Sum of the square roots of the first n natural numbers

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SUMMARY

The discussion focuses on calculating the sum of the square roots of the first n natural numbers. It concludes that an exact closed-form solution does not exist, but approximations can be made using integral calculus. The integral approximation of \(\frac{2}{3} n^{\frac{3}{2}}\) is highlighted as a superior method compared to a simple integer approximation formula. Participants agree that while approximations are necessary, they provide a reasonable estimate for practical purposes.

PREREQUISITES
  • Understanding of integral calculus
  • Familiarity with square root functions
  • Basic knowledge of approximation methods
  • Ability to compute powers, specifically \(n^{3/2}\)
NEXT STEPS
  • Research integral approximation techniques in calculus
  • Explore numerical methods for estimating square roots
  • Learn about error bounds in numerical approximations
  • Investigate advanced mathematical software for symbolic computation
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Mathematicians, students studying calculus, software developers implementing mathematical functions, and anyone interested in numerical methods for approximating mathematical sums.

Amith2006
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Is there a way to find the,"Sum of the square roots of the first n natural numbers"?
 
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I don't think you could do it exactly. You could approximate it by the integral of \sqrt{x}, and get a bound on the error.
 
As StatusX says I'm pretty sure there's no way to do in exactly in closed-form. If you don't have a way to calculate square roots at all (ie. you're doing it without a calculator and don't want to go through an approximation method), then a simple integer approximation would be

\frac{2}{3}\lfloor \sqrt{n} \rfloor^3 - \frac{1}{2}\lfloor \sqrt{n} \rfloor^2 - \frac{1}{6} \lfloor \sqrt{n} \rfloor + \lfloor \sqrt{n} \rfloor(n-\lfloor \sqrt{n} \rfloor^2),

but it's not very good. The integral approximation \frac{2}{3} n^{\frac{3}{2}} is much better, but you have to be able to compute n^{3/2}
(2/3 \lfloor n^{3/2}\rfloor is also better than the one I gave above though).
 
Last edited:

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