Mathematical Tools for Physics

11—Numerical Analysis 333

Substitute this into the equation fory(0):

y(0) =

∑N

k=1

αk

∑∞

n=0

(−kh)n

y(n)(0)

n!

+h

∑N

k=1

βk

∑∞

n=0

(−kh)n

y(n+1)(0)

n!

.

This should be an identity to as high an order as possible. The coefficient ofh^0 gives

1 =

∑N

k=1

αk. (40)

The next orders are

0 =

∑

k

αk(−kh) +h

∑

k

βk

0 =

∑

k

1

2

αk(−kh)^2 +h

∑

k

βk(−kh)

..

. (41)

N= 1is Euler’s method again.
N= 2gives
α 1 +α 2 = 1
α 1 + 4α 2 = 2(β 1 + 2β 2 )

α 1 + 2α 2 =β 1 +β 2

α 1 + 8α 2 = 3(β 1 + 4β 2 ).

The solution of these equations is

α 1 =− 4 α 2 = +5 β 1 = +4 β 2 = +2

y(0) =− 4 y(−h) + 5y(− 2 h) +h

[

4 y′(−h) + 2y′(− 2 h)

]

. (42)

To start this algorithm off, two pieces of information are needed: the values ofyat−hand at− 2 h. This is in
contrast to Runge-Kutta, which needs only one point.