Problem 4: Taylor polynomials and errors.

> f := x -> exp(x^2 - 1) - 1;

a )

> taylor(f(x),x=1,3);

Hence

> P := x -> 3*(x-1)^2 + 2*(x-1) + 0;

It is not necessary to simplify this, but in case you wish to:

> simplify(P(x));

Just for fun, let's compare:

> plot([f(x),P(x)], x=-1..1);

Notice that the approximation is good near +1, but not so good near -1. Part c) uses Lagrange's formula:

c )

> diff(f(x), x$3)/3!;

In Lagrange's formula, the x here is replaced by c, and the whole thing is multiplied by (-1-(1))^3.

The e-factors are never bigger than 1, so the magnitude of this error is smaller than

max |2 c + 4 c^3/3| for c between -1 and 1. Kindergarten calculus tells us this value will occur when c=+1 or c = -1 (end points ), because the derivative of this expression is not 0 for -1 < c < 1. Therefore the error is less in magnitude than

> (2 + 4/3)*2^3;

Evidently, the Lagrange form of the error is more than 20 . WIth more work, we could do a little better, but this is not required on the test