. While I solved the problem (given below), the sender said that the source solution arrived at the conclusion that points $A$, $O$ and $E$ are collinear and $OC \perp AE$, so $\displaystyle \angle OCB = \angle OCD = \frac{45^{\circ}}{2}=22.5^{\circ}$. While I agree with points being collinear and $OC \perp AE$ but that cannot obviously be the reason for the angles being equal. In fact the solution does not depend on them being equal as we can see. I am seeking help in establishing $\angle OCB = \angle OCD$ if that is indeed true, which I cannot see how one can conclude based on what is given.

My solution: Say, $\angle OCB = \theta$. Then, $\angle ACB = \angle OCD = (45^{\circ}-\theta)$ and $\angle DCE = \theta$.

Segment $AB= \displaystyle r^2 \left[\frac{\pi}{4}-\theta-\sin(45^{\circ}-\theta)\cos(45^{\circ}-\theta)\right]$ Segment $DE= \displaystyle r^2 \left[\theta-\sin \theta \cos \theta\right]$

$\triangle OBC = r^2 \sin \theta \cos \theta$ $\triangle ODC = r^2 \sin(45^{\circ}-\theta)\cos(45^{\circ}-\theta)$

Section $BOD = \dfrac{\pi}{4} r^2$

Adding all of the above, total area in green $= \dfrac{\pi}{2} r^2$. So the red area has to be the same too.

In addition to my question on $OC$ being bisector of $\angle BCD$, let me also know if any of you have a simpler solution.

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rowanstree-blog · 7 years

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Day #167

Today I started with a math class on the graphs of the trigonometric functions, sine, cosine and tangent. I started the lesson of with a video on the graph of the sinusoidal function, y=sin(x). I started with a question asking what the domain of the sinusoidal function was. The domain is all sets of valuable inputs for the function. I used the unit circle and a y, theta graph to wrote and plot the sinusoidal equation. Once I did all of that, I concluded that the domain of this function is (-1,1) including -1 and 1. I then watched a video on how to construct the graph of y=tan(x). I concluded that the graph would have a vertical asymptote every five radians and approach positive snd negative infinity while doing so. Considering the tangent of theta is the slope of the terminal ray, I just had to take the sin of theta over the cosine of theta to find out the tangent of several pi radians. I then watched a video on intersection points of y=sin(x) and y=cos(x). In this video, I wrote terms of the table in terms of radians for both the cosine of theta and the sine of theta on the unit circle and graphed accordingly. After that, I watched a video on basic trigonometric entities involving symmetry. In the lesson, I drew several terminal rays in all four quadrants of a graph in a unit circle. I then wrote down which sine and cosine functions were equivalent and then followed it up with a video on tangent identities which is the proportion of a sin function over its equivalent cosine function. Next, I watched a video on sine and cosine identities involving symmetry in which I took random sine and cosine functions and determined if they were equal to each other. I then watched a video on tangent identities periodicity. In this video, I did pretty much the same thing except I determined if its slope, or tangent was equal to 1/2. After all of that, I took an Algebra 1 class on the Quadratic equation sample problems as a refresher. I then took a French lesson on Duolingo followed by a Chemistry class on how to fill in an ICE table according to the equilibrium constant expression of a reaction. I stands for initial concentration which should be given in order to calculate E which stands for equilibrium concentration which can be found after adding or subtracting I and C which stands for change in concentration. First, you must write all of your initial concentrations If your equilibrium constant is larger than or equal to 10^4, then your products are favored. If Kc is less than or equal to 10^-2, then your reactants are favored. This information can help you figure out which side is equal to zero which will later determine how you fill in your ICE table. Depending on which side is favored, you must add or subtract the molecules stoichiometric coefficient of x. Add, subtract and solve for x to find out your equilibrium constant. After that, I took a space science class on a simulation video of Earth’s tilt and rotation which determines its season. This information can contribute to the reasoning behind why the equator doesn't have seasons. It is because it is never turned away from the sun, like us hence our four seasons as we turn towards and away from the sun. I then took an environmental science class on various studies launched and questions asked on the transgenic contamination of Mexican maize. Even though there have been both false positives and false negatives in order to protect and expose the firms causing this transgenic contamination. It is true that transgenic contamination is becoming more and more abundant and the effects are still unknown. I then did a quick exercise followed by an hour long lunch. After that, I took a women’s history class on Mary Harris “Mother” Jones who was a labor organizer who played an outspoken role in the world that was mislead by her appearance. She was a passionate and resilient woman that fought for what she believed in and due to that, became known as “the most dangerous woman in America.” I then took an economics class on Allocatice efficiency which is when the marginal cost is equal to the marginal benefit and I did an example that showed this both literally and graphically. Next, I took a grammar class on commas and introductory elements which are followed by a comma and are placed in the beginning of a sentence. After that, I read some of “The Beauty Myth” on how markets rely so much on feminine appearance more that the actual product to “appeal” to the reader while they are actually making them more self conscious.

