MA 1505 Tutorial 3: Taylor Series

The Taylor Series of f(x) at the point x_{0} is

f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(x_{0})}{n!} (x-x_{0})^{n}.

e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n!}

\ln(1+x)=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{n}}{n}

\sin x= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n-1}}{(2n-1)!}

\cos x =\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n}}{(2n)!}

\frac{1}{1-x} =\sum_{n=0}^{\infty} x^{n}

Question 1. Let S=\sum_{n=0}^{\infty} \frac{1}{n!(n+2)} . Calculate the value of S.

Solution.

Method (i).

S=\sum_{n=0}^{\infty} \frac{1}{n!(n+2)}

= \sum_{n=0}^{\infty} \frac{n+1}{(n+2)!}

= \sum_{n=0}^{\infty} \frac{(n+2)-1}{(n+2)!}

= \sum_{n=0}^{\infty} (\frac{1}{(n+1)!}-\frac{1}{(n+2)!})

= 1

Method (ii). Integrate the Taylor series of xe^{x} to show that S=1.

The Taylor series of x e^{x} is \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} . Take the integration of the function on the interval [0,1], we get

\int_{0}^{1} xe^{x} dx

=\int_{0}^{1} \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!} dx

= \sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{x^{n+1}}{n!} dx

= \sum_{n=0}^{\infty} \frac{1}{n!(n+2)}=S .

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

Method (iii). Differentiate the Taylor series of (e^{x}-1)/x.

The Taylor series of f(x)= (e^{x}-1)/x is \sum_{n=1}^{\infty} \frac{x^{n-1}}{n!} . Differentiate f(x) and get f^{'}(x)= \sum_{n=2}^{\infty} \frac{x^{n-2}}{(n-2)!n} . Moreover, f^{'}(x)= \frac{e^{x}x-(e^{x}-1)}{x^{2}} and f^{'}(1)=1=S.

Method (iv). Assume the function f(x)=\sum_{n=0}^{\infty} x^{n}/(n!(n+2)). This implies f(0)=0. Assume

g(x)=\int_{0}^{x}f(t)dt= \sum_{n=0}^{\infty} \int_{0}^{x} \frac{t^{n}}{n!(n+2)}dt = \sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+2)!} = \frac{1}{x} \sum_{n=0}^{\infty} \frac{x^{n+2}}{(n+2)!} = \frac{1}{x}(e^{x}-1-x).

Since f(x)=g^{'}(x), we get f(x) = x^{-1}(e^{x}-1)-x^{-2}(e^{x}-1-x). That means f(1)=1.

Method (v). Assume the function f(x)=\sum_{n=0}^{\infty} x^{n}/(n!(n+2)).

f(x)= \sum_{n=0}^{\infty} \frac{x^{n}}{(n+1)!} - \frac{x^{n}}{(n+2)!} = x^{-1}\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)!} - x^{-2}\sum_{n=0}^{\infty} \frac{x^{n+2}}{(n+2)!} = x^{-1} (e^{x}-1) - x^{-2}(e^{x}-1-x). Therefore, f(1)=1.

Remark. There is a similar problem: calculate \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!(n+2)}. Answer is 1-2e^{-1}.

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

\frac{1}{2} \int_{0}^{1} t^{n-1}(1-t)^{2} dt= \frac{1}{n(n+1)(n+2)}

and calculate the value of the summation

S=\frac{1}{1\cdot 2 \cdot 3} + \frac{1}{3\cdot 4 \cdot 5} + \frac{1}{5\cdot 6\cdot 7} + \frac{1}{7\cdot 8 \cdot 9}+.... .

Solution. 

\frac{1}{2}\int_{0}^{1} t^{n-1}(1-t)^{2}dt

= \frac{1}{2} \int_{0}^{1} (t^{n+1}-2t^{n}+t^{n-1}) dt

= \frac{1}{2} (\frac{1}{n+2}-\frac{2}{n+1}+\frac{1}{n})

= \frac{1}{n(n+1)(n+2)} .

