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Author: Subject: Calculus! For beginners, with a ‘no theorems’ approach!
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[*] posted on 30-4-2016 at 11:42


Oh bugger.

I remember that formula (vaguely) from school, but did not recognise the derivative as a Quadratic equation.

Heigh-ho.

Got it half right, which is pretty good for a Saturday !




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[*] posted on 3-5-2016 at 05:45
Minimum distance problem:


A planner for Vulture Chemicals plots a grid on a map and determines that VC’s four most important customers are located at A(1, 5), B(0, 0), C(8, 0) and D(4,9) where units are in 100 miles. At what point W(x, y) should a warehouse be located in order to minimize the sum of the distances from W to A, B, C, and D?

Distance optimisation 2.png - 6kB

Preamble:

In a Cartesian coordinate system the distance between two points:
$$(x_1,y_1),(x_2,y_2)$$
... can be calculated as follows:
$$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$
For this problem it's easier to use the squares of the distances:
$$d^2=(x_1-x_2)^2+(y_1-y_2)^2$$
If S is the sum of the squares of the distances for all four customers to the planned warehouse:
$$S=\Sigma d^2$$
Then:
$$S(x,y)=[(x-1)^2+(y-5)^2]+(x^2+y^2)+[(x-8)^2+y^2]+[(x-4)^2+(y-9)^2]$$
By means of the partial derivatives fx and fy, find the minimum for S.

(This is the last optimisation problem. Next: famous second order DVs)




[Edited on 3-5-2016 by blogfast25]




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[*] posted on 3-5-2016 at 05:50


What you do is not equivalent. Minimizing the sum of distances in general need not give the same result as minimizing the sum of the squares of distances. It may be that in this particular case the answer is the same (I did not do any math on this), but in general this certainly is not the case.



The art of wondering makes life worth living...
Want to wonder? Look at https://woelen.homescience.net
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[*] posted on 3-5-2016 at 07:32


Is this the S(x,y) = 0 thing ?

If so,
$$S(x,y)=[(x-1)^2+(y-5)^2]+(x^2+y^2)+[(x-8)^2+y^2]+[(x-4)^2+(y-9)^2]$$
$$=x^2-2x+1+y^2-10y+25+x^2+y^2+x^2-16x+64+y^2+x^2-8x+16+y^2-18y+81$$
$$=4x^2+4y^2-26x-28y+187$$
$$f_x = 8x-26, f_y = 8y-28$$
$$f_x = 0, x=\frac {26}{8} = 3.25$$
$$f_y = 0, y=\frac {28}{8} = 3.5$$
$$\therefore S(x,y) = 0 = S(3.25, 3.5)$$

It looks like point D should be (9, 4) where it is shown on the plot, rather than (4, 9).

@woelen: i think the excercise was made more interesting by having an actual application.

Whether it works or not is a moot point, especially if it's me doing the calculations !




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[*] posted on 3-5-2016 at 08:16


@woelen; I'll address your point a little later on. Thanks!

@aga:

Correct!

But your final statement is not:

$$S(x,y) = 0 = S(3.25, 3.5)$$

The sum of the distances is far from zero. But fx=fy=0, for (3.25,3.5).

The D point was badly drawn, but the (4,9) coordinates were correct and as intended.

It's also a little quicker if you use the chain rule on terms like:
$$(x-4)^2$$
$$f_x=2(x-4) \times f_x(x-4)=2(x-4) \times 1=2(x-4)$$
Still, as they say: 'whatever works for you'.

[Edited on 3-5-2016 by blogfast25]




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[*] posted on 3-5-2016 at 08:29


Quote: Originally posted by woelen  
What you do is not equivalent. Minimizing the sum of distances in general need not give the same result as minimizing the sum of the squares of distances. It may be that in this particular case the answer is the same (I did not do any math on this), but in general this certainly is not the case.


You're correct.

If:
$$d_i=\sqrt{(x-x_i)^2+(y-y_1)^2}$$
Then:
$$S=\Sigma d_i=\Sigma_i \sqrt{(x-x_i)^2+(y-y_1)^2}$$
And:
$$f_x=\Sigma_i \frac{(x-x_i)}{\sqrt{(x-x_i)^2+(y-y_1)^2}}$$
$$f_y=\Sigma_i \frac{(y-y_i)}{\sqrt{(x-x_i)^2+(y-y_1)^2}}$$
Then that would not give the same results as we computed.

