A fifth degree equation in two variables: a clever solution

Question:

Verify the identity: (2xy+(x^{2}-2y^{2}))^{5}+(2xy-(x^{2}-2y^{2}))^{5}=(2xy+(x^{2}+2y^{2})i)^{5}+(2xy-(x^{2}+2y^{2})i)^{5}

let us observe first that each of the fifth degree expression is just a quadratic in two variables x and y. Let us say the above identity to be verified is:

P_{1}+P_{2}=P_{3}+P_{4}

Method I:

Use binomial expansion. It is a very longish tedious method.

Method II:

Factorize each of the quadratic expressions P_{1}, P_{2}, P_{3}, P_{4} using quadratic formula method (what is known in India as Sridhar Acharya’s method):

Now fill in the above details.

You will conclude very happily that :

The above identity is transformed to :

P_{1}=(x+y+\sqrt{3}y)^{5}(x+y-\sqrt{3}y)^{5}

P_{2}=(-1)^{5}(x-y-\sqrt{3}y)^{5}(x-y+\sqrt{3}y)^{5}

P_{3}=(i^{2}(x-y-\sqrt{3}y)(x-y+\sqrt{3}y))^{5}

P_{4}=((-i^{2})(x+y+\sqrt{3}y)(x-y-\sqrt{3}y))^{5}

You will find that P_{1}=P_{4} and P_{2}=P_{4}

Hence, it is verified that the given identity P_{1}+P_{2}=P_{3}+P_{4}. QED.

Regards,
Nalin Pithwa.

Set Theory, Relations, Functions: preliminaries: part 10: more tutorial problems for practice

Problem 1:

Prove that a function f is 1-1 iff f^{-1}(f(A))=A for all A \subset X. Given that f: X \longrightarrow Y.

Problem 2:

Prove that a function if is onto iff f(f^{-1}(C))=C for all C \subset Y. Given that f: X \longrightarrow Y.

Problem 3:

(a) How many functions are there from a non-empty set S into \phi\?

(b) How many functions are there from \phi into an arbitrary set S?

(c) Show that the notation \{ X_{i} \}_{i \in I} implicitly involves the notion of a function.

Problem 4:

Let f: X \longrightarrow Y be a function, let A \subset X, B \subset X, C \subset Y and D \subset Y. Prove that

i) f(A \bigcap B) \subset f(A) \bigcap f(B)

ii) f^{-1}(f(A)) \supset A

iii) f(f^{-1}(C)) \subset C

Problem 5:

Let I be a non-empty set and for each i \in I, let X_{i} be a set. Prove that

(a) for any set B, we have B \bigcap \bigcup_{i \in I}X_{i}=\bigcup_{i \in I}(B \bigcap X_{i})

(b) if each X_{i} is a subset of a given set S, then (\bigcup_{i \in I}X_{i})^{'}=\bigcap_{i \in I}(X_{i})^{'} where the prime indicates complement.

Problem 6:

Let A, B, C be subsets of a set S. Prove the following statements:

(i) A- (B-C)=(A-B)\bigcup(A \bigcap B \bigcap C)

(ii) (A-B) \times C=(A \times C)-(B \times C)

🙂 🙂 🙂

Nalin Pithwa

A not so easy quadratic equation problem for RMO or IITJEE maths

Question:

Suppose that we are given a monic quadratic polynomial p(t). Prove that for any integer n, there exists an integer k such that p(n)p(n+1)=p(k).

🙂 🙂 🙂

Solution:

Let p(t)=t^{2}+bt+c and which implies p(t+1)=(t+1)^{2}+b(t+1)+c. We want p(n)p(n+1)=p(k). By trial and error, we get k=t(t+1)+bt+c.

By the way, we could have gotten the same solution by method of undetermined coefficients. But that would also need intelligent guess-works.

Nalin Pithwa

PS: I will post the solution after some time. Meanwhile, please try.

Set Theory, Relations, Functions: Preliminaries: Part IX: (tutorial problems)

Reference: Introductory Real Analysis, Kolmogorov and Fomin, Dover Publications.

