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Chapter 10: Problem 6

What condition must be met by a function \(f\) for it to have a Taylor series centered at \(a ?\)

Short Answer

Expert verified
Answer: The necessary and sufficient condition for a function to have a Taylor series centered at a point a is that the remainder term R_n(x) converges to 0 as n approaches infinity.

Step by step solution

01

Understanding Taylor series

A Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. In general, a Taylor series for a function \(f\) centered at \(a\) can be written as: \[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n\] where \(f^{(n)}(a)\) is the \(n\)-th derivative of \(f\) evaluated at \(a\), and \(n!\) is the factorial of \(n\).
02

Knowing about convergence

For a function \(f\) to have a Taylor series centered at \(a\), the series must converge to the function. That is, the difference between the function and its Taylor series representation should become smaller and smaller as more terms are included in the series.
03

Introducing the remainder term

To determine if the Taylor series converges to the function \(f\), we introduce a remainder term, \(R_n(x)\), defined as the difference between the function and the Taylor polynomial of degree \(n\), that is: \[R_n(x) = f(x) - \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k\]
04

Condition for convergence

A necessary and sufficient condition for a function \(f\) to have a Taylor series centered at \(a\) is that the remainder term \(R_n(x)\) converges to \(0\) as \(n \to \infty\). Mathematically, we can express this condition as: \[\lim_{n\to\infty} R_n(x) = 0\]
05

Conclusion

In conclusion, for a function \(f\) to have a Taylor series centered at \(a\), the necessary and sufficient condition is that the remainder term \(R_n(x)\) converges to \(0\) as \(n \to \infty\). This condition ensures that the Taylor series converges to the function, and the difference between the function and its Taylor series representation becomes smaller as more terms are included in the series.

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