How to Find This without Calculator?

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Discussion Overview

The discussion revolves around methods to calculate ##2^{1.2}## without the use of a calculator. Participants explore various mathematical techniques, approximations, and reasoning approaches to achieve this calculation.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant suggests that ##2^{1.2}## can be expressed as ##2\sqrt[5]{2}##, noting the irrationality of ##\sqrt[5]{2}## complicates the calculation.
  • Another proposes using Taylor expansion around ##x=1## for an approximation of ##2^{1.2}##.
  • A different approach involves using logarithms, where if ##y=2^{1.2}##, then ##\log(y)=1.2 \log(2)##, and suggests looking up logarithms in a table or using a slide rule.
  • One participant discusses Newton's method for finding ##\sqrt[5]{2}##, initially reporting rapid convergence but later retracts, indicating that the method is not as calculator-friendly as initially thought.
  • A detailed hand calculation using Newton's method is provided, showing step-by-step estimates for ##\sqrt[5]{2}## and arriving at an approximation for ##2^{1.2}##.
  • Another participant presents an alternative method using the approximation of ##2^{0.2}## through the exponential function and logarithms, achieving a high degree of accuracy without extensive calculations.

Areas of Agreement / Disagreement

Participants present multiple competing methods and viewpoints without reaching a consensus on a single approach. The discussion remains unresolved regarding the best method to calculate ##2^{1.2}## without a calculator.

Contextual Notes

Some methods rely on approximations and assumptions about logarithmic values, while others involve iterative calculations that may not be straightforward without computational tools. The accuracy of various approaches is also contingent on the precision of constants used.

basty
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How do I find ##2^{1.2}## without using calculator?
 
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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.
 
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}##...
 
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!
 
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
 
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] <br /> = \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
 
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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