Question

Taylor's theorem with cauchy's form of remainder proof

Answer 1

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Taylor's theorem with Cauchy's form of remainder proof..
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Answer 2

Proof for Taylor's theorem with Cauchy's form of remainder:

Statement: If f is a function defined in [a,b] such that

(i.) f ,f¹, f¹¹.....fⁿ⁻¹ are continuous in [a,b]

(ii.) fⁿ(x) exists in (a,b) then there exists at least one real number θ ∈ (0,1), then show that f(b) = f(a) + (b-a) f¹(a) + (b-a)²/2 f¹¹(a) +.....+ (b-a)ⁿ⁻¹/(n-1) fⁿ⁻¹(a) + (b-a)  { (b-a)ⁿ⁻¹ (1-θ)ⁿ⁻¹ }/n-1 fⁿ{a+θ(b-a)}

Proof:

Consider a function ∅ defined as

⇒ Φ(x) = f(x) + (b-x) f¹(x) + (b-x)²/2 f¹¹(x) +.....+ (b-x)ⁿ⁻¹/(n-1) fⁿ⁻¹(x) + (b-x) A

Where, A = constant so that Φ(a) = Φ(b)

⇒ Φ(a) = f(a) + (b-a) f¹(a) + (b-a)²/2 f¹¹(a) +.....+ (b-a)ⁿ⁻¹/(n-1) fⁿ⁻¹(a) + (b-a) A

⇒ Φ(b) = f(b) + (b-b) f¹(x) + (b-b)²/2 f¹¹(b) +.....+ (b-b)ⁿ⁻¹/(n-1) fⁿ⁻¹(b) + (b-b) A

⇒ Φ(b) = f(b)

⇒ Φ(a) = f(a) + (b-a) f¹(a) + (b-a)²/2 f¹¹(a) +.....+ (b-a)ⁿ⁻¹/(n-1) fⁿ⁻¹(a) + (b-a) A =  f(b)

Now, given that

1.) Φ(x) is continuous in [a,b]

2.) Φ(x) is derivable in (a,b)

3.) Φ(a) = Φ(b)

So, we can apply Rolles theorem

⇒ c ∈ (a,b) then Φ¹(c) = 0

⇒ Φ¹(x) = f¹(x) + { -f¹(x) + ( b-x )f¹¹(x) } + {-( b-x )f¹¹(x) + ( -b-x )²/2 f¹¹(x) }+......+ { -( b-x )ⁿ⁻²/n-2 fⁿ⁻¹(x) +-( b-x )ⁿ⁻¹/n-1 fⁿ(x) }

⇒ Φ¹(x) = -( b-x )ⁿ⁻¹/n-1 fⁿ(x)

⇒ Φ¹(a+θ(b-a)) = (b-(a+θ(b-a))ⁿ⁻¹/n-1 fⁿ{a+θ(b-a)}

⇒ Φ¹(a+θ(b-a)) = { b-a-θb+θa }ⁿ⁻¹/n-1 fⁿ{a+θ(b-a)}

⇒ Φ¹(a+θ(b-a)) = { (b-a) (1-θ) }ⁿ⁻¹/n-1 fⁿ{a+θ(b-a)}

⇒ Φ¹(a+θ(b-a)) =  { (b-a)ⁿ⁻¹ (1-θ)ⁿ⁻¹ }/n-1 fⁿ{a+θ(b-a)}

Therefore A =   { (b-a)ⁿ⁻¹ (1-θ)ⁿ⁻¹ }/n-1 fⁿ{a+θ(b-a)}

⇒ f(b) = f(a) + (b-a) f¹(a) + (b-a)²/2 f¹¹(a) +.....+ (b-a)ⁿ⁻¹/(n-1) fⁿ⁻¹(a) + (b-a) A

⇒ f(b) = f(a) + (b-a) f¹(a) + (b-a)²/2 f¹¹(a) +.....+ (b-a)ⁿ⁻¹/(n-1) fⁿ⁻¹(a) + (b-a)  { (b-a)ⁿ⁻¹ (1-θ)ⁿ⁻¹ }/n-1 fⁿ{a+θ(b-a)} where 0<θ<1

Hence, Taylor's theorem with cauchy's form of remainder proved.

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