How do you prove that every analytic function has a power series representation?

 How do you prove that every analytic function has a power series representation?💝💝

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To prove that every analytic function has a power series representation, we utilize the definition and properties of analytic functions, along with some fundamental results from complex analysis. Here's a step-by-step outline of the proof:

Definitions and Key Concepts

1. Analytic Function: A function $𝑓$ is said to be analytic at a point ${𝑧}_0$ if it is differentiable at ${𝑧}_0$ and in some neighborhood around ${𝑧}_0$.

2. Power Series Representation: We want to show that if $𝑓$ is analytic at ${𝑧}_0$, then there exists a power series

   $$𝑓(𝑧)=\sum _{𝑛=0}^{\mathrm{\infty }}{𝑎}_{𝑛}(𝑧-{𝑧}_0{)}^{𝑛}$$

   that converges to $𝑓(𝑧)$ in some neighborhood around ${𝑧}_0$.

Steps in the Proof

Step 1: Taylor Series Expansion

The Taylor series of a function $𝑓$ at ${𝑧}_0$ is given by:

$$𝑓(𝑧)=\sum _{𝑛=0}^{\mathrm{\infty }}\frac{{𝑓}^{(𝑛)}({𝑧}_0)}{𝑛!}(𝑧-{𝑧}_0{)}^{𝑛}$$

1. that converges to $𝑓(𝑧)$ in some neighborhood around ${𝑧}_0$.

Steps in the Proof

Step 1: Taylor Series Expansion

The Taylor series of a function $𝑓$ at ${𝑧}_0$ is given by:

$$𝑓(𝑧)=\sum _{𝑛=0}^{\mathrm{\infty }}\frac{{𝑓}^{(𝑛)}({𝑧}_0)}{𝑛!}(𝑧-{𝑧}_0{)}^{𝑛}$$

where ${𝑓}^{(𝑛)}({𝑧}_0)$ denotes the $𝑛$-th derivative of $𝑓$ at ${𝑧}_0$.

Step 2: Analyticity and Derivatives

Since $𝑓$ is analytic at ${𝑧}_0$, it is infinitely differentiable at ${𝑧}_0$. Therefore, all the derivatives ${𝑓}^{(𝑛)}({𝑧}_0)$ exist.

Step 3: Existence of the Taylor Series

By the definition of an analytic function, $𝑓$ is not just differentiable at ${𝑧}_0$, but in some neighborhood around ${𝑧}_0$. Hence, we can write the function $𝑓$ as:

$$𝑓(𝑧)=\sum _{𝑛=0}^{\mathrm{\infty }}\frac{{𝑓}^{(𝑛)}({𝑧}_0)}{𝑛!}(𝑧-{𝑧}_0{)}^{𝑛}$$

Step 4: Convergence of the Power Series

The power series representation ${\sum }_{𝑛=0}^{\mathrm{\infty }}\frac{{𝑓}^{(𝑛)}({𝑧}_0)}{𝑛!}(𝑧-{𝑧}_0{)}^{𝑛}$ converges to $𝑓(𝑧)$ in some neighborhood around ${𝑧}_0$. This follows from the fact that $𝑓$ is analytic at ${𝑧}_0$, meaning that $𝑓$ can be represented as a power series in some disk $\mathrm{\mid }𝑧-{𝑧}_0\mathrm{\mid }<𝑅$ where $𝑅$ is the radius of convergence.

Step 5: Uniqueness of the Power Series

The coefficients of the power series are uniquely determined by the derivatives of $𝑓$ at ${𝑧}_0$. Specifically, ${𝑎}_{𝑛}=\frac{{𝑓}^{(𝑛)}({𝑧}_0)}{𝑛!}$. This ensures that the series is not only convergent but also the unique representation of $𝑓$ in the neighborhood around ${𝑧}_0$.

Conclusion

By the definition of analyticity, the fact that the function is infinitely differentiable, and the construction of the Taylor series, we have shown that every analytic function $𝑓$ can be expressed as a power series:

$$𝑓(𝑧)=\sum _{𝑛=0}^{\mathrm{\infty }}{𝑎}_{𝑛}(𝑧-{𝑧}_0{)}^{𝑛}$$

where ${𝑎}_{𝑛}=\frac{{𝑓}^{(𝑛)}({𝑧}_0)}{𝑛!}$. This series converges to $𝑓(𝑧)$ in some neighborhood around ${𝑧}_0$. Thus, every analytic function has a power series representation.🙏🙏