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devphilamaths · 3 years

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Solve cos(pi/2+theta) | cos(pi/2 +x) | cos pi/2 + x formula, Find value cos pi by 2 + x Hi friends.. In this tutorial Find the Value of cos(pi/2+theta), cos(pi/2 +x)cos pi/2 + x formula Exact Value of cos pi/2 + x Find the value of cos pi by 2 +...

#Solve cos(pi/2+theta)#cos(pi/2 +x)#cos pi/2 + x formula #Find value cos pi by 2 + x #Exact Value of cos pi/2 + x

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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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devphilamaths · 3 years

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Solve cosec(pi/2+theta) | cosec pi/2 + x formula, csc (pi/2 +x) | Find value cosec pi by 2 + x

#Solve cosec(pi/2+theta)#cosec pi/2 + x formula #csc (pi/2 +x)#Find value cosec pi by 2 + x #math #12th class #formula #sin #cos #ten #cosec #sec #cot

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devphilamaths · 3 years

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Cos(π/2-x) =sinx solve trigonometric identities | Cos(pi/2-theta) =sintheta solve 12th Class Math

Cos(π/2-x) =sinx

#Cos(π/2-x) =sinx #Cos(π/2-x) =sinx solve trigonometric identities #Cos(pi/2-theta) =sintheta solve 12th Class Math #Cos(pi/2-theta) =sintheta solve #=sinx

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devphilamaths · 3 years

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sin(π/2-x) =cosx solve trigonometric identities | sin(pi/2-theta) =sintheta

sin(π/2-x) =cosx

sin(π/2-x)

sin

12th IIT JEE

#sin(π/2-x) =cosx #sin(π/2-x) =cosx solve #sin(π/2-x) =cosx solve trigonometric identities #sin(pi/2-theta) =sintheta #sin(pi/2-theta)#sin #sin(π/2-x)

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mathematicianadda · 4 years

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Simplify $\tan^{-1} ( \frac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}} )$ with trigonometric substitution https://ift.tt/eA8V8J

I will explain my approach, help me with the last step please! $$ \tan^{-1} {\left(\frac {x - \sqrt {1-x^2}}{x + \sqrt {1-x^2}}\right)}$$

substituting x = $\sin \theta$ (as learnt from book) and solving 1-$\sin^2 \theta$ = $\cos^2 \theta$ $$ \tan^{-1} {\left(\frac {\sin \theta - |\cos \theta|}{\sin \theta + |\cos \theta| }\right)}$$

For solving modulus, it was important to determine range of $\theta$ , therefore I defined it (as it is my variable,i can define it my way) for [-$\pi$/2 , $\pi$/2] so that sine covers all values from $-1$ to $1$ (as , $ -1 \le x \le 1 \,$ , from domain ) and $\cos \theta$ is positive , and hence $|\cos \theta| = \cos \theta$.

$$ \tan^{-1} {\left(\frac {\sin \theta - \cos \theta}{\sin \theta + \cos \theta }\right)}$$ = dividing by $\cos \theta$ $$ \tan^{-1} {\left(\frac {\tan \theta - 1}{\tan\theta + 1 }\right)}$$

= by formula of $\tan (\theta - \pi/4)$ $$ \tan^{-1}( \tan{\left(\theta - \pi/4\right)})$$

That's where I am stuck ,as according to the identity,$\quad$ $tan^{-1} ( \tan \alpha) = \alpha$ $\quad$ only when $\, -\pi/2 <\alpha < \pi/2$ . But here $$ -3\pi/4 \le \,(\theta-\pi/4) \, \le \pi/4 $$ Therefore, I am not going to get ($ \,\theta - \pi/4 $) out of the expression. What i get will be based on that graph of $\bf {\tan^{-1} (\tan x)}$ . $$ (\theta - \pi/4) +\pi \,$$ for $\,-3\pi/4 \le \, (\theta -\pi/4) \, < -\pi/2 \,\,$ and

$$\theta -\pi/4$$ for $\,-\pi/2 < \, (\theta -\pi/4) \, \le \pi/4 \,\,$

My teacher just cancelled arctan and tan and wrote $\theta - \pi/4$ and he didn't even include that modulus function over $\cos \theta$.

