[SOLVED] nth root in a simple calculator

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May 16th 2010, 05:02 AM
rohankg88

[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'.

add 1

square the result 12 times...

Any idea why i get the correct answer ?
May 16th 2010, 05:08 AM
Prove It

Quote:

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'.

add 1

square the result 12 times...

Any idea why i get the correct answer ?

Why not just evaluate $x^{\frac{1}{n}}$ ?
May 16th 2010, 05:39 AM
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 ??
May 16th 2010, 09:08 AM
roninpro

I attempted to do this to compute $3^{1/4}$ , but it didn't work. Can you give an example?
May 16th 2010, 10:56 AM
CaptainBlack

Quote:

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'.

add 1

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
May 16th 2010, 12:33 PM
awkward

Quote:

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.
May 16th 2010, 07:37 PM
CaptainBlack

Quote:

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
November 29th 2012, 10:09 PM
Pavitra

Re: [SOLVED] nth root in a simple calculator

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