Math Mystery: Why Is \sqrt{2}^{4} Not 4?

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Discussion Overview

The discussion revolves around the mathematical expression \(\sqrt{2}^{4}\) and the discrepancies observed when calculating it using a calculator. Participants explore the implications of numerical precision and the methods calculators use for computation, focusing on the theoretical and practical aspects of square roots and their powers.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why \(\sqrt{2}^{4}\) does not yield exactly 4 on their calculator, noting the result is very close but not equal.
  • Another participant asserts that the result is indeed 4, attributing the discrepancy to the calculator's method of computation.
  • A different participant explains that calculators may use a Taylor series expansion for evaluating functions, which can lead to approximations.
  • Another participant challenges the use of Taylor series in calculators, suggesting that other algorithms like CORDIC or pseudo-division are more commonly employed for square root calculations.
  • One participant emphasizes that calculators work with finite decimal precision, which affects the accuracy of the result when raising the approximate value of \(\sqrt{2}\) to the fourth power.

Areas of Agreement / Disagreement

Participants express differing views on the methods calculators use and the implications of numerical precision. There is no consensus on the exact nature of the calculations or the validity of the approximations presented.

Contextual Notes

The discussion highlights limitations related to numerical precision and the dependence on the specific algorithms used by calculators, which are not fully resolved within the thread.

AznBoi
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Isn't \sqrt{2}^{4} equal to 4?? How come when I plug this into my calculator it gives me the number:3.999999999996 , which is very close to 4 but isn't?? Is there something wrong with the settings of my calculator?
 
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It is 4. This is just the all to do with the way the calculator works it out... If you can try
Pi/6=ans=x
x-x^3/3!+x^5/5!-x^7/7!+x^9/9!...
It would say: .4999999999999999999999999
Instead of .5
 
it is because the calculator evaluates the root function using a taylor series expansion which looks like what above poster wrote out. the calculator approximates the answer
 
Well, the Taylor series are not used by pocket calculators or microprocessors for calculating trigonometric function or other elementary functions. But, indeed, any numerical calclation has a finite precision.

For trigonometric functions, the CORDIC algorithm is mostly used: http://en.wikipedia.org/wiki/CORDIC .
For the square root, a generalisation of the same algorithm can also be used.
But the most used algorithm for the square root seems to be the "pseudo-division": http://www.jacques-laporte.org/Meggitt_62.pdf .

Of course, the Taylor series has numerous applications, including for numerical computation of some functions. But very often for numerical application the convergence can be improved by using other more specific methods. A famous book by http://www.math.sfu.ca/~cbm/aands/" give many of these methods.
 
Last edited by a moderator:
It doesn't matter what method the calculator uses to evaluate square roots.

The reason is that the calculator only works out sqrt(2) to a finite number of decimal places. When you raise that approximate value to the fourth power, it does not equal 4 exactly.

Sqrt(2) = 1.41421...

If you calculator only stored numbers to 3 decimal places, 1.414^4 = 3.9976...
To 4 decimal places, 1.4142^4 = 3.9998...
etc.
 

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