Why does the value of tolerance stop at n=2 (third value of the iteration) within the while loop?

I am using a while loop to determine the taylor expansion of cos(x), I am trying to work out how many iterations (with the result of each iteration) it takes to reach the tolerance value (tol) for a given value of x (in this case sin(pi/5)) and tolerance of exp(-7).

When I run the for loop, I get 3 values output (n=0,1,2) but not a third value, as the answer for e suggests that I should get to n=3, yet my code seems to stop as soon as I reach n=2. However, the output value is still greater than the allowed tolerance, so I am unsure why the while loop does not complete another iteration (to be within the tolerance value).

clear;clc;
tol = exp(-7);
x = sin(pi/5);
target = cos(x);
counter = 1;
n(counter) = 0;
result(counter) = (((-1)^n(counter)) * (x^(2*n(counter)))/(factorial(2*n(counter))));
while abs(result(counter)-target) > tol
    counter = counter + 1;
    n(counter) = n(counter-1) + 1;
    result(counter) = result(counter-1) + (((-1)^n(counter)) * (x^(2*n(counter)))/(factorial(2*n(counter))))
end

[Aquatris]

The output value is not greater than the allowed tolerance since the difference between result(counter) and target becomes less than the tol.

Can you explain what you mean by "as the answer for e suggests that I should get to n=3"?

clear;clc;
tol = exp(-7);
x = sin(pi/5);
target = cos(x);
counter = 1;
n(counter) = 0;
result(counter) = (((-1)^n(counter)) * (x^(2*n(counter)))/(factorial(2*n(counter))));
while abs(result(counter)-target) > tol
    counter = counter + 1;
    n(counter) = n(counter-1) + 1;
    result(counter) = result(counter-1) + (((-1)^n(counter)) * (x^(2*n(counter)))/(factorial(2*n(counter))));
end
n

n = 1×3

0 1 2

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[result(end) target]

ans = 1×2

0.8322 0.8322

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abs(result(end)-target)<tol

ans = logical
   1

Answer 1

[Star Strider]

The loop appears to be working correctly.

Adding ‘Check_Convergence’ and examiining the results demonstrates this —

clear;clc;
tol = exp(-7)

tol = 9.1188e-04

x = sin(pi/5);
target = cos(x)

target = 0.8322

counter = 1;
n(counter) = 0;
result(counter) = (((-1)^n(counter)) * (x^(2*n(counter)))/(factorial(2*n(counter))));
while abs(result(counter)-target) > tol
    counter = counter + 1;
    n(counter) = n(counter-1) + 1
    result(counter) = result(counter-1) + (((-1)^n(counter)) * (x^(2*n(counter)))/(factorial(2*n(counter))))
    Check_Convergence = abs(result(counter)-target)
end

n = 1×2

0 1

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result = 1×2

1.0000 0.8273

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Check_Convergence = 0.0049

n = 1×3

0 1 2

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result = 1×3

1.0000 0.8273 0.8322

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Check_Convergence = 5.6925e-05

The calculation meets the criterion after the second iteration, and the loop then stops.

.

[Asher]

Fantastic thank you!

[Star Strider]

As always, my pleasure!

Answer 2

[Torsten]

Maybe you mean

tol = 1e-7

instead of

tol = exp(-7)

?

However: It's correct that MATLAB quits the while-loop after three values:

tol = exp(-7)

tol = 9.1188e-04

x = sin(pi/5);
target = cos(x);
counter = 1;
n(counter) = 0;
result(counter) = (((-1)^n(counter)) * (x^(2*n(counter)))/(factorial(2*n(counter))));
while abs(result(counter)-target) > tol
    counter = counter + 1;
    n(counter) = n(counter-1) + 1;
    result(counter) = result(counter-1) + (((-1)^n(counter)) * (x^(2*n(counter)))/(factorial(2*n(counter))));
end
abs(result(counter)-target)

ans = 5.6925e-05

[Asher]

Wonderful thank you!