Monthly Archives: February 2017

Matchstick Problem

Determination is like a match stick !!! 🙂

Singapore Maths Tuition

matchstick quiz

Translation: How do we move only 1 matchstick to make the equation valid?

(“Not Equals” sign $latex neq$ is not allowed)

This is a really tricky question… Hint: Need to think in Chinese, this is a Chinese joke 😛

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Motivational Books for Students (Educational)

The Motivational Books recommended in the above post seems to be very popular with readers on this site. Many readers (presumably parents) have bought the books! Buy a Christmas present for yourself this Xmas. 🙂

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Limits that arise frequently

We continue our presentation of basic stuff from Calculus and Analytic Geometry, G B Thomas and Finney, Ninth Edition. My express purpose in presenting these few proofs is to emphasize that Calculus, is not just a recipe of calculation techniques. Or, even, a bit further, math is not just about calculation. I have a feeling that such thinking nurtured/developed at a young age, (while preparing for IITJEE Math, for example) makes one razor sharp.

We verify a few famous limits.

Formula 1:

If |x|<1, \lim_{n \rightarrow \infty}x^{n}=0

We need to show that to each \in >0 there corresponds an integer N so large that |x^{n}|<\in for all n greater than N. Since \in^{1/n}\rightarrow 1, while |x|<1. there exists an integer N for which \in^{1/n}>|x|. In other words,

|x^{N}|=|x|^{N}<\in. Call this (I).

This is the integer we seek because, if |x|<1, then

|x^{n}|<|x^{N}| for all n>N. Call this (II).

Combining I and II produces |x^{n}|<\in for all n>N, concluding the proof.

Formula II:

For any number x, \lim_{n \rightarrow \infty}(1+\frac{x}{n})^{n}=e^{x}.

Let a_{n}=(1+\frac{x}{n})^{n}. Then, \ln {a_{n}}=\ln{(1+\frac{x}{n})^{n}}=n\ln{(1+\frac{x}{n})}\rightarrow x,

as we can see by the following application of l’Hopital’s rule, in which we differentiate with respect to n:

\lim_{n \rightarrow \infty}n\ln{(1+\frac{x}{n})}=\lim_{n \rightarrow \infty}\frac{\ln{(1+x/n)}}{1/n}, which in turn equals

\lim_{n \rightarrow \infty}\frac{(\frac{1}{1+x/n}).(-\frac{x}{n^{2}})}{-1/n^{2}}=\lim_{n \rightarrow \infty}\frac{x}{1+x/n}=x.

Now, let us apply the following theorem with f(x)=e^{x} to the above:

(a theorem for calculating limits of sequences) the continuous function theorem for sequences:

Let a_{n} be a sequence of real numbers. If \{a_{n}\} be a sequence of real numbers. If a_{n} \rightarrow L and if f is a function that is continu0us at L and defined at all a_{n}, then f(a_{n}) \rightarrow f(L).

So, in this particular proof, we get the following:

(1+\frac{x}{n})^{n}=a_{n}=e^{\ln{a_{n}}}\rightarrow e^{x}.

Formula 3:

For any number x, \lim_{n \rightarrow \infty}\frac{x^{n}}{n!}=0

Since -\frac{|x|^{n}}{n!} \leq \frac{x^{n}}{n!} \leq \frac{|x|^{n}}{n!},

all we need to show is that \frac{|x|^{n}}{n!} \rightarrow 0. We can then apply the Sandwich Theorem for Sequences (Let \{a_{n}\}, \{b_{n}\} and \{c_{n}\} be sequences of real numbers. if a_{n}\leq b_{n}\leq c_{n} holds for all n beyond some index N, and if \lim_{n\rightarrow \infty}a_{n}=\lim_{n\rightarrow \infty}c_{n}=L,, then \lim_{n\rightarrow \infty}b_{n}=L also) to  conclude that \frac{x^{n}}{n!} \rightarrow 0.

