Numerical Methods Problem Solution

Can i please get help with this question? I am new to MATLAB and do not know how to go about solving this in MATLAB, maybe i would be able to solve subsequest problems when i see the steps to this question. Thanks.

[Dyuman Joshi]

How would you solve that question with pen and paper? Do you any knowledge of programming (irrelevant of the programming language)?

[Chijioke Nkwocha]

Solving with pen and paper, i have to, then i will input value for to for ti in the equation t(i+1) to get t1 and i will keep doing this until i get to t25. As i compute the above, i will substitute the values for t0 to t25 to get pi.

I do not have knowledge of programming, just trying to learn. Writing codes for iteration has been a challenge.

[Dyuman Joshi]

Fair enough, I'll give you an outline of the code so you can work with it. In case you have any queries, do mention.

[Chijioke Nkwocha]

I was able to come up with a code for the aforementioned problem, but i doubt it is right. Any idea to make it better?

Answer 1

[Dyuman Joshi]

In MATLAB, indexing start with 1 and not 0 (as it does in some programming languages).

%defining the initial values
t0 = 1/sqrt(3);

t = (sqrt(t0^2+1)-1)/t0;
pi_approx = (sqrt(t^2+1)-1)/t;

for i=1:24
    t(i+1) = (sqrt(t(i)^2+1)-1)/t(i);
    pi_approx(i+1) = 6*2^i*t(i+1);
end

%converting to table
y=array2table([t;pi_approx]', 'VariableNames',{'t','pi_approx'});

[Chijioke Nkwocha]

In the code above, the "for loop" calculates starting from t2, meanwhile, t1 has not been calculated previously.

[Dyuman Joshi]

Do you want t1 to be calculated exclusively? If so, I can modify the code.

The code I have written calculates the corresponding value but it is not allocated to t1, because I treated t0 as t1.

(Reason - "In MATLAB, indexing start with 1 and not 0")

[Chijioke Nkwocha]

Yes, t1 has to be calculated and the corresponding pi also.

[Dyuman Joshi]

I have edited my answer, check it.

[Chijioke Nkwocha]

Thanks.