If you are not familiar with continuous functions or closed intervals, start here.
If you would like a review of derivatives, start here.
Theorem: Taylor's Formula
Let f(x) be a continuous function over the closed interval [a,b] that has an (n+1) derivative which is referenced by f(n+1)(x) where n is a positive integer.
Then:
f(b) = f(a) + f'(a)(b-a) + ... + [f(n)(a)/n!](b-a)n + [f(n+1)(z)/(n+1)!](b-a)n+1)
for some number z which lies between a and b.
Proof:
(1) Let H = f(b) - f(a) - f'(a)(b-a) - [f''(a)/2!](b-a)2 - ... - [f(n)(a)/n!](b-a)n
(2) Let K = H/(b-a)n+1
(3) Let g(x) be a function such that:
g(x) = f(b) - f(x) - f'(x)(b-x) -[f''(x)/2!](b-x)2 - ... - [f(n)(x)/n!](b-x)n - K(b-x)n+1
(4) We can see that g(x) is a continuous function since:
(a) f(x) is continuous over [a,b] by the given.
(b) f(n) is a continuous function since by the given we know that f(x) has an (n+1) derivative and since f(x) is differentiable at x, then it is continuous at x (see here)
(c) (b-x)n is continuous over [a,b] since f(x) is continuous since:
Let h(x)=b-x
h(x) is continuous since it is the addition of two continuous functions i(x)=b and j(x)=-x [See here for proof of the Addition Law]
(b-x)n is continuous by the Multiplication Law [See here for proof of the Multiplication Law]
(b-x)n = (b - x)*(b-x)*... since n is a positive integer.
(d) f(b) is continuous since it is a constant. (See here)
(e) Each f(n)(x)/n! is continuous since (-1/n!) can be thought of as a constant function h(x)=(-1/n!) and the multiplication of two continuous functions is itself continuous (see here)
(f) Finally, g(x) is continuous because the addition of a set of continuous functions is itself continuous (see here)
(5) We can see that g(a) = 0 since:
g(a) = H - H/(b-a)n+1*(b-a)n+1 = H - H = 0
(6) We can also see that g(b) = 0 since:
g(b) = f(b) - f(b) - f'(b)(b-b) - [f''(b)/2!](b-b)2 - ... - [f(n)(b)/n!](b-b)n - K(b-b)n+1
(7) By Rolle's Theorem, since g(x) is continuous on [a,b], we know that there exists a value z such that z ∈ [a,b] and g'(z) = 0. [See here for Rolle's Theorem]
(8) If we differentiate on g(x), we get:
g'(x) = -f'(x) + f'(x) -f(2)(x)(b-x) + f(2)(x)(b-x) - (1/2!)f(3)(x)(b-x)2 + (1/2!)f(3)(x)(b-x)2 - (1/3!)f(4)(x)(b-x)3 + ... + [1/(n-1)!]f(n)(x)(b-x)n-1 - (1/n!)f(n+1)(x)(b-x)n + (n+1)K(b-x)n since:
(a) g(x) = f(b) - f(x) - f'(x)(b-x) -[f''(x)/2!](b-x)2 - ... - [f(n)(x)/n!](b-x)n - K(b-x)n+1
(b) By Lemma 3 here, we can differentiate each individual product in the sum.
(c) f(b) is a constant so d/dx(f(b)) = 0 [See here for Constant Rule]
(d) d/dx(-f(x)) = -f'(x) [See here for details]
(e) d/dx[-f'(x)(b-x)] = -f''(x)(b-x) - f'(x)(-1) = f'(x) -f''(x)(b-x) [See here for the Product Rule]
(f) d/dx[(-f''(x)/2!)(b-x)2] = (-f(3)(x)/2!)(b-x)2 + [-f''(x)/2!](2)(b-x)(-1)] =
= (-f(3)/2!)(b-x)2 + f''(x)(b-x) [See here for Generalized Power Rule]
(g) d/dx([-f(n)(x)/n!](b-x)n) = (-f(n+1)(x)/n!)(b-x)n + [-f(n)(n)/n!](n)(b-x)n-1*(-1) =
= (-f(n+1)(x)/n!)(b-x)n +[ f(n)/(n-1)!](b-x)n-1
(h) Since K is a constant
d/dx(-K(b-x)n+1) = (n+1)(-K)(b-x)n(-1) = (n+1)K(b-x)n
(9) We see that most of the terms cancel out so that we get:
g'(x) = (n+1)K(b-x)n - (1/n!)f(n+1)(x)(b-x)n
(10) Applying the fact that g'(z) = 0 from step #7 gives us:
g'(z) = (n+1)K(b-z)n - (1/n!)f(n+1)(z)(b-z)n = 0
(11) We can divide both sides by (b-z)n to get:
(n+1)K - (1/n!)f(n+1)(z) = 0
(12) Now we can rearrange (#11) to get:
K = [(1/n!)f(n+1)(z)]/(n+1) = [f(n+1)(z)]/(n+1)!
(13) Now, using (#12) and (#3) with x=a, we get:
g(a) = 0 = f(b) - f(a) - f'(a)(b-a) - [f''(a)/2!](b-a)2 - ... - [f(n)(a)/n!](b-a)n - [(b-a)n+1][f(n+1)(z)/(n+1)!]
(14) Now, if we move over all the elements after f(b), we get:
f(b) = f(a) + f'(a)(b-a) + [f''(a)/2!](b-a)2 + ... + [f(n)(a)/n!](b-a)n + [(b-a)n+1][f(n+1)(z)/(n+1)!]
QED
References
- Edwards & Penny, Calculus and Analytic Geometry
7 comments:
A question (rather a dumb one, I guess), what is the need for proving g(x) is a continuous function?
Hi Jagadeesh,
We need to show that g(x) is a continuous function in order to apply Rolle's Theorem which only applies to continuous functions.
-Larry
Wow, thanks. I still don't get it really good, but at least I was able to read through the proof.
Bit of nitpicking about a part that confused me:
(8) (e):
(e) d/dx[-f'(x)(b-x)] = -f'''(x)(b-x) - f'(x)(-1) = f'(x) -f''(x)(b-x) [See here for the Product Rule]
You accidentally have written the third derivative of f(x) instead of second:
It should be:
(e) d/dx[-f'(x)(b-x)] = -f''(x)(b-x) - f'(x)(-1) = f'(x) -f''(x)(b-x) [See here for the Product Rule]
Hi Rauni,
Thanks for your comment. I'll try to simplify the explanation of this proof at a future time.
In the mean time, I've fixed the typo you found (nitpicking encouraged) :-)
-Larry
This is a nice proof, thanks. I see a minor typo on 8a, the "n+1" at the end should be an exponent.
This is difficult to understand. The proof given in wikipedia was simpler.
A silly question but how can one know that imaginary numbers can be applied in lemma 2? Is there a way of proving it?
Anyway your blog is amazing. just amazing.
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