Sums and Approximations In our example, where f(a)= In(1+r), we have f"(a)=-1/(1 +a)2, which means f"(o)=-1. This gives the quadratic approximation In(1 +r)aI-22/2. Lets add this to our plot 1.5 y=In(1+a) y=x-x2/2 2 Taylor's Theorem says that one can continue in this way, getting progressively better approximations by incorporating higher-order derivatives Theorem(Taylors theorem). Suppose that f: R-R is(n 1)-times differentiable on the interval0,x小Then f(x)=f(0)+f(0)x+ f(0)r2 f(m)(0)xn2,fan+1)(x) (n+1)! for some a∈[0,x The final term describes the"error"in an n-term Taylor approximation; we'l return to that later. Here are the leading terms in the Taylor series for some functions that often C=1+x+a2++4Sums and Approximations 11 In our example, where f(x) = ln(1 + x), we have f��(x) = −1/(1 + x)2, which means f��(0) = −1. This gives the quadratic approximation ln(1 + x) ≈ x − x2/2. Lets add this to our plot: 6 1.5 y = x y = ln(1 + x) 1 0.5 y = x − x2/2 0 - 0 1 2 3 Taylor’s Theorem says that one can continue in this way, getting progressively better approximations by incorporating higher­order derivatives. Theorem (Taylor’s theorem). Suppose that f : R R is (n + 1)­times differentiable on the interval [0, x]. Then → 2 n f(n+1)(z)xn+1 f��(0)x f(n) (0)x f(x) = f(0) + f� (0)x + + . . . + + 2! n! (n + 1)! for some z ∈ [0, x]. The final term describes the “error” in an n­term Taylor approximation; we’ll return to that later. Here are the leading terms in the Taylor series for some functions that often arise: 2 3 4 x x x x e = 1 + x + + + + . . . 2! 3! 4! 2 3 4 5 x x x x ln(1 + x) = x − + 5 − . . . 2 + 3 − 4 x x x3 5x √ 4 1 + x = 1 + 2 − 8 2 + + . . . 16 − 128