## Maclaurin Series Terms and Convergence or Divergence

This is one of the HW questions I’m trying to solve

$$f(x)=cos(frac{x}{x+1})$$

Find the first four terms of this Maclaurin Series…

I tried using $$sum_{n=0}^infty (-1)^n frac{x^{2n}}{2n!} = cosx$$ & $$sum_{n=0}^infty (-1)^n x^n= frac{1}{1+x}$$ but no luck! I can’t properly substitute the derivative of the series in the Maclaurin Series.

Also related to this question is its second part$$sum_{} (n)^p (cos (frac{1}{n-1})-cos(frac{1}{n})$$
for what values of “P” it converges or diverges.
I would really appreciate it if someone could type in a detailed solution so I can improve my understanding of this topic.

## simplifying expressions – Remove specific terms from Mathematica output

I have a very complicated output from Mathematica – which may run to several pages – and I want to keep only some of the terms.

The expression I am working with is a double summation. It results in (several pages) of output, but many of the terms include variables raised to a power where the power is a function of $$n$$. For example:

``````$$begin{equation} a^4b^2(2a^3+a^7-4abc(6a-9)+a^{2n+6}left(8+9bc-6a^{2n-8}right)) end{equation}$$
``````

In my problem, $$n$$ is very large, such that I want to be able to ignore any individual terms involving this power because they become negligible. The expression above would become:

``````$$begin{equation} a^4 b^2 (2a^3 + a^7 - 4abc(6a-9) ) end{equation}$$
``````

Any assistance gratefully received – I have about 5 pages of output to sift through otherwise!

I have looked at this post

and have tried the delete with function suggested, but
it removes all of the terms in the above example.

