## Friday, March 10, 2006

### 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

A question (rather a dumb one, I guess), what is the need for proving g(x) is a continuous function?

Larry Freeman said...

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

Lilleke said...

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]

Larry Freeman said...

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

Greg Sidebottom said...

This is a nice proof, thanks. I see a minor typo on 8a, the "n+1" at the end should be an exponent.

Sanju said...

This is difficult to understand. The proof given in wikipedia was simpler.

Unknown said...

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.