Question

Let S be the set of all continuous functions f: [-1,1]→\(\R\) satisfying the following three conditions:
(i) f is infinitely differentiable on the open interval (-1,1),
(ii) the Taylor series
\(f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+...\) of f at 0 converges to f(x) for each x ∈ (-1,1),
(iii) \(f(\frac{1}{n})=0\ \text{for all}\ n\isin\N\)
Then which of the following is/are true?

• A. f(0) = 0 for every f ∈ S.
• B. \(f'(\frac{1}{2})=0\) for every \(f\isin S\).
• C. There exists \(f\isin S\) such that \(f'(\frac{1}{2})\ne0\)
• D. There exists f ∈ S such that f (x) ≠ 0 for some x ∈ [-1,1].

Answer 1

The Correct Option is A, B

To determine which options are correct, let's examine the conditions and conclusions step by step: 

1. \(S\) is the set of all continuous functions \(f: [-1,1] \to \mathbb{R}\) that satisfy the following:
   • (i) \(f\) is infinitely differentiable on the open interval \((-1,1)\).
   • (ii) The Taylor series \(f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \ldots\) converges to \(f(x)\) for each \(x \in (-1,1)\).
   • (iii) \(f\left(\frac{1}{n}\right) = 0\) for all \(n \in \mathbb{N}\).
2. From condition (ii), we know that the Taylor series converges to the function \(f(x)\) within \((-1,1)\). The points \(\frac{1}{n}\) (for \(n \in \mathbb{N}\)) are within this interval, thus the series converges to 0 at these points due to condition (iii).
3. Let's analyze each option:

Option 1: \(f(0) = 0\) for every \(f \in S\).

Since the Taylor series expansion converges to the function, and since \(f\left(\frac{1}{n}\right)=0\) for all \(n\), the only way \(f(x)\) can remain zero at those infinitely approaching points is if \(f(x) = 0\) for all \(x\). Therefore, \(f(0) = 0\) must be true. Thus, this option is correct.

Option 2: \(f'\left(\frac{1}{2}\right) = 0\) for every \(f \in S\).

Since the derivative at all points of accumulation must evaluate to zero (due to how conditions are set up given restrictions from condition (iii)), this option is correct.

Option 3: There exists \(f \in S\) such that \(f'\left(\frac{1}{2}\right) \neq 0\).

This is incorrect because if there were such a function, it would violate the convergence setup and the rules applied in option 2.

Option 4: There exists \(f \in S\) such that \(f(x) \neq 0\) for some \(x \in [-1,1]\).

This is incorrect because the Taylor series already constrains \(f(x)\) to 0.

Thus, the correct statements are: "f(0) = 0 for every f ∈ S." and "f'(\(\frac{1}{2}\))=0 for every f ∈ S."