Category Archives: MA 1505 Mathematics I

MA 1505 Tutorial 9 and 10: Line Integral and Green’s Formula

Line integral of a scalar field:

Assume f: U \subseteq \mathbb{R}^{n} \rightarrow \mathbb{R} is a smooth function,

where U is the domain of f.

\int_{C} f ds =\int_{a}^{b} f(r(t)) \cdot ||r^{'}(t)|| dt

where r:[a,b] \rightarrow C is a smooth curve.

Line integral of a vector field:

Assume \mathbf{F}: U \subseteq \mathbb{R}^{n} \rightarrow \mathbb{R}^{n} is a smooth vector function,

\int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{a}^{b} \mathbf{F}( \mathbf{r}(t)) \cdot \mathbf{r}^{'}(t) dt

where r:[a,b] \rightarrow C  is a smooth curve.

Green’s Formula: 

Assume \bold{F}=(P,Q) is a vector field,

\oint_{\partial D} \bold{F}\cdot d\bold{s}=\oint_{\partial D} Pdx + Qdy = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{ \partial P}{\partial y}) dxdy,

where D is the domain and \partial D denotes the boundary of D. The orientation of \partial D satisfies the left hand rule. That means if you walk along the boundary of D, the domain D must be on your left.

429px-Green's-theorem-simple-region.svg

D is a simply connected region with boundary consisting four boundaries C_{1}, C_{2}, C_{3}, C_{4}, the orientation is counterclockwise.

macroscopic_microscopic_circulationmacroscopic_microscopic_circulation_hole

In the first graph, \partial D which denotes the boundary of D has only one closed curve C and the orientation of C is counterclockwise. However, in the second graph, \partial D contains two curves, i.e. the blue one and the red one. The orientation on the blue one which is the outer boundary of D is counterclockwise, the orientation on the red one which in the inner boundary of D is clockwise. That means if you walk along the boundary of D, the domain D must be on your left. This is the left hand rule.

Corollary of Green’s Theorem

Assume D is a domain in the plane, Area(D) denotes the area of D, then the area can be calculated from the following formulas:

Area(D)=\iint_{D} dxdy

= \oint_{\partial D} xdy = \oint_{\partial D} -ydx =\oint_{\partial D} (-\frac{y}{2} dx +\frac{x}{2} dy)

Fundamental Theorem of Line Integral:

\int_{C}\nabla f \cdot d\bold{r}=f(\text{terminal point})-f(\text{initial point}),

where C denotes a curve from initial point to terminal point and f is a scalar field.

Conservative Vector Field:

1. \bold{F}=(P,Q,R) is called a conservative vector field, if there exists a scalar field f such that \nabla f=\bold{F}. It is equivalent to these conditions:

\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}, \frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y}, \frac{\partial R}{\partial x}=\frac{\partial P}{\partial z}.

2. \bold{F}=(P,Q) is called a conservative vector field, if  there exists a scalar field f such that \nabla f=\bold{F}. It is equivalent the condition \frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}.

Path Independence:

A key property of a conservative vector field is that its integral along a path depends only on the endpoints of that path, not the particular route taken.

For example, if \bold{F}=(P,Q) or \bold{F}=(P,Q,R) is a conservative vector field, then the value of the line integral I=\int_{C} \bold{F}\cdot d\bold{r} depends only on the initial point and terminal point of the curve C. That means if \bold{F} is a conservative vector field, the curves C_{1} and C_{2} have the same initial points and terminal points, then these two line integrals are equal: \int_{C_{1}} \bold{F}\cdot d\bold{r}=\int_{C_{2}}\bold{F}\cdot d\bold{r}. For this reason, a line integral of a conservative vector field is called path independent.

Question 1. For each non-zero constant a>0, let C_{a} denote the curve y=a \sin x, where 0 \leq x \leq \pi. Let

I(a)= \int_{C_{a}} (1+y^{3})dx + (2x+y)dy

Find the minimum value of I(a) in the domain a>0.

Solution.

Method (i).  Use the definition of line integration.

Since y=a \sin x , 0 \leq x \leq \pi , dy= a \cos x dx ,

I(a)= \int_{0}^{\pi} (1+a^{3} \sin^{3} x) dx + (2x+ a \sin x) a \cos x dx

= \int_{0}^{\pi} ( 1+ a^{3} \sin^{3} x + 2a x \cos x + a^{2} \sin x \cos x) dx.

