**Prove that **

**Proof:**

To prove the above is the same as proving

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

which equals , which in turn equals

, that is,

.

Thus, the proof.

More later,

Nalin Pithwa

Mathematics demystified

February 28, 2015 – 4:10 pm

**Prove that **

**Proof:**

To prove the above is the same as proving

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

which equals , which in turn equals

, that is,

.

Thus, the proof.

More later,

Nalin Pithwa

February 26, 2015 – 2:55 pm

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

**Example. Let and let for **

.** Compute .**

**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, for all . But, why? We reveal the secret with trigonometric substitution; that is, we define

with such that . It is clear that if is a real number, such an is unique, because

is a bijection. Because , we can rewrite the given condition as

, which is

by the addition and subtraction formulae. Consequently,

, or (because tan has a period of ). In any case, it is not difficult to see that for some integer k. Therefore, , that is, the sequence

has a period 4, implying that .

More later,

Nalin Pithwa

February 21, 2015 – 12:54 pm

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

February 18, 2015 – 7:07 pm

**Question: **Prove that for any complex number z:

or

**Solution: Suppose by way of contradiction that**

and

Setting with yields .

We obtain

and

and consequently,

and .

Summing these inequalities implies which is a contradiction.

More later,

Nalin Pithwa

February 15, 2015 – 8:21 pm

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

February 10, 2015 – 7:48 pm

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

.

**Proof.**

Let X be an n-set and fix an element x of X, and let Y denote the set . 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 , then B is a -subset of Y and hence, can be chosen in ways, while in the second case, A is itself a k-subset of Y and hence, can be chosen in 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:

which is obtained by letting in the binomial theorem. By making the substitution in the above expression, we obtain:

Stated in other words, the above identity tells us that a set of order n has subsets in all. Again, the susbstitution yields

.

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

,

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

.

Hence, the LHS reduces to the alternating sum

using the binomial theorem. Sim1ilarly, to find , 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

February 3, 2015 – 12:14 am

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

.

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

is divisible by and also find the quotient.

Please do send your questions, comments, suggestions …

More later…

Nalin Pithwa