https://colleenyoung.wordpress.com/2017/11/05/math-is-fun/

With thanks and regards to Colleen Young.

Mathematics demystified

November 5, 2017 – 5:36 pm

https://colleenyoung.wordpress.com/2017/11/05/math-is-fun/

With thanks and regards to Colleen Young.

November 28, 2016 – 6:13 pm

(Non-linear) equations involving three or more variables can only be solved in special cases. We shall here consider some of the most useful methods of solution.

**Example 1.**

Solve the following system of equations:

Equation I

Equation II

Equation III

**Solution I:**

From II and III, we get

Plug in ; then this equation becomes .

Also, from I, we get

hence, we obtain , .Thus,we have

, ; and , . Hence, the solutions are

; ; ;

or,

; ; .

**Example 2:**

Solve the following system of equations:

Equation I

Equation II

Equation III

**Solution: 2:**

Write u, v, w for , , respectively; thus

, , …..call this equations A

Multiplying these equations together, we have

. Hence, .

Combining this result with each of the equations in A, we have

; or, ; therefore, we get two sets of linear equations, as follows:

for which the solution set is: ; and the other set is:

for which the solution set if .

**Example 3:**

Solve the following system of equations:

…call this equation (1)

…call this equation (2)

…call this equation (3)

Subtracting (2) from (1), we get the following:

, that is, …call this equation (4).

Similarly, from (1) and (3), we get the following:

…call this equation (5).

Hence, from equations (4) and (5), by division, we get the following:

, hence,

Substituting in Equation (3), we obtain .

From (2), we get the following:

Solving these homogeneous equations (*hint: *put , where m is parameter to be found), we get the following:

, ; and therefore,

or, ; ; and therefore, .

**Example 4:**

Solve the following system of equations:

**Solution 4:**

Multiply the equations by y, z, and x respectively, and add; then,

…call this equation I

Multiply the equations by z, x and y respectively; and add; then

…call this equation II

From (I) and (II), by cross multiplication,

, suppose.

Substitute in any one of the given equations; then,

; hence, we get the following solution:

.

More stuff in the pipeline,

Nalin Pithwa

November 14, 2016 – 12:45 pm

In order to find values of expressions involving a, b, c when these quantities are connected by the equation , we might employ the substitution:

If, however, the expressions involve a, b, c symmetrically the method exhibited by the following example is preferable:

**Example: **

If , show that

**Solution:**

We have identically

, where , , .

Hence, using the condition given,

Taking logarithms and equating the coefficients of , we have , which is equal to the coefficient of in the expansion of , which is equal to the coefficient of in the following:

By putting we obtain

, and hence, we get

, and the required result follows at once.

More stuff to follow…

*Nalin Pithwa*

November 7, 2016 – 11:43 am

Many identities can be readily established by making use of the properties of the cube roots of unity; as usual these will be denoted by .

**Problem 1:**

Show that

**Solution 1:**

The expression, E, on the left vanishes when ; hence, it must contain as a factor.

Putting , we have

Hence, E contains as a factor; and similarly, we may show that it contains as a factor; that is, E is divisible by

, or .

Further, E being of seven, and of five dimensions, the remaining factor must be of the form , thus,

.

Putting , we have ; putting , we have , and hence, ;

.

**Problem 2:**

Show that the product of and can be put in the form .

**Solution 2:**

The product

By taking these six factors in the pairs ,

,

and ,

we obtain the partial products:

, , and

where , , and

Thus, the product

which, in turn, equals

More esoteric algebraic miscellany is planned for you!

Nalin Pithwa

October 27, 2016 – 4:14 pm

**Problem 1:**

Prove that .

**Proof I:**

First carry out the following steps:

(a) Is the given expression symmetric?

(b) Is the given expression alternating?

(c) Is the given expression cyclic?

(d) Is the given expression homogeneous?

So here goes the proof:

Denote the expression by E. Clearly, the substitutions , give us . Hence, two of the factors are . Similarly, the substitution give us . So the other factor is .

The degree of the given expression E is 5. E is symmetric w.r.t. a and b; it is homogeneous. The factors we found above are . So where F(x,y) ought to be of degree 2, homogeneous and symmetric. So, let ; thus, we have

. Now, find k, A, and B by the undetermined coefficients.

Hence, you will get .

**Problem 2:**

Find the factors of .

**Solution 2:**

First carry out the following steps:

(a) Is the given expression symmetric w.r.t. a and b; b and c; c and a?

(b) Is the given expression alternating w.r.t. a and b; b and c; c and a?

(c) Is the given expression cyclic w.r.t. a, b and c?

(d) Is the given expression homogeneous and if so, what is the degree of each expression?

Denote the expression by E; then, E is a function of a which vanishes when , , , so the following are certainly factors of E: , and hence, E contains as a factor.

Now note that E is of fourth degree and symmetric, so the remaining factor is of the form : , where M is a coefficient to be determined.

So, we have now: , and you can easily find that , and hence,

More mathematical miscellany for IITJEE mathematics to follow!

Nalin Pithwa

October 22, 2016 – 2:51 pm

**Example 1:**

Find the roots of , given that one root is the negative of the another.

**Solution 1:**

If the roots are a, b, and c, we have , say. So, using Viete’s relations,

.

Hence, and . Hence, the roots are .

**Example 2:**

Let , with such that a and have the same sign. Show that the equation cannot have both roots in the interval .

**Solution 2:**

Let be roots of the given quadratic equation. We have

which in turn equals

.

If both belong to then each term of the sum will be negative, which is a contraction. Hence, the proof.

**Example 3:**

Consider all lines which meet the graph of in four distinct points, say . Then, show that is independent of the line and find its value.

**Solution 3:**

Let be any line which intersects the graph at , where . Then, are the roots of

.

(Note that are distinct as would imply ).

The above equation reduces to an equation of degree 4, namely,

. Hence, by Viete’s relations,

.

**Example 4:**

The product of two of the four roots of is 24. Find k.

**Solution 4:**

Let the given equation be written as , and let the roots of the equation be with . Now,

, so . Also,

with , . Comparing coefficients of and x we get and . This gives . Comparing coefficients of , .

**Example 5:**

If and are roots of , where p and q are integers with , then show that

(i) is an integer. ()

(ii) is an integer divisible by q. ().

**Solution 5:**

Since are the roots of , we get

—– Equation A

—– Equation B

Note that . For , multiplying this equation by , we get . Similarly, . Hence,

Eqn C

Also, for , Equation D

1.By equation A and D, and are both integers. Hence, by Eqn C, it follows by induction on n that is an integer for .

2. Since , equation D shows that . Further, and so . Hence, by C, it follows by induction on n that for .

**Example 6:**

Find all integers a such that the equation has three integer roots.

**Solution 6:**

Let the integer roots of the given equation be . Then,

, . *Let this be Equation I.*

Hence, . *Let this be Equation II.*

So, and so the solution of *Equation II are essentially the following:*

—- Call this ***

—- Call this *****

Both these sets satisfy *Equation I. *Hence, the required values of a are corresponding to the roots in (***) and (*****) respectively.

**Example 7:**

Find the remainder when is divided by .

**Solution 7:**

Dividing by , we get , we get

.

Put or . Hence,

Using Binomial theorem, we get

, which in turn equals

*Call this Equation @@@*

Now, equating coefficients of we get

, , .

Solving these equations, we get

, ,

Hence, the remainder is

More algebraic stuff in the pipeline!!

Nalin Pithwa