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

S=\frac{1}{2} \int_{0}^{1} (1+t^{2}+t^{4}+t^{6}+...)(1-t)^{2}dt

= \frac{1}{2} \int_{0}^{1} \frac{1}{1-t^{2}} (1-t)^{2} dt

= \frac{1}{2} \int_{0}^{1} \frac{1-t}{1+t} dt

= \frac{1}{2} \int_{0}^{1} (\frac{2}{1+t}-1)dt

= \frac{1}{2}( 2\ln(1+t)-t)_{t=0}^{t=1}

= \ln 2 -\frac{1}{2} .

Method (ii).

Since \frac{1}{n(n+1)(n+2)}= \frac{1}{2}(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}) ,

S=\sum_{ odd} \frac{1}{n(n+1)(n+2)}

= \frac{1}{2} \sum_{odd} ( \frac{1}{n}- \frac{2}{n+1}+\frac{1}{n+2})

= \frac{1}{2} ( \frac{1}{1}-\frac{2}{2}+\frac{1}{3}+ \frac{1}{3}-\frac{2}{4}+\frac{1}{5}+ \frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{7}-\frac{2}{8}+\frac{1}{9}+...)

= \frac{1}{2} ( \frac{1}{1} + 2( -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-...))

= \frac{1}{2} ( 1 + 2 (\ln 2-1))

= \ln 2 -\frac{1}{2} .

Here we use the Taylor series of \ln(1+x)= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^{n}}{n} and \ln 2= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....

Question 3. Assume \zeta(k)=1+\frac{1}{2^{k}} + \frac{1}{3^{k}} + ... = \sum_{m=1}^{\infty} \frac{1}{m^{k}}.

Prove

\sum_{k=2}^{\infty} (\zeta(k)-1)=1.

\sum_{k=1}^{\infty} (\zeta(2k)-1)=3/4.

Proof.

\sum_{k=2}^{\infty} (\zeta(k)-1)

= \sum_{k=2}^{\infty} \sum_{m=2}^{\infty} \frac{1}{m^{k}}

= \sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \frac{1}{m^{k}}

= \sum_{m=2}^{\infty} \frac{1}{(m-1)m}

= \sum_{m=2}^{\infty} ( \frac{1}{m-1} - \frac{1}{m})

= 1.

\sum_{k=1}^{\infty} ( \zeta(2k)-1)

= \sum_{k=1}^{\infty} \sum_{m=2}^{\infty} \frac{1}{m^{2k}}

= \sum_{m=2}^{\infty} \sum_{k=1}^{\infty} \frac{1}{m^{2k}}

= \sum_{m=2}^{\infty} \frac{1}{m^{2}-1}

= \sum_{m=2}^{\infty} \frac{1}{2} ( \frac{1}{m-1}-\frac{1}{m+1})

= \frac{1}{2}(1+\frac{1}{2})

= \frac{3}{4}.

Question 4. Calculate the summation S= \sum_{k=1}^{\infty} \frac{(-1)^{k} k}{4k^{2}-1}.

Solution. 

S=\sum_{k=1}^{\infty} \frac{(-1)^{k} k}{4k^{2}-1}

=\frac{1}{4} \sum_{k=1}^{\infty} (-1)^{k} ( \frac{1}{2k-1} +\frac{1}{2k+1})

= \frac{1}{4} \sum_{k=1}^{\infty} ( \frac{(-1)^{k}}{2k-1} - \frac{(-1)^{k+1}}{2k+1})

= \frac{1}{4} \cdot \frac{-1}{2-1} = - \frac{1}{4}.

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 x=r \cos(\theta), y=r \sin(\theta), where r\in (0,\infty) and \theta \in [0, 2\pi).

\iint_{D} f(x,y) dxdy= \iint_{D^{'}} f(r \cos \theta, r \sin \theta) r dr d\theta

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

I= \int_{-\infty}^{\infty} e^{-x^{2}}dx.

Solution.

Method (i).

I=\int_{-\infty}^{\infty} e^{-x^{2}} dx= \int_{-\infty}^{\infty} e^{-y^{2}}dy .