The first step in showing that |x|^{n}/n! \rightarrow 0 is to choose an integer M>|x|, so that (|x|/M)<1. Now, let us the rule (formula 1, mentioned above), so we conclude that:(|x|/M)^{n}\rightarrow 0. We then restrict our attention to values of n>M. For these values of n, we can write:

\frac{|x|^{n}}{n!}=\frac{|x|^{n}}{1.2 \ldots M.(M+1)(M+2)\ldots n}, where there are (n-M) factors in the expression (M+1)(M+2)\ldots n, and

the RHS in the above expression is \leq \frac{|x|^{n}}{M!M^{n-M}}=\frac{|x|^{n}M^{M}}{M!M^{n}}=\frac{M^{M}}{M!}(\frac{|x|}{M})^{n}. Thus,

0\leq \frac{|x|^{n}}{n!}\leq \frac{M^{M}}{M!}(\frac{|x|}{M})^{n}. Now, the constant \frac{M^{M}}{M!} does not change as n increases. Thus, the Sandwich theorem tells us that \frac{|x|^{n}}{n!} \rightarrow 0 because (\frac{|x|}{M})^{n}\rightarrow 0.

That’s all, folks !!


Nalin Pithwa.

Education gives strength: Anand Kumar tells Maoist youngsters

I reproduce this highly inspirational news from The DNA, Mumbai, Thursday, February 23, 2017: 

Raipur: Super 30 founder Anand Kumar has exhorted the youth in Maoist areas to shun the gun and embrace pen to script a new tale of peace and prosperity through real empowerment which, he believes, only education can ensure.

Kumar was addressing students at the education city in Jawanga village of Chhattisgarh’s insurgency-hit Dantewada district at a programme on Tuesday.

State Chief Minister Raman Singh and Rajya Sabha MP Dr. Subhash Chandra were also present on the occasion.

Stressing that education was the only way to bridge the gap in the society, Kumar said, “Education can lend you strength in the real sense. It can bring about generational change.”

Technology should also be increasingly used to make quality education accessible to all, which is another prerequisite for an egalitarian society and tackling poverty effectively said Kumar, who runs a residential and free-of-cost “Super 30” programme for the last 15 years for talented students from the underprivileged sections.

Narrating his “Super 30” experience, he  gave examples of students from the most underprivileged sections and how they reached the Indian Institutes of Technology (IITs) through their hard work.

“All the problems in the world originate basically because of four-five reasons, like illiteracy, ignorance, poverty, lack of opportunities, unemployment, mindless violence etc. And, if we look at them closely, all have their genesis in lack of education. It is the lack of education that breeds inferiority complex and later manifests itself in different ways,” Kumar said.

Chief Minister asked the students to set a goal in their life and work hard with honesty to achieve it.

“Bastar is changing rapidly. The students (of education city) will play a key role in the development of Bastar. The day when maximum students of Bastar will sit on higher administration posts, that day will be memorable for me,” he said.

“With the efforts of the government, several children of Bastar have qualified for IIT, engineering and medical colleges. These children are not only future of Chattisgarh but the country too,” he said.

The Chief Minister also released Halbi-Gondi-English dictionary for students. –PTI.

Hats off to Mr. Anand Kumar and his “Super 30”!

From Nalin Pithwa

Are kids no longer learning multiplication tables in school?

This topic is very close to my heart and head. I had written a remark about this in an earlier blog article. 

Today, I suddenly found some “echoes” or “resemblances” to my views. Kindly let me re-blog, or share the opinions about this from today’s EETimes, authored by Max Maxfield, Designline Editor. The URL is pasted below:;

Thanks Max and EETimes !

From Nalin Pithwa.

Francis Su, Retiring President of MAA, “Mathematics for human flourishing”

Cauchy’s Mean Value Theorem and the Stronger Form of l’Hopital’s Rule

Reference: Thomas, Finney, 9th edition, Calculus and Analytic Geometry.

Continuing our previous discussion of “theoretical” calculus or “rigorous” calculus, I am reproducing below the proof of the finite limit case of the stronger form of l’Hopital’s Rule :

L’Hopital’s Rule (Stronger Form):

Suppose that


and that the functions f and g are both differentiable on an open interval (a,b) that contains the point x_{0}. Suppose also that g^{'} \neq 0 at every point in (a,b) except possibly at x_{0}. Then,

\lim_{x \rightarrow x_{0}}\frac{f(x)}{g(x)}=\lim_{x \rightarrow x_{0}}\frac{f^{x}}{g^{x}} ….call this equation I,

provided the limit on the right exists.

