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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.
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
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
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
Here we use the Taylor series of and .
Question 3. Assume
Question 4. Calculate the summation
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
Since , we get
Assume y=sx, we get
Question 2. Calculate the value of
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
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
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
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
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,
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
Assume , for all ,
Take the summation at the both sides,
Question 1. Assume is a function and . satisfies the partial differential equation
(1) is a function of y.
(2) is a function of x.
(3) The solution of has the form for some function and .
(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 .
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
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
Method (i) Use derivatives.
Since has continuous partial derivative of all orders,
This implies i.e.
Method (ii). Use integration.
Comparing them, we know and where and are constants.
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
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:
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
Assume , we get
From Parserval’s identity, we know
Assume , we get
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Theorem 1. Real Koebe Principle