Since

\int_{0}^{\pi} (a^{3} \sin^{3} x) dx = \frac{4}{3} a^{3},

\int_{0}^{\pi} (2a x \cos x) dx =-4a,

\int_{0}^{\pi} (a^{2} \sin x \cos x) dx =0,

we get

I(a)= \pi + \frac{4}{3} a^{3} - 4a on a>0 .

I^{'}(a) = 4a^{2}-4 .

The minimum value is taken at a=1, the I(1)= \pi -\frac{8}{3}.

Method (ii). Use Green’s Formula.

Consider the domain D bounded by y= a\sin x and x=0, P(x,y)=1+y^{3} and Q(x,y)= 2x+y.

i.e.

D: 0\leq x \leq \pi, 0\leq y \leq a \sin x.

From Green’s Formula, pay attention to the orientation,

\iint_{D} ( 2-3y^{2}) dxdy = -I(a) + \int_{0}^{\pi} dx

Therefore,

I(a) = \pi - \iint_{D} ( 2- 3y^{2}) dxdy

= \pi - \int_{0}^{\pi} \int_{0}^{a \sin x} (2-3y^{2}) dy dx

= \pi - \int_{0}^{\pi} ( 2a \sin x - a^{3} \sin^{3} x) dx

= \pi - 4a + \frac{4}{3} a^{3}.

The derivative of I(a) is I^{'}(a) = 4 a^{2}-4, the minimum value is taken at a=1, and I(1)= \pi-\frac{8}{3}.

Question 2. Prove the area of the disc with radius R is \pi R^{2}.

Solution. 

Method (i). Definition of Integration.

Area=4 \int_{0}^{R} \sqrt{R^{2}-x^{2}} dx

= 4R^{2} \int_{0}^{\frac{\pi}{2}} cos^{2}\theta d \theta  (x=\sin \theta)

= 4R^{2} \int_{0}^{\frac{\pi}{2}} \frac{1+ \cos (2\theta)}{2} d\theta

= \pi R^{2}.

Method (ii). Green Formula

Area(D)=\iint_{D} dxdy

= \oint_{\partial D} xdy = \oint_{\partial D} -ydx =\oint_{\partial D} (-\frac{y}{2} dx +\frac{x}{2} dy)

For D=\{ x^{2}+y^{2}\leq R^{2} \}, on \partial D, x=\cos \theta, y=\sin\theta, where \theta \in [0, 2\pi).

Area(D)= \oint_{\partial D} x dy

= \int_{0}^{2\pi} R \cos \theta \cdot R \cos \theta d\theta

= \pi R^{2}.

Question 3. Prove the area of the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1 is \pi ab.

Solution.  It is similar to Question 2, and x=a \cos \theta, y= b \cos \theta.

Question 4.  Calculate

\int_{C} 2xy dx + (x^{2}+z)dy +y dz

where C consists two line segments: C_{1} from (0,0,0) to (1,0,2), and C_{2} from (1,0,2) to (3,4,1).

Solution. 

Method (i).  From the definition of line integral,

C_{1}: r_{1}(t)=(t,0,2t), C_{2}: r_{2}(t)=(1+2t, 4t, 2-t).

\int_{C_{1}} 2xy dx + (x^{2}+z)dy +y dz=0

\int_{C_{2}} 2xy dx + (x^{2}+z)dy +y dz

=\int_{0}^{1}( 48 t^{2} + 24 t +12) dt

=40.

Method (ii). Check the vector field \mathbf{F}=(2xy, x^{2}+z, y) is a conservative vector field. Since

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

\frac{\partial}{\partial x}(y)=\frac{\partial}{\partial z}(2xy)=0

\frac{\partial}{\partial y}(y)=\frac{\partial}{\partial z}(x^{2}+z)=1

Therefore, \mathbf{F} is a conservative vector field, and we can assume \nabla f=\mathbf{F} , i.e. f_{x}=2xy, f_{y}=x^{2}+z, f_{z}=y. Hence,

f(x,y,z)=x^{2}y+yz+K for some constant K.

Therefore, the answer is f(3,4,1)-f(0,0,0)=40.

Question 5. Evaluate

\oint_{C} (x^{5}-y^{5}) dx + (x^{5}+y^{5}) dy,

where C denotes the boundary with positive orientation of the region between the circles x^{2}+y^{2}=a^{2} and x^{2}+y^{2}=b^{2} with 0<a<b.

Solution.

Method (i). The definition of the line integral.

