Gaussian Integral, the Jacobian and the Gamma Function

To evaluate the Gaussian Integral requires the use of double integrals and the Jacobian for the change of variables from rectangular to polar coordinates. The following YouTube video gives a clear account of how this integral is evaluated:

The following video deals with the gamma function for value 1/2 (=1/2!) and the use of the above integral technique to evaluate it.

The final video is by the author of the second video and is an introduction to the gamma function. I'm just skimming the surface of things here but it's a start.

The Wallis Formula for Pi and the Dawson Function

I spent quite some time watching this video on the derivation of the Wallis product and practised until I could reproduce it without any external assistance. A crucial part of the solution relies on integration by parts to set up a reduction formula for the integral of \( sin(x)^n\). Here is the very well-presented and easy to understand video:

Note: the following discussion centres on integration by parts and is not related to the Wallis function.

I sometimes practise using integration by parts to solve integrals that I think of and last night my mind fell on a deceptively easy-looking integral, namely \(e^{x^2}\). The graph of this function is well-behaved and I thought that there would be an easy solution but try as I might I couldn't find it. Reluctantly, I checked first Symbolab and then WolframAlpha to find out how it could be done. Here's what the former had to say:

What on earth is this F(x) that just appears out of nowhere? WolframAlpha offered the same solution but had accompanying documentation that described F(x) as the Dawson integral defined by:

There is a quite comprehensive article about the Dawson integral or Dawson Function, as it's alternatively called, on Wikipedia but it's largely incomprehensible to me at the moment. Maybe I can come to terms with it later. 

The Harmonic Series

Here's an interesting problem: an ant traverses a circle with a circumference of one metre at a rate of one centimetre per second. After each second, the circumference of the circle increases by one metre. Will the ant ever return to its starting point?

Let's consider the matter. In the first second, the ant traverses 1% of the circumference; in the second second, it traverses 1/2%; in the third second, 1/3% and so on. The cumulative distance covered is given by the sum of:

1 + 1/2 + 1/3 + 1/4 + 1/5 + ... 

This is the harmonic series and it is divergent, meaning that the sum is continuously increasing and never reaches any upper limit. Contrast this to a convergent geometric series, the sum of whose terms approaches ever closer to 2 as more terms are added:

1 + 1/2 + 1/4 + 1/8 + 1/16 + ...

In the case of the harmonic series, the sum will certainly surpass the 100% required for the ant to return to its starting point. It may take a long time but it will happen. In fact OEIS A004080 tells us that it would take the following number of seconds:

15092688622113788323693563264538101449859497

This of course represents more than 4.78 x 10^23 years which is far in excess of the age of the universe! The following screenshot is taken from the WolframAlpha article about the harmonic series.

Taken from http://mathworld.wolfram.com/HarmonicSeries.html

This Numberphile YouTube video was the inspiration for this post.

Reversible Digits Problem for Mother and Child

My daughter turns 36 on November 23rd and her mother, who is born on the same day, turns 63. I was struck by the reversible digits and immediately wondered whether this had happened before. A little investigation revealed that it had: at 25 and 52, at 14 and 41 and (if allowing leading zeroes) at 03 and 30. In fact it happens every 11 years. I wondered whether this was true for every mother and child. It's not: such reversibility of digits can only happen if the mother's age when she gives birth is a multiple of 9 e.g. 9, 18, 27, 36 etc. Here is my proof of this assertion. 9 is a little young of course but it has happened I think.

Let \(n\) be the age of the mother when the child is born.

Suppose mother and child turn \(x \, y\) and \(y \,x\) years old in a certain year.

Now the mother's age of \(x \, y\) has numerical value of \(10x+y\).

Similarly the child's age of \(y \, x\) has a numerical value of \(10y+x\).

We know that \( (10x+y)-(10y+x) =n\).

This means that \(9(x-y)=n\) where \(x>y\).

\(x\) and \(y\) have integer values from 0 to 9 and thus \(n\) must be a multiple of 9.

Suppose \(n=9\), this means \(x-y=1\) and there are various solutions.

Let's consider \(x=1\) and \(y=0\), giving the mother's age as 10 and the child's as 01.

Next consider \(x=2\) and \(y=1\), giving the mother's age as 21 and the child's as 12.

Next consider \(x=3\) and \(y=2\), giving the mother's age as 32 and the child's as 23.

We see can see clearly when the ages of mother and child share the same digits and why these match ups occur every 11 years.

ADDENDUM:  June 19th 2022

Here is some SageMath code that will spit out the above data (change the offset from 27 to some other number and experiment):