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Math Help - [SOLVED] nth root in a simple calculator

  1. #1
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    Post [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 ?
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  2. #2
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    Quote Originally Posted by rohankg88 View Post
    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}}?
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  3. #3
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    Post 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 ??
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  4. #4
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    I attempted to do this to compute 3^{1/4}, but it didn't work. Can you give an example?
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  5. #5
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    Quote Originally Posted by rohankg88 View Post
    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
    Last edited by CaptainBlack; May 16th 2010 at 11:08 AM.
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  6. #6
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    Quote Originally Posted by CaptainBlack View Post
    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.
    Last edited by awkward; May 16th 2010 at 12:35 PM. Reason: Slow to catch on
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by awkward View Post
    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
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    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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