Therefore

I^{2}=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} dxdy

= \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^{2}} r dr d\theta

= 2\pi \int_{0}^{\infty} e^{-r^{2}}r dr

= 2\pi \frac{1}{2} e^{-r^{2}}|_{r=0}^{r=\infty}

= \pi .

Hence I=\sqrt{\pi} .

Method (ii).

Since I=\int_{-\infty}^{\infty} e^{-x^{2}}dx = 2\int_{0}^{\infty} e^{-x^{2}}dx = 2\int_{0}^{\infty} e^{-y^{2}}dy , we get

I^{2}=\int_{0}^{\infty} \int_{0}^{\infty} e^{-x^{2}-y^{2}} dy dx

Assume y=sx, we get

I^{2}=4 \int_{0}^{\infty} \int_{0}^{\infty} e^{-x^{2}(1+s^{2})} x ds dx

=4 \int_{0}^{\infty} \int_{0}^{\infty} e^{-(1+s^{2})x^{2}} x dx ds

=4 \int_{0}^{\infty} \frac{1}{2(1+s^{2})} ds

=4 \cdot \frac{1}{2} \arctan s|_{s=0}^{s= \infty}

= \pi

Therefore, I=\sqrt{\pi}

Question 2. Calculate the value of

\lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} \int_{x}^{t} \sin{y^{2}} dy dx}{t^{4}}.

Solution.

Method (i). Leibniz Integration Rule.

\frac{d}{d\theta} ( \int_{a(\theta)}^{b(\theta)} f(x,\theta)dx)

= \int_{a(\theta)}^{b(\theta)} f_{\theta}(x,\theta) dx + f(b(\theta), \theta)\cdot b^{'}(\theta) - f(a(\theta),\theta) \cdot a^{'}(\theta)

Here f_{\theta}(x,\theta) denotes the partial derivative of f(x, \theta) with respect to the variable \theta.

In the question, assume G(x,t)=\int_{x}^{t} \sin{y^{2}} dy .

Making use of L’Hospital Rule, we have

\lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} \int_{x}^{t} \sin{y^{2}} dy dx}{t^{4}}

= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} G(x,t)dx}{t^{4}}

= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} G_{t}(x,t)dx+ G(t,t)\cdot 1 - G(0,t)\cdot 0}{ 4 t^{3}}

= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} \sin{t^{2}}dx}{4t^{3}}

= \lim_{t\rightarrow 0^{+}} \frac{ t \sin{t^{2}}}{4t^{3}}= \frac{1}{4}

Method (ii). Reverse the order of integration.

The integration domain is 0\leq x \leq t and x \leq y \leq t. It is same as 0\leq y \leq t and 0\leq x\leq y.

Answer= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} \int_{0}^{y} \sin{y^{2}} dxdy}{t^{4}}

= \lim_{t\rightarrow 0^{+}} \frac{\int_{0}^{t} y \sin{y^{2}}dy}{t^{4}}

= \lim_{t\rightarrow 0^{+}} \frac{ t \sin{t^{2}}}{4t^{3}}

=\frac{1}{4} .

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

\iint_{R} (4-x^{2}-y^{2})dxdy.

Solution. 

Since we want to find the largest possible value, then we must guarantee that on the region R, the function f(x,y)=4-x^{2}-y^{2} is non-negative. That means the region R is 4-x^{2}-y^{2}\geq 0. i.e. x^{2}+y^{2}\leq 4. Therefore, we should calculate the integration

\iint_{x^{2}+y^{2}\leq 4} (4-x^{2}-y^{2}) dxdy

= \int_{0}^{2\pi} \int_{0}^{2} (4-r^{2})r dr d\theta

= 2\pi \int_{0}^{2} (4r-r^{3})dr

= 8\pi

Question 4. I \subseteq \mathbb{R} is a real interval, calculate the maximum value of

\int_{I} (1-x^{2}) dx.

Solution.

To calculate the maximum value of the integration, the maximal interval I=[-1,1]. Therefore, the maximum value of the integration is

\int_{-1}^{1} (1-x^{2}) dx = \frac{4}{3}.

Qustion 5. Calculate the multiple integration

\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} e^{x^{2}+y^{2}} dy dx.

Solution.

Method (i).  Use the polar coordinate.

