User:EGM6341.s11.TEAM1.HW3

Find
Solve for n, such that Eq 1.1 is relevant.

Solution
Therefore, the below inequality is proved.

$$\displaystyle \Rightarrow \left|E_{4}^{L}(\frac{7\pi}{8})\right| = 1.480782755480525e-003 \leq \frac{\left|q_{5}(\frac{7\pi}{8})\right|}{5!} = 8.171616062246945e-003 $$

Define the error of Taylor series at nth order as,

Substituting known values into the equations yields,

Our condition is satisfied when n=6,


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$$\displaystyle \begin{align} &\therefore \text{When n=6,} \quad E_6^T < E_4^L \\ &\left | f \left ( \frac{7 \pi\ }{8} \right ) - f_6^T \left ( \frac{7 \pi\ }{8} \right ) \right | = \left | E_6^T \left ( \frac{7 \pi\ }{8} \right ) \right | \le \ \left | f \left ( \frac{7 \pi\ }{8} \right ) - f_4^L \left ( \frac{7 \pi\ }{8} \right ) \right | = \left | e_4^L \left ( \frac{7 \pi\ }{8} \right ) \right | \end{align} $$
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 Result table table and Matlab code 

Matlab code

Result

EGM6341.s11.TEAM1.Yoon 00:00, 10 February 2011 (UTC)

Find
Prove Eq2.1

Solution
Let $$x_{0},x_{1},...,x_{n} \ $$ be distinct real numbers, and let $$f \ $$ be a given real valued function with $$n+1 \ $$ continuous derivatives on the interval $$I_{t}=\left \{ t,x_{0},...,x_{n} \right \},$$ with $$t \ $$ some given number. Assume $$t \ $$ does not equal any node point, and then define

The function $$ G(x) \ $$ is $$n+1 \ $$ times continously differentiable on the interval $$I_{t} \ $$, as are $$E(x) \ $$ and $$\Psi (x) \ $$.

Author--EGM6341.s11.TEAM1.Chiu 01:14, 15 February 2011 (UTC)

Find
1) Using Eq3.1 plot 3 figures simliar to those on p14.2 using uniform discretization $$ \left \{ x_0,...,x_4 \right \} $$ and select $$ t=+5, x=0.75 \ $$ and for $$ i=2 \ $$, plot $$ l_{2,4}(.) \ $$. Find $$ e_4^L(f,t) \ $$ and $$ e_4^L(f,x) \ $$ 2) Using Eq3.2 plot 3 figures simliar to those on p14.2 using uniform discretization $$ \left \{ x_0,...,x_8 \right \}  $$ and select $$ t=+4.5, x=3 \ $$ and for $$ i=3 \ $$, plot $$ l_{3,8}(x) \ $$. Find $$ e_8^L(f,t) \ $$ and $$ e_8^L(f,x) \ $$

Solution
Part 1

From lecture 9, page 9-2, equation 1 :

From lecture 11, page 11-3, equation 3 :



Using the following error definition:

The error at $$t=5 \ $$ and $$x=0.75 \ $$ is

 Matlab code: 

Part 2

Similarly,



and the error at $$t = 4.5 \ $$ and $$x = 3 \ $$ is

 Matlab code: 

Author Egm6341.s11.team1.arm 17:03, 13 February 2011 (UTC)

Find
1) Perform a literature search to find history and applications (if any) of the bifolium curve. 2) Find the area in one leaf (folium) to $$ 10^{-6} \ $$ accuracy using the following methods:

2.1) Compute using Trapezoidal Theorem 2.2) Compute using Simpson's Theorem 2.3) Area equal to sum of triangles

Solution
Part 1) Background on bifolium curve Information on the history and origin of the bifolium curve is sparse. The earliest found references to the curve originate in papers written by Isaac Newton in the 1690s . The curve was later included in several indexes of quartic curves in the late 1800s . The few sources that could be found that discussed the curve were either mathematics articles or books and did not mention any possible applications of the bifolium curve.

