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A friend of mine came upon the following problem.

Solve for $a$ the equation $\displaystyle\int_{0}^a \left(3^{\frac{1}{3} \left( x^3 - 3x \right)}- 1\right)\, dx = 0$.

By typing the problem into either wolfram, maple, or matlab it is easily solvable. Assume for a minute you want to solve this by using pen paper, and a very basic calculator. multiplication, division, exponents. No iterative functions, no integrals etc.

How would one most easily solve this problem ? I tried using Simpsons 3/8 combined with newton. But alas, the computations were not accurate enough. I would like to obtain a precision of 3 digits. Meaning 2.23 or better =)

Here is the "exact" value

2.227185349853887802594276227693823508660419768745303460008020665494240139789974722809139563708277641

I tried using newton/rhapson

$$ N(a) = a - \dfrac{f(a)}{f^{\,\prime\!}(a)} $$

Where $$ f(a) = \int_{0}^a \left(3^{\frac{1}{3} \left( x^3 - 3x \right)}- 1\right)\, dx $$

My first guess was $a=2$ by looking at the graph. For estimating $f(2)$ I used simpsons 3/8 rule.

http://i44.tinypic.com/2vj9xrn.png

Thanks in advance.

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What is the top limit of this integral? –  Aleksey Pichugin Nov 6 '11 at 14:43
8  
Are you trying to solve for $a$? In that case the only answer is $a=0$, since $3^z$ is positive for all real $z$. –  Craig Nov 6 '11 at 14:44
2  
@Craig, thanks for your comment, I was beginning to believe I had slipped into a parallel universe... –  Did Nov 6 '11 at 14:47
2  
@N3buchadnezzar: While you are at it, could you please replace your numerical solution with the correct value of $a\approx 2.227185$. –  Aleksey Pichugin Nov 6 '11 at 15:01
1  
Are you sure the $-1$ is inside the integral? –  J. M. Nov 6 '11 at 15:01

1 Answer 1

I would use the bisection method. It is clear that $f(\sqrt3)<0$. From the inequality $e^t-1\ge t$ it follows easely that $f(\sqrt6)>0$. So the solution lies between $\sqrt3=1.732\dots$ and $\sqrt6=2.449\dots$ The first approximation will be $a_1=(\sqrt3+\sqrt6)/2=2.090\dots$ What is the advantage of this method? We do not need to compute the value of $f(a_1)$. It is enough to be able to decide if $f(a_1)<0$ or $f(a_1)>0$, and this does not need as much precission. Of course nothing comes for free: more iterations will be needed.

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