# 3rd order Runga Kutta method agrees with Taylor Series up to terms of order $h^3$?

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 $h^3$?

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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.

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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).

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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