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

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 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.

|log{xx_{1}}| + |log{xx_{2}}| + …+ |log{xx_{n}}| + |log{x/x_{1}}| + |log{x/x_{2}}| + …+|log{x/x_{n}}|= |log{x_{1}}+ log{x_{2}}+ ….+log{x_{n}}|

Solve the following :

Find all positive real numbers x, x_{1}, x_{2}, \ldots, x_{n} such that

|\log{xx_{1}}|+|\log{xx_{2}}| + \ldots + |\log{xx_{n}}| + |\log{\frac{x}{x_{1}}}| + |\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}|= |\log{x_{1}}+ \log{x_{2}}+\log{x_{3}}+ \ldots + \log{x_{n}}|
…let us say this is given equality A

Solution:

Use the following inequality: |a-b| \leq |a| + |b| with equality iff ab \leq 0

So, we observe that : |\log{xx_{1}}|+|\log{\frac{x}{x_{1}}}| \geq |\log{xx_{1}}-\log{\frac{x}{x_{1}}}| = |\log{x_{1}^{2}}|=2 |\log{x_{1}}|,

Hence, LHS of the given equality is greater than or equal to:

2(|\log{x_{1}}|+|\log{x_{2}}|+|\log{x_{3}}|+ \ldots + |\log{x_{n}}|)

Now, let us consider the RHS of the given equality A:

we have to use the following standard result:

|\pm a_{} \pm a_{2} \pm a_{3} \ldots \pm a_{n}| \leq |a_{1}|+|a_{2}| + \ldots + |a_{n}|

So, applying the above to RHS of A:

|\log{x_{1}}+\log{x_{2}}+\ldots + \log{x_{n}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|.

But, RHS is equal to LHS as given in A:

That is, |\log{xx_{1}}|+|\log{xx_{2}}|+ \ldots + |\log{xx_{n}}| +|\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}|+ \ldots + |\log{\frac{x}{x_{n}}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|

Now, just a few steps before we proved that LHS is also greater than or equal to : That is,

|\log{xx_{1}}|+|\log{xx_{2}}|+\ldots + |\log{xx_{n}}|+ |\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}| \geq 2(|\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|)

The above two inequalities are like the following: x \leq y and x \geq 2y; so what is the conclusion? The first inequality means x2y or x=2y; clearly it means the only valid solution is x=2y.

Using the above brief result, we have here:

|\log{x_{1}}|+|\log{x_{2}}|+ \ldots +|\log{x_{n}}| =2(|\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|)

Hence, we get |\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|=0, which in turn means that (by applying the definition of absolute value):

|\log{x_{1}}|=|\log{x_{2}}|= \ldots =|\log{x_{n}}|, which implies that x_{1}=x_{2}= \ldots  x_{n}=1.

Substituting these values in the given logarithmic absolute value equation, we get:

n \times |\log{x}|+ n \times |\log{x}|=0, that is 2n \times |\log{x}|=0, and as n \neq 0, this implies that |\log{x}|=0 which in turn means x=1 also.

You and your research or you and your studies for competitive math exams

You and your research ( You and your studies) : By Richard Hamming, AT and T, Bell Labs mathematician;