Let $f(z) = \sum_{j=0}^{\infty}a_jz^j$ be the Maclaurin expansion of a fnction $f(z)$ analytic at the origin. Prove each of the following statements.

Question

Let $f(z) = \sum_{j=0}^{\infty}a_jz^j$ be the Maclaurin expansion of a function $f(z)$ analytic at the origin. Prove each of the following statements.

$(a)$ $\sum_{j=0}^{\infty}a_jz^{2j}$ is the Maclaurin expansion of $g(z) : = f(z^2)$

$(b)$ $\sum_{j=0}^{\infty}a_jz^{m+j}$ is the Maclaurin expansion of $H(z) : = z^mf(z)$

Hint: You may use the uniqueness of Taylor series( the following theorem) when solving the questions above.

Theorem: If $\sum_{j=0}^{\infty}a_j(z-z_0)^{j}$ converges to $f(z)$ in some circular neighborhood of $z_0$ (that is, the radius of its circle of convergence is nonzero), then

$$a_j = \dfrac{f^{(j)}(z_0)}{j!} $$ where $j = 0,1,2,....$

Consequently $\sum_{j=0}^{\infty}a_j(z-z_0)^{j}$ is the Taylor expansion of $f(z)$ around $z_0$

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I don't understand how should I use the Theorem to solve the problem, any help would be appreciated

Answer 1

Hint: The theorem says that if a power series converges to the function, then the power series is uniquely defined. So all you need to do is show that the power series given converge to the functions given, assuming that the original series converges to $f(x)$. Of course you need to show that $f(z^2)$ and $z^m f(z)$ are analytic at $0$ to make the proof air-tight but that shouldn't be hard if you have theorems about making new analytic functions out of old analytic functions.

Answer 2

Hint: First show that the sum converges to the given function in a circular neighborhood of $0$ by using the following equations. $$\sum^\infty_{j=0}a_jz^{2j}=\sum^\infty_{j=0}a_j(z^{2}-0)^{j}$$ And $$\sum^\infty_{j=0}a_jz^{m+j}=\sum^\infty_{j=0}a_jz^{m}z^{j}=z^m\sum^\infty_{j=0}a_jz^{j}$$