# Official Samsung Galaxy J4 SM-J400M DS Stock Rom

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Official Samsung Galaxy J4 SM-J400M DS Stock Rom

Samsung Galaxy J4 SM-J400M Official Firmware Download/Install/Flash. This phone uses a new version of Android, version 7 (Nougat). It was first officially released on Android 4.0.4 KitKat and later it was updated to Android 6.0.1 Marshmallow.

Galaxy J series

Category:Smartphones
Category:Samsung mobile phones
Category:Android (operating system) devicesQ:

what is the correct power series for a function $f$ defined on the upper half plane such that $f(z) = \frac{\sin z}{z}$?

I am trying to find the series expansion of the function $f(z) = \frac{\sin z}{z}$ at $z = 0$. I used the power series centered at $\infty$ to get the result below.
Is this the right way to proceed? And what would be the right power series for the same function when $z$ is in the upper half plane?
$f(z) = \frac{\sin z}{z}$
Given the relation: $f(z) = \frac{\sin z}{z}$,
we can find that:
$f(z) = \sum_{n = 0}^{\infty} (-1)^n\frac{(z)^{2n+1}}{(2n+1)!}$
$f(z) = \sum_{n = 0}^{\infty} (-1)^n\frac{z^{2n}}{(2n)!}$

A:

Using the $f(z)=\sin z$ power series, we have the following relation:
$$\sin z = \sum_{n=0}^\infty (-1)^n \frac{z^{2n}}{(2n)!}$$
From this, we can use the relation:
$$\frac{\sin z}{z} = -\sum_{n=0}^\infty (-1)^n \frac{z^{2n-1}}{(2n-1)!}$$
This tells us that we should write the following power series for $f$:
f(z) https://biodiversidad.gt/portal/checklists/checklist.php?clid=4293

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