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# Daily Archives: October 12, 2013

# MA 1505 Tutorial 3: Taylor Series

The Taylor Series of f(x) at the point is

**Question 1. **Let . Calculate the value of S.

**Solution.**

**Method (i).**

**Method (ii). **Integrate the Taylor series of to show that S=1.

The Taylor series of is . Take the integration of the function on the interval [0,1], we get

.

The left hand side equals to 1 from integration by parts.

**Method (iii).** Differentiate the Taylor series of .

The Taylor series of is . Differentiate f(x) and get . Moreover, and .

**Method (iv). **Assume the function This implies f(0)=0. Assume

Since we get That means f(1)=1.

**Method (v).** Assume the function

Therefore, f(1)=1.

**Remark. **There is a similar problem: calculate Answer is

**Question 2.** Let n be a positive integer. Prove that

and calculate the value of the summation

.

**Solution. **

.

To calculate the value of S, there are two methods.

**Method (i). **The summation of S, n is only taken odd numbers. From the first step, we know the summation

.

**Method (ii).**

Since ,

.

Here we use the Taylor series of and .

**Question 3. **Assume

Prove

**Proof.**

**Question 4. **Calculate the summation

**Solution. **

# MA 1505 Tutorial 7: Integration of Two Variables Functions

In the tutorial 7, we will learn to calculate the integration of two variables, reverse the order of integration and polar coordinate.

The formulas of polar coordinate are , , where and .

**Question 1.** The application of polar coordinate. Calculate the value of

**Solution.**

**Method (i).**

.

Therefore

.

Hence .

**Method (ii).**

Since , we get

Assume y=sx, we get

Therefore,

**Question 2.** Calculate the value of

**Solution.**

**Method (i).** Leibniz Integration Rule.

Here denotes the partial derivative of with respect to the variable .

In the question, assume .

Making use of L’Hospital Rule, we have

**Method (ii).** Reverse the order of integration.

The integration domain is and . It is same as and .

.

**Question 3.** MA1505 2010-2011 Semester 2, Question 6(b).

Let R be a region of xy-plane, find the largest possible value of the integration

**Solution. **

Since we want to find the largest possible value, then we must guarantee that on the region R, the function is non-negative. That means the region R is . i.e. . Therefore, we should calculate the integration

**Question 4. ** is a real interval, calculate the maximum value of

**Solution.**

To calculate the maximum value of the integration, the maximal interval Therefore, the maximum value of the integration is

**Qustion 5. **Calculate the multiple integration

**Solution.**

**Method (i). **Use the polar coordinate.

**Method (ii). **Make the substitution , then

The region is and

That is equivalent to and

The integration is

# MA 1505 Tutorial 6: Partial Derivatives and Directional Derivative

In the tutorial, we will learn the partial derivatives for multiple variable functions.

Assume is a two variable function, then we use the notations to describe the partial derivatives of

denotes the partial derivative of f under the variable x.

denotes the partial derivative of f under the variable y.

Similarly, we can also define the second derivative of

,

.

Assume is a unit vector, i.e. its length is 1. If is at the point p, then we can define the **directional derivative** of at point p as

**Theorem 1. Geometric mean is not larger than Arithmetic mean.**

For n positive real numbers ,

“=” if and only if

**Theorem 2. Cauchy’s Inequality.**

For 2n real numbers ,

“=” if and only if

**Proof.**

**Method (i). **Construct a non-negative function f(x) with respect to variable x

Consider the equation f(x)=0, there are only two possibilities: one is the equation f(x)=0 has only one root, the other one is the function has no real roots. Therefore,

Hence,

Moreover, if “=”, then f(x)=0 has only one root , i.e. for all , . That means

By the way, the solution of is and

**Method (ii). **Since , we know

for all

Assume , for all ,

Take the summation at the both sides,

**Question 1.** Assume is a function and . satisfies the partial differential equation

Prove

(1) is a function of y.

(2) is a function of x.

(3) The solution of has the form for some function and .

Proof.

(1) **Method (i) **Make use of derivative.

First, we know . Second, take the partial derivative of the function with respect to the variable x. That means,

from the partial differential equation. Therefore, the function is independent of the variable x. i.e. the function is a function of variable y.

**Method (ii) **Make use of integration.

Since , , then we take the integration of x at the both sides,

, the left hand side is , the right hand side is for some function That means, . and is a function of .

(2) is similar to (1).

(3) From part (1), we know is a function of y. Assume . Take the integration of y at the both sides, we have

for some function for some functions and

**Question 2. **Assume and are non-negative. Find the maximum value of .

**Solution.**

**Method (i)**. Langrange’s Method.

Take three partial derivatives of g,

Solve these three equations, we get and , therefore the maximum value is taken at and

**Method (ii)**. Change to one variable function.

Since L+K=150, we can define the one variable function

The derivative of

The critical point is The maximal value of g(L) is taken at

**Method (iii)**. Mathematical Olympic Method.

Use the fact that the geometric mean is not larger than the arithmetic mean.

.

The maximum value is taken at i.e.

**Question 3.** Assume and are non-negative variables. . Find the maximum value of

**Solution.**

**Method (i).** Langrange’s Method

Take four partial derivatives of the critical point is taken at i.e. the maximum value of f(x,y,z) is taken at

**Method (ii)** Math Olympic Method

The maximum value is taken at , i.e.

**Question 4. ** 2012 Exam MA1505 Semester 1, Question 3(a)

Assume has continuous partial derivatives of all orders, if

Find the value of the constant

**Solution.**

**Method (i) **Use derivatives.

Since has continuous partial derivative of all orders,

Since and

we have

This implies i.e.

**Method (ii). **Use integration.

Comparing them, we know and where and are constants.

Therefore

# MA 1505 Tutorial 5: Fourier Series

In this tutorial, we will learn how to calculate the Fourier series of periodic functions.

Assume is a periodic function with period , i.e. for all . The Fourier Series of is defined as where

for all

for all

**Theorem 1. ** If satisfies Lipchitz condition on , then

**Theorem 2. Parseval’s Identity.**

**Question 1. **Assume for all and on

What is the value of

**Solution. **From Theorem 1, on . Therefore, and . Hence,

**Question 2. **Prove these identities:

**Solution.**

Choose the function on and f(x) is a periodic function with period .

Use the formulas of and , we can prove that the Fourier series of is

From Theorem 1, take , then

Therefore, .

Assume , we get

.

Therefore .

From **Parserval’s identity**, we know

Therefore .

Assume , we get

Therefore, .

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**Theorem 1. **Real Koebe Principle