\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} e^{x^{2}+y^{2}} dydx

= \int_{0}^{\pi/2} \int_{0}^{1} e^{r^{2}} r dr d\theta

= \frac{\pi}{2} \int_{0}^{1} e^{r^{2}} r dr

= \frac{\pi}{2} (\frac{e^{r^{2}}}{2}) |_{r=0}^{r=1}

= \frac{\pi}{4}(e-1).

Method (ii). Make the substitution y=sx, then dy=x ds.

The region is 0\leq x \leq 1 and 0\leq s \leq \sqrt{1-x^{2}}/x.

That is equivalent to 0 \leq s \leq \infty and 0 \leq x \leq 1/\sqrt{1+s^{2}}.

The integration is

\int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}/x} e^{x^{2}+s^{2}x^{2}} xds dx

= \int_{0}^{\infty} \int_{0}^{1/\sqrt{1+s^{2}}} e^{(1+s^{2})x^{2}} x dx ds

= \int_{0}^{\infty} (\frac{1}{2(1+s^{2})} e^{(1+s^{2})x^{2}} |_{x=0}^{x=1/\sqrt{1+s^{2}}}) ds

= \int_{0}^{\infty} \frac{e-1}{2(1+s^{2})}ds

= \frac{e-1}{2} \arctan s|_{s=0}^{s=\infty}

= \frac{\pi}{4} (e-1).

MA 1505 Tutorial 6: Partial Derivatives and Directional Derivative

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

Assume z=f(x,y) is a two variable function, then we use the notations to describe the partial derivatives of f(x,y).

f_{x}=\frac{\partial f}{\partial x} denotes the partial derivative of f under the variable x.

f_{y}=\frac{\partial f}{\partial y} denotes the partial derivative of f under the variable y.

Similarly, we can also define the second derivative of f(x,y).

f_{xx}=\frac{\partial^{2} f}{\partial x^{2}} ,

f_{xy}=f_{yx}=\frac{\partial^{2}f}{\partial x\partial y}=\frac{\partial^{2}f}{\partial y\partial x}  \text{ if } f(x,y) \text{ is a } C^{2} \text{ function.}

f_{yy}=\frac{\partial^{2} f}{\partial y^{2}} .

Assume u=(a,b) is a unit vector, i.e. its length is 1. If f(x,y) is C^{1} at the point p, then we can define the directional derivative of f(x,y) at point p as

f_{x}(p) a + f_{y}(p) b

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

For n positive real numbers a_{1}, a_{2}, ..., a_{n} ,

(a_{1}...a_{n})^{\frac{1}{n}} \leq \frac{a_{1}+...+a_{n}}{n}

“=” if and only if a_{1}=a_{2}=...=a_{n}.

Theorem 2. Cauchy’s Inequality.

For 2n real numbers a_{1},..., a_{n}, b_{1},...,b_{n} ,

(a_{1}b_{1}+...+a_{n}b_{n})^{2}\leq (a_{1}^{2}+...+a_{n}^{2})(b_{1}^{2}+...+b_{n}^{2})

“=” if and only if \frac{a_{1}}{b_{1}}=...=\frac{a_{n}}{b_{n}}.

Proof.

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

f(x)= \sum_{i=1}^{n}(a_{i}x-b_{i})^{2}= (\sum_{i=1}^{n} a_{i}^{2}) x^{2} - 2(\sum_{i=1}^{n} a_{i}b_{i}) x + (\sum_{i=1}^{n} b_{i}^{2}).

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,

\Delta=4(\sum_{i=1}^{n} a_{i}b_{i})^{2} - 4 ( \sum_{i=1}^{n} a_{i}^{2}) \cdot ( \sum_{i=1}^{n} b_{i}^{2}) \leq 0.

Hence, (a_{1}b_{1}+...+a_{n}b_{n})^{2}\leq (a_{1}^{2}+...+a_{n}^{2})(b_{1}^{2}+...+b_{n}^{2}).

Moreover, if “=”, then f(x)=0 has only one root x_{0}, i.e. for all 1\leq i \leq n, a_{i}x_{0}-b_{i}=0. That means

\frac{a_{1}}{b_{1}} =...=\frac{a_{n}}{b_{n}}.

