Monthly Archives: February 2018

Education System

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This blog post has nothing to do with mathematics but just wanted to vent out my emotions.

I know that my opinions regarding the education system don’t matter since there always have been smarter people (i.e. people scoring more than me) around me in my home, school and college (and according to this system, only the opinions of top scorers matter). But, since WordPress allows me to express my opinions, here are the few comics which are in sync with my opinions:

ind111

© Bill Watterson

Testing

couldn’t find the creator of this comic

av

© Awantha Artigala

bill

© Bill Watterson

I don’t think there is any solution to this problem since there are so many human beings on earth (i.e large variety of minds…).

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Listening Maths

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Earlier I told about new mediums now available to enjoy maths. You can find a list of YouTube channels, to begin with here. And a list of dedicated math podcasts here.

In this post, I want to point out some episodes from the  podcast “In Our Time” which discuss mathematics:

Apart from them, another podcast worth listening to is “The Wizard of Mathematics, Srinivasa Ramanujan” by Prof. Srinivas Kotyada on FM Gold:

A topic I wanted to discuss for long time

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If you are an average maths undergraduate student (like me), you might have ended up in a situation of choosing between “just completing the degree by somehow passing the courses without caring about the grades” and “repeating a course/taking fewer courses so as to pass all courses with nice grades only”. Following is a nice discussion from Reddit:

A sequence I didn’t like

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Following is a problem I encountered many times in my high school olympiads, but was never able to solve it. Hence didn’t like it.

What is the 100th term in the sequence 1,2,2,3,3,3,4,4,4,4,5,\ldots?

Following is a quick solution:

In this post, I will discuss the solution given in The Green Book of Mathematical Problems (problem 14).

Determine a function f(n) such that the n^{th} term of the sequence 1,2,2,3,3,3,4,4,4,4,5,\ldots is given by \lfloor f(n)\rfloor.

Let’s denote the n^{th} number of the sequence by a_n, i.e. a_n=\lfloor f(n)\rfloor. The integer m first occurs in the sequence when each of the integers from 1 to m-1 have already appeared 1 to m-1 times, respectively. Hence, if a_n=m then
n = [1 + 2 + 3 + \ldots + (m-1)]+1 +\ell = \frac{m(m-1)}{2} + 1 + \ell
for \ell = 0,1,2,\ldots, m-1.

Hence we have:
\displaystyle{0\leq n - \frac{m(m-1)}{2} - 1 \leq m-1}

\displaystyle{\Rightarrow \frac{m^2-m+2}{2}\leq n \leq \frac{m^2+m}{2}}

\displaystyle{\Rightarrow m^2-m+2\leq 2n \leq m^2+m}

\displaystyle{\Rightarrow \left(m-\frac{1}{2}\right)^2+\frac{7}{4}\leq 2n \leq \left(m+\frac{1}{2}\right)^2-\frac{1}{4}}

\displaystyle{\Rightarrow (2m-1)^2+7\leq 8n \leq (2m+1)^2 - 1}

\displaystyle{\Rightarrow (2m-1)^2 \leq 8n - 7 \leq (2m+1)^2 - 8}

\displaystyle{\Rightarrow (2m-1)^2 \leq 8n - 7 < (2m+1)^2}

\displaystyle{\Rightarrow 2m-1 \leq \sqrt{8n-7}<2m+1}

\displaystyle{\Rightarrow m \leq \frac{1+\sqrt{8n-7}}{2} < m+1}

\displaystyle{\Rightarrow m=\left\lfloor \frac{1+\sqrt{8n-7}}{2} \right\rfloor}

Hence we have a_n = \left\lfloor \frac{1+\sqrt{8n-7}}{2} \right\rfloor. Thus, we have

\displaystyle{\boxed{f(n) =\frac{1+\sqrt{8n-7}}{2}}}

Now compared to the earlier solution obtained by observing the pattern, one might ask “Is there is a better formula?”. For that, you might also look at the discussion at Math.SE.