SOLUTION: procedure to solve cube root of 35 with out calculator
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Question 1024132: procedure to solve cube root of 35 with out calculator
Found 2 solutions by fractalier, solver91311:
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
There are a number of algorithms that will show you...here is one link...
http://www.mathpath.org/Algor/cuberoot/algor.cube.root.htm
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
There are a couple of ways to go about it.
You can just approximate.
3 is too small, because 3^3 = 27
4 is too large, because 4^3 = 64
Interpolate: 64 - 27 is 37 and 35 - 27 is 8, so use 3 8/37 or a little less than 3 and 1/4.
Multiply out 3.25^3 and you'll find out that is a little small. So pick something a little larger and try again. Keep it up until you get as close as you like.
You can also use Newton-Raphson:
You want to solve the equation:
Newton-Raphson says that if you make a guess, 
, then 
is a closer guess if
}{f'(x_o)})
So for the cube root of 
)
So, since 
for your problem and 3 is an integer lower bound on the desired value, you can use 3 as your first guess:
)
Doing the arithmetic:
If you calculate
^3)
You will find that you are a little on the low side, so perform Newton-Raphson again using 
as the initial guess.
\ -\ \frac{1}{3}\left(\left(\frac{89}{27}\right)\ -\ \frac{35}{\left(\frac{89}{27}\right)^2}\right))
Yes, the arithmetic is as ugly as a mud fence, but it is all rational numbers so you can do it old school without a calculator.
The good news is that, given an appropriate first guess, Newton-Raphson generally converges rapidly for well-behaved functions. In this particular case, if you do the calculation above, your approximation will be correct to the third decimal place; that is your answer will be on the high side by only 0.0002 or so.
John
My calculator said it, I believe it, that settles it
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