Homework # 4

4.3
The following infinite series can be used to approximate e^x:

Use the Taylor series to estimate f(x) = e^(-x) at xi+1 = 1 for xi = 0.2. Employ the zero- , first-, second-, and third- order versions.

(a) This Maclaurin series expansion is a special case of the taylor series when a = 0 because the Taylor series is the same as the Maclaurin series if a = 0, which in this case is x_i.

(b) This code prints out the first four iterations of e^(-x):

Output:  

0.8187    0.1637    0.4257    0.3559

-122.5417   55.4917  -15.7217    3.2686

Matlab:

%f(x) = exp(-x)

%(i+1) = 1

%(i) = 0.2

xi = 0.2;

xi1 = 1;

true = 0.3679;

sol(1) = exp(-xi);

sol(2) = sol(1) -  exp(-xi)*(xi1-xi);

sol(3) = sol(2) + (exp(-xi))/(factorial(2))*(xi1-xi)^2;

sol(4) = sol(3) - (exp(-xi))/(factorial(3))*(xi1-xi)^3;

for iter = 1:4

    error(iter) = ((true - sol(iter)) / true) * 100; 

    

    

    

end

sol()

error()

4.5
Use zero- through third- order Taylor series expansions to predict f(3) for:

using a base point at x = 1. Compute the true percent relative error for each approximation.

For this Problem I used the taylor series to expand out the function f(x) and found the true result after the third order approximation

Output: 

(Error) 111.1913   85.9206   36.1011         0

Matlab:

%f(x) = 25x^3 - 6x^2 + 7x -88

true = 25*(3)^3 - 6*(3)^2 +7*(3) -88;

x = 3;

a = 1;

sol(1) = 25*a^3 - 6*a^2 + 7*a -88;

sol(2) = 75*a^2 - 12*a +7;

sol(3) = 6*(25*a - 2);

sol(4) = 150;

%sol(1) + sol(2)*(x-a) + (sol(3)/factorial(2))*(x-a)^2 + (sol(4)/factorial(3))*(x-a)^3

Solution = sol(1);

Error(1) = ((true - Solution)/true)*100;

for iter = 2:4

    

Solution = Solution + (sol(iter)/factorial(iter-1))*(x-a)^(iter-1);     

Error(iter) = ((true - Solution)/true)*100; 

    

    

end

Solution

Error()

4.8

To solve this problem I simply used the values given and solved the equation to get the two error values

Output:

Error range 20 : 1.2058    1.5425

Error range 40 : 1.0598    1.7351 

Matlab:

A = 0.15;

e = 0.90;

T = 650;

errt= 20;

errt2 = 40;

sigma = 5.67 * 10^(-8);

Sol = A*e*sigma*(T-errt)^4

Sol = A*e*sigma*(T+errt)^4

Sol2 = A*e*sigma*(T-errt2)^4

Sol2 = A*e*sigma*(T+errt2)^4
  

4.12

For these problems I used the formula: 

and derived each formula to get the condition numbers for each equation.
(a)

=  158.115



(b)

=  -10



(c)

 =  0




(d)

  =  -0.000499917....




(e)

 =  -10001




4.17

This Problem was fairly simple, I just plugged in the values given and computed the angle values based of the two error values

OutPut : 

33.3081  34.6676

Matlab:

%v_e/v_0 = 2;

%alpha = 0.25

%asind  = arc sin in degrees

angle = asind((1 + 0.25)*sqrt(1-(0.25/(1+0.25))*(2)^2));

% range = 

Solution = angle-(angle*.02)

Solution = angle+(angle*.02)