I found the exercise somewhere, made it a little harder but didn't stop to think. Grrr. :mad: :)

[Edited on 3-5-2016 by blogfast25]




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[*] posted on 3-5-2016 at 09:27


Quote: Originally posted by blogfast25  
I found the exercise somewhere, made it a little harder but didn't stop to think. Grrr. :mad: :)

Teaching is a 2 way thing.

Maybe a little has rubbed off ;)




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[*] posted on 3-5-2016 at 11:29


Quote: Originally posted by aga  

Maybe a little has rubbed off ;)


Don't get me worried... :(




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[*] posted on 3-5-2016 at 11:33
Famous second order DEs:


Newtonian equations of motion:

Consider an object of mass m on which a number of forces act, shown below are 3 forces:

Newton's second law.png - 8kB

Then Newton’s Second Law tells us:
$$\Sigma_i \vec{F_i}=m\vec{a}$$
Worded: 'the vector sum of the acting forces equals the mass times the acceleration vector'

Analytically, this is usually solved by using the property that motions in the x, y and z directions are completely independent of each other. This allows to break down the vector problem into three independent non-vector equations:
$$\Sigma_i F_{x,i}=ma_x$$
$$\Sigma_i F_{y,i}=ma_y$$
$$\Sigma_i F_{z,i}=ma_z$$

From now on we’ll drop the indices x, y and z because it is usually very clear from the context of the problem which axis is meant. Also we’ll call:
$$\Sigma_i F_i=F$$
Where F is the sum of the force components along the intended axis, often also called the net force (along that axis).
So for a given axis we can now simply write:
$$ma=F$$
Also:
$$m\frac{dv}{dt}=F$$
And (for an x-axis):
$$m\frac{d^2x}{dt^2}=F$$
Specific examples:

1. F is constant:

Real world example: a free falling body (without drag).
$$m\frac{dv}{dt}=F$$
Integrate with boundary conditions t=0, v=v<sub>0</sub>, x=x<sub>0</sub>:
$$\int_{v_0}^v dv=\int_0^t \frac{F}{m}dt$$
$$\implies v(t)=v_0+\frac{F}{m}t$$
$$v(t)=\frac{dx(t)}{dt}=v_0+\frac{F}{m}t$$
Integrate again:
$$\int_{x_0}^xdx=\int_0^t(v_0+\frac{F}{m}t)dt$$
$$\implies x(t)=x_0+v_0t+\frac{F}{2m}t^2$$
2. F is a function of v:
$$ma=f(v)$$
$$m\frac{dv}{dt}=f(v)$$
Integrate:
$$m\int \frac {dv}{f(v)}=\int dt$$
Alternatively:
$$mx''(t)=f[x'(t)]$$
Real world example: projectile with drag. We derived such a problem higher up:
$$ma=mg-F_{drag}$$
Specifically:
$$v’=9.8-\frac{5v}{50}=9.8-\frac{v}{10}$$
We solved this by twice integrating, to obtain x(t).

But it can also be solved by writing it as:
$$\frac{d^2x}{dt^2}=9.8-\frac{1}{10}\frac{dx}{dt}$$
Or:
$$x''+0.1x'-9.8=0$$
The boundary conditions were:
$$t=0, x(0)=0, x'(0)=-10$$
I used Wolfram Alpha's DSolve function for this and obtained:
$$x(t)=98t+1080e^{-0.1t}-1080$$
Which is also what we found before.

[Edited on 3-5-2016 by blogfast25]




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[*] posted on 3-5-2016 at 11:48


In
$$\Sigma_i \vec{F_i}=m\vec{a}$$
what does the i signify ?

The arrows mean Vectors, although quite how they work is a mystery.

A bit of utoob showed a tail-to-tip example which made sense in isolation (vector addition).

Putting them into an equation, Fwooosh ! Straight over head.

Edit:

It is nice to see that Wolfram got it right as well ;)

Another thing: why is an accurate depiction of my brain being used in the example image ?


[Edited on 3-5-2016 by aga]




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[*] posted on 3-5-2016 at 12:20


Quote: Originally posted by aga  
In
$$\Sigma_i \vec{F_i}=m\vec{a}$$
what does the i signify ?

The arrows mean Vectors, although quite how they work is a mystery.