The proof of the stronger form of l’Hopital’s Rule is based on Cauchy’s Mean Value Theorem, a mean value theorem that involves two functions instead of one. We prove Cauchy’s theorem first and then show how it leads to l’Hopital’s Rule. 

Cauchy’s Mean Value Theorem:

Suppose that the functions f and g are continuous on [a,b] and differentiable throughout (a,b) and suppose also that g^{'} \neq 0 throughout (a,b). Then there exists a number c in (a,b) at which

\frac{f^{'}(c)}{g^{'}(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}…call this II.

The ordinary Mean Value Theorem is the case where g(x)=x.

Proof of Cauchy’s Mean Value Theorem:

We apply the Mean Value Theorem twice. First we use it to show that g(a) \neq g(b). For if g(b) did equal to g(a), then the Mean Value Theorem would give:

g^{'}(c)=\frac{g(b)-g(a)}{b-a}=0 for some c between a and b. This cannot happen because g^{'}(x) \neq 0 in (a,b).

We next apply the Mean Value Theorem to the function:

F(x) = f(x)-f(a)-\frac{f(b)-f(a)}{g(b)-g(a)}[g(x)-g(a)].

This function is continuous and differentiable where f and g are, and F(b) = F(a)=0. Therefore, there is a number c between a and b for which F^{'}(c)=0. In terms of f and g, this says:

F^{'}(c) = f^{'}(c)-\frac{f(b)-f(a)}{g(b)-g(a)}[g^{'}(c)]=0, or

\frac{f^{'}(c)}{g^{'}(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}, which is II above. QED.

Proof of the Stronger Form of l’Hopital’s Rule:

We first prove I for the case x \rightarrow x_{o}^{+}. The method needs no  change to apply to x \rightarrow x_{0}^{-}, and the combination of those two cases establishes the result.

Suppose that x lies to the right of x_{o}. Then, g^{'}(x) \neq 0 and we can apply the Cauchy’s Mean Value Theorem to the closed interval from x_{0} to x. This produces a number c between x_{0} and x such that \frac{f^{'}(c)}{g^{'}(c)}=\frac{f(x)-f(x_{0})}{g(x)-g(x_{0})}.

But, f(x_{0})=g(x_{0})=0 so that \frac{f^{'}(c)}{g^{'}(c)}=\frac{f(x)}{g(x)}.

As x approaches x_{0}, c approaches x_{0} because it lies between x and x_{0}. Therefore, \lim_{x \rightarrow x_{0}^{+}}\frac{f(x)}{g(x)}=\lim_{x \rightarrow x_{0}^{+}}\frac{f^{'}(c)}{g^{'}(c)}=\lim_{x \rightarrow x_{0}^{+}}\frac{f^{'}(x)}{g^{'}(x)}.

This establishes l’Hopital’s Rule for the case where x approaches x_{0} from above. The case where x approaches x_{0} from below is proved by applying Cauchy’s Mean Value Theorem to the closed interval [x,x_{0}], where x< x_{0}QED.

A remark on memory power, concentration and good math

I have come across several maths teachers (and principals and of course, parents) in Mumbai and Bangalore, who  claim that students don’t need to learn by heart, by practising writing daily multiplication tables from 1 to 30 at a tender age. Whereas this is the way I had learnt these multiplication tables. The argument forwarded is that there is now the influence of Western culture (meaning USA and UK) where students don’t learn any multiplication tables because this is the age of electronics/IT/calculators.

I have seen that this leads to severe dependence on calculators from a young age even for simple numerical problems; and in math syllabii/exams of schools of India, school kids make “silly” mistakes, which is a euphemism of numerical mistakes. In my opinion, a high numerical ability is a sign of a good IQ. Of course, we have all heard of Shakuntala Devi’s feats. So, also the feats of John von Neumann.

