Question 1024132
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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:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x^3\ -\ 35\ =\ 0]


Newton-Raphson says that if you make a guess, *[tex \Large x_o], then *[tex \Large x_1] is a closer guess if


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ x_o\ -\ \frac{f(x_o)}{f'(x_o)}]


So for the cube root of *[tex \Large a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ x_o\ -\ \frac{x_o^3\ -\ a}{3x_o^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ x_o\ -\ \frac{1}{3}\left(x_o\ -\ \frac{a}{x_o^2}\right)]


So, since *[tex \Large a = 35] for your problem and 3 is an integer lower bound on the desired value, you can use 3 as your first guess:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ 3\ -\ \frac{1}{3}\left(3\ -\ \frac{35}{9}\right)]


Doing the arithmetic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_1\ =\ \frac{89}{27}]


If you calculate


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{89}{27}\right)^3]


You will find that you are a little on the low side, so perform Newton-Raphson again using *[tex \Large x_1\ =\ \frac{89}{27}] as the initial guess.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x_2\ =\ \left(\frac{89}{27}\right)\ -\ \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
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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