Problem 1:

Prove that if A \bigcup B=A and A \bigcap B=A, then A=B.

Problem 2:

Show that in general (A-B)\bigcup B \neq A.

Problem 3:

Let A = \{ 2,4, \ldots, 2n, \ldots\} and B= \{ 3,6,\ldots, 3n, \ldots\}. Find A \bigcap B and A - B.

Problem 4:

Prove that (a) (A-B)\bigcap (C)=(A \bigcap C)-(B \bigcap C)

Prove that (b) A \Delta B = (A \bigcup B)-(A \bigcap B)

Problem 5:

Prove that \bigcup_{a}A_{\alpha}-\bigcup_{a}B_{\alpha}=\bigcup_{\alpha}(A_{\alpha}-B_{\alpha})

Problem 6:

Let A_{n} be the set of all positive integers divisible by n. Find the sets (i) \bigcup_{n=2}^{\infty}A_{n} (ii) \bigcap_{n=2}^{\infty}A_{n}.

Problem 7:

Find (i) \bigcup_{n=1}^{\infty}[n+\frac{1}{n}, n - \frac{1}{n}] (ii) \bigcap_{n=1}^{\infty}(a-\frac{1}{n},b+\frac{1}{n})

Problem 8:

Let A_{\alpha} be the set of points lying on the curve y=\frac{1}{x^{\alpha}} where (0<x<\infty). What is \bigcap_{\alpha \geq 1}A_{\alpha}?

Problem 9:

Let y=f(x) = <x> for all real x, where <x> is the fractional part of x. Prove that every closed interval of length 1 has the same image under f. What is the image? Is f one-to-one? What is the pre-image of the interval \frac{1}{4} \leq y \leq \frac{3}{4}? Partition the real line into classes of points with the same image.

Problem 10:

Given a set M, let \mathscr{R} be the set of all ordered pairs on the form (a,a) with a \in M, and let aRb if and only if (a,b) \in \mathscr{R}. Interpret the relation R.

Problem 11:

Give an example of a binary relation which is:

  • Reflexive and symmetric, but not transitive.
  • Reflexive, but neither symmetric nor transitive.
  • Symmetric, but neither reflexive nor transitive.
  • Transitive, but neither reflexive nor symmetric.

We will continue later, 🙂 🙂 🙂

PS: The above problem set, in my opinion, will be very useful to candidates appearing for the Chennai Mathematical Institute Entrance Exam also.

Nalin Pithwa

 

 

A quadratic equation question for pRMO or preRMO

Question:

Find the necessary and sufficient condition that the quadratic equation ax^{2}+bx+c=0 where a \neq 0 has one root which is the square of the other.

Solution:

Let the two roots of the given quadratic equation ax^{2}+bx+c=0, with a \neq 0 be \alpha and \beta such that \beta = \alpha^{2}.

Then, we know \alpha+\beta=-\frac{b}{a} and \alpha\beta=\frac{c}{a} so that \alpha+\alpha^{2}=-\frac{b}{a} and \alpha^{3}=\frac{c}{a}. From the latter relation, we get that \alpha = (\frac{c}{a})^{\frac{1}{3}}. Substituting this in the first relation of sum of roots, we get the following necessary and sufficient condition:

(\frac{c}{a})^{\frac{1}{3}} + (\frac{c}{a})^{\frac{2}{3}} = -\frac{b}{a}.

The above is the desired solution.

🙂 🙂 🙂

Nalin Pithwa

Set Theory, Relations, Functions: Preliminaries: part VIIIA

(We continue from part VII of the same blog article series with same reference text).

Theorem 4:

A set M can be partitioned into classes by a relation R (acting as a criterion for assigning two elements to the same class) if and only R is an equivalence relation on M.

Proof of Theorem 4:

Every partition of M determines a binary relation on M, where aRb means that “a belongs to the same class as b.” It is then obvious that R must be reflexive, symmetric and transitive, that is, R is an equivalence relation on M.

