## oracle: strange behavior of pl / sql: the last character is deleted in the Select statement, but the actual data has a different length

We have a table with the name of a column. `A` with type `nvarchar(23)`.

the following query will always return `23` which means that the actual length of all records is `23`.

``````select length(trim(req.A)), count(*)
from tableName req
group by length(trim(req.A));
``````
```|length(trim(req.A))|count(*)|
------------------------------
|23                 |1006    |
```

But when we select from this table with this query, it behaves differently and it seems that the last character is always removed in the Gridview result in the pl / sql developer.

``````select LENGTHB(req.A) lenb, length(req.A) len, req.* from tableName req
where req.A = 'NHBBBBB1398052635902235'; -- Note to the equal sign and the last charactar (5) of the where clause
``````

the result is:

```|lenb|len|          A           |
---------------------------------
|46  |23 |NHBBBBB139805263590223|
```

How can you see the last character (`5`) is deleted in the selected result.

Can you explain what happens? Is this related to pl / sql configurations? How to solve this?

## blocksize – What is the actual size of the Bitcoin block? 2mb or 1mb?

The limit on the block size excluding witness data is 10,000,000 bytes.

There is no limit on the block size, including witness data. Instead, there is a limit on the weight of the block at 40,000,000 units of weight. Depending on the composition of the block, 40,000,000 WU can correspond to anything from 10,000,000 bytes to about 40,000,000 bytes (including witness data).

But a block with 4000000 WU is "full", regardless of whether it is 1 MB or 3.8 MB of data.

## Actual analysis: limit of the function composed in isolated points

I have some doubts about the previous claim. In particular, I am not sure if it is correct if $$b$$ it is an isolated point of the domain of $$f$$.

I have tried to find the counterexample below. Is my counterexample wrong? And if not, how can we solve the previous claim so that it is correct?

Counterexample Define $$f: left {2 right } rightarrow mathbb {R}$$ by $$f left (2 right) = 3$$.

Define $$g: mathbb {R} rightarrow mathbb {R}$$ by $$g left (x right) = x + 1$$

So

• $$lim_ {x rightarrow1} g left (x right) = 2$$
• $$f left (x right)$$ is continuous in $$2$$ (why $$2$$ it is an isolated point of the domain of $$f$$)
• $$f left ( lim_ {x rightarrow1} g left (x right) right) = f left (2 right) = 3$$.

But nevertheless, $$lim_ {x rightarrow1} f left (g left (x right) right)$$ make
does not exist. (The composite function $$f circ g$$ is not defined for any $$x neq 1$$.)

And so, contrary to the previous statement, $$lim_ {x rightarrow1} f left (g left (x right) right) neq f left ( lim_ {x rightarrow1} g left (x right) right). PS$$

## Actual analysis – Explanation of the test: "Each convergent sequence is limited"

I am looking at this test:

Leave $$(a_n) _ {n in mathbb {N}}$$ be a sequence with $$a_n a$$. Choose $$varepsilon = 1$$, then there is a $$N in mathbb {N}$$ such that $$| a_n-a | < varepsilon$$ for all $$n geq N$$.
$$| a_n | = | (a_n-a) + a | leq | a_n-a | + | to | <1+ | to | quad text {for all} n geq N$$

Choose $$A: = max {1+ | to |, | a_1 |, …, | a_ {N-1} | }$$. It turns out that $$| a_n | leq A$$ for all $$n in mathbb {N}$$

I do not understand that $$A: = max {1+ | to |, | a_1 |, …, | a_ {N-1} | }$$ means or why one would choose $$A$$ how that. That is a problem that I always face with different tests.

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## Actual analysis: each interval \$ I subset mathbb {r} \$ is connected. [Proof clarification]

I struggled to understand a part of the next test.

Topological test that each Interval \$ I subset mathbb {R} \$ is connected

Definition: A topological space is connected if, and only if, it can not be divided into two
subsets that are not empty, open and separate, or, similarly, if the empty set and the complete set are the
Only subsets that are open and closed at the same time.

Test. Suppose $$I = A cup B$$ Y $$A cap B = emptyset$$, $$A$$ Y $$B$$ They are not empty and they are open in the subspace-topology of $$I subset mathbb {R}$$. Choose $$a in A$$ Y $$b in B$$ and suppose $$a . Leave $$s: = mathrm {inf} {x in B ~ | ~ a . Then in each neighborhood of $$s$$ there are points of $$B$$ (due to the definition of infimum), but also of $$A$$then if not $$s = a$$, so $$a and the open intervall $$(a, s)$$ It is entirely in $$A$$. Y so $$s$$ it can not be an internal point of $$A$$ neither $$B$$, but this is a contradiction to the property that both $$A$$ Y $$B$$ be open and $$s in A cup B$$.

My problems
I'm struggling with bold parts. Probably because I am more familiar with metric spaces than with topological spaces. I know that $$forall epsilon> 0, exist and in B: s + epsilon> y$$ And this implies that each ball open. $$beta (s, epsilon)$$ contains a point of $$B$$. Furthermore, if $$a = s in A$$ clearly $$beta (s, epsilon)$$ contains a point of $$A$$otherwise, if $$a so $$(a, s) subset A$$ which also means that $$beta (s, epsilon)$$ contain a point $$A$$. If this argument is correct, then my problem is the translation of these ideas of open balls and distances to the topological framework. I have the intuition that he is trying to show that $$s$$ It is a limit point of $$B$$ in $$I$$. As $$A$$ Y $$B$$ coordinated $$I$$, $$B ^ c = A$$. So, $$s$$ can never be (an inner point) either in $$A$$ or in $$B$$.

I am trying to provide a test in the context of a topological space. Can someone clarify my doubts?

