Parametric Integration is an Integration technique that was popularized by Richard Feynman but was known since Leibinz’s times. But this technique rarely gets discussed beyond a niche set of problems mostly in graduate school in the context of Contour Integration.
A while ago, having become obsessed with this technique I wrote this note on applying it to Laplace transform problems and it is now public for everyone to take a look.
( Link to notes on Google Drive )
I would be open to your suggestions, comments and improvements on it as well. Cheers!
When one is solving problems on the two dimensional plane and you are using polar coordinates, it is always a challenge to remember what the velocity/acceleration in the radial and angular directions () are. Here’s one failsafe way using complex numbers that made things really easy :
From the above expression, we can obtain and
From this we can obtain and with absolute ease.
Something that I realized only after a mechanics course in college was done and dusted but nevertheless a really cool and interesting place where complex numbers come in handy!
So, I read this post on the the area of the sine curve some time ago and in the bottom was this equally amazing comment :
Diameter of the circle/ The distance covered along the x axis starting from and ending up at .
And therefore by the same logic, it is extremely intuitive to see why:
Because if a dude starts at and ends at , the effective distance that he covers is 0.
If you still have trouble understanding, follow the blue point in the above gif and hopefully things become clearer.
In this post, let’s derive a general solution for the Laplacian in Spherical Coordinates. In future posts, we shall look at the application of this equation in the context of Fluids and Quantum Mechanics.
The Laplacian in Spherical coordinates in its ultimate glory is written as follows:
To solve it we use the method of separation of variables.
Plugging in the value of into the Laplacian, we get that :
Dividing throughout by and multiplying throughout by , further simplifies into:
It can be observed that the first expression in the differential equation is merely a function of and the remaining a function of and only. Therefore, we equate the first expression to be and the second to be . The reason for choosing the peculiar value of is explained in another post.
The first expression in (1) the Euler-Cauchy equation in .
The general solution of this has been in discussed in a previous post and it can be written as:
The second expression in (1) takes the form as follows:
The following observation can be made similar to the previous analysis
The first expression in the above equation (2) is the Associated Legendre Differential equation.
The general solution to this differential equation can be given as:
The solution to the second term in the equation (2) is a trivial one:
Therefore the general solution to the Laplacian in Spherical coordinates is given by:
Many a times it is not discussed as to How the Taylor/Maclaurin series came to be in its current form. This short snippet is all about it.
Let us assume that some function can be written as a power series expansion. i.e
We are left with the task of finding out the coefficients of the power series expansion.
Substitution x = 0, we obtain the value of .
Lets differentiate wrt x.
Evaluating at x =0 , we get
That’s it we have found all the coefficient values, the only thing left to do is to plug it back into the power series expression:
The above series expanded about the point x = 0 is called as the ‘Maclaurin Series’. The same underlying principle can be extended for expanding about any other point as well i.e ‘Taylor Series’.
When you are working with Spherical harmonics, then the Legendre Differential Equation does not appear in its natural form i.e
Instead, it appears in this form:
It seems daunting but the above is the same as the LDE. We can arrive at it by taking and proceeding as follows:
Now, applying chain rule, we obtain that
Now simplifying the above expression, we obtain that:
Plugging in the values of and into the Legendre Differential Equation,
Now if we do some algebra and simplify the trigonometric identities, we will arrive at the following expression for the Legendre Differential Equation:
If we take the solution for the LDE as , then the solution to the LDE in the above form is merely .