# [SOLVED] nth root in a simple calculator

#### rohankg88

I came across this trick to find n th root of any number on a calculator that does not have nth root function or logs !! problem is I can't figure out why it works ...

To find : nth root of x.

take sqrt of 'x' 12 times. (i.e. take sqrt (x) then again take sqrt (x) ... 12 times)

subtract 1 from this ans.

divide by 'n'.

square the result 12 times...

Any idea why i get the correct answer ?

#### Prove It

MHF Helper
I came across this trick to find n th root of any number on a calculator that does not have nth root function or logs !! problem is I can't figure out why it works ...

To find : nth root of x.

take sqrt of 'x' 12 times. (i.e. take sqrt (x) then again take sqrt (x) ... 12 times)

subtract 1 from this ans.

divide by 'n'.

square the result 12 times...

Any idea why i get the correct answer ?
Why not just evaluate $$\displaystyle x^{\frac{1}{n}}$$?

#### rohankg88

Curiosity...

In simple calculators there's no function for x ^ 1/n.

I just wanted to know why this method works... whats the math behind it ??

#### roninpro

I attempted to do this to compute $$\displaystyle 3^{1/4}$$, but it didn't work. Can you give an example?

#### CaptainBlack

MHF Hall of Fame
I came across this trick to find n th root of any number on a calculator that does not have nth root function or logs !! problem is I can't figure out why it works ...

To find : nth root of x.

take sqrt of 'x' 12 times. (i.e. take sqrt (x) then again take sqrt (x) ... 12 times)

subtract 1 from this ans.

divide by 'n'.

square the result 12 times...

Any idea why i get the correct answer ?
The approximation you give is:

$$\displaystyle x^{1/n}\approx \left( 1+\frac{x^{1/4096}-1}{n}\right)^{4096}$$

This is using the result that for large $$\displaystyle m$$:

$$\displaystyle e^x \approx \left(1+\frac{x}{m}\right)^m$$

so:

$$\displaystyle x^{1/n}=e^{\ln(x)/n}\approx \left( 1+\frac{\ln(x)}{n.m}\right)^m$$

and presumably a suitable approximation for $$\displaystyle \ln(x^{1/4096})\approx x^{1/4096}-1$$.

I will leave the detail to others to work out

CB

Last edited:
• General, awkward and AllanCuz

#### awkward

MHF Hall of Honor
The approximation you give is:

$$\displaystyle x^{1/n}\approx \left( 1+\frac{x^{1/4096}-1}{n}\right)^{4096}$$

This is using the result that for large $$\displaystyle m$$:

$$\displaystyle e^x \approx \left(1+\frac{x}{m}\right)^m$$

so:

$$\displaystyle x^{1/n}=e^{\ln(x)/n}\approx \left( 1+\frac{\ln(x)}{n.m}\right)^m$$

and presumably a suitable approximation for $$\displaystyle \ln(x^{1/4096})\approx x^{1/4096}-1$$.

I will leave the detail to others to work out

CB
The final approximation you need is

$$\displaystyle \frac{\ln(x)}{m} \approx x^{1/m}-1$$ for large $$\displaystyle m$$.

To show this, start by observing that
$$\displaystyle t^{1/m} \approx 1$$ for large $$\displaystyle m$$,
so
$$\displaystyle \frac{1}{t} \approx t^{1/m-1}$$.
Now integrate both sides of the equation from 1 to x.

Last edited:

#### CaptainBlack

MHF Hall of Fame
The final approximation you need is

$$\displaystyle \frac{\ln(x)}{m} \approx x^{1/m}-1$$ for large $$\displaystyle m$$.

To show this, start by observing that
$$\displaystyle t^{1/m} \approx 1$$ for large $$\displaystyle m$$,
so
$$\displaystyle \frac{1}{t} \approx t^{1/m-1}$$.
Now integrate both sides of the equation from 1 to x.
If you had not noticed that is what I gave (may be I gave the impression that that is what I wanted, if so that is the wrong impression).

CB

#### Pavitra

if u hv understood the reason behind this, could u please share it with me too???