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Calculus- Early Transcendentals, 2021a

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186 Applications of Derivatives<br />

5.4.3 Taylor Polynomials<br />

We can go beyond first order derivatives to create polynomials approximating a function as closely as we<br />

wish, these are called Taylor Polynomials.<br />

While our linear approximation L(x)= f ′ (a)(x − a)+ f (a) at a point a was a polynomial of degree 1<br />

such that both L(a)= f (a) and L ′ (a)= f ′ (a), we can now form a polynomial<br />

T n (x)=a 0 + a 1 (x − a)+a 2 (x − a) 2 + a 3 (x − a) 3 + ···+ a n (x − a) n<br />

which has the same first n derivatives at x = a as the function f .<br />

By successively computing the derivatives of T n , we obtain:<br />

a 0 = f (a)= f (a)<br />

0!<br />

a 1 = f ′ (a)<br />

1!<br />

a 2 = f ′′ (a)<br />

2!<br />

···<br />

a k = f (k) (a)<br />

k!<br />

···a n = f (n) (a)<br />

n!<br />

where f (k) (x) is the k th derivative of f (x), andn! = n(n − 1)(n − 2)...(2)(1), referred to as factorial<br />

notation.<br />

Here is an example.<br />

Example 5.31: Approximate e using Taylor Polynomials<br />

Approximate e x using Taylor polynomials at a = 0, and use this to approximate e.<br />

Solution. Inthiscaseweusethefunction f (x)=e x at a = 0, and therefore<br />

Since all derivatives f (k) (x)=e x ,weget:<br />

T n (x)=a 0 + a 1 x + a x x 2 + a 3 x 3 + ...+ a n x n<br />

Thus<br />

a 0 = f (0)=1<br />

a 1 = f ′ (0)<br />

1!<br />

= 1<br />

a 2 = f ′′ (0)<br />

2!<br />

=<br />

2!<br />

1<br />

a 3 = f ′′′ (0)<br />

3!<br />

=<br />

3!<br />

1<br />

···<br />

a k = f (k) (0)<br />

k!<br />

=<br />

k!<br />

1<br />

···<br />

a n = f (n) (0)<br />

n!<br />

=<br />

n!<br />

1<br />

T 1 (x)=1 + x = L(x)<br />

T 2 (x)=1 + x + x2<br />

2!<br />

T 3 (x)=1 + x + x2<br />

2! + x3<br />

3!

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