By the way, the solution of ax^{2}+bx+c=0 is \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} and \Delta=b^{2}-4ac.

Method (ii). Since a^{2}+b^{2}\geq 2 ab, we know

ab\leq \frac{1}{2}(\lambda^{2} a^{2}+ b^{2}/\lambda^{2}) for all \lambda \neq 0.

Assume \lambda^{2}=\sqrt{(\sum_{i=1}^{n}b_{i}^{2})/(\sum_{i=1}^{n} a_{i}^{2})} , for all 1\leq i \leq n,

a_{i}b_{i}\leq \frac{1}{2} (\lambda^{2}a_{i}^{2}+b_{i}^{2}/\lambda^{2})

Take the summation at the both sides,

\sum_{i=1}^{n} a_{i}b_{i} \leq \frac{1}{2}( \lambda^{2}\sum_{i=1}^{n}a_{i}^{2} + (\sum_{i=1}^{n} b_{i}^{2})/ \lambda^{2})= \sqrt{(\sum_{i=1}^{n} a_{i}^{2}) \cdot (\sum_{i=1}^{n}b_{i}^{2})}.

Question 1. Assume u(x,y) is a C^{2} function and u>0 . u(x,y) satisfies the partial differential equation u u_{xy}= u_{x}u_{y}.

Prove

(1) \frac{\partial \ln u}{\partial y} is a function of y.

(2) \frac{\partial \ln u}{\partial x} is a function of x.

(3) The solution of u(x,y) has the form u(x,y)=f(x) g(y) for some function f(x) and g(y) .

Proof.

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

First, we know \frac{\partial \ln u}{\partial y}=\frac{u_{y}}{u} . Second, take the partial derivative of the function with respect to the variable x. That means,

\frac{\partial }{\partial x} (\frac{u_{y}}{u})= \frac{u_{xy}u- u_{x}u_{y}}{u^{2}}=0 from the partial differential equation. Therefore, the function \frac{u_{y}}{u} is independent of the variable x. i.e. the function is a function of variable y.

Method (ii) Make use of integration.

Since u u_{xy}=u_{x}u_{y} , \frac{u_{x}}{u}=\frac{u_{xy}}{u_{y}} , then we take the integration of x at the both sides,

\int \frac{u_{x}}{u} dx =\int \frac{u_{xy}}{u_{y}} dx , the left hand side is \ln u, the right hand side is \ln |u_{y}| + h_{1}(y) for some function h(y). That means, \frac{|u_{y}|}{u}=e^{-h_{1}(y)} . and \frac{u_{y}}{u} is a function of y .

(2) is similar to (1).

(3) From part (1), we know \frac{\partial \ln u }{\partial y} is a function of y. Assume \frac{\partial \ln u }{\partial y} = h_{2}(y). Take the integration of y at the both sides, we have

\ln u= \int h_{2}(y) dy + h_{3}(x) for some function h_{3}(x) . u = e^{\int h_{2}(y) dy} \cdot e^{h_{3}(x)} = g(y) \cdot f(x) for some functions f(x) and g(y).

Question 2. Assume L+K=150,  L and K are non-negative. Find the maximum value of f(L,K)=50 L^{0.4} K^{0.6} .

Solution.

Method (i). Langrange’s Method.

g(L,K,\lambda)=f(L,K)-\lambda(L+K-150)=50L^{\frac{2}{5}}K^{\frac{3}{5}}-\lambda(L+K-150).

Take three partial derivatives of g,

\frac{\partial g}{\partial \lambda} = -(L+K-150)=0

\frac{\partial g}{\partial L} = 50 \cdot \frac{2}{5} L^{-\frac{3}{5}}K^{\frac{3}{5}} - \lambda

\frac{\partial g}{\partial K}= 50 \cdot \frac{3}{5} L^{\frac{2}{5}} K^{-\frac{2}{5}} - \lambda=0

Solve these three equations, we get 2K=3L and L+K=150 , therefore the maximum value is taken at L=60 and K=90.

Method (ii). Change to one variable function.