So what will be the exact answer because if everyone decide $\theta$ as per they like then there will not be a finite answer. Everyone will have their own answers and in each answer they have multiple cases as I just discussed above.

So please help me, very hopefully I signed up in stackexchange!

Found Solution :-

I was confused because I was thinking that there can be many solutions differing person to person, but even if you choose any value of $\theta$ , you are going to get two solutions which are in the asked question above. The problem resolves when we write $\theta$ in terms of $sin^{-1} x$ as then we would not simply write like $$ \theta = \sin^{-1} x $$ we would write an equation,$$ \sin^{-1} x = \sin^{-1} (\sin \theta)$$, now if $\theta$ is not in range of $-\pi/2$ and $\,\pi/2$ , then there would be some constant in $\pi$ (like , $\pi/4 , 2\pi$ etc. we would have to add or subtract according to the graph of 'sin inverse sin' and when we would put that value of $\theta$ , we would end with the solutions as answered by people. (I write the answer in this edit to help anyone who will reach here after searching web , thanks to everyone for answers)

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mathematicianadda · 4 years

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Integrate a weighted Bessel function over the unit disk https://ift.tt/eA8V8J

I would like to evaluate a complex-valued integral of the form

$$ I_e = \int_0^1 x e^{iax} J_0(b \sqrt{1-x^2}) dx $$

where $a$ and $b$ are positive real numbers and $J_0(z)$ is the Bessel function of the first kind. I am particularly interested in the special case of $a=(c+1)b$ for $c \ll 1$.

The task boils down to evaluating two real-valued integrals

$$ I_s = \int_0^1 x \sin(ax) J_0(b \sqrt{1-x^2}) dx $$ $$ I_c = \int_0^1 x \cos(ax) J_0(b \sqrt{1-x^2}) dx $$

The integral with the sine has a simple form given by Gradshteyn and Ryzhik (6.738.1) which, after simplification, becomes

$$ I_s =\sqrt{\frac{a^2}{a^2 + b^2}}j_1(\sqrt{a^2 + b^2}) $$

where $j_1(z)$ is the spherical Bessel function of the first kind.

I am not exactly sure how this expression was derived. Perhaps it holds a clue. I tried substituting the integral form of the Bessel function and integrating analytically but did not get very far.

By symmetry, I naively expected the integral involving the cosine to be proportional to the spherical Bessel function of the second kind $y_1(z)$ (and thus, the complex-valued integral to be proportional to the spherical Hankel function of the second kind), but that does not appear to be the case.

$$ I_c \approx -\sqrt{\frac{a^2}{a^2 + b^2}}y_1(\sqrt{a^2 + b^2}) $$

The agreement for the special case of $a=b$ is not too terrible, but something is going wrong for small values of $b$. If I take the difference between the true value of the integral and my guess, it is proportional to another Bessel function, but the arguments are crazy and it does not make any sense.

Now, if I assume that it is proportional to the spherical Bessel function of the first kind instead, and compute the ratio between the true value of the integral and my guess, I get the classic graph of the cotangent.

For $b \gg 1$, it appears

$$ I_c \approx \cot(\sqrt{a^2 + b^2} + \phi)) \sqrt{\frac{a^2}{a^2 + b^2}}j_1(\sqrt{a^2 + b^2} + \phi) $$

In practice, this is hardly useful, since the period of the cotangent would have to correspond to zeros of the spherical Bessel function, and these are hard to compute. Additionally, $j_1(z)$ and the actual $I_c$ appear to be slightly off-phase.

Of course, ideally, I would like to solve the problem analytically, but I am not sure how (using a series expansion, perhaps?). I would appreciate any tips or guidance.

Thank you!

Edit: by transforming using $x=\sin{\theta}$, there is an almost perfect match in Gradshteyn and Ryzhik (6.688.2), except that the leading term in my case is $\sin{2\theta}$ rather than $\sin{\theta}$. Additionally, it gives a result in terms of the spherical Bessel function of the first kind rather than the second kind, so it does not explain my empirical approximation.

$$ I_c = \int_0^{\pi/2} \frac{1}{2} \sin{2\theta} \cos(a \cos{\theta}) J_0(b \sin{\theta}) d\theta $$

Using the representation given above, we can expand $\cos(a \cos{\theta})$ in a series, which does give an analytic solution, but it converges rather poorly and does not appear to contain any spherical Bessel terms (just the regular Bessel functions). I am yet to find an expansion in terms of spherical Bessel functions.

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