Taylor Series (4/7/06) We have seen that if f(x) is a function for which we can compute all of its derivatives (i.e., first derivative f '(x), second derivative.

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Presentation transcript:

Taylor Series (4/7/06) We have seen that if f(x) is a function for which we can compute all of its derivatives (i.e., first derivative f '(x), second derivative f (2) (x), third derivative f (3) (x), and so on), then we can write down a representation of f as a power series in terms of the values of those derivatives at the center of the interval of convergence. This is called the Taylor Series for f(x).

Taylor Series centered at 0 (aka Maclaurin Series) We get that, on the interval of convergence, f(x) = f(0) + f '(0) x + f (2) (0) x 2 / 2! +… + f (n) (0) x n / n! +… = To estimate f, we can use the first so many terms (as a calculator does!).

Examples of series centered at 0 Since if f (x) = e x, then for every n, f (n) (x) = e x also, so f (n) (0) = 1. Hence the Taylor Series for e x is very simple: e x = 1 + x + x 2 /2! + … + x n /n! + …, which converges everywhere. Show that the Taylor series for sin(x) is sin(x) = x – x 3 /3! + x 5 /5! – x 7 /7! + …. By computing the derivatives of sin at 0.

Taylor Series centered at a If it is easier (or more useful) to evaluate f ‘s derivatives at a point a rather than zero, then we simply center the power series at a and evaluate the derivative at a: f(x) = f(a) + f '(a) (x-a) + f (2) (a) (x-a) 2 / 2! + …+ f (n) (a) (x-a) n / n! +…

Examples of series centered at 1 The natural log function isn’t even defined at 0, so we can’t center a series for it there. Show, computing derivatives evaluated at 1, that a Taylor series for ln(x) is ln(x) = (x-1) - (x-1) 2 /2 + (x-1) 3 /3 - (x-1) 4 /4 + … (Note that this only converges on 0 < x  2) Now try it for the square root function. These derivatives at 1 are a bit trickier!

Assignment For Monday, read Section through page 767 and do Exercises 3, 5, 7, 11, 14, & 15. Wednesday will be an optional Q & A session. Friday (4/14) is Test #2.