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How to Find This without Calculator?

  1. Aug 27, 2015 #1
    How do I find ##2^{1.2}## without using calculator?
     
  2. jcsd
  3. Aug 27, 2015 #2

    andrewkirk

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    Since it's equal to ##2\sqrt[5]{2}## and ##\sqrt[5]{2}## is irrational you won't be able to.
     
  4. Aug 27, 2015 #3

    Ssnow

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    You can use Taylor expansion in ##1## for a good approximation...

    ## 2^{x} \approx 2+2\ln{2}(x-1)+(\ln{2})^{2}(x-1)^{2}+\cdots ##

    putting ##x=\frac{12}{10}##...
     
  5. Aug 27, 2015 #4

    HallsofIvy

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    Or: if [itex]y= 2^{1.2}[/itex] then [itex]log(y)= 1.2 log(2)[/itex] and you can look up the logarithms in a logarithm table.

    Or use a slide rule!
     
  6. Aug 27, 2015 #5

    jasonRF

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    Newton's method is a great way to find ##\sqrt[5]{2}##.
    https://en.wikipedia.org/wiki/Newton's_method
    This converges very quickly, using 1 as my first guess, the 3rd iteration gave a fractional error of ##3\times 10^{-5}##, the 4th gave fractional error of ##1\times 10^{-9}##.

    EDIT: brain freeze! This is not so calculator friendly after all - you need to compute 4th powers of decimals to make this work. You would want a calculator for this. Ooops!

    jason
     
  7. Aug 27, 2015 #6

    jasonRF

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    This was nagging me on my drive home - so here is a hand calculation using Newtons method for ##
    \sqrt[5]{2}##
    . If ## s_n## is the ## n^{th}## estimate of the root, then I get the following for the ##p^{th}## root of ##x##:
    [tex]
    s_{n+1} = \frac{1}{p}\left[(p-1) s_n + \frac{x}{s_n^{p-1}} \right]
    [/tex]
    For us p=5, x=2, so,
    [tex]
    s_{n+1} = \frac{1}{5}\left[4 s_n + \frac{2}{s_n^{4}} \right]
    [/tex]

    if I use ## s_0 = 1##, then ##s_1 = 1.2##. For ##s_2## I get
    [tex]
    s_{2} = \frac{1}{5}\left[ 4.8 + \frac{2}{2.0736} \right] = \frac{1}{5}\left[ 4.8 + \frac{1}{1.0368} \right] \approx \frac{1}{5}\left[ 4.8 + 1 - 0.0368 \right]
    = \frac{5.7632}{5} \approx 1.15
    [/tex]

    So I get ##2^{1.2} \approx 2.30 ##. Better than 1% accuracy in this case....

    jason
     
  8. Aug 27, 2015 #7

    mfb

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    I would use this approach:

    21.2 = 2*20.2

    ##2^{0.2} = e^{0.2 \ln 2} \approx e^{0.2 \cdot 0.7} = e^{0.14} \approx 1+0.14##
    2*1.14 = 2.28, better than 1% accuracy and the above calculation works without pen and paper if you know ln(2) ≈ 0.7.

    Better: ln(2)=0.693 or 1% smaller than 0.7, and we can add the second order for the exponential (using 0.14 ≈ 1/7) => 20.2≈1+0.14-0.0014 + 1/2*1/72 = 1.1486 because 1/72 is about 1/50 and 1/2*1/50=0.01.
    2*1.1486 = 2.2972 - an error of just 0.01%.
     
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