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somebody help me please! I've gone over the irritation of approx_sqrt and the only way I've got each value of the function while using a calculator it was if I used y=2

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Suleman
Suleman 2023-2-20
编辑： John D'Errico 2023-2-22
Ran in:
format long g
approx_sqrt(2)
y =
1.5
y =
1.41666666666667
y =
1.41421568627451
ans =
1.41421568627451
function y=approx_sqrt(x)
y=x;
while abs(y^2-x)>0.001*x
y=(x/y+y)/2
end
end
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Suleman
Suleman 2023-2-21
@John D'Errico
while abs(y^2-x)>0.001*x
so using 2 in the y place is the only way i got those values
y =
1.5
y =
1.41666666666667
y =
1.41421568627451
ans =
1.41421568627451
when i used the scientific calcator to further understand the function. So i'm asking is what i did correct! Because i tried everthing to get those values after hours of rewatching this video Lesson 6.2 while-loops in MATLAB - YouTube
8:42 and that's the time stamp if you wanna know what i struggled with. after i got the result for when x=2 i used to for the after i tried to followed the video's explanation i could the same the numbers the prgramming was giving until i used 2 for the y. So is what i did right? check the video please at the time stamp 8:42 if i failed to explain clearly. thank in advance.

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回答（1 个）

John D'Errico
John D'Errico 2023-2-22
编辑：John D'Errico 2023-2-22
Ran in:
The site seems to want an answer to your question, and we just answered you in the comments. Oh well.
But seriously, your solution is correct. This is indeed Newton's method, as applied to the square root. For example, if I run your code, we see this (actually, I changed the congergence tolerance so it would iterate at least one more time, but the code is the same):
format long g
approx_sqrt(3)
y =
2
y =
1.75
y =
1.73214285714286
y =
1.73205081001473
y =
1.73205080756888
ans =
1.73205080756888
sqrt(3)
ans =
1.73205080756888
What do we know about Newton's method? It has quadratic convergence near a single root. One thing that tells us is as we get near to the solution, you should expect to see double the number of correct digits with each iteration.
So the first iteration started out at y=2. Then we see 1.75, so 2 significant digits are correct. The next iteration gave us 1.732, so with 4 correct digits. The next iteration has the approximation at 1.73205081, so the firsy 8 digits are correct. And then the final iteration gave us essentially full double precision accuracy.
function y=approx_sqrt(x)
y=x;
while abs(y^2-x)>1e-14*x
y=(x/y+y)/2
end
end
Again, this is exactly how that method should work. It did exactly what I would expect. (And the equations you used are correct.)

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