0
$\begingroup$

Recall the fourth order Runge-Kutta method: $$x_{n+1} = x_n + \frac16(k_1 + 2k_2 + 2k_3 + k_4)$$ Apply it, with $h = 1$ to the initial value problem in the previous question to obtain a better approximation to $\sqrt2$. Provide a numerical answer rounded to five decimal places.

I need a hint solving this equation. Please help!

$\endgroup$
  • $\begingroup$ What, exactly, is "the previous question"? And what have you tried? $\endgroup$ – Arthur Jan 4 '16 at 10:52
  • $\begingroup$ Thank you for the edit. I do not understand, where to insert h. i tried k4 = hf(xn + k3, tn + h) $\endgroup$ – Spacemoose Jan 4 '16 at 10:54
  • $\begingroup$ The first problem is this one: math.stackexchange.com/questions/1599366/… $\endgroup$ – Spacemoose Jan 4 '16 at 13:01
1
$\begingroup$

The differential equation in question is $$ \frac{dx}{dt} = \frac{1}{2x}, \quad x(1) = 1 $$ which has general solution $x(t) = \sqrt{t+C}$, and with the given initial value, we get the specific solution $x(t) = \sqrt t$. That means that starting at $t_0 = 1$, and using some numerical method with $h = 1$, we should get to an approximation of $\sqrt2$ in one step. We are told to use Runge-Kutta of fourth order, and that is what we're going to do.

Gathering what we're given, both from the problem itself and the definition of fourth-order Runge-Kutta, we have $$ f(t, x) = \frac 1{2x}\\ x_0 = 1\\ t_0 = 0\\ x_{n+1} = x_n + \frac16(k_1 + 2k_2 + 2k_3 + k_4)\\ k_1 = f(t_n, x_n)\\ k_2 = f(t_n + \frac12, x_n + \frac{k_1}{2})\\ k_3 = f(t_n + \frac12, x_n + \frac{k_2}{2})\\ k_4 = f(t_n + 1, x_n + k_3) $$ (Note that there are supposed to be quite a lot of $h$ strewn in here and there as well. Since we're told that $h = 1$, I've skipped those.)

The solution to our problem is $x_1$. So we just start calculating: $$ k_1 = f(t_0, x_0) = \frac{1}{2x_0} = \frac12\\ k_2 = f(t_0 + \frac12, x_0 + \frac{k_1}{2}) = \frac{1}{2(x_0 + k_1/2)} = \frac{1}{2(1 + 1/4)} = \frac25\\ k_3 = f(t_0 + \frac12, x_0 + \frac{k_2}{2}) = \frac1{2(x_0 + k_2/2)} = \frac{1}{2(1 + 1/5)} = \frac{5}{12}\\ k_4 = f(t_0 + 1, x_0 + k_3) = \frac{1}{2(x_0 + k_3)} = \frac{1}{2(1 + \frac{5}{12})} = \frac{6}{17} $$ and with this we can enter everything into the final expression: $$ x_1 = x_0 + \frac16(k_1 + 2k_2 + 2k_3 + k_4)\\ = 1 + \frac16\left(\frac12 + \frac45 + \frac56 + \frac6{17}\right)\\ = \frac{1082}{765} \approx 1.414379084967320261437 $$

$\endgroup$
  • $\begingroup$ You are such a boss O.O $\endgroup$ – Spacemoose Jan 4 '16 at 16:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.