Here is a small sample of the Mathematica Output I am trying to simplify:

``````((1 - a)^2 (1 - b)^2 (1 -
c)^2 (a^(4 + 2 n) (-1 + b^2) (b - c)^2 (-1 + b c) (-1 + c^2) +
a^(6 + 2 n) b (-1 + b^2) (b - c)^2 c (-1 + b c) (-1 + c^2) -
a^(5 + 2 n) (-1 + b^2) (b - c)^2 (-1 + c^2) (-c + b^2 c +
b (-1 + c^2)) +
2 a^(2 + n)
b c (-b c^(5 + n) + b^3 c^(5 + n) + b^(5 + n) c (-1 + c^2) +
c^(2 + n) (-1 + c^2) - b^4 c^(2 + n) (-1 + c^2) +
b^(2 + n) (-1 + c^4) - b^(4 + n) (-1 + c^4)) +
2 a^(6 + n)
b c (-b c^(1 + n) + b^3 c^(1 + n) + b^(1 + n) c (-1 + c^2) -
c^(2 + n) (-1 + c^2) + b^4 c^(2 + n) (-1 + c^2) -
b^(2 + n) (-1 + c^4) + b^(4 + n) (-1 + c^4)) +
2 a^(5 +
n) (-c^(3 + n) - b^2 c^(3 + n) + b^4 c^(3 + n) +
b^6 c^(3 + n) + b^(3 + n) (-1 + c^2) (1 + c^2)^2 +
b^(6 + n) c (-1 + c^4) - b c^(2 + n) (-1 + c^4) +
b^5 c^(2 + n) (-1 + c^4) + b^(2 + n) (c - c^5)) -
2 a^(3 +
n) (c^(5 + n) + b^2 c^(5 + n) - b^4 c^(5 + n) - b^6 c^(5 + n) -
b^(5 + n) (-1 + c^2) (1 + c^2)^2 + b^(6 + n) c (-1 + c^4) -
b c^(2 + n) (-1 + c^4) + b^5 c^(2 + n) (-1 + c^4) +
b^(2 + n) (c - c^5)) +
a^3 (-b^(5 + 2 n) (-1 + c^2) (1 + c^2)^2 -
2 b^3 (-1 + (a b)^n) (-1 + c^2) (1 + c^2)^2 +
b^(4 + 2 n) c (-1 + c^4) + b^(6 + 2 n) c (-1 + c^4) -
b c^2 (1 + c^2) (-2 + c^2 + c^(2 + 2 n)) +
b^5 (1 + c^2) (1 - 2 c^2 + c^(4 + 2 n)) +
c^3 (-2 + c^2 + c^(2 + 2 n) + 2 (a c)^n) -
b^6 c (1 + c^(4 + 2 n) + 2 c^2 (-1 + (a c)^n)) +
b^2 c (2 - c^4 + c^(4 + 2 n) + 2 c^2 (-1 + (a c)^n)) -
b^4 c (-1 + 2 c^4 + c^(4 + 2 n) + 2 c^2 (-1 + (a c)^n))) +
a^5 (c^3 - c^(5 + 2 n) + 2 c^5 (a c)^n - 2 b^6 c^5 (a c)^n +
b^(4 + 2 n) c (-1 + c^2) - 2 b^(2 + n) c^(3 + n) (-1 + c^2) +
2 b^(6 + n) c^(3 + n) (-1 + c^2) + b^(5 + 2 n) (-1 + c^4) -
2 b^(3 + n) c^(2 + n) (-1 + c^4) +
2 b^(5 + n) c^(2 + n) (-1 + c^4) +
b^(6 + 2 n) (c + c^3 - 2 c^5) +
b c^2 (-1 + c^2) (1 + c^(2 + 2 n)) +
b^3 (1 + c^2) (1 - 2 c^2 + c^(4 + 2 n)) -
2 b^5 (-(a b)^n - (1 + (a b)^n) c^2 + (a b)^n c^4 + (a b)^
n c^6 + c^(6 + 2 n)) +
b^4 (c - 2 c^3 + c^(5 + 2 n) - 2 c^5 (a c)^n) +
b^2 c (-1 - c^2 + 2 c^4 (1 + (a c)^n))) +
a^6 b c (2 b^(4 + n) c^(2 + n) + 2 b^(2 + n) c^(4 + n) -
2 (b c)^(4 + n) - b^(4 + 2 n) (-1 + c^2) +
b^(5 + 2 n) c (-1 + c^2) - 2 b^5 (a b)^n c (-1 + c^2) +
c^2 (-1 + c^(2 + 2 n)) +
b^3 c (-1 + c^(4 + 2 n) - 2 c^4 (a c)^n) -
b c (-2 + c^2 + c^(4 + 2 n) - 2 c^4 (a c)^n) -
b^2 (1 + c^(4 + 2 n) + 2 c^2 (-1 + (b c)^n))) +
a^4 (c^2 - 2 c^4 + 2 b^(6 + n) c^(2 + n) +
2 b^(5 + n) c^(3 + n) + 2 b^(3 + n) c^(5 + n) +
2 b^(2 + n) c^(6 + n) + c^(4 + 2 n) +
b^(5 + 2 n) c^3 (-1 + c^2) + b^(4 + 2 n) (1 + c^2 - 2 c^4) +
b^(6 + 2 n) c^2 (-1 + c^4) + b c (-2 + c^2 + c^4) -
2 b^4 (1 + c^2 - 3 c^4 + c^(4 + 2 n)) +
b^6 c^2 (-1 + c^(4 + 2 n) - 2 c^4 (b c)^n) +
b^5 (c - 2 c^3 + c^(5 + 2 n) - 2 c^5 (b c)^n) -
b^3 c (-1 + 2 c^4 + c^(4 + 2 n) + 2 c^2 (-1 + (b c)^n)) -
b^2 (-1 + 2 c^4 + c^6 - c^(4 + 2 n) + c^(6 + 2 n) +
2 c^2 (-1 + (b c)^n))) -
a^2 (2 b^(6 + n) c^(2 + n) + 2 b^(5 + n) c^(3 + n) +
2 b^(3 + n) c^(5 + n) + 2 b^(2 + n) c^(6 + n) +
b^(5 + 2 n) (c - c^3) - b^(4 + 2 n) (-1 + c^4) +
b^(6 + 2 n) c^2 (-2 + c^2 + c^4) + c^4 (-1 + c^(2 n)) +
b c^3 (-2 + c^2 + c^(2 + 2 n)) -
2 b^2 c^2 (-3 + (a b)^n - (-1 + (a b)^n) c^2 + c^4 + c^(
4 + 2 n) + (a c)^n + (b c)^n) +
b^6 c^2 (-2 + c^2 + c^(4 + 2 n) - 2 c^4 (b c)^n) +
b^4 (-1 + 2 c^4 + c^6 - c^(4 + 2 n) + c^(6 + 2 n) +
2 c^2 (-1 + (a c)^n)) -
b^3 c (2 - c^4 + c^(4 + 2 n) + 2 c^2 (-1 + (b c)^n)) +
b^5 (c + c^3 - 2 c^5 (1 + (b c)^n))) -
b^2 c^2 (-b^(2 + 2 n) (-1 + c^2) + b^(3 + 2 n) c (-1 + c^2) +
c^2 (-1 + c^(2 n)) - b^2 (1 - 2 c^2 + c^(2 + 2 n)) +
b^3 c (-1 + c^(2 + 2 n) + 2 (b c)^n - 2 c^2 (b c)^n) +
b c (2 - c^(2 + 2 n) - 2 (b c)^n + c^2 (-1 + 2 (b c)^n))) +
a b c (2 b^(2 + n) c^(5 + n) - 2 b^(4 + n) c^(5 + n) +
b^(5 + 2 n) c^2 (-1 + c^2) -
2 b^(5 + n) c^(2 + n) (-1 + c^2) + b^(4 + 2 n) c (-1 + c^4) +
b^3 (-2 + c^2 + c^4) - b^(3 + 2 n) (-2 + c^2 + c^4) +
2 c^3 (-1 + c^(2 n)) + b^5 c^2 (-1 + c^(2 + 2 n)) +
b^4 c (-1 + c^2 - c^(2 + 2 n) + c^(4 + 2 n) + 2 (b c)^n) -
b^2 c (-2 - c^2 + c^4 + c^(2 + 2 n) + c^(4 + 2 n) +
2 (b c)^n) +
b c^2 (2 - c^(2 + 2 n) - 2 (b c)^n +
c^2 (-1 + 2 (b c)^n)))))/((-1 + a^2) (a - b)^2 (-1 +
a b) (-1 + b^2) (a - c)^2 (b - c)^2 (-1 + a c) (-1 + b c) (-1 +
c^2))
``````

