Insights on Differentiable Functions & Expansions

1. SOME GENERAL TEHOREMS ON DIFFERENTIABLE FUNCTIONS & EXPANSIONS a) Rolle's TheoremIf a function (x) is (i) continuous in closed interval [a, b] (ii) differentiable in open interval (a, b) and (iii) f(a) = f(b) ; then there exists atleast one real numbersuch that f '(c) = 0. (b) Lagrange's Mean Value Theorem.If a function f(x) is (i) continuous in closed interval [a, a + h] (ii) differentiable in open interval (a, a + h) then there exists atleast one real number θ; 0< θ<1, such that f(a + h) = f(a) + hf '(a + θh). 1 Taylor's Theorem with Lagrange's Form of Remainder after 'n' Terms If a function f(x) is such that (i) f(x), f '(x) f "(x).....f n-1(x) are continuous in closed interval [a, a+ h] (ii) f n(x) exists in open interval (a, a + h). then there exists atleast one real number θ; 0< θ<1 such that Proof.Consider the function where A is a constant to be chosen so that F(a) = F(a+h)

2. Now Putting these values in (2), we have To apply Rolle's theorem on F(x) : (i) f(x), f '(x), f "(x) ......., f n-1(x) are continuous on the closed interval [a, a+h] ........ (given) and (a+h-x), (a+h-x)2, (a+h-x)3, ............ (a+h-x)nbeing polynomials are continuous on the closed interval [a, a+h]. Also the algebraic sum of continuous functions is continuous. therefore, F(x) is continuous in the closed interval [a, a+h]. (ii) As f n (x) exists in the open interval (a, a+h) therefore, f(x), f '(x), f "(x) ........., f n-1(x) are all derivable in the open interval (a, a+h). Also (a+h-x), (a+h-x)2, ........... (a+h-x)n being polynomials are derivable in the open interval (a, a+h) therefore,F(x) is derivable in the open interval (a, a+h). (iii) F(a) = F(a+h) F(x) satisfies all the three conditions of Rolle's theorem in [a, a+h]. Hence there exists atleast one real number θ; 0< θ<1, such that F'(a+ θh) = 0

3. Differentiating (1) w.r.t. x, we get F'(x) = f'(x) Putting x= a + θh, we get Putting x=+ θh, we get But which is the required expression.

4. Cor. Maclaurin's theorem with Lagrangbe's form of remainder Putting a = 0, h = x in in Taylor's theorem, we have which is Maclaurin's theorem with Lagrange's form of remainder. Maclaurin's Theorem in full form : If a function f (x) is such that (i) f(x), f '(x), f "(x), ........... f n-1(x) are continuous in closed interval [0, x] (ii) f n(x) exists in open interval (0, x), then there exists atleast one real number θ; 0< θ<1, such that Another form of Taylor's Theorem with Lagrange's form of reminder Let b=a+h i.e., h=b-a and c=a+ θh=a+ θ(b-a) and Taylor's Theorem with Cauchy's Form of Remainder If a function f(x) is such that (i) f(x), f '(x), f "(x), ....f n-1(x) are all continuous in the closed interval [a, a+h] (ii) f n(x) exists in open interval (a, a+h) then there exists atleast one real number θ; 0<θ<1, such that

5. Proof : Let where A is a constant such that F(a) = F(a+h). Putting x = a in (1), we have Putting x = a + h in (1), we have F(a + h) = f(a+h) + 0 + 0 + .......... = f(a+h) Now F(a+h) = F(a) Then from (2) and (3), we have Now, (i) f(x), f '(x), f "(x) ............, f n-1(x) are continuous on the closed interval [a, a+h) ........... (given) and (a+h-x), (a+h-x)2, (a+h-x)3, ................ (a+h-x)nbeing polynomials are continuous on the closed interval [a, a+h]. Also the algebraic sum of continuous functions is continuous Therefore, F(x) is continuous in the closed interval [a, a + h]. (ii) As f n(x) exists in the open interval (a, a + h) Therefore, f(x), f '(x),f "(x) ......., f n-1(x) are all derivable in the open interval (a, a+h). Also (a+h-x), (a+h-x)2, ....... (a+h-x)nbeing polynomials are derivable in the open interval (a, a+h) Therefore, F(x) is derivable in the open interval (a, a+h). (iii) F(a) = F(a+h) Therefore, F(x) satisfies all the three conditions of Rolle's theorem in [a, a+h]. Hence there exists atleast one real number θ; 0<θ<1, such that F'(a+θh) = 0 Differentiating both sides of (1) w.r.t. x,

6. Putting x = a + θh, we get

7. Putting this value of A in (4), we get

8. Cor. Maclaurin’s Theorem with Cauchy’s form of remainder. Putting a = 0, h =x in the above theorem, we have Whichis the Maclaurin’s theorem with Cauchy’s form of remainder. is called Cauchy’s form of remainder afer n terms in Maclaurins’ expansion of f(x). This theorem can be stated as : If a function f(x) is such that (i) (ii) Then there exists atleast one real number