https://gowers.wordpress.com/2008/07/30/recognising-countable-sets/

Thanks Dr. Gowers’. These are invaluable insights into basics. Thanks for giving so much of your time.

https://gowers.wordpress.com/2008/07/30/recognising-countable-sets/

Thanks Dr. Gowers’. These are invaluable insights into basics. Thanks for giving so much of your time.

**Reference: Nordic Mathematical Contest 1987, R. Todev:**

**Question:**

Let f be a function, defined for natural numbers, that is strictly increasing, such that values of the function are also natural numbers and which satisfies the conditions and for all natural numbers m and n. Define the smallest possible value of a.

**Solution:**

Since, is a function satisfying the conditions of the problem, the smallest possible a is at most 4. Assume that . It is easy to prove by induction that for all . So, taking into account that f is strictly increasing, we get

as well as .

So, we arrive at . But, this is not possible, since is an integer. So, .

Cheers,

Nalin Pithwa.

**Question 1:**

If , find the least value of .

**Solution 1:**

Let

will be minimum when will be minimum.

Now, ….call this equation I.

Hence, z will be maximum when is maximum but is the product of two factors whose sum is .

Hence, will be maximum when both these factors are equal, that is, when

. From equation I, maximum value of . Hence, the least value of .

**Some basics related to maximum and minimum:**

*Basic 1:*

Let a and b be two positive quantities, S their sum and P their product; then, from the identity:

, we have

and .

Hence, if S is given, P is greatest when ; and if P is given, S is least when . That is, *if the sum of two positive quantities is given, their product is greatest when they are equal; and, if the product of two positive quantities is given, their sum is least when they are equal.*

*Basic 2:*

*To find the greatest value of a product the sum of whose factors is constant.*

*Solution 2:*

Let there be n factors , and suppose that their sum is constant and equal to s.

Consider the product , and suppose that a and b are any two unequal factors. If we replace the two unequal factors a and b by the two equal factors , the product is increased, while the sum remains unaltered; hence, *so long as the product contains two unequal factors it can be increased without altering the sum of the factors; *therefore, the product is greatest when all the factors are equal. In this case, the value of each of the n factors is , and the greatest value of the product is , or .

*Corollary to Basic 2:*

If are unequal, ;

that is, .

By an extension of the meaning of the terms *arithmetic mean *and *geometric mean*, this result is usually stated as follows: *the arithmetic mean of any number of positive quantities is greater than the geometric mean.*

*Basic 3:*

*To find the greatest value of * *when * *is constant; **m,n, p, ….being positive integers.*

*Solution to Basic 3:*

Since m,n,p, …are constants, the expression will be greatest when is greatest. But, this last expression is the product of factors whose sum is , or , and therefor constant. Hence, will be greatest when the factors are all equal, that is, when

Thus, the greatest value is .

**Some examples using the above techniques:**

**Example 1:**

Show that where r is any real number.

**Solution 1:**

Since

Hence,

that is, , which is the desired result.

**Example 2:**

Find the greatest value of for any real value of x numerically less than a.

**Solution 2:**

The given expression is greatest when is greatest; but, the sum of the factors of this expression is , that is, ; hence, is greatest when , that is, . Thus, the greatest value is .

*Some remarks/observations:*

The determination of **maximum **and **minimum** values may often be more simply effected by the solution of a quadratic equation than by the foregoing methods. For example:

*Question:*

Divide an odd integer into two integral parts whose product is a maximum.

*Answer:*

Let an odd integer be represented as ; the two parts by x and ; and the product by y; then ; hence,

but the quantity under the radical sign must be positive, and therefore y cannot be greater than , or, ; and since y is integral its greatest value must be ; in which case , or n; thus, the two parts are n and .

**Sometimes we may use the following method:**

Find the minimum value of .

**Solution:**

Put ; then the expression

which in turn equals

.

Hence, the expression is a a minimum when the square term is zero; that is when .

Thus, the minimum value is , and the corresponding value of x is .

**Problems for Practice:**

- Find the greatest value of x in order that may be greater than .
- Find the minimum value of , and the maximum value of .
- Show that and .
- Find the maximum value of when x lies between 7 and -2.
- Find the minimum value of .

More later,

Nalin Pithwa.

**Question:**

If , , be the roots of the cubic equation . Prove that the equation in y whose roots are is obtained by the transformation . Hence, form the equation with above roots.

**Solution:**

Given that , , are the roots of the equation:

…call this equation I.

