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To start,
![$\displaystyle \int^1_0 f(x)dx = \int^1_0 f(x)B_0(x)dx$](img20.png) |
(2.10) |
It is also obvious that
so that,
![$\displaystyle \int^1_0 f(x)dx= f(1)B_1(1)-f(0)B_1(0) - \int^1_0 f'(x)B_1(x)dx$](img22.png) |
(2.11) |
Carrying on and seeing that
and
, we would find,
![$\displaystyle \int^1_0f(x)dx = \frac{1}{2}[f(1)+f(0)] - \sum^{q}_{p=1} \frac{1}...
...}[f^{(2p-1)}(1)-f^{(2p-1)}(0)] + \frac{1}{(2q)!}\int^1_0 f^{(2q)}(x)B_{2q}(x)dx$](img25.png) |
(2.12) |
The transform
,
,etcetera, yields,
root
2006-09-15