Error in a code which computes sqrt(a)

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Anastasia Kyriakou
Anastasia Kyriakou 2020-2-25
回答： Samatha Aleti 2020-2-28
One of the methods to compute sqer (a) , a>0 is
X(n+1) = (a + (X(n)*X(n-1))/(X(n)+X(n-1)), n = 1, 2, …, with X0=1 and X1=a (That is, it is known that lim n-> infin of Xn = sqrt(a)
Write a function [sqa, nitr] = mySqrt(a) which implements this calculation. The function should use a while-loop, terminate when the difference between Xn+1 and Xn becomes smaller than eps(10*a), and output Xn+1 in sqa and the value of n at which the while-loopwas terminated in nitr. Test your function for a = 103041
I have written this but it does not work
function [sqa, nitr] = mySqrt (a)
%[sqa, nitr] = mySqrt (a)
% computes square root of a
% sqa = a;
sqaprev = a;
nitr = 0;
X(n+1) = (a + (X(n)*X(n-1))/(X(n)+X(n-1))); %find the second term
sqa= X(n+1)
while abs (sqaprev-sqa) >= eps (10*a)
sqaprev = sqa;
sqa = (1/2) *(sqaprev+ (a/sqaprev));
nitr = nitr + 1;
end %while
end
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Anastasia Kyriakou
Anastasia Kyriakou 2020-2-25
do i have to write X(n+1) in terms of sqa etc?
Adam
Adam 2020-2-25
When you have a sequence in which the next term is defined in terms of the previous one you have to initialise the first term somehow, otherwise there is no previous term for the first case, to get the whole sequence going. In your case your next term is based on two previous terms, which is why you are given the first two terms of the sequence in the question so that you can then loop round increasing n and referrinig to the previous two terms, which will exist if you insert them as e.g.
X = [ 0 a ];

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回答（1 个）

Samatha Aleti
Samatha Aleti 2020-2-28
You may need to initialize the variables "sqaprev", "sqa" as follows:
function [sqa, nitr] = mySqrt (a)
sqaprev = 0;
sqa= a;
nitr = 0;
while abs (sqaprev-sqa) >= eps (10*a)
sqaprev = sqa;
sqa = (1/2) *(sqaprev+ (a/sqaprev));
nitr = nitr + 1;
end
end

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