Part 2.1) Computation with Trapizoidal method and its error  Trapizoidal method for Bifolium(n=16) Calculated summation of the blue areas and then substracted the summation of the red areas

Required-Research - 3 equations(methods) of bipolium curve

1) Polar equation(Which is given in the lecture note
 * $$\displaystyle \rho = a \sin^2 \theta cos \theta$$


 * advantage for our purpose:
 * - Computation is Simpler and faster then the Cartesian bifolium equation method.


 * disadvantage for our purpose:
 * - Even if we can change coordinate system from Polar to Cartesian, we cannot get exact even distances for range of [a,b]
 * - For example coordinate changing functions(cart2pol,pol2cart) provided from MathWork are just using Jacobian(Gradient) to change coordinate from Polar to Cartesian.

2) x,y Expicit polar equation
 * $$\displaystyle x=a \sin^3 \theta \cos \theta$$
 * $$\displaystyle y=a\sin^2 \theta \cos^2 \theta$$


 * Not necessary to change coordinate however x,y are calculated based on theta. Therefore it still have same x,y coordinates with the method 1.
 * has Same advantages and disadvantages compared to Cartesian equation

3) Cartesian coordinate equation


 * $$\displaystyle f= -2yx^2 + y^4 + 2x^2 y^2 + x^4$$


 * We can calculate x,y coordinate exactly how we wanted. however, to calculate each coordinate we need to seperate upper point and lower point from one polynomial(with power 4).
 * need to tranform the equation to polinomials(with power 4) corresponding to each x points and calculate each polynomials to get roots(two values are imaginary and the rest are for upper y value and lower y value).
 * disadvantage: Computation time is tediously long since symbolic is used to calculate each polynomials

 4) Exact Area of bifolium(We only consider rightside of folium)


 * When $$\displaystyle \rho = 4a \sin^2 \theta \cos \theta ,$$ the area enclosed is,
 * Therefore, our exact area of one leaf in bifolium is $$\displaystyle \frac{\pi}{16} $$

 5) Used algorithem for solution
 * 1. From the Cartesian bifolium equation, switch symbolic x to constant x'(devided into regular intervals) and generate polynomials
 * 2. Compute each polynomials to get roots and remove imaginary roots
 * 3. Now we have y1 and y2 corresponding to one x'(one is for upper y and the other is for lower y)
 * 4. Solve for area of each trapezoids and substract from the exact area to get error

MATLAB Code

Part2.2)Computation with Simpson's method and its error

MATLAB Code

The error of Simpson's method goes to convergence point(about 0.0418487209446289)

Part 2.3)Area=Summation of Triangles Compare two figures down below

required

Test Result Trial condition: n=16, used Trapezoidal method to find out approximation area of Bifolium

Matlab code

EGM6341.s11.TEAM1.Yoon 00:00, 10 February 2011 (UTC)

Find
Prove that Eq5.1 is equal to $$\begin{align}e^{(n+1)}(x)=f^{(n+1)}(x)- \cancelto{0}{f_n^{L(n+1)}(x)} \end{align}$$ using Eq5.2 thru Eq5.4.

Solution
Suppose that $$ f(x) \ $$ is at leat $$ n+1 \ $$ times differentiable function. From Eq.5.3, we can confirm that;

Author --EGM6341.s11.TEAM1.Sujin 23:26, 13 February 2011 (UTC)

Find
Prove Eq 6.1

Solution
Thus,

Author --EGM6341.s11.TEAM1.Sujin 23:26, 13 February 2011 (UTC)

Find
Show steps of integration of Eq7.1 to obatin Eq7.2

Solution
Knowing $$\displaystyle (x-b) \le 0 ,$$ $$ \left | (x-b) \right \vert\ $$ must equal $$ b-x \ $$ to be positive plug into Eq7.1 and begin to solve: Solving the right side of the Eq7.3: Integrate A part first to get: Evaluate from a to b: Use common denominators and reduce: Use a common denominator of 6 and reduce further:

Now, plug Eq7.8 back into Eq7.4: Multiply and reduce to:

