Home » Articles « Back to article

PROOF OF TAYLOR'S THEOREM

bladeX -rise of a new sun

Blazing goIITian

Posted on
17 Jul 2009 15:21:10 IST

Posts: 1249

17 Jul 2009 15:21:10 IST

Like 0 people liked this

3318

PROOF OF TAYLOR'S THEOREM

Mathematics , Integral Calculus

Proof: Taylor's theorem in one variable

[edit] Integral version

We first prove Taylor's theorem with the integral remainder term.^[4]

The fundamental theorem of calculus states that

$int_a^x , f'(t) , dt=f(x)-f(a),$

which can be rearranged to:

$f(x)=f(a)+ int_a^x , f'(t) , dt.$

Now we can see that an application of Integration by parts yields:

$egin{align}f(x) &= f(a)+xf'(x)-af'(a)-int_a^x , tf''(t) , dt &= f(a)+int_a^x , xf''(t) ,dt+xf'(a)-af'(a)-int_a^x , tf''(t) , dt &= f(a)+(x-a)f'(a)+int_a^x , (x-t)f''(t) , dt.end{align}$

The first equation is arrived at by letting $u=f'(t),$ and dv = dt; the second equation by noting that $int_a^x , xf''(t) ,dt = xf'(x)-xf'(a)$ ; the third just factors out some common terms.

Another application yields:

$f(x)=f(a)+(x-a)f'(a)+ rac 1 2 (x-a)^2f''(a) + rac 1 2 int_a^x , (x-t)^2f'''(t) , dt.$

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

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

$f(x) = f(a)+ rac{f'(a)}{1!}(x - a)+ cdots+ rac{f^{(n)}(a)}{n!}(x - a)^n+ int_a^x rac{f^{(n+1)} (t)}{n!} (x - t)^n , dt. qquad(*)$

We can 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

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

${} = - left[ rac{f^{(n+1)} (t)}{(n+1)n!} (x - t)^{n+1} ight]_a^x + int_a^x rac{f^{(n+2)} (t)}{(n+1)n!} (x - t)^{n+1} , dt$

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

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

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

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

where $ξ$ is some number from the interval [a, x]. The last integral can be solved immediately, which leads to

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

More generally, for any function G(t), the mean value theorem asserts the existence of $ξ$ in the interval [a, x] satisfying

$R_n = int_a^x rac{f^{(n+1)} (t)}{n!} (x - t)^n rac{G'(t)}{G'(t)}, dt = rac{f^{(n+1)}(xi)}{n!} (x-xi)^n rac{1}{G'(xi)} int_a^x G'(t) , dt$

$= rac{f^{(n+1)}(xi)}{n!} (x-xi)^n cdot rac{G(x)-G(a)}{G'(xi)}.$

R_nis the remainder :

SOURCE : WIKIPEDIA

Share this article on:

Comments (1)

swapnil joshi

Blazing goIITian

Joined: 10 Apr 2009 12:36:21 IST

Posts: 466

18 Jul 2009 10:41:20 IST

Like 0 people liked this

this is easy

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team

Free Sign Up!

Proof: Taylor's theorem in one variable

[edit] Integral version

Comments (1)

Quick Reply

Reply

Preparing for IIT-JEE ?

Arihant Revision Package for IIT JEE - Books, Practice Tests + Rank Predictor

@ INR 1,995/-

For Quick Info