Taylor's theorem

In calculus, Taylor's theorem, named after the mathematician Brook Taylor, who stated it in 1712, gives the approximation of a differentiable function near a point by a polynomial whose coefficients depend only on the derivatives of the function at that point.

Proof

We first prove Taylor's theorem with the integral remainder term.

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The fundamental theorem of calculus states that

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:f(x) = f(a) + int_a^x (x-t)^0 , f'(t) , dt.

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This proves the theorem for n = 0.

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Integration by parts yields the case n = 1:

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:f(x) = f(a) +f'(a),(x-a)+int_a^x (x-t)^1 , f(t) , dt.

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By repeating this process, we may derive Taylor's theorem for higher values of n.

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This can be formalized by applying the technique of induction. So, suppose that Taylor's theorem holds for a particular n, that is, suppose that

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:

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f(x) = f(a)

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+ rac{f'(a)}{1!}(x - a)

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+ cdots

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+ rac{f^{(n)}(a)}{n!}(x - a)^n

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+ int_a^x rac{f^{(n+1)} (t)}{n!} (x - t)^n , dt. qquad(*)

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We can again rewrite the integral using integration by parts. An antiderivative of (x − t)n as a function of t is given by −(x−t)n+1 / (n + 1), so

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: int_a^x rac{f^{(n+1)} (t)}{n!} (x - t)^n , dt

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:: {} = - left_a^x + int_a^x rac{f^{(n+2)} (t)}{(n+1)n!} (x - t)^{n+1} , dt

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:: {} = rac{f^{(n+1)} (a)}{(n+1)!} (x - a)^{n+1} + int_a^x rac{f^{(n+2)} (t)}{(n+1)!} (x - t)^{n+1} , dt.

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Substituting this in (*) proves Taylor's theorem for n + 1, and hence for all nonnegative integers n.

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The remainder term in the Lagrange form can be derived by the mean value theorem in the following way:

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:

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R_n = int_a^x rac{f^{(n+1)} (t)}{n!} (x - t)^n , dt =f^{(n+1)}(xi) int_a^x rac{(x - t)^n }{n!} , dt.

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The last integral can be solved immediately, which leads to

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:

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R_n = rac{f^{(n+1)}(xi)}{(n+1)!} (x-a)^{n+1}.

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