A bit of utoob showed a tail-to-tip example which made sense in isolation (vector addition).

Putting them into an equation, Fwooosh ! Straight over head.

Edit:

It is nice to see that Wolfram got it right as well ;)

Another thing: why is an accurate depiction of my brain being used in the example image ?


i is the index (number) of a force, as in:
$$F_1,F_2,...,F_i,...,F_n$$

Vector calculation:

Vectors.png - 4kB

Right: an example of two vectors being added up.

Left: F would be the resultant vector from adding up all vectors on the brain:

$$\Sigma_i \vec{F_i}=\vec{F}$$
The acceleration vector

$$\vec{a}$$

... is obtained by:

$$\vec{a}=\frac{\vec{F}}{m}$$

Both F and a vectors thus are parallel and point in the same direction, they're simply different in size.

But as explained, usually we kind of avoid all that vectoring by looking at the x,y,z force components.

I'll post a simple example.

[Edited on 3-5-2016 by blogfast25]




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[*] posted on 3-5-2016 at 12:33


So in
$$\Sigma_i \vec{F_i}=\vec{F}$$
the meaning of
$$\vec{F}$$

is : "The sum of all i nvolved vectors"

Figuring out the Words is almost as hard as the maths.




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[*] posted on 3-5-2016 at 12:52


Quote: Originally posted by aga  
So in
$$\Sigma_i \vec{F_i}=\vec{F}$$
the meaning of
$$\vec{F}$$

is : "The sum of all i involved vectors"

Figuring out the Words is almost as hard as the maths.


Yes!

Now hold your fire for 5 mins.




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[*] posted on 3-5-2016 at 13:10
Object sliding down an incline:


Consider an object of mass m sliding down a smooth incline (angle theta), without friction or drag:

Ball on an incline.png - 5kB

Carefully note the choice of x and y-axis, please.

What are the forces and accelerations in play?

The weight mg is decomposed into an x and a y-component.

y-axis:

$$F_y=-mg\cos \theta$$

This force is countered by a reaction force FN provided by the incline, so that:

$$F_N-mg\cos\theta=0$$
There's no NET force acting on the object in the y-direction.
$$\implies a_y=0$$

No acceleration in the y-direction.

x-axis:

$$F_x=mg\sin \theta$$

In the absence of friction or drag, Fx is also the ONLY force acting on the object along the x-axis, so:

$$ma_x=mg\sin\theta$$
$$\implies a_x=g\sin \theta$$

The object will accelerate along the x-axis with constant acceleration:
$$a_x=g\sin \theta$$
<hr>
Fun with a shopping trolley and two toddlers!

You're pushing a shopping trolley with force F1. Toddler 1 want to pull the trolley towards the chocolates with force F3, while toddler 2 want to pull it to the toys, with force F2:

Shopping trolley.png - 12kB

On the right we add up the force vectors, first F3 + F1, then add to that F2, to get the resultant ('net') force FR.

The trolley will experience an acceleration vector a:

$$\vec{a}=\frac{\vec{F_R}}{m}$$

[Edited on 4-5-2016 by blogfast25]




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[*] posted on 4-5-2016 at 10:25


Could the Distribution centre problem not be solved with vector maths ?

I.e. the distance being the Force, and the angle being the other bit of the vector.

I guess it'd get complicated finding a common solution, unless that's what vector derviatives are all about.




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[*] posted on 4-5-2016 at 11:05


Quote: Originally posted by aga  
Could the Distribution centre problem not be solved with vector maths ?



Let think about that.

Coming up, another example of a Newtonian equation of motion.




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[*] posted on 4-5-2016 at 11:26


I started thinking about working from the Centre and solving for each of the vector forces, then realised that the Centre is the actual answer.

Amazingly, over 10 years ago i posed this exact question on a freelance programmer's website to get a way to calculate the position of a mobile phone tower based on GPS co-ordinates with TA values in order to map where all the mobile towers are.

TA=Timing Advance = 1 for every 550m your phone is away from the tower.

It's used for keeping phones at differing distances away from the tower in synch with the tower's timing chain.

The idea was to implement a non-GPS based positioning system based on Known tower sites.

Cheap too, as the GSM units do not even need a SIM card (=no contract) to get the TA information (phones MUST connect with no SIM to do 999/112 calls).

Never did get any useful solution, then they went and released the tower locations on a public website.

