How to calculate the cube root of value in theoretical?
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So the function you calculate for something like the rrth root is
⎛⎝x1210−1r+1⎞⎠210(x1210−1r+1)210
Well 210210 is just being used here as a suitable large number which doesn't kill the precision of your calculator. Let's use nn for 210210 and set yn=n(x−−√n−1)yn=n(xn−1).
Then we are calculating (for large nn, so near the limit)
fn(x)=(1+ynrn)nfn(x)=(1+ynrn)n
Now limn→∞yn=lnxlimn→∞yn=lnx. And the limit of the function fnfn (for constant yy) is eyreyr(these are both standard limits, though the second is better known than the first) and we have
elnxr=elnx1r=x−−√relnxr=elnx1r=xr
The skill in getting the best precision for this method is in taking the square root (and then squaring back again) enough times to get close to the limit, without losing precision by dropping too many significant digits.
⎛⎝x1210−1r+1⎞⎠210(x1210−1r+1)210
Well 210210 is just being used here as a suitable large number which doesn't kill the precision of your calculator. Let's use nn for 210210 and set yn=n(x−−√n−1)yn=n(xn−1).
Then we are calculating (for large nn, so near the limit)
fn(x)=(1+ynrn)nfn(x)=(1+ynrn)n
Now limn→∞yn=lnxlimn→∞yn=lnx. And the limit of the function fnfn (for constant yy) is eyreyr(these are both standard limits, though the second is better known than the first) and we have
elnxr=elnx1r=x−−√relnxr=elnx1r=xr
The skill in getting the best precision for this method is in taking the square root (and then squaring back again) enough times to get close to the limit, without losing precision by dropping too many significant digits.
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