## Sort terms of attributes by date modified or as per user selection, in product backed of woocommerce

Sort terms of attributes by date modified or as per user selection, Right now it is sorted alphabetically when we select any terms from the box in a product backed of woocommerce

## Consider 3 sequence 2n,3n,5n make combination of these 3 sequence into y sequence.then how find terms in increasing weight of number?

2n=2,4,6,8,10,12,14,16,…
3n=3,6,9,12,15,18,21,27,..
5n=5,10,15,20,25,30,35,40,45,..
Combination of sequence-
(2n,3n,5n)=y=2,3,4,5,6,8,9,10,12,14,15,16,18,20,21,22,24,25,…
How find terms in y in increasing order(weight)?
But we easily find combination of 2 sequence. let combination of (2n,3n)then terms in that seq. (2n,3n,2n,3n,2n,3n,2n,3n,….)but problem in combination of more than 2 sequence.is there formula?

## database design – Representing credit terms with single-entry accounting?

I have a very simple `transactions` table which represents single-entry accounting in my app. Accounting in my app is not complicated, and I don’t want to go the full dual-entry bookkeeping route, which I’ve done before and is overkill for this specific app.

The table has a `type` enum (`DEBIT`/`CREDIT`), and an `amount` column.

The table represents the clients “balance”. They can currently only pre-pay:

``````| id | description    | type   | amount |
|----+----------------+--------+--------|
| 1  | pre-payment    | CREDIT | 100.00 |
| 2  | signup bonus   | CREDIT | 25.00  |
| 3  | campaign spend | DEBIT  | 110.00 |
``````

Let’s say I want to allow “credit terms” where people can “post pay” after their campaign has run.

Is there a simple way to represent credit using single-entry accounting? I understand that’s one of the great benefits of dual-entry, but I’m hoping there is a somewhat established simple way. These transactions are only a few per month so I don’t need an overly cumbersome dual-entry system.

The problem is that…

``````CREDIT  \$100,   credit-deposit
DEBIT   \$99.95  campaign-spend
???     \$99.95  payment
``````

I guess the problem is that I need to credit their balance so their account shows a positive balance (so they can spend it). But when their account hits \$0 I also need to collect the actual money which is a transaction in itself.

I’m thinking maybe I can somehow have an ‘accounts_recievable’ boolean? And then I can tally that separately?

## taxonomy terms – Is there anyway create relationship for filter criteria which has vocabularies( state, city and location) in views – Drupal 8

list of available Vocabularies in my content type.

• Country(Term: US)
• State (Terms: CA , CO, DE.. etc)
• City (Terms:Fremont, Irwin, Los Angeles, Monument etc..)
• Location(Terms: All loacations)

I’m trying to create a view for published content result which uses above vocabularies. I need to expose filters to the users where I added above vocabulary fields. How to create a relationship to select particular state->city-> location
(For ex: State(CA)->City(Fremont)->location(xxx-fremont))

