Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the initial value problem: $$y(0) = 1, y ′ (t) = λy(t)$$

Using that the solution is $y(t) = e^{λt}$, write out a Taylor series for $y(t_{i+1})$ about $y(t_i)$ up to terms of order $h^4$ (note: use that $h = t_{i+1} − t_i$).

Write out what the RK32 method would be for this problem. Show that it agrees with the Taylor Series expansion up to terms of order $h^3$.

I have determined the Taylor series for $y(t_{i+1})$ about $y(t_i)$ as

$$y(t_i) + \lambda y(t_i)h + \frac{1}{2}\lambda^2 y(t_i)h^2 + \frac{1}{3!}\lambda^3 y(t_i)h^3 + \cdots$$

But I am not able to write out the equation in RK3-2 form

RK3-2 is defined as

$k_1 = f(t_i, y_i)$

$k_2 = f(t_i + \frac{1}{2}h, y_i + \frac{1}{2}hK_1)$

$k_3 = f(t_i + h, y_i - hk_1 + 2hK_2)$

$y_{i+1} = y_i + h(\frac{1}{6}k_1 + \frac{4}{6}k_2 +\frac{1}{6}k_3)$

How can I show that the RK3-2 method agrees with the Taylor expansion up to terms of order $h^3$?

share|cite|improve this question
up vote 5 down vote accepted

In your case $f(t,y)=\lambda y$ so

$k_1 = \lambda y_i$

$k_2 = \lambda (y_i+\frac{1}{2}h \lambda y_i)$

$k_3 = \lambda (y_i-h\lambda y_i+2h \lambda (y_i+\frac{1}{2}h \lambda y_i))$

Expand the formula for $y_{i+1}$ and you'll get exactly the Taylor expansion in $y_i$ up to order 3.

share|cite|improve this answer

Let's compute what $k_i$ are in terms of $y$ and $t$:

$$k_1 = \lambda y(t_i)$$ $$k_2 = \lambda \left(y\left(t_i\right)+\frac{h}{2} k_1\right) = \lambda \left(y\left(t_i\right)+\frac{h}{2}\lambda y\left(t_i\right)\right)$$ $$\vdots$$

Plug these into your RK formula $y_{n+1} = y_n +\sum_{i=1}^k b_ik_i$ and you should be able to show agreement (it's a bunch of tedious algebra, mostly).

share|cite|improve this answer
This agrees with Beni's answer, who got it right much faster than me! – Emily Nov 16 '12 at 17:31
But in $k_2$, the $t_i$ term is actually $t_i + \frac{1}{2}h$. So why have you got $y(t_i)$ instead of $y(t_i + \frac{1}{2}h)$? – sonicboom Nov 16 '12 at 17:34
No, you only replace $t$ with $t_i+\frac{1}{2}h$ when $t$ appears explicitly. In $k_2$, you replace $y$ with $y_i+\frac{1}{2}hk_1$. The important thing to note is you replace it with a $y_i$, and $y_i = y(t_i)$. – Emily Nov 16 '12 at 17:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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