Question

In this problem you will utilize a for loop to compute the an approximation of the value of using the Leibniz's formula for Pi. The formula uses a summation. Follow the instructions correctly and read the questions thoroughly. 1. Set up the initial number of iteration i to obtain six iterations. 2. The approximate value of it can be given by the following formula: Approximated = Enzo 2n+1 Where n is an integer starting at zero indicating the iterations. The initial value of the sum which is equal to zero. 3. Use the above equation and the for loop to compute the first five approximations of the value of . Save the approximate values in a vector. Hint: create an index value j. 4. Using and the Percent difference formula calculate how close your approximation was to the real value for the fourth iteration. Display the percent difference and approximation using disp(). Hint: This does not require a loop. %Difference = Approximated- - x 100% (Approximated+7)
Please complete using matlab

Answer 1

Ans:-

Code:-

%number of iteration i
i = 5;
%from n = 0 to 5
n = 0:i;
%each terms whose sum is approx of pi
S = 4.*((-1).^n)./(2.*n.+1);
%first approx equal to first term of sum
A(1) = S(1);
%rest of approximation equal to sum of previous approximation and next term
for j = 2:5%loop for next approximations
A(j) = A(j-1)+S(j);%sum of previous approx and next term
end
%find the difference
diff = abs(A.-pi)./((A.+pi)./2) .*100;
disp(diff);%display the difference %

lebiniz. mx number of iteration i 2 i = 5; 3 from n = 0 to 5 4 n = 0:i; Peach terms whose sum is approx of pi 6 S = 4.*((-1). n)./12.*n.+1); 7 first approx equal to first term of sum 8 A(1) = S(1); 9 frest of approximation equal to sum of previous approximation and next term 10 for j - 2:5*loop for next approximations 11 A()) = A (-1) +S()); $sum of previous approx and next term 12 Lend 13 find the difference 14 diff = abs (A.-pi)./((A.+pi)./2) .*100; disp (diff); $display the difference $

Output:-

>> lebiniz 24.0397 16.3535 9.8384 8.1617 6.1127

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