Conversely, let R be an equivalence relation on M, and let K_{a} be the set of all elements x \in M such that xRa (clearly, a \in K_{a}, since R is reflexive). Then, two classes K_{a} and K_{b} are either identical or disjoint. In fact, suppose that an element c belongs to both K_{a} and K_{b}, so that cRa and cRb. But by symmetry of R, being an equivalence relation, we can infer that aRc also and, further by transitivity, we say that aRb. If now, x \in K_{a} then we have xRa and hence, xRb (since we already have aRb and using transitivity).

Similarly, we can prove that x \in K_{b} implies that x \in K_{a}.

Therefore, K_{a}=K_{b} if K_{a} and K_{b} have an element in common. Therefore, the distinct sets K_{a} form a partition of M into classes.

QED.

Remark:

Because of theorem 4, one often talks about the decomposition of a set M into equivalence classes.

There is an intimate connection between mappings and partitions into classes, as illustrated by the following examples:

Example 1:

Let f be a mapping of a set A into a set B and partition A into sets, each consisting of all elements with the same image b=f(a) \in B. This gives a partition of A into classes. For example, suppose f projects the xy-plane onto the x-axis by mapping the point (x,y) into the point (x,0). Then, the preimages of the points of the x-axis are vertical lines, and the representation of the plane as the union of these lines is the decomposition into classes corresponding to f.

Example 2:

Given any partition of a set A into classes, let B be the set of these classes and associate each element a \in A with the class (that is, element of B) to which it belongs. This gives a mapping of A into B. For example, suppose we partition three-dimensional space into classes by assigning to the same class all points which are equidistant from the origin of coordinates. Then, every class is a sphere of a certain radius. The set of all these classes can be identified with the set of points on the half-line [0, \infty) each point corresponding to a possible value of the radius. In this sense, the decomposition of 3-dimensional space into concentric spheres corresponds to the mapping of space into the half-line [0,\infty).

Example 3:

Suppose that we assign all real numbers with the same fractional part to the same class. Then, the mapping corresponding to this partition has the effect of “winding” the real line onto a circle of unit circumference. (Note: The largest integer \leq x is called the integral part of x, denoted by [x], and the quantity x -[x] is called the fractional part of x).

In the next blog article, let us consider a tutorial problem set based on last two blogs of this series.

🙂 🙂 🙂

Nalin Pithwa

 

 

 

 

 

A quadratic and trigonometry combo question: RMO and IITJEE maths coaching

Question:

Given that \tan {A} and \tan {B} are the roots of the quadratic equation x^{2}+px+q=0, find the value of

\sin^{2}{(A+B)}+ p \sin{(A+B)}\cos{(A+B)} + q\cos^{2}{(A+B)}

Solution:

Let \alpha=\tan{A} and \beta=\tan{B} be the two roots of the given quadratic equation: x^{2}+px+q=0

By Viete’s relations between roots and coefficients:

\alpha+\beta=\tan{A}+\tan{B}=-p and \alpha \beta = \tan{A}\tan{B}=q but we also know that \tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\frac{-p}{1-q}=\frac{p}{q-1}

Now, let us call E=\sin^{2}{(A+B)}+p\sin{(A+B)\cos{(A+B)}}+\cos^{2}{(A+B)} which in turn is same as

\cos^{2}{(A+B)}(\tan^{2}{(A+B)}+p\tan{(A+B)}+q)

We have already determined \tan{(A+B)} in terms of p and q above.

Now, again note that \sin^{2}{\theta}+\cos^{2}{\theta}=1 which in turn gives us that \tan^{2}{\theta}+1=\sec^{2}{\theta} so we get:

\sec^{2}{(A+B)}=1+\tan^{2}{(A+B)}=1+\frac{p^{2}}{(q-1)^{2}}=\frac{p^{2}+(q-1)^{2}}{(q-1)^{2}} so that

\cos^{2}{(A+B)}=\frac{1}{\sec^{2}{(A+B)}}=\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}}

Hence, the given expression E becomes:

(\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}})(\frac{p^{2}}{(q-1)^{2}}+\frac{p^{2}}{q-1}+q), which is the desired solution.

🙂 🙂 🙂

Nalin Pithwa.