Good morning friends,

We define a function of multivariables to be differentable using the norm.
$f: \mathbb{R^n} \rightarrow X, X \subset \mathbb{R^m}$ is differentiable at the point $x_0 \in \mathbb{R^n}$ if
$ lim_{h\rightarrow 0 , h,0\in \mathbb{R^n}} \frac{||f(x_0+h)-f(x_0)-L(h)||}{||h||}= 0 \in \mathbb{R^1}$
If for example $X = \mathbb{R^1} $ then we have in the fraction above a reel number, since both $ f, L$ will be building in the reels. Apparently, as i have seen in some notes, the norm then becomes merly an absolute power. What i do not understand, is how to transform the norm f…

Hello, i am having trouble understanding this rigoriously.
let $f: [a,b]\times[c,d] \rightarrow \mathbb{R} $ continious and let the partial derivative $\frac{\partial f }{\partial y} $ exist and also be continious then we define the function
$ F(y) := \int_a^bf(x,y)dx$ and we want to show that F is differentiable with the derivative $F'(y) = \int_a^b\frac{\partial f}{\partial y}(x,y)dx$
Proving it is written that :

Because $\frac{\partial f}{\partial y}$ has an anti derivative, it is for $y \in [c,d]: \int_c^y\frac{\partial f}{\partial \xi} d\xi= f(x,y)-f(x,c)$ thus $F(y) = \int_a^b[\int_c…

Okay. So, proving that $g$ is entire isn't too bad. Let $f(z) = \sum_{n=0}^{\infty} a_n z^n$ be the power series expansion of $f$ at $z=0$. I think it's true that no matter what branch cut of the square root we take, $(\sqrt{z})^2 = z$ (this is all sort of weird...what does $\sqrt{~~~~}$ actually refer to?). Since $f$ is even, $a_n = 0$ whenever $n$ is odd. Hence,

$$f(z) = \sum_{n=0}^{\infty} a_{2n} (z^2)^n$$
and therefore

$$g(z) = f(\sqrt{z}) = \sum_{n=0}^{\infty} a_{2n} [(\sqrt{z})^2]^n = \sum_{n=0}^{\infty} a_{2n} z^n$$