[SOLVED] nth root in a simple calculator

May 2010
2
0
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 ?
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
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 \(\displaystyle x^{\frac{1}{n}}\)?
 
May 2010
2
0
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 ??
 
Nov 2009
485
184
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
Nov 2005
14,972
5,271
someplace
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:

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

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:

awkward

MHF Hall of Honor
Mar 2008
934
409
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
Nov 2005
14,972
5,271
someplace
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
 
Nov 2012
1
0
India
if u hv understood the reason behind this, could u please share it with me too???