Beautiful Proofs(#3): Area under a sine curve !

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 :

Screenshot from 2017-06-07 00:19:11

\sum sin(\theta)d\theta =   Diameter of the circle/ The distance covered along the x axis starting from 0 and ending up at \pi.

And therefore by the same logic, it is extremely intuitive to see why:

\int\limits_{0}^{2\pi} sin/cos(x) dx = 0

Because if a dude starts at 0 and ends at 0/ 2\pi/ 4\pi \hdots, the effective distance that he covers is 0.

Circle_cos_sin.gif

If you still have trouble understanding, follow the blue point in the above gif and hopefully things become clearer.

 

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Why is the area under one hump of a sine curve exactly 2?

Girls' Angle

blog_073013_02

I was talking with a student recently who told me that he always found the fact that $latex int_0^{pi} sin x , dx = 2$ amazing. “How is it that the area under one hump of the sine curve comes out exactly 2?” He asked me if there is an easy way to see that, or is it something you just have to discover by doing the computation.

If you’ve wondered about this too, perhaps you’ll find the following of interest.

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A strange operator

In a previous post on using the Feynman’s trick for Discrete calculus, I used a very strange operator ( \triangledown ). And whose function is the following :

\triangledown n^{\underline{k}} = \frac{n^{\underline{k+1}}}{k+1}

What is this operator? Well, to be quite frank I am not sure of the name, but I used it as an analogy to Integration. i.e

\int x^{n} = \frac{x^{n+1}}{n+1} + C

What are the properties of this operator ? Let’s use the known fact that n^{\underline{k+1}} = (n-k) n^{\underline{k}}

\triangledown n^{\underline{k}} = \frac{n^{\underline{k+1}}}{k+1}

\triangledown n^{\underline{k}} = \frac{(n-k) n^{\underline{k}}}{k+1}

And applying the operator twice yields:

\triangledown^2 n^{\underline{k}} = \frac{n^{\underline{k+2}}}{(k+1)(k+2)}

\triangledown^2 n^{\underline{k}} = \frac{(n-k-1) n^{\underline{k+1}}}{(k+1)(k+2)}

\triangledown^2 n^{\underline{k}} = \frac{(n-k-1)(n-k) n^{\underline{k}}}{(k+1)(k+2)}

We can clearly see a pattern emerging from this already, applying the operator once more :

\triangledown^3 n^{\underline{k}} = \frac{(n-k-2)(n-k-1)(n-k) n^{\underline{k}}}{(k+1)(k+2)(k+3)}

\vdots

Or in general, the operator that has the characteristic prescribed in the previous post is the following:

\triangledown^m n^{\underline{k}} = \frac{n^{\underline{k+m}}}{(k+m)^{\underline{m}}} n^{\underline{k}}

If you guys are aware of the name of this operator, do ping me !

Inverse of an Infinite matrix

\begin{bmatrix} 0 & 1 & 0 & 0 & 0 & 0 & \hdots \\ 0 & 0 & 2 & 0 & 0 & 0 & \hdots \\ 0 & 0 & 0 & 3 & 0 & 0 & \hdots \\ 0 & 0 & 0 & 0 & 4 & 0 & \hdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \end{bmatrix}

What is the inverse of the above matrix ? I would strongly suggest that you think about the above matrix and what its inverse would look like before you read through.

On the face of it, it is indeed startling to even think of an inverse of an infinite dimensional matrix. But the only reason why this matrix seems weird is because I have presented it out of context.

You see, the popular name of the matrix is the Differentiation Matrix and is commonly denoted as D.

The differentiation matrix is a beautiful matrix and we will discuss all about it in some other post, but in the this post lets talk about its inverse. The inverse of the differentiation matrix is ( as you might have guessed ) is the Integration Matrix (I^*)

I^* = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & \hdots \\ 1 & 0 & 0 & 0 & 0 & 0 & \hdots \\ 0 & \frac{1}{2} & 0 & 0 & 0 & 0 & \hdots \\ 0 & 0 & \frac{1}{3} & 0 & 0 & 0 & \hdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \end{bmatrix}

And it can be easily verified that DI^* = I, where I is the Identity matrix.

Lesson learned: Infinite dimensional matrices can have inverses. 😀

 

 

On the beauty of Parametric Integration and the Gamma function

Parametric integration is one such technique that once you are made aware of it, you will never for the love of god forget it. Let me demonstrate :

Now this integral might seem familiar to many of you and to evaluate it is rather simple as well.

\int\limits_0^{\infty} e^{-sx} dx = \frac{1}{s}

Knowing this you can do lots of crazy stuff. Lets differentiate this expression wrt to the parameter in the integral – s (Hence the name parametric integration ). i.e

\frac{d}{ds}\int\limits_0^{\infty} e^{-sx} dx = \frac{d}{ds}\left(\frac{1}{s}\right)

\int\limits_0^{\infty} x e^{-sx} dx = \frac{1}{s^2}

Look at that, by simple differentiation we have obtained the expression for another integral. How cool is that! It gets even better.
Lets differentiate it once more:

\int\limits_0^{\infty} x^2 e^{-sx} dx = \frac{2*1}{s^3}

\int\limits_0^{\infty} x^3 e^{-sx} dx = \frac{3*2*1}{s^4}

\vdots

If you keep on differentiating the expression n times, one gets this :

\int\limits_0^{\infty} x^n e^{-sx} dx = \frac{n!}{s^{n+1}}

Now substituting the value of s to be 1, we obtain the following integral expression for the factorial. This is known as the gamma function.

\int\limits_0^{\infty} x^n e^{-x} dx = n! = \Gamma(n+1)

There are lots of ways to derive the above expression for the gamma function, but parametric integration is in my opinion the most subtle way to arrive at it. 😀

Source: http://www.maa.org/sites/default/files/268948443847.pdf