[CST-2] NA2 page 10

[Ashish Jain]

> Our supervisor just said that it's just an example of "a" Chebyshev
> approximation...but I'm really not sure how to get it!  I've tried for
> hours (and given up!!!) - anyone with a better explaination, please tell
> us...
>
> > - how is the error from truncating the Taylor series after x^6
> > calculated?
>
> Since we are looking at the interval [-1, 1] we can say that the error in
> truncating the Taylor series after x^6 is the maximum value of the next
> terms x^7 and x^8.  i.e. 1/factorial(7) + 1/factorial(8) ~ 0.00023.
> (Note: we can ignore all the other terms from x^9 and onwards as they will
> make no difference, since we are considering numbers to 5 decimal places
> only.)



The remainder when you truncated a Taylor series is given by
|R(x)|=(f^n(Xi)* x^n)/n!  which in this case (interval [-1,1]) reduces to
1/7! (Could also consider 1/8! for better accuracy)

To economize the series you represent it as

P5(x)=P6(x)-a(6)T(6)(x)

After subsituting you'll get a relation in degree 5 giving you the
polynomial P5 in x of degree 5. The line on the notes can be found as by
subsituting

T(0) for 1
T(1) for x
1/2(T(2)+1) for x^2  ( from T(2)=2x^2 -1 )

and so on.... in the expression.

The P5 obtained can then further be economize to get a polynomial P4 and
the error thus obtained is smaller than mere truncating the Taylor's
series.



Ashish