# Thread: [SOLVED] nth root in a simple calculator

1. ## [SOLVED] nth root in a simple calculator

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 ?

2. Originally Posted by 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 ?
Why not just evaluate $x^{\frac{1}{n}}$?

3. ## 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 ??

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

5. Originally Posted by 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 ?
The approximation you give is:

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

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

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

so:

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

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

I will leave the detail to others to work out

CB

6. Originally Posted by CaptainBlack
The approximation you give is:

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

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

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

so:

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

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

I will leave the detail to others to work out

CB
The final approximation you need is

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

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

7. Originally Posted by awkward
The final approximation you need is

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

To show this, start by observing that
$t^{1/m} \approx 1$ for large $m$,
so
$\frac{1}{t} \approx t^{1/m-1}$.
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

8. ## Re: [SOLVED] nth root in a simple calculator

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

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# how to calculate 20th root of any number in normal calculator

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