Square root
Author
Ok, I''ve asked countless math teachers, professors, and majors, the same thing: How do you get the sqaure root of a number without a calculator, and without guessing? I want to make an algorithm for a simple calculator program. Obviously, it''s possible. Calculators can do it, a desktop can definately do it. But nobody knows how to do it! It''s always "they taught us that a long time ago, but I don''t remember. They don''t teach that anymore. It must''ve been the invention of the calculator..."
If anybody knows how I could do this, LET ME KNOW, PLEASE!!!!!
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Try Newtons Iteration:
square root
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square root
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For sqrt(x) do the iteration y = (y + x/y)/2 to approximate sqrt(x), with some start value for y.
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If you do the algebra...
sqrt(x) = y for some number y.
=> x = y^2
=> log(x) = log(y^2), where log is the natural logarithm.
=> log(x) = 2 * log(y)
=> log(y) = log(x) / 2
=> y = exp(log(x)/2) (or e^(log(x)/2)), where exp is the exponential function.
This would be extremely slow of course, but exp and log are defined in math.h.
There is another method explained here:
http://forum.swarthmore.edu/dr.math/problems/steve.8.6.96.html
sqrt(x) = y for some number y.
=> x = y^2
=> log(x) = log(y^2), where log is the natural logarithm.
=> log(x) = 2 * log(y)
=> log(y) = log(x) / 2
=> y = exp(log(x)/2) (or e^(log(x)/2)), where exp is the exponential function.
This would be extremely slow of course, but exp and log are defined in math.h.
There is another method explained here:
http://forum.swarthmore.edu/dr.math/problems/steve.8.6.96.html
E.g. if we want to calculate sqrt(3), we take e.g. 1 as start value, we get
1. y = (1 + 3/1)/2 = 2
2. y = (2 + 3/2)/2 = 7/4
3. y = (7/4 + 3/(7/4))/2 = ........
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Edited by - ga on June 20, 2000 8:33:56 AM
1. y = (1 + 3/1)/2 = 2
2. y = (2 + 3/2)/2 = 7/4
3. y = (7/4 + 3/(7/4))/2 = ........
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Edited by - ga on June 20, 2000 8:33:56 AM
To calculate any sqrt stuff try this:
y = exp(ln(base)/power);
or
exp(number, 1/base)
[Sorry, Psycho, but after a several conversation with
some people i am no further sure whether this is correct]
-------
pseudo-programmer, webmaster @
digital impulse
Edited by - Wolti on June 22, 2000 9:36:30 AM
y = exp(ln(base)/power);
or
exp(number, 1/base)
[Sorry, Psycho, but after a several conversation with
some people i am no further sure whether this is correct]
-------
pseudo-programmer, webmaster @
digital impulse
Edited by - Wolti on June 22, 2000 9:36:30 AM
Author
Thanks for your help, everybody! I appreciate it. Anyone else with suggestions, feel free to add to the thread. I need as many suggestions as I can get.
The easiest way to do is by x^(1/2) because:
x = y^2
ln(x) = ln(y^2)
ln(x) = ln(y) * 2
ln(x) / 2 = ln(y)
ln(x) * 1/2 = ln(y)
ln(x^(1/2)) = ln(y)
x^(1/2) = y
BTW, isn''t sqrt in the math.h to??
x = y^2
ln(x) = ln(y^2)
ln(x) = ln(y) * 2
ln(x) / 2 = ln(y)
ln(x) * 1/2 = ln(y)
ln(x^(1/2)) = ln(y)
x^(1/2) = y
BTW, isn''t sqrt in the math.h to??
x^(1/2) isnt really a solution.
Like... you can write x^2 as x*x, and you can figure that out with using normal multiplication. But..how do you write x^(1/2) the only way is the sqrt(x)...which is back where you started.
Its true. I mean, x^(1/2) == sqrt(x), but it doesnt really help you figure it out with a pencil and paper.
sorry to be picky
Like... you can write x^2 as x*x, and you can figure that out with using normal multiplication. But..how do you write x^(1/2) the only way is the sqrt(x)...which is back where you started.
Its true. I mean, x^(1/2) == sqrt(x), but it doesnt really help you figure it out with a pencil and paper.
sorry to be picky
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