Homework # 3

3.2

Convert the base - 8 numbers: 71,563 and 3.14 to base 10

To convert the numbers to base ten you must multiply each digit in the number by 8 to the power of its decimal place:

71,563 = 7*(8^4) + 1*(8^3) + 5*(8^2) + 6*(8^1)+ 3*(8^0) =
               7*(4096) + 1*(512) + 5*(64) + 6*(8)+ 3*(1) = 29,555

3.14 = 3*1 + 1*0.125 + 4*0.015625 = 3.1875

3.5

Write an algorithm that computes f(n) = 10,000 from i = 1 to 10,000


Matlab:

sum = 0;

for iter = 1 : 10000

    

    sum = sum + 1/(iter^4) ;   

    

    

end

sum2 = 1.0823;

for iter = 10000 : 1

    

    sum2 = 1/(iter^4) - sum2 ;   

    

    

end

sum

sum2

3.6
Evaluate e^(-5)
Using both of the following equations:

and compare with the true value of 6.737947 x 10^(-3).

Matlab:

function ANS = P36(iter)

% initilizes first values

x = 5;

equation1(1) = 1;

equation2(1) = 1;

equation2de(1) = 1;

Err1(1) = (exp(-5) - 1)/(exp(-5))*1;

Err2(1) = (exp(-5) - 1)/(exp(-5))*1;

%loop to calculate each value based on input values

for loop = 1:iter

   

   %retrieves each term for the iterations

   term = x^loop/factorial(loop); 

   % flips the sign

   equation2de(loop+1) = equation2de (loop) +term;

   equation2(loop + 1) = 1/equation2de(loop+1);

   if (mod(loop,2) == 1)

       term = -term;

   end

   %saves each value to the matrix

   equation1(loop+1) = equation1(loop) + term;

   Errvalue1 = (exp(-5) - equation1(loop))/(exp(-5))*1;

   Err1(loop+1) =  Errvalue1 ;

   Errvalue2 = (exp(-5) - equation2(loop))/(exp(-5))*1;

   Err2(loop+1) =  Errvalue2 ;

   

end

%prints out values and graphs

equation1

equation2

Err1

Err2

plot ((0:iter),equation1,'black');

hold on;

plot ((0:iter),equation2,'red');

plot ((0:iter),Err1,'blue');

plot ((0:iter),Err2,'green');

end

Plot:

Black is Equation 1

Red is Equation 2

Percent error:

Equation values: