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# Cube root.

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Were given this "iterative formula", by our teacher, to calculate the cube root of 25 to an accuracy of at least 10 decimal places: a(n+1) =[A/a^2 + 2*a]/3, where n =0, 1, 2, 3....etc, A =25 and a=initial guess. But, I'm having problems getting the right answer. Any help would be great. Thank you.

Feb 7, 2019

#1
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Do you understand what an "iterative formula" is? It means that the answer you get from your first "guess" is fed back into the formula to get your 2nd result, which is fed back to get the 3rd result and so on.
Here is my attempt with the initial guess of "a" = 2.5. I calculated the first five to 25 decimal places and the 6th to 50 decimal places.
1-3
2- 2.925925925925925925925926 to 25 decimal places.
3- 2.924018982396379181082487 .........................................
4- 2.924017738213395471233655.........................................
5- 2.924017738212866065506787..........................................
6-2.924017738212866065506787360137922778530498635101.                                                                                                                                                                           This 6th iteration gives you about 50 accurate decimal places. By the way, the 4th iteration gives you 11 accurate decimal places.

Feb 7, 2019
edited by Guest  Feb 7, 2019
#2
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OK....let's guess that the cube root is 2

First iteration  = [ 25/ 2^2 + 2*2 ] / 3 =   3.4166666666

Second iteration

[ 25/ (3.4166666666)^2 + 2 (3.4166666666) ] / 3    = 2.9916385749

Third iteration

[ 25 / ( 2.9916385749)^2 + 2( 2.9916385749) ] / 3  =   2.9255346747

Fourth iteration

[ 25 / (2.9255346747)^2 + 2 (2.9255346747) ] / 3   =   2.9240185246

Fifth iteration

[ 25 / (2.9240185246)^2 + 2(2.9240185246) ] / 3   =  2.9240177382

Sixth iteration

[ 25 / (2.9240177382)^2 + 2(2.9240177382) ] / 3   =   2.9240177382

And they agree to 10 decimal places

[Note....that the closer our first guess is to the true value...the fewer iterations that may be necessary....]

Feb 7, 2019