Author EGM6341.s11.TEAM1.WILKS 06:29, 13 February 2011 (UTC)EGM6341.s11.TEAM1.WILKS

Given
where $$ h:=\frac{(b-a)}{2} \ $$ or $$ (b-a)=2h \ $$

Find
Show the steps of integrationto prove Eq8.1 equals Eq8.2

Solution
Because we are looking to determine the integral of an absolue valued expression, we need to ensure that as we integrate over the interval $$[a, b],$$ that we only sum together positive values. The question is, how can we accomplish this? If we take the expression as is and integrate over the interval $$[a, b],$$ we can see immediately that due to the even distance between each zero root, the definite integral will result in zero. Choosing dummy values $$a=1$$ and $$b=2,$$ and plotting using MATLAB, we can see below by observation why the definite integral is zero. The positive and negative areas are the same and therefore, eliminate each other in the definite integral and results in telling us something useless.

Therefore, after expanding the polynomial into its cubic form we write the integral out, breaking its limits of integration into to integrals. Because we have expressed the integral in its cubic form, it can be easily integrated. Now we can go ahead and integrate each expression, saving the application of the limit values for the next step. Now applying the limits to each term we arrive at the following expression. The terms that we want to group together are highlighted in matching colors. Wwe need to gather all the terms by common factors and powers. Simplifying the last term we can see that it cancels itself out. Now rewriting we have the following. Now let's factor each of the terms respectively. First we will expand each group. Collecting common terms we have: Now, notice the first term is a forth order polynomial. These are not usually easily factor, but here we see that the power count up and down just like we would see in the binomial expansion, showing symmetry in both $$a$$ and $$b.$$ Therefore, because we know our solution ahead of time, we can see that the factor $$(b-a)$$ is our ultimate goal among all three terms. So, we develop an algorithm to solve for the factor$$(b-a)^2$$ in the fourth order polynomial of the first term. We look for the square of the factor because we are dealing with a fourth order polynomial. Therefore, we are looking for something like the following. In order for there to be a number which exists that renders the above argument true, we must demand some logical conditions which arise from the distributive property of real numbers. Lets call our unknown $$y$$. The products which sum to give the $$-6a^2b^2$$ according to the distributive rule are $$7b^2a^2+7a^2b^2-2ab(y)$$. Therefore, if there is a number that exists in the real number field for which the expression in Eq.8.12 is true we require the following. Solving for $$y$$ we find: Therefore, we know the factors in Eq.12 do exist and our unknown number is $$10ab.$$ Carrying through with factoring the rest of the common terms out (they are highlighted in blue so the reader may follow them through the equations) we have: Using the rule for the difference of 2 squares in the $$2^{nd}$$ term of Eq.8.15 we have: Now we can factor out the common factor $$(-1)(b-a)^2.$$ Find a common denominator and collect like terms.

And therefore, we have arrived at the following solution.

Author Egm6341.s11.team-1.langpm 07:33, 16 February 2011 (UTC)

Signatures
Solved problem 1 -- EGM6341.s11.TEAM1.Yoon 00:00, 10 February 2011 (UTC)

Solved problem 2 -- EGM6341.s11.TEAM1.Chiu 01:14, 15 February 2011 (UTC)

Solved problem 3 -- Egm6341.s11.team1.arm 17:04, 13 February 2011 (UTC)

Solved problem 4 -- EGM6341.s11.TEAM1.Yoon 00:00, 10 February 2011 (UTC)

Solved problem 5 --EGM6341.s11.TEAM1.Sujin 23:28, 13 February 2011 (UTC)

Solved problem 6 --EGM6341.s11.TEAM1.Sujin 23:28, 13 February 2011 (UTC)

Solved problem 7 -- EGM6341.s11.TEAM1.WILKS 03:47, 13 February 2011 (UTC)EGM6341.s11.TEAM1.WILKS

Solved problem 8 --Egm6341.s11.team-1.langpm 10:25, 16 February 2011 (UTC)