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

g(L)=f(L,150-L)=50 L^{\frac{2}{5}}(150-L)^{\frac{3}{5}}.

The derivative of g^{'}(L)=50 \cdot (\frac{2}{5} L^{-\frac{3}{5}}(150-L)^{\frac{3}{5}} - L^{\frac{2}{5}}\frac{3}{5}(150-L)^{-\frac{2}{5}}).

The critical point is L=60. The maximal value of g(L) is taken at L=60, K=90.

Method (iii). Mathematical Olympic Method.

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

f(L,K)=50 L^{\frac{1}{5}}L^{\frac{1}{5}} K^{\frac{1}{5}} K^{\frac{1}{5}} K^{\frac{1}{5}}

= \frac{50}{3^{\frac{2}{5}} 2^{\frac{3}{5}}} (3L)^{\frac{1}{5}} (3L)^{\frac{1}{5}} (2K)^{\frac{1}{5}} (2K)^{\frac{1}{5}} (2K)^{\frac{1}{5}}

\leq \frac{50}{3^{\frac{3}{5}} 2^{\frac{2}{5}}} \frac{3L+3L+2K+2K+2K}{5}

= \frac{50}{3^{\frac{2}{5}} 2^{\frac{3}{5}}} \frac{6(L+K)}{5}

= \frac{50}{3^{\frac{2}{5}} 2^{\frac{3}{5}}} \frac{6\cdot 150}{5} .

The maximum value is taken at 3L=2K. i.e. L=60, K=90.

Question 3. Assume 24x+18y+12z=144 and x, y, z are non-negative variables. f(x,y,z)=18 x^{2} y z . Find the maximum value of f(x,y,z).

Solution.

Method (i). Langrange’s Method

g(x,y,z,\lambda)=f(x,y,z)-\lambda(24x+18y+12z-144) = 18 x^{2} y z-\lambda(24x+18y+12z-144)

Take four partial derivatives of g, the critical point is taken at x=z=1.5y. i.e. the maximum value of f(x,y,z) is taken at x=z=3, y=2.

Method (ii) Math Olympic Method

f(x,y,z)=18 x \cdot x \cdot y \cdot z

= \frac{18}{12*12*18*12} (12x) \cdot (12x) \cdot (18y) \cdot (12z)

\leq \frac{18}{12*12*18*12} (\frac{12x+12x+18y+12z}{4})^{4}

= \frac{18}{12*12*18*12} (\frac{144}{4})^{4}

The maximum value is taken at 12x=12x=18y=12z, i.e. x=z=3, y=2.

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

Assume f(x,y) has continuous partial derivatives of all orders, if

\nabla f = (xy^{2}+kx^{2}y+x^{3}) \textbf{i} + (x^{3}+x^{2}y+y^{2}) \textbf{j},

Find the value of the constant k.

Solution.

Method (i) Use derivatives.

Since f has continuous partial derivative of all orders, f_{xy}= f_{yx}.

Since f_{x}= xy^{2}+kx^{2}y+x^{3} and f_{y}=x^{3}+x^{2}y+y^{2},

we have

\frac{\partial}{\partial y} (xy^{2}+kx^{2}y+x^{3})= \frac{\partial}{\partial x} (x^{3}+x^{2}y+y^{2})

This implies 2xy+kx^{2}=3x^{2}+2xy. i.e. k=3.

Method (ii). Use integration.

f_{x}=xy^{2}+kx^{2}y+x^{3} \Rightarrow f(x,y)=\frac{1}{2}x^{2}y^{2} + \frac{k}{3}x^{3}y + \frac{1}{4}x^{4} + h_{1}(y),

f_{y}=x^{3}+x^{2}y+y^{2} \Rightarrow f(x,y)=x^{3}y+\frac{1}{2}x^{2}y^{2}+\frac{1}{3}y^{3}+h_{2}(x).

Comparing them, we know k=3, h_{1}(y)=\frac{1}{3}y^{3}+C_{1} and h_{2}(x)=\frac{1}{4}x^{4}+C_{2}, where C_{1} and C_{2} are constants.

Therefore k=3.