On circle x^{2}+y^{2}=b^{2}, it is counterclockwise, x= b \cos \theta \text{ and } y= b \sin \theta,  \theta is from 0 to 2\pi.

On circle x^{2}+y^{2}=a^{2}, it is clockwise, x=a \cos \theta \text{ and } y= a \sin \theta, \theta is from 2\pi to 0.

On the circle C_{b}: x^{2}+y^{2}=b^{2} ,

\oint_{C_{b}} (x^{5}-y^{5})dx + ( x^{5}+y^{5}) dy

= b^{6} \int_{0}^{2\pi} ( \cos^{5} \theta - \sin^{5} \theta) \cdot ( - \sin \theta) + ( \cos^{5} \theta + \sin^{5} \theta) \cdot \cos \theta ) d \theta

= b^{6} \int_{0}^{2\pi} ( \cos^{6} \theta + \sin^{6} \theta) - \sin \theta \cos \theta ( \cos^{4} \theta - \sin^{4}\theta ) d\theta

= b^{6} \int_{0}^{2\pi} ( 1- 3\sin^{2} \theta \cos^{2}\theta - \frac{1}{2} \sin 2\theta \cos 2\theta ) d\theta

= b^{6} \int_{0}^{2\pi} ( \frac{5}{8} -\frac{3}{8} \cos 4\theta -\frac{1}{4} \sin 4 \theta ) d\theta

= b^{6} \cdot \frac{5}{4} \cdot \pi.

Pay attention to the orientation, we get the answer is

\frac{5 \pi}{4} ( b^{6}-a^{6}).

Method (ii). Green’s Theorem.

\oint_{C} (x^{5}-y^{5}) dx + (x^{5}+y^{5}) dy,

= \iint_{D}( 5 x^{4}+ 5 y^{4}) dxdy

= 5 \int_{0}^{2\pi} \int_{a}^{b} r^{5}( \cos^{4} \theta + \sin^{4} \theta) dr d \theta

= 5 \int_{a}^{b} r^{5} dr \cdot \int_{0}^{2 \pi} ( \cos^{4} \theta + \sin^{4}\theta) d\theta

= \frac{5( b^{6}-a^{6})}{6} \int_{0}^{2\pi} (1-2\sin^{2}\theta \cos^{2} \theta) d\theta

= \frac{5(b^{6}-a^{6})}{6} \cdot \frac{3 \pi}{2}

= \frac{5}{4} \pi ( b^{6}-a^{6}).

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MA 1505 Tutorial 8: Surface Area and Volume

Assume z=z(x,y) is a surface on \mathbb{R}^{3}, the domain R is the projection of the surface z=z(x,y)  on xy-plane. Then the area of the surface is

\iint_{R} \sqrt{1+z_{x}^{2}+z_{y}^{2}} dxdy,

where z_{x} and z_{y} are partial derivatives of z=z(x,y) with respect to the variable x and y respectively.

If a surface is z=z(x,y)\geq 0 and the projection of it on xy-plane is R, then the volume bounded by xy-plane and the surface z=z(x,y) is

\iint_{R} z(x,y) dxdy.

Theorem 1. 

The surface area of the sphere with radius R is 4\pi R^{2}.

The volume of the sphere with radius R is \frac{4}{3} \pi R^{3}.

Proof.  The equation of the sphere with radius R is x^{2}+y^{2}+z^{2}=R^{2}.

First we calculate the surface area of sphere.

Assume R=\{ (x,y): x^{2}+y^{2} \leq R^{2} \}, the function z=z(x,y)=\sqrt{ R^{2}-x^{2}-y^{2}}.

Then

z_{x}=(-x)/ \sqrt{R^{2}-x^{2}-y^{2}},

z_{y}=(-y)/ \sqrt{R^{2}-x^{2}-y^{2}}.

Therefore

\sqrt{1+z_{x}^{2}+z_{y}^{2}} =R/ \sqrt{R^{2}-x^{2}-y^{2}}.

The surface area of half-sphere is

\iint_{x^{2}+y^{2}\leq R^{2}} \frac{R}{\sqrt{R^{2}-x^{2}-y^{2}}} dxdy

= \int_{0}^{2\pi} \int_{0}^{R} \frac{Rr}{\sqrt{ R^{2}-r^{2}}} dr d\theta

= 2 \pi R \int_{0}^{R} \frac{r}{\sqrt{R^{2}-r^{2}}} dr

= 2 \pi R^{2}

Hence, the total surface area of the sphere with radius R is 4\pi R^{2}.

