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Taylor’s Formula November 14, 2007

Posted by jagadeeshbp in Uncategorized.
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Let f(x) be a function continous over the closed interval, [a,b] such thatf(x) has (n+1)^{th} derivative, which is written as f^{(n+1)}(x); then:

f(b) = f(a) + f'(a) (b-a) + ... + \frac{f^{(n)}(a)}{n!} (b-a)^n + \frac{f^{(n+1)}(z)}{(n+1)!} (b-a)^{(n+1)}

PROOF:

Let

H = f(b) - f(a) - f'(a) (b-a) - .. - \frac{f^{(n)}(a) }{n!}(b-a)^n

Also,

K = \frac{H}{(b-a)^(n+1)}

Consider a polynomial,

g(x) = f(b) - f(x) - f'(x)(b-x) - f''(x)(b-x)^2... - \frac{f^{(n)}(x)}{n!}(b-x)^n - K(b-x)^{n+1}

We can see that,

g(a) = H - K(b-a)^{n+1} = 0

Similarly,

g(b) = f(b) - f(b) - f'(b)(b-b) - ... - \frac{f^{(n)}(b) }{n!}(b-b)^{n} - K(b-b)^{n+1} = 0

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