Taylor's Formula

In today's blog, I will go over Taylor's Formula. This is a theorem that can be used to establish the Taylor Series and the MacLaurin Series which I use in my proof of Euler's Identity.

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