Contemporaneously Google released Google Maps which blew away all the vector mapping companies.




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[*] posted on 4-5-2016 at 11:28


3. F is a function of x:

Real world example: Simple Harmonic Oscillator.

SHO.png - 4kB

A massless Hookean spring with spring constant k is anchored at A and attached to an object of mass m. Ignore all friction or drag.

Initially the spring is neither extended nor compressed but at t=0 we move the object to x=x0, then release it. The restoring force exerted by the spring is kx and the Newtonian equation is:

$$ma_x=-kx(t)$$
$$a_x=\frac{d^2x(t)}{dt^2}=x''(x)$$
$$mx''(t)+kx(t)=0$$
Or in 'shorthand':
$$mx''+kx=0$$
We assume the solution to be of the general form:
$$x=C_2\cos C_1 x$$
So:
$$x'=-C_2C_1 \sin C_1 x$$
$$x''=-C_2C_1^2\cos C_1 x$$
Plug it all into the DV:
$$-mC_2C_1^2\cos C_1 x+kC_2\cos C_1 x=0$$
C2cosC1x drops out and we get:
$$mC_1^2=k$$
$$C_1=\sqrt{\frac{k}{m}}$$
Now with the boundary condition:
$$t=0, x=x_0$$
$$\implies C_2=x_0\cos 0=x_0$$
So:
$$x(t)=x_0\cos \Big(\sqrt{\frac{k}{m}}t\Big)$$
C1 is often called the angular velocity, omega:
$$\omega=\sqrt{\frac{k}{m}}$$
The frequency of oscillation is:
$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$$
The period of oscillation is:
$$T=\frac1f$$
<hr>
Would it matter if we rotate the whole set up, axis included, clockwise by 90 degrees? The object would then be hanging off the spring.

Let's see. The Newtonian equation now is:

$$ma=mg-kx(t)$$
$$mx''(t)+kx(t)-mg=0$$
$$mx''(t)+k[x(t)-\frac{mg}{k}]=0$$
Substitute:
$$u=x(t)-\frac{mg}{k}$$
$$u'=x'(t)$$
$$u''=x''(t)$$
$$mu''+ku=0$$
The derivation is the same as above and yields:
$$x(t)=\Big(x_0-\frac{mg}{k}\Big)\cos \Big(\sqrt{\frac{k}{m}}t\Big)$$
So gravity doesn't affect angular velocity, frequency or period of oscillation. Only amplitude is affected.

[Edited on 4-5-2016 by blogfast25]




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[*] posted on 4-5-2016 at 11:35


Wow !

I assumed Gravity would have an effect on the frequency, although that may be due to naturally shoving friction in there.

[Edited on 4-5-2016 by aga]




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[*] posted on 4-5-2016 at 13:35


Quote: Originally posted by aga  

Amazingly, over 10 years ago i posed this exact question on a freelance programmer's website to get a way to calculate the position of a mobile phone tower based on GPS co-ordinates with TA values in order to map where all the mobile towers are.



There's a numerical solution to the minimum distance problem.

Create a Cartesian x,y grid (shown in blue) and calculate the sum of distances for each x,y point. A visual inspection (or Excel triage) will show you where you are near to the optimum (x,y) point. Repeat with a finer mesh x,y grid in that area, for greater precision.

Distance optimisation 3.png - 5kB

Quote: Originally posted by aga  
Wow !

I assumed Gravity would have an effect on the frequency, although that may be due to naturally shoving friction in there.

[Edited on 4-5-2016 by aga]


There are numerous applications of SHOs (see e.g. pendulum clocks) and some with engineered friction, like a car's shock absorbers which provides a critically damped SHO:

http://hyperphysics.phy-astr.gsu.edu/hbase/oscda2.html

In such a device the initial perturbation x0 is absorbed and then the system returns to x=0 as soon as possible.

The idealised DV is:

$$mx''+kx-g=\epsilon x'$$

... because the friction force is:

$$F_F=\epsilon v=\epsilon x'$$


[Edited on 5-5-2016 by blogfast25]




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[*] posted on 5-5-2016 at 11:55


I imagined the solution would be simpler in polar co-ordinates, seeing as the Location and Distance are the 'known' elements.

Perhaps that's why it never got figured out properly.