## taxonomy terms – How to create and save users in a loop?

Hello i want to create or update users in a loop. Updating is no problem with user_load but i can’t create user with taxonomy field i think my user_save statement is wrong.

`````` // Create new User
if(empty(\$uid)){
\$new_user= array(
'name' => 'wirdfit'.\$row->id,
'mail' => 'wirdfit_'.\$row->id.'@wirdfit.lbt',
'status' => 1,
'access' => REQUEST_TIME,
'roles' => array(),
'is_new' => TRUE,
// Set Custom field
'field_id' => array(LANGUAGE_NONE => array(array('value' => \$row->id))),
'field_type' => array(LANGUAGE_NONE => array(array('value'=>wirdfit)))
);
// Set Taxonomy
\$types = explode(',', \$row->zustzliche_suchkriterien);
for ( \$i = 0; \$i < count (\$types); \$i++){
\$terms = taxonomy_get_term_by_name(\$types(\$i));
\$tid = key(\$terms);
if (\$tid){
\$edit = array(
'field_zusaetzliche_suchkriterien' => array(
'und' => array(
\$i => array(
'tid' => \$tid))));
}
}
user_save(null, \$new_user, \$edit);
}
``````

\$new_user field zusaetzliche_suchkriterien empty
\$edit not empty
is user_save statement wrong?
user_save(null, \$new_user, \$edit);

## bitcoin core – In realistic and practical terms, how exactly should we solve the mega problem of sky-high and slow transaction fees?

I just paid \$20 in transaction fee to move 0.1 BTC from one wallet.dat to another on my very same Bitcoin Core installation.

It hurts every time. I feel like I’ve wasted a damn fortune over the years in just transaction fees.

I’ve just recently come to the horrible realization that nobody is going to use my service with these large (and slow) transactions, and especially now that few are willing to part from any of their satoshis to actually buy stuff and services with their digital gold.

I’ve tried to get Lightning set up, but it’s incomprehensible to me. I neither understand how to send it nor how to accept it in my (custom) payment system which already supports Bitcoin.

I just can’t seem to find the energy to research and implement Lightning. It’s as if I have already wasted all of my “energy container”, so to speak.

It doesn’t help to hear about these various “sidechains” being worked on. I need something concrete which is actually supported and trusted — not vague promises of future improvements.

Could somebody offer some kind of promising development such as a new feature in Bitcoin itself (thus supported by Bitcoin Core and the other big wallets/clients) which obsoletes all these “side-projects” and somehow enables rapid and near-free micro transactions without involving weird “solutions”?

Or is this fundamentally impossible using Bitcoin? It has to be both this slow and expensive? Because if so, and if they keep making all these stupid “altcoins” which nobody is ever going to use when there’s almost as many of them as human beings on this planet, we’re never going to get anywhere.

I don’t care if somecoin has instant payments and zero fees, if nobody uses and supports it, and it’s being run by some sketchy scammer as a centralized project. I truly don’t understand what they are waiting for, the ones with all the technical skills to make this happen. Why was this not part of Bitcoin from the start?

Don’t they realize that being able to send literally 1 satoshi to somebody for free and instantly would change everything? I fear every single time I have to pay using Bitcoin, because it feels like half my stash is gone instantly due to the damn fees.

## taxonomy terms – How do I manage breadcrumbs when the node uses multiple vocabularies?

I have a blog in Drupal 8 and it uses the Categories vocabulary to display breadcrumbs as Home > Blog > (Category) > Post title.

I added taxonomies from the Tags vocabulary, which means every post can have one category, and multiple tags. Since I added the Tags vocabulary, it has taken over the breadcrumbs on individual blog posts, which are now Home > Blog > (Tag) > Post title.

Is it possible to control which vocabulary should have priority when it comes to displaying breadcrumbs?
Is it possible to control which vocabulary is displayed on blog posts based on where the user came from? For example, if the user is on the individual Tag page (Home > Blog > (Tag)) and clicks on a blog post belonging to that tag, the blog post should display the breadcrumbs using the Tags vocabulary (Home > Blog > (Tag) > Post title). However, if the user is on the individual Category page (Home > Blog > (Category)) and clicks on a blog post belonging to that category, the blog post should display the breadcrumbs using the Category vocabulary (Home > Blog > (Category) > Post title).

## ordinary differential equations – Formula for the derivative of finite power series in reversed order of terms.

I wanted to solve the polar part in Schrödinger’s wave equation for the H-atom by direct substitution of functions of form:-
$$Theta_{lm}(theta) = a_{lm} sin^{|m|}theta sum_{r≥0}^{r≤(l-|m|)/2}(-1)^rb_r cos^{l-|m|-2r}theta$$
The $$a$$‘s are normalisation constants, no problem there. However, the problem of determining the $$b$$‘s ultimately drops down to finding the first and second derivatives of the polynomial in $$z=cos theta$$:
$$P(z)=sum_{r≥0}^{r≤(l-|m|)/2} (-1)^rb_r z^{l-|m|-2r}$$
Which is a finite power series written in decreasing order of powers. I couldn’t find a formula for so (well sometimes I get that dumb), but I think it does exist, maybe some reference book or website. I emphasize that what I’m doing is right the reverse of Frobenius-method. Thanks.