## Monthly Archives: February 2015

### Trigonometry puzzle

Prove that $1-\cot 23\deg=\frac{2}{1-\cot 22\deg}$

Proof:

To prove the above is the same as proving $(1-\cot 23\deg)(1-\cot 22\deg)=2$

Indeed, by the addition and subtraction formulae, we obtain:

$(1-\cot 23\deg)(1-\cot 22\deg)=(1-\frac{\cos 23\deg}{\sin 23\deg})(1-\frac{\cos 22\deg}{\sin 22\deg})$

which equals $(\frac{\sin 23\deg}-\cos 23\deg{\sin 23\deg})(\frac{\sin 22\deg-\cos 22\deg}{\sin 22\deg})$, which in turn equals

$(\frac{\sqrt{2}\sin(23\deg-45\deg)}{\sin 23\deg-\sin 22\deg})(\frac{\sqrt{2}\sin(22\deg-45\deg)}{1})$, that is,

$\frac{2\sin(-22\deg)\sin (-23\deg)}{\sin(23\deg)\sin (22\deg)}=2$.

Thus, the proof.

More later,

Nalin Pithwa

### Sequences and series or trigonometry problem or both?

The IITJEE Mains are to be held on April 4 2015.  So, let’s focus on some worthwhile basic problems.

Example. Let $x_{0}=2003$ and let $x_{n+1}=\frac {1+x_{n}}{1-x_{n}}$ for

$n \geq 1$. Compute $x_{2004}$.

Solution:

At the first sight, you might think it’s a plain manipulation of recurring sequences. But, wait…let’s see…

With a little algebraic computation, we can show that this sequence has a period of 4; that is, $x_{n+4}=x_{n}$ for all $n \geq 1$. But, why? We reveal the secret with trigonometric substitution; that is, we define

$\alpha_{n}$ with $-90 \deg < \alpha_{n} < 90 \deg$ such that $x_{n}=\tan \alpha_{n}$. It is clear that if $x_{n}$ is a real number, such an $\alpha_{n}$ is unique, because

$\tan: (-90 \deg, 90 \deg) \rightarrow \Re$ is a bijection. Because $1=\tan 45 \deg$, we can rewrite the given condition as

$\tan \alpha_{n+1}=\frac{\tan 45\deg + \tan \alpha_{n}}{1-\tan 45\deg\tan\alpha_{n}}$, which is

$\tan(45\deg+\alpha_{n})$ by the addition and subtraction formulae. Consequently,

$\alpha_{n+1}=45\deg+\alpha{n}$, or $\alpha_{n+1}=45\deg+\alpha_{n}-180\deg$ (because tan has a period of $180\deg$). In any case, it is not difficult to see that $\alpha_{n+4}=\alpha_{n}+k.180\deg$ for some integer k. Therefore, $x_{n+4}=\tan{\alpha_{n}}=x_{n}$, that is, the sequence

$\{x_{n}\}_{n \geq 0}$ has a period 4, implying that $x_{2004}=x_{0}=2003$.

More later,

Nalin Pithwa

### A very brief note on the history of Calculus

Calculus is one of the triumphs of the human mind.It emerged from investigations into such basic questions as finding areas, lengths and volumes. In the third century BC, Archimedes determined the area under the arc of a parabola. In the early seventeenth century, Fermat and Descartes studied the problem of finding tangents to curves. But, the subject really came to life in the hands of Newton and Leibniz in the late seventeenth century. In particular, they showed that the geometric problems of finding the areas of planar regions and of finding the tangents to plane curves are intimately related to one another. In subsequent decades,  the subject developed further through the work of several mathematicians, most notably, Euler, Cauchy, Riemann and Weierstrass.

Today, calculus occupies a central place in mathematics and is an essential component of undergraduate education. It has an immense number of applications both within and outside mathematics. Judged by the sheer variety of the concepts and results it has generated, calculus can be rightly viewed as a fountainhead of ideas and disciplines in mathematics.

The history of calculus can greatly aid and enrich the understanding of the subject. We encourage you to look at the internet for the same.