I think that it works “both” ways: learning multiplication tables by practice at a tender age develops a child concentration powers, and if a kid has enough concentration from a tender age, he can mug up/practise multiplication tables. This might also lead to “average” concentration span in kids, and if I may venture further, it will not give rise to ADHD.. ..I know this might sound simplistic solution….

I refer the reader to the child hood of Paul Halmos, famous Hungarian born, American mathematician, and one of the finest expositors of math. In his autobiography, “I want to be a mathematician”, Paul Halmos had written: “When my cousin and I were 8  and 9, our grandfather used to race us against one another in the multiplication of three-digit numbers in our heads, of course — no paper and pencil. I neither loved nor hated the game. I liked to win, and I liked getting the coin that was awarded to the winner. (Is that when I turned pro?). We were both equally good.”

My own dad had told me that he had learnt the fractional tables also, meaning, the tables of one by four, one by two, three by four, etc. and he told me that as a CA he knew all the volumes of Income Tax Law by Nani Palkhivala and Kanga!!

🙂 🙂 🙂

Nalin Pithwa.

Who rode the horse?

One day a young man and an older man left the village for the city, one on horse, one in a car. Soon it was apparent that if the older man had ridden three times as far as he had, he would have half as far to ride as he had, and if the young man had ridden half as far as he had, he would have three times as far to ride as he had.

Who rode the horse? 🙂 🙂 🙂

Nalin Pithwa.

PS: I hope you liked this puzzle ! Reference: The Moscow Puzzles, Boris A. Kordemsky, Amazon India,

To live your best life, do mathematics

I am essentially re-blogging this from Singapore Maths Tuition; of course, I like math v much…and would practise it even after a hard day’s labour job ! 🙂

To Live Your Best Life, Do Mathematics

The Sandwich Theorem or Squeeze Play Theorem

It helps to think about the core concepts of Calculus from a young age, if you want to develop your expertise or talents further in math, pure or applied, engineering or mathematical sciences. At a tangible level, it helps you attack more or many questions of the IIT JEE Advanced Mathematics. Let us see if you like the following proof, or can absorb/digest it:

Reference: Calculus and Analytic Geometry by Thomas and Finney, 9th edition.

The Sandwich Theorem:

Suppose that g(x) \leq f(x) \leq h(x) for all x in some open interval containing c, except possibly at x=c itself. Suppose also that \lim_{x \rightarrow c}g(x)= \lim_{x \rightarrow c}h(x)=L. Then, \lim_{x \rightarrow c}f(x)=c.

Proof for Right Hand Limits:

Suppose \lim_{x \rightarrow c^{+}}g(x)=\lim_{x \rightarrow c^{+}}h(x)=L. Then, for any \in >0, there exists a \delta >0 such that for all x, the inequality c<x<c+\delta implies L-\in<g(x)<L+\in and L-\in<h(x)<L+\in ….call this (I)

These inequalities combine with the inequality g(x) \leq f(x) \leq h(x) to give

L-\in <g(x) \leq f(x) \leq h(x)<L+\in

L-\in <f(x)<L+\in

-\in <f(x)-L<\in….call this (II)

Therefore, for all x, the inequality c<x<c+\delta implies |f(x)-L|<\in. …call this (III)

Proof for LeftHand Limits:

Suppose \lim_{x \rightarrow c^{-}} g(x)=\lim_{x \rightarrow c^{-}}=L. Then, for \in >0 there exists a \delta >0 such that for all x, the inequality c-\delta <x<c implies L-\in<g(x)<L+\in and L-\in<h(x)<L+\in …call this (IV).

We conclude as before that for all x, c-\delta <x<c implies |f(x)-L|<\in.

Proof for Two sided Limits:

If \lim_{x \rightarrow c}g(x) = \lim_{x \rightarrow c}h(x)=L, then g(x) and h(x) both approach L as x \rightarrow c^{+} and as x \rightarrow c^{-} so \lim_{x \rightarrow c^{+}}f(x)=L and \lim_{x \rightarrow c^{-}}f(x)=L. Hence, \lim_{x \rightarrow c}f(x)=L. QED.

Let me know your feedback on such stuff,

Nalin Pithwa