c. Proof that, if a function \( f(x) \) has a power series representation centered at \( x=a \) that converges to \( f(x) \) on some open interval containing \( a \), then this power series a Taylor series for \( f(x) \) at \( x=a \).

To prove that if a function \( f(x) \) has a power series representation centered at \( x = a \) that converges to \( f(x) \) on some open interval containing \( a \), then this power series is a Taylor series for \( f(x) \) at \( x = a \), we will follow these steps: 1. **Definition of Power Series**: A power series centered at \( x = a \) can be expressed as: \[ P(x) = \sum_{n=0}^{\infty} c_n (x - a)^n \] where \( c_n \) are the coefficients of the series. 2. **Convergence**: We are given that this power series converges to \( f(x) \) on some open interval \( I \) containing \( a \). This means that for all \( x \) in \( I \), \[ f(x) = \sum_{n=0}^{\infty} c_n (x - a)^n. \] 3. **Taylor Series Definition**: The Taylor series of \( f(x) \) at \( x = a \) is given by: \[ T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n. \] Our goal is to show that \( P(x) = T(x) \) for all \( x \) in the interval \( I \). 4. **Derivatives of \( f(x) \)**: Since \( f(x) \) is represented by a power series that converges to \( f(x) \), we can differentiate \( f(x) \) term by term within the interval of convergence. Specifically, we can compute the \( n \)-th derivative of \( f(x) \) at \( x = a \): \[ f^{(n)}(x) = \sum_{k=0}^{\infty} c_k \frac{d^n}{dx^n} (x - a)^k. \] The \( n \)-th derivative of \( (x - a)^k \) is given by: \[ \frac{d^n}{dx^n} (x - a)^k = \begin{cases} 0 & \text{if } k < n \\ \frac{k!}{(k-n)!} (x - a)^{k-n} & \text{if } k \geq n \end{cases} \] Therefore, we can write: \[ f^{(n)}(a) = \sum_{k=n}^{\infty} c_k \frac{k!}{(k-n)!} (a - a)^{k-n} = c_n. \] 5. **Coefficients of the Taylor Series**: From the above, we find that: \[ f^{(n)}(a) = c_n. \] This means that the coefficients \( c_n \) of the power series \( P(x) \) are exactly the coefficients of the Taylor series \( T(x) \). 6. **Conclusion**: Since the power series \( P(x) \) converges to \( f(x) \) and has coefficients that match those of the Taylor series \( T(x) \), we conclude that: \[ P(x) = T(x) \quad \text{for all } x \text{ in the interval } I. \] Thus, the power series representation of \( f(x) \) is indeed the Taylor series for \( f(x) \) at \( x = a \). Therefore, we have shown that if a function \( f(x) \) has a power series representation centered at \( x = a \) that converges to \( f(x) \) on some open interval containing \( a \), then this power series is a Taylor series for \( f(x) \) at \( x = a \).