More later,

Nalin Pithwa

### Complex numbers

Question: Prove that for any complex number z:

$|z+1| \geq \frac {1}{\sqrt {2}}$ or $|z^{2}+1| \geq 1$

Solution: Suppose by way of contradiction that

$|1+z| < \frac {1}{\sqrt {2}}$ and $|1+z^{2}| <1$

Setting $z=a+ib$ with $a,b \in R$ yields $z^{2}=a^{2}-b^{2}+i2ab$.

We obtain

$(1+a^{2}-b^{2})^{2}+4a^{2}b^{2}<1$ and $(1+a)^{2}+b^{2}<1/2$

and consequently,

$(a^{2}+b^{2})^{2}+2(a^{2}-b^{2})<0$ and $2(a^{2}+b^{2})+4a+1<0$.

Summing these inequalities implies $(a^{2}+b^{2})^{2}+(2a+1)^{2}>0$ which is a contradiction.

More later,

Nalin Pithwa

### Elementary geometry problem

Show that the only regular figures which may be fitted together so as to form a plane surface are (a) equilateral triangles (b) squares (c) regular hexagons.

More later…

Nalin Pithwa

### Pascal’s identity

Theorem. (Pascal’s Identity) Let n and k be positive integers. Then,

${n \choose k} = {{n-1}\choose k}+{{n-1} \choose {k-1}}$.

Proof.

Let X be an n-set and fix an element x of X, and let Y denote the set $X-{x}$. For any k-combination (i.e., a k-subset) A of X, either x is in A or x is not in A. In the first case, if B is the set $A-{x}$, then B is a ${k-1}$ -subset of Y and hence, can be chosen in ${n-1} \choose {k-1}$ ways, while in the second case, A is itself a k-subset of Y and hence, can be chosen in ${n-1} \choose k$ ways. The proof is complete by invoking the addition principle. QED.

It is also easy to see that Pascal’s identity follows from the binomial theorem. (if we have proved the latter without using the former as ,we did). Pascal’s identity gives rise to the famous Pascal’s triangle, initial portion of which is drawn below. Each entry is obtained from the two entries directly above it as given in Pascal’s identity. This obtains all the binomial coefficients where n runs from 1 to 6 and the horizontal lines correspond to a fixed value of n. Note that Pascal triangle is an infinite triangle, only a finite portion of this triangle (from 1 to 6) is shown in the figure below/attached. A more familiar form of the binomial theorem is as follows:

$(1+x)^{n}=\sum_{k=0}^{n}{n \choose k}x^{k}$

which is obtained by letting $y=1$ in the binomial theorem. By making the substitution $x=1$ in the above expression, we obtain:

$\sum_{k=0}^{n}{n \choose k}=2^{n}$

Stated in other words, the above identity tells us that a set of order n has $2^{n}$ subsets in all. Again, the susbstitution $x=-1$ yields

$\sum_{k} {n \choose {2k}}=\sum_{k}{n \choose {2k+1}}$.

Thus, in any set the number of subsets of odd order is the same as the number of subsets of even order.

A large number of identities involving binomial coefficients are actually proved either using a combinatorial identity or a known polynomial expression such as the binomial theorem and manipulating it. For example, to prove that

${n \choose 1}-2{n \choose 2}+3{n \choose 3}- \ldots +(-1)^{n-1}n {n \choose k}=0$,

we observe that each summand on the left hand side (ignoring the sign) has the form :

$k {n \choose k}= n {{n-1}\choose {k-1}}$.

Hence, the LHS reduces to the alternating sum

$n \times \sum_{i=0}^{n-1}(-1)^{i}{{n-1} \choose i}=0$

using the binomial theorem. Sim1ilarly, to find $n \sum \frac {1}{k+1}{n \choose k}$, we take the familiar form of the binomial theorem and integrate both sides (as polynomials in x) w.r.t. x from 0 to 1.

More later…

Nalin Pithwa

### yet another important algebric identity

Several blogs ago,  I had suggested the use of a powerful fundamental algebraic identity:

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.

Now, this identity becomes a basic applied tool in Algebra and Trigonometry and Number theory, perhaps.

A salient feature of mathematics is that mathematicians always try to generalize a technique or concept.

Can you generalize the above identity as follows (a proof is required please):

Let there be quantities a,b,c,d…then prove that

$a^{3}+b^{3}+c^{3}+d^{3}+\ldots -3(abc+bcd+abd+\ldots)$ is divisible by $(a+b+c+d+ \ldots)$ and also  find the quotient.