Calculating the nth root in simple calculator?
Recently during my physics class that how to take a cube root in a simple calculator.
Follow the steps given below
Step1. Press the square root button 13 times
Step2. subtract 1 from it
Step3. divide by 3,
Step4. add 1 to it
Step5. Press these keys x,= 13 times and you will get the cube root of the number.
On asking these to my maths teacher , in addition to these he told me that how to take out nth root [if we want to take out 7th root of 128 which is (128)^1/7 =2 ] it can be also done using a simple calculator.
We have to repeat the same process but in step 3 instead of dividing by 3 we have to divide by the number n that is nth root [ if we want to take 7th root of 128 which is (128)^1/7 then in third step you have to divide it by 7 ] rest of the method remains same.
So by this in simple calculator also nth root can be found. So I want to know why it is happening so ?
$\endgroup$ 33 Answers
$\begingroup$So the function you calculate for something like the $r$th root is $$\left(\frac{x^{\frac 1{2^{10}}}-1}r+1\right)^{2^{10}}$$Well $2^{10}$ is just being used here as a suitable large number which doesn't kill the precision of your calculator. Let's use $n$ for $2^{10}$ and set $y_n=n(\sqrt [n]x-1)$.
Then we are calculating (for large $n$, so near the limit) $$f_n(x)=\left(1+\frac {y_n}{rn}\right)^n$$
Now $\lim_{n\to \infty} y_n=\ln x$. And the limit of the function $f_n$ (for constant $y$) is $e^{\frac yr}$ (these are both standard limits, though the second is better known than the first) and we have $$e^{\frac {\ln x}r}=e^{\ln x^{\frac 1r}}=\sqrt[r]x$$
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.
$\endgroup$ $\begingroup$For numbers very close to 1, the $nth$ root function has a slope of $1/n$ - that is, the $nth$ root of $1+x$ is approximately $1+x/n$, when x is very small. Your teacher was using repeated roots to get the number into a range where this is true, then repeated squaring to get the result back up to its correct range. These functions (square root and square) don't affect the cube root calculation provided they are matched.
The trick is limited by the precision on your calculator, but sounds like it would work quite often, at least for cube roots. I wouldn't trust it for important work though - there are plenty of calculating tools available.
$\endgroup$ $\begingroup$If $x$ is the input then the operations in the instructions give
$$\begin{align}\left(\frac{x^{2^{-10}}-1}{3}+1\right)^{2^{10}}&=\left[\left(1+\frac{x^{2^{-10}}-1}{3}\right)^{\frac{3}{x^{2^{-10}}-1}}\right]^{\frac{x^{2^{-10}}-1}{3}2^{10}}\end{align}$$
When $10\to\infty$ (the $10$ is playing the role here of a number that is assumed to be large enough) the part in brackets tends to $e$. On the other hand, the exponent $$\frac{x^{2^{-10}}-1}{3\cdot 2^{-10}}\to\frac{1}{3}\ln(x)=\ln(\sqrt[3]{x})$$
So, the first expression tends to $e^{\ln(\sqrt[3]{x})}=\sqrt[3]{x}$.
The same argument works when you replace $3$ by other numbers.
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