MA 1505 Tutorial 5: Fourier Series

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

Assume f(x) is a periodic function with period 2\pi, i.e. f(x)=f(x+2\pi) for all x \in \mathbb{R} . The Fourier Series of f(x) is defined as a_{0}+\sum_{n=1}^{\infty} (a_{n} \cos(nx) +b_{n} \sin(nx)), where

a_{0}= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) dx,

a_{n}= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx for all n\geq 1,

b_{n}= \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx for all n\geq 1,

Theorem 1.  If f(x) satisfies Lipchitz condition on (-\pi, \pi) , then

f(x) =a_{0}+\sum_{n=1}^{\infty} (a_{n} \cos(nx) +b_{n} \sin(nx)).

Theorem 2. Parseval’s Identity.

\frac{1}{\pi} \int_{-\pi}^{\pi} |f(x)|^{2} dx= 2a_{0}^{2}+ \sum_{n=1}^{\infty} (a_{n}^{2}+b_{n}^{2}).

Question 1. Assume f(x)=f(x+2\pi) for all x\in \mathbb{R} and f(x)=1505+1506x+1507x^{2}+1508x^{3} on [-\pi, \pi).

What is the value of a_{0}+\sum_{n=1}^{\infty}a_{n} ?

Solution. From Theorem 1, f(x)=a_{0}+\sum_{n=1}^{\infty} (a_{n} \cos(nx) +b_{n} \sin(nx)) on (-\pi, \pi) . Therefore, f(0)=a_{0}+\sum_{n=1}^{\infty}a_{n} and f(0)=1505 . Hence, a_{0}+\sum_{n=1}^{\infty}a_{n}=1505.

Question 2. Prove these identities:

\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{2}}=\frac{\pi^{2}}{8}

\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}

\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{4}}=\frac{\pi^{4}}{96}

\sum_{n=1}^{\infty} \frac{1}{n^{4}}=\frac{\pi^{4}}{90}

Solution.

Choose the function f(x)=|x| on (-\pi, \pi) and f(x) is a periodic function with period 2\pi.

Use the formulas of a_{n} and b_{n}, we can prove that the Fourier series of f(x)=|x| is

\frac{\pi}{2} + \sum_{n=1}^{\infty} \frac{2((-1)^{n}-1)}{\pi}  \cdot \frac{cos(nx)}{n^{2}}

From Theorem 1, take x=0, then

0= \frac{\pi}{2} + \sum_{n=1}^{\infty} \frac{2((-1)^{n}-1)}{n^{2} \pi}  = \frac{\pi}{2} + \sum_{m=1}^{\infty} \frac{-4}{(2m-1)^{2}\pi}  = \frac{\pi}{2} - \frac{4}{\pi} \sum_{m=1}^{\infty} \frac{1}{(2m-1)^{2}}

Therefore, \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{2}}=\frac{\pi^{2}}{8} .

Assume S=\sum_{n=1}^{\infty} \frac{1}{n^{2}} , we get

S=\sum_{odd} \frac{1}{n^{2}} + \sum_{even} \frac{1}{n^{2}}  = \frac{\pi^{2}}{8} + \frac{1}{4} S .

Therefore S=\frac{\pi^{2}}{6} .

From Parserval’s identity, we know

\frac{2\pi^{2}}{3}= \frac{1}{\pi} \int_{-\pi}^{\pi} x^{2}dx  = 2\cdot (\frac{\pi}{2})^{2} + \sum_{n=1}^{\infty} \frac{4((-1)^{n}-1)^{2}}{\pi^{2}\cdot n^{4}}  = \frac{\pi^{2}}{2} + \sum_{m=1}^{\infty} \frac{16}{\pi^{2} (2m-1)^{4}}

Therefore \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{4}} = \frac{\pi^{4}}{96} .

Assume S=\sum_{n=1}^{\infty} \frac{1}{n^{4}} , we get

S=\sum_{odd} \frac{1}{n^{4}} + \sum_{even} \frac{1}{n^{4}}  = \frac{\pi^{4}}{96} + \frac{1}{16} S

Therefore, S=\frac{\pi^{4}}{90} .