By relationships between roots and co-efficients, (Viete’s relations), we get

and , and

Now, , that is,

, or …call this equation II.

Subtracting Equation II from Equation I, we get

since

which is the required transformation.

Now, , that is,

Putting this value of x in Equation I, we get

, that is,

, which is the required equation.

Cheers,

Nalin Pithwa.

Some classic examples are presented below to illustrate the use of mathematical induction to prove inequalities:

**Example 1:**

Prove the inequality for all positive integers n.

**Proof 1:**

Let be the proposition that .

*Basis step:*

is true, because . This completes the basis step.

*Inductive step:*

We first assume the inductive hypothesis that is true for the positive integer k. That is, the inductive hypothesis is the statement that . To complete the inductive step, we need to show that if is true, then , which is the statement that is also true. That is, we need to show that if , then . To show that this conditional statement is true for the positive integer k, we first add 1 to both sides of and then note that . This tells us that

.

This shows that is true; namely, that , based on the assumption that is true. The induction step is complete.

Therefore, because we have completed both the basis step and the inductive step, by the principle of mathematical induction we have shown that is true for all positive integers n. QED.

**Example 2:**

Prove that for every positive integer n with . (Note that this inequality is false for ).

**Proof 2:**

Let be the proposition that

*Basis Step:*

To prove the inequality for requires that the basis step be . Note that is true because .

*Inductive Step:*

For the inductive step, we assume that is true for the positive integer k with . That is, we assume that with . We must show that under this hypothesis, is also also true. That is, we must show that if for the positive integer , then . We have

(by definition of exposition)

which (by the inductive hypothesis)

which because

which equals (by definition of factorial function).

This shows that is true when is true. This completes the inductive step of the proof.

We have completed the basis step and the inductive step. Hence, by mathematical induction is true for all integers n with . That is, we have proved that is true for all integers n with . QED.

*An important inequality for the sum of the reciprocals of a set of positive integers will be proved in the example below:*

**Example 3:**

**An inequality for Harmonic Numbers**

The harmonic numbers are defined by .

For instance, . Use mathematical induction to show that , whenever n is a nonnegative integer.

**Proof 3:**

To carry out the proof, let be the proposition that .

*Basis Step:*

is true because .

*Inductive Step:*

The inductive hypothesis is the statement that is true, that is, , where k is a nonnegative integer. We must show that if is true, then , which states that , is also true. So, assuming the inductive hypothesis, it follows that

(by the definition of harmonic number)

which equals (by the definition of the harmonic number)

(by the inductive hypothesis)

(because there are terms each greater than or equal to )

(canceling a common factor of in second term), which in turn, equals .

This establishes the inductive step of the proof.

We have completed the basis step and the inductive step. Thus, by mathematical induction is true for all nonnegative integers. That is, the inequality for the harmonic number is valid for all non-negative integers n. QED.

**Remark:**

The inequality established here shows that the **harmonic series** is a divergent infinite series. This is an important example in the study of infinite series.

More later,

Nalin Pithwa

**Reference: Discrete Mathematics and its Applications by Kenneth H. Rosen, Seventh Edition.**

This is one of the prime utilities of trigonometry —- to calculate heights and distances, called as surveying in civil engineering.

**1. **From the extremities of a horizontal base line AB, whose length is 1 km, the bearings of the foot C of a tower are observed and it is found that is 56 degrees and 23 minutes and is 47 degrees and 15 minutes, and the elevation of the tower from A is 9 degrees and 25 minutes, find the height of the tower.

**2. **A man in a balloon observes that the angle of depression of an object on the ground bearing due north is 33 degrees; the balloon drifts 3 km due west and the angle of depression is now found to be 21 degrees. Find the height of the balloon.

**3. **A tower PN stands on level ground. A base AB is measured at right angles to AN, the points A, B, and N being in the same horizontal plane, and the angles PAN and PBN are found to be and respectively. Prove that the height of the tower is

AB

If AB is 100m, degrees and degrees, calculate the height.

**4. **At each end of a horizontal base of length 2a, it is found that the angular height of a certain peak is and that at the middle point it is . Prove that the vertical height of the peak is

**5. **To find the distance from A to P, a distance AB of 1 km. is measured for a convenient direction. At A the angle PAB is found to be 41 degrees 18 min and at B, the angle PBA is found to be 114 degrees and 38 min. What is the required distance to the nearest metre?

More later,

Nalin Pithwa