## Actual analysis: Prove that \$ vert vert vert vert cdot vert vert vert \$ is equivalent to \$ vert vert cdot vert vert_ {1} \$

Leave $$vert vert vert cdot vert vert vert$$ be a norm in $$ell ^ {1}$$ with the following properties:

$$1.$$ $$( ell ^ {1}, vert vert vert cdot vert vert)$$ is the Banach space

$$2.$$ for all $$x in ell ^ {1}$$: $$vert vert x vert vert _ { infty} leq vert vert vert x vert vert vert$$

Show, using the closed graph theorem that $$vert vert vert cdot vert vert vert$$ It is equivalent to $$vert vert cdot vert vert_ {1}$$.

My idea:

Define $$J: ( ell ^ {1}, vert vert vert cdot vert vert vert) a ( ell ^ {1}, vert vert cdot vert vert_ {1}) , x mapsto x$$

I need to show that $$J$$ It is a closed operator, as $$( ell ^ {1}, vert vert vert cdot vert vert)$$ Y $$( ell ^ {1}, vert vert cdot vert vert_ {1})$$ They are already banach.

Then leave $$(x ^ {n}) n subseteq ( ell ^ {1}, vert vert vert cdot vert vert vert)$$ where $$x ^ {n} xrightarrow {n a infty} x$$ Y $$exists and in ( ell ^ {1}, vert vert cdot vert vert_ {1})$$ so that $$Tx ^ {n} xrightarrow {n to infty} and$$. Now, how can I show that? $$Tx = y$$

And then, how am I going to show that $$( ell ^ {1}, vert vert vert cdot vert vert)$$ is closed?

## Actual analysis – Test: Riemann Sum error is decreasing for a particular function

Leave $$n$$ be a fixed integer and define $$f (x) = frac {1} {n} sum_ {k = 1} ^ {n} left ( frac {(k / n) ^ x-1} {x} right)$$ Y $$g (x) = int_ {0} ^ {1} frac {t ^ x-1} {x} dt$$ for $$x> 0$$.

Test it $$f (x) -g (x)$$ is decreasing in $$x in (0,1)$$ for any fixed $$n.$$

//

My thoughts: Note that $$f (x) = frac {1} {n} sum_ {k = 1} ^ {n} left ( frac {(k / n) ^ x-1} {x} right) rightarrow int_ {0} ^ {1} frac {t ^ x-1} {x} dt$$ as $$n rightarrow infty$$ why $$f (x)$$, for any fixed $$n$$ Y $$x,$$ is the correct sum of Riemann $$h (t) = frac {t ^ x-1} {x}$$ in $$[0,1]$$. Therefore, we wish to show that the difference between a Riemann sum and the integral decreases in $$x in (0,1).$$

As $$h (t)$$ it is negative and it is increasing $$t in (0.1)$$ for any fixed $$x> 0$$, $$f (x) -g (x)$$ it is positive for any $$x> 0$$. In addition, you can see graphically that $$x$$ increases from $$0$$ to $$1$$, $$h (t) = frac {t ^ x-1} {x}$$ vertical contracts towards $$x$$-axis, so it seems intuitive that the error of the Riemann sum decreases.

With these observations out of the way, I could not find any clear way to rigorously prove the desired statement. Of course, we can try to differentiate ourselves. $$f (x) -g (x)$$ and show that the result is always negative in $$(0,1),$$ But the resulting derivative is quite complicated.

Any help or new ideas would be appreciated. =)

## 7 – How to align the views of the pager URL with the actual page number?

When you create pager in views (D7), you get urls with query parameters like `page = X` default. But this X does not match the actual page number.

`page = 1` is generated for `two` Link in the pager and so on.

Is there any way to make them equal without too much Kung Fu code or did I miss any checkbox?

## Actual analysis: if \$ A \$ is a subset of \$ (V, |. |) \$, where \$ V \$ is a vector space. Let \$ overline {A} \$ close. Show \$ overline {x + A} = x + overline {A} \$

Yes $$A$$ it is a subset of $$(V, |. |)$$, where $$V$$ It is a vector space. Leave $$overline {A}$$ denotes closure. Show that yes $$x in V$$ so $$overline {x + A} = x + overline {A}$$.

Tried:

$$( Rightarrow)$$
To consider $$and in overline {x + A} = (x + A) cup (x + A) & # 39;$$ where $$(x + A) & # 39;$$ It is the set of limit points.
$$Rightarrow and in x + A$$ or $$y in (x + A) & # 39;$$

Scenario 1: $$Rightarrow and in x + overline {A}$$ (in which case we have finished).

Scenario 2: $$y in x & # 39;$$ Y $$y in A & # 39;$$.

Continuing with Scenario 2:
As $$y in x & # 39;$$ but $$x & # 39;$$ it's just an element that means $$y = x = overline {x}$$. As $$y in A & # 39;$$ this means $$and in overline {A}$$.

Thus $$y in x + overline {A}$$

$$( Leftarrow)$$

leave $$y in x + overline {A}$$

$$Rightarrow y = x$$ Y $$y in A cup A & # 39;$$

$$Rightarrow and in x + A$$ Y $$y in x + A & # 39;$$

Scenario 1: $$y in x + A$$. If this is the case we have finished because $$x + A subset overline {x + A}$$.

Scenario 2: $$y in x + A & # 39;$$. This means $$y = x$$ Y $$y in A & # 39;$$. Meaning $$y = x = overline {x}$$ Y $$y in A & # 39;$$ medium $$and in overline {A}$$. Which it involves $$y in overline {x + A}$$.

Commentary: I may be thinking too much about something, but I really do not think this is the right way to respond to this problem. I never used