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[*] posted on 5-5-2016 at 12:36


Quote: Originally posted by aga  
I imagined the solution would be simpler in polar co-ordinates, seeing as the Location and Distance are the 'known' elements.

Perhaps that's why it never got figured out properly.


Polar coordinates wouldn't help at all here. And distance is definitely an unknown element here...

More about the facility location problem, here: it's hard and has been from about 1860! :D

[Edited on 5-5-2016 by blogfast25]




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[*] posted on 5-5-2016 at 12:46


Sorry, i was thinking about the mobile phone towers problem there.

The TA value is basically a discrete integer representing a 550m increment per step : basically the distance to the the tower with crappy granularity.

So the two known values would be GPS co-odinates of the A,B,C,D locations, each with a known TA (=crappy granuarity distance) from the central point = (The tower).

Problem was to calculate the GPS co-ordinates of that central point from a set of n GPS readings and measured TA values.

The geometry kind of suggested that at least 3 readings from varied positions would be needed to get an actual 'fix'.

Edit: these were the considered cases :

GPS.gif - 6kB

'X' marks the tower spot.

A,B, C are the known positions, the radius from each being the known distance to X.

F is the false positive, it being the same distance from A and B.

[Edited on 5-5-2016 by aga]




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[*] posted on 6-5-2016 at 08:14
The Schrödinger equation (SE):


No mini-Pantheon of famous second order DEs can be complete without one of the most successful physics equations ever. Here it is again, time-independent and for one dimension only:
$$-\frac{\hbar^2}{2m}.\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x)$$
Now let’s see if with our new-found calculus ability we can analytically solve the simplest of all quantum problems: a particle in a one dimensional box (Pi1DB, see also the QP thread):

Schrodinger box.png - 3kB

Mathematically:
$$-\frac{L}{2} \leq x\leq +\frac{L}{2}\:\text{then}\: V=V_0$$
$$|x| > \frac{L}{2} \:\text{then}\: V=+\infty$$
So the SE becomes:
$$-\frac{\hbar^2}{2m}.\frac{d^2\psi(x)}{dx^2}+V_0\psi(x)=E\psi(x)$$
Or:
$$-\frac{\hbar^2}{2m}.\frac{d^2\psi(x)}{dx^2}=(E-V_0)\psi(x)$$
So we’re looking for a function that twice derived kind of, sort of looks like the negative of itself. That’s quite reminiscent of:
$$(\sin x)'=\cos x$$
$$(\cos x)'=-\sin x$$
Or:
$$-(\sin x)''=\sin x$$
So we tentatively propose a solution of the form:
$$\psi=A\sin kx$$
$$\psi'=Ak\cos kx$$
$$\psi''=-Ak^2\sin kx$$
Plug it all into the SE:
$$-\frac{\hbar^2}{2m}(-Ak^2\sin kx)=(E-V_0)A\sin kx$$
$$\frac{\hbar^2}{2m}k^2=(E-V_0)$$
So it seems to be a fit but what are k and A?

We know that:
$$x=+\frac{L}{2}\implies \psi=0$$
$$\implies A\sin \Big(k\frac{L}{2}\Big)=0$$
And because:
$$A \neq 0$$
$$\implies \frac{kL}{2}=2\pi n$$
$$n=1,2,3,...$$
So:
$$k=\frac{4\pi n}{L}$$
$$\psi=A\sin \Big(\frac{4\pi n}{L}x\Big)$$
Also:
$$E_n=V_0+\frac{8\pi^2\hbar^2n^2}{mL^2}$$
To find A we need to apply the Normalisation condition:
$$\int_{-L/2}^{+L/2}\psi^2dx=1$$
$$\int_{-L/2}^{+L/2}\big[A\sin \Big(\frac{4\pi n}{L}x\Big)\Big]^2dx=1$$
It’s a banal integral which I’ll leave for readers to check, which yields:
$$A=\sqrt{\frac{2}{L}}$$
$$\psi_n(x)=\sqrt{\frac{2}{L}}\sin \Big(\frac{4\pi n}{L}x\Big)$$

First 8 wave functions (not to scale):

wave functions.png - 66kB

[Edited on 6-5-2016 by blogfast25]

[Edited on 6-5-2016 by blogfast25]




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[*] posted on 6-5-2016 at 08:59


This thread suddenly passed my level of knowledge, haha. I just took my AP Calculus AB test for university credit yesterday. Good stuff.



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