Form 6
L17-18 Polynomials (1)
L19-20 Polynomial (2)
L21 Polynomials (3)
L22-24 Inequality (I)- (AM-GM and Cauchy Schwarz)
L25-26 Inequality (II)- (Using Calculus)
L27 Inequality (III) – (Hard problems)
L28 Sequences (I) – (Recurrence Relations)
L29-30 Sequences (II) – (Gauss’s Arithmetic-Geometric Mean and Euler’s number – e)
L31-32 Functional Equations [I will cover basic knowledge about continuity/differentiability needed for this]

*The topics here may change, and I am open to suggestions of my current and prospective students.

Given a real number . Find the greatest area of triangle ABC with circum-radius

Then we have .

Multiplying both sides by yields:

Since , we must have divides

Now, from multiplying both sides by and argue similarly yields divides Hence the theorem.

Suppose is a polynomial of degree . Suppose further that

(i) for all and

(ii)

Find the leading coefficient of and the degree of

I won’t discuss the solution I presented in class here. Here are two innovative ways suggested by the two students.

1) Using Taylor series. One of the students based his solution using the Taylor’s formula. The ingenious thought is that he centered the expansion at Namely, he writes the expansion as:

This method yields the correct answer and I am very impressed.

2) Using the Mean Value Theorem. Another student suggested using the Mean-Value theorem, but he could not finish the proof. I gave this idea some thought and eventually made it to work. Here is my solution whose idea is inspired by my student.

From Differentiating both sides times gives:

for all

Now applying the mean value theorem to yields:

, where

This gives:

Now note that is a polynomial with infinitely many distinct roots, namely, . [Note that they are indeed distinct because the intervals are disjoint. Therefore, by the identity theorem.

Hence, we have for all This immediately yields the degree of is , and that the leading coefficient can be found by integrating times, which yields the answer to be

For example, integrating once yields: , integrating twice yields

Likewise, integrating 101 times yields:

By Viete's theorem, we have (after simplification) ), and since

(ii) This part is actually easy, just applying the previous parts. However, one must pay attention to note (and you have to explicitly state this fact) that are not equal to or Since dividing by something potentially zero is a fatal error in HKAL.

Book 4, 02IQ11.
(a) (ii), (Method 1) Here is a better way to present the solution. (to avoid adding the extra line explaining why might not be good enough.)
for all This would imply is strictly increasing for all since f has no local max or local min.

(method 2): Suppose it has more than one real root, then all roots are real since complex roots come in pairs. Let the roots be Then by Viete’s theorem, we have , and Now, This means , which is absurd.

The rest are as discussed in class.

Solve the following matrix equation where

Page 27
(i)
Hence

Note that we have implicitly used the condition to avoid being

(ii) The sum equals

Again note that we have implicitly used the condition to avoid being

Exercise: try to do the problem on page 27 without using the previous parts. This time, use the theory of roots of unity.

Lesson 2

Page 50

(a) (i) , where . Hence the roots are simply , for

(ii) Next, we have

Thus

Now,

Finally, since dividing both sides by yields the desired factorization for at once.

(b)(i)

(b)(ii) Let , we have and

Adding the subtracting the two equations yield

, and

Plugging this in (a)(ii) yields

Dividing both sides by 2i yields the result at once.

(b)(iii) Dividing both sides by and taking limit as yields

by b(ii).

Now, using yields

Finally, for since for We have

(c) From (a)(ii), setting z = -1 yields

Using yields

Since for , we have

Therefore, is a cubic in with leading coefficient

1(b) Note that and have two identical rows, and therefore by properties of determinants, we have By Factor theorem, we have are factors of Since is a cubic, we must have , where is some root to be determined, and being the leading coefficient. By part (a), we have Hence as claimed.

1(c) Since the coefficient of of is zero, we have the sum of roots is zero. This implies , and therefore we have as claimed.

2) Suppose is a polynomial. Then , but , which implies Hence we have which implies . i.e. is constant, which is absurd. Hence is not a polynomial.