Is there any equivalent of Taylor/Maclaurin series of $ln(1+x)$ for $|x| > 1$

I happened to come across Taylor series and Maclaurin series recently, but everywhere I read about the expansion for $ ln(1+x)$ , it was stated that the approximation is valid for $ -1 < x < 1$ .

I understand that the bounds for $ x$ are because the series doesn’t converge for $ |x|>1$ , but is there any equivalent of this series for the value of $ |x|$ as greater than $ 1$ ?

Please note that I am not asking if we can compute for $ |x|>1$ or not, as that can be done by computing for $ \frac 1x$ , which will then lie between $ -1$ and $ 1$ .

Also, I’m quite new to all this, so new that today was the day I read the name ‘Maclaurin’ for the first time. So any answers understandable with high school mathematics are highly appreciated.

Thanks!

$M \leq X$. Is it true that $M^*$ is a subspace of $X^*$ where $*$ denotes the dual space?

Let $ X$ be a Banach space and $ M$ be a closed subspace i.e. $ M \leq X$ .

Is it true that $ M^*$ is a subspace of $ X^*$ where $ *$ denotes the dual space?

What I have tried:

I know by Hahn Banach theorem that every $ f \in M^*$ has an extension $ \hat f \in X^*$ such that $ \hat f |_M =f$ and the operator norm doesn’t change i.e. $ \|\hat f\| =\|f\|$ .

I found that such an extension can be chosen to be linear if $ X$ is a Hilbert space here, hence my claim might not hold in the general case.

Any help is appreciated.

Find harmonic of order $3$ for $2\pi$ periodic funciton $f(x) = \pi -|x|$


Let $ f$ be a $ 2\pi$ periodic function such that $ f(x) = \pi -|x|$ in $ [-\pi,\pi]$ . Find its harmonic of order $ 3$

I’m learning about orthogonal families of polynmials and discrete polynomial approximations by fourier functions. I’ve came upon this exercise, but I don’t know what an harmonic of order $ 3$ means.

What will be $\lim\limits_{x \rightarrow 0^+} \left ( 2 \sin {\frac {1} {x}} + \sqrt x \sin {\frac {1} {x}} \right )^x.$

Evaluate

$ $ \lim\limits_{x \rightarrow 0^+} \left ( 2 \sin {\frac {1} {x}} + \sqrt x \sin {\frac {1} {x}} \right )^x.$ $

I tried by taking log but it wouldn’t work because there are infinitely many points in any neighbourhood of $ 0$ where $ \ln \left ( \sin {\frac {1} {x}} \right )$ doesn’t exist. How to overcome this situation?

Any help will be highly appreciated.Thank you very much for your valuable time.

How to update the apt-check count of “X packages can be updated” message

Is there a command to update the apt-check results so I can use bash script to check them?

I run upgrade and it shows 0 but apt-check still shows 12 packages can be updated. How often is this suppose to update on its own? I have numerous servers. Some seem to update instantly. Others go days still showing updates are available when they aren’t.

I have read that rebooting can reset it but thats silly. Why would you need to reboot to refresh a simple message especially when most updates do NOT require a reboot.

# apt-get update && apt-get upgrade 0 upgraded, 0 newly installed, 0 to remove # /usr/lib/update-notifier/apt-check --human-readable 12 packages can be updated. 9 updates are security updates.