Second, we calculate the volume of the sphere with radius R.

V= 2\iint_{x^{2}+y^{2} \leq R^{2}} \sqrt{R^{2}-x^{2}-y^{2}} dx dy

= 2 \int_{0}^{2\pi} \int_{0}^{R} \sqrt{R^{2}-r^{2}}\cdot r dr d\theta

= 2 \cdot 2\pi \cdot \int_{0}^{R} \sqrt{R^{2}-r^{2}}\cdot r dr

= \frac{4}{3} \pi R^{3}.

Theorem 2. The volume of the ellipsoid \frac{x^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}}=1 is \frac{4}{3}\pi abc.

Proof.

The upper bound of the volume is

z= c\cdot \sqrt{1-\frac{x^{2}}{a^{2}}- \frac{y^{2}}{b^{2}}}.

The lower bound of the volume is

z=- c\cdot \sqrt{1-\frac{x^{2}}{a^{2}}- \frac{y^{2}}{b^{2}}}.

Assume R=\{\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} \leq 1 \}, the volume of the ellipsoid is

V= 2 \iint_{R} c \cdot \sqrt{1-\frac{x^{2}}{a^{2}}- \frac{y^{2}}{b^{2}}} dxdy

= 2 c \int_{0}^{2\pi} \int_{0}^{1} \sqrt{1-r^{2}} \cdot (r \cdot a \cdot b) dr d\theta

where we use the substitution x=a \cdot r \cos \theta and y=b\cdot r \sin \theta, the determinant of Jacobian matrix is a\cdot b\cdot r.

Therefore, the value equals to

2abc \cdot (2\pi) \int_{0}^{1} \sqrt{1-r^{2}} r dr = \frac{4}{3} \pi abc.

MA 1505 Tutorial 2: Integration

L.Hospital Rule: if the ratio is \infty/\infty or 0/0, then we can use the L.Hospital Rule to calculate the limit. Precisely, the L.Hospital Rule is 

\lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \frac{f^{'}(x)}{g^{'}(x)},

where a is a finite real number or infinity.

If f(x) is a continuous function, then F(x)= \int_{a}^{x} f(t) dt is a differentiable function and its derivative F^{'}(x)=f(x).

General 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)

Question 1. Calculate the value S=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x+\cos x}dx .

Solution. 

S=\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x +\cos x }dx

= \int_{0}^{\frac{\pi}{2}} \frac{\cos(\frac{\pi}{2}-t)}{\sin(\frac{\pi}{2}-t) + \cos( \frac{\pi}{2}-t)} dt

= \int_{0}^{\frac{\pi}{2}} \frac{\sin t}{\cos t+\sin t}dt

Therefore

2S=S+S

= \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x+\cos x} + \frac{\sin x}{\cos x+\sin x} dx

= \int_{0}^{\frac{\pi}{2}} 1 dx= \frac{\pi}{2}

Hence, S=\frac{\pi}{4} .

Question 2. Calculate the value S=\int_{0}^{\frac{\pi}{4}} \ln(1+\tan{x}) dx .

Solution. 

S=\int_{0}^{\frac{\pi}{4}} \ln(1+\tan x) dx

= \int_{0}^{\frac{\pi}{4}} \ln(1+\tan(\frac{\pi}{4}-x)) dx

= \int_{0}^{\frac{\pi}{4}} \ln(\frac{2}{1+\tan x})dx

= \frac{\pi \ln2}{4} - \int_{0}^{\frac{\pi}{4}} \ln(1+\tan x) dx

= \frac{\pi \ln2}{4} - S

Therefore, S=\frac{\pi \ln2}{8} .

Question 3. 

S=\int_{0}^{1} \frac{x^{4}}{x^{4}+(1-x)^{4}}dx

Solution.

S=\int_{0}^{1} \frac{x^{4}}{x^{4}+(1-x)^{4}}dx

= \int_{0}^{1} \frac{(1-x)^{4}}{x^{4}+(1-x)^{4}}dx

= 1- \int_{0}^{1} \frac{x^{4}}{x^{4}+(1-x)^{4}}dx

Therefore, S=0.5.

Question 4. 

\lim_{x\rightarrow 0} \frac{\sin x}{x}=1.

\lim_{x\rightarrow \infty} x\tan\frac{1}{x} = \lim_{y\rightarrow 0}\frac{\tan y}{y}=1, where y=1/x.

\lim_{x\rightarrow \infty} \frac{\ln x}{x^{a}}=0, where a>0.

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.