Question

Let f be a function such that f and all its derivatives exist in an open interval I containing c, then the Taylor’s series of f at x = c is given by :

• A. $f(c) + \frac{f'(c)}{1!}x + \frac{f''(c)}{2!}x^2 + \dots$
• B. $f(c) - \frac{f'(c)}{1!}(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \dots$
• C. $f(c) + \frac{f'(c)}{1!}(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \dots$
• D. $f(c) - \frac{f'(c)}{1!}x + \frac{f''(c)}{2!}x^2 - \dots$

Hint:

To remember the formula, make sure every term of the expansion is consistent in its indices: the $n^{\text{th}}$ derivative, the $n!$ denominator, and the power $(x-c)^n$ must all share the same index $n$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A Taylor series is an infinite sum of terms expressed in terms of the function's derivatives at a single point $c$.
Detailed Explanation:
The Taylor series representation of a real or complex-valued function $f(x)$ that is infinitely differentiable at a point $c$ is: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n \] Let us expand this summation term by term:
- For $n = 0$: $\frac{f^{(0)}(c)}{0!} (x-c)^0 = f(c)$
- For $n = 1$: $\frac{f^{(1)}(c)}{1!} (x-c)^1 = \frac{f'(c)}{1!} (x-c)$
- For $n = 2$: $\frac{f^{(2)}(c)}{2!} (x-c)^2 = \frac{f''(c)}{2!} (x-c)^2$
Summing these terms: \[ f(x) = f(c) + \frac{f'(c)}{1!}(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \dots \] This matches the formulation in Option (C).
Note: When $c = 0$, this series is specifically referred to as the Maclaurin series.

Step 2: Final Answer:
The Taylor's series expression matches Option (C).