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 Mathematics I

MA 1505 Tutorial 7: Integration of Two Variables Functions

October 12, 2013 zr9558 Leave a comment

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 Mathematics I

MA 1505 Tutorial 6: Partial Derivatives and Directional Derivative

October 12, 2013 zr9558 Leave a comment

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 Mathematics I

MA 1505 Tutorial 5: Fourier Series

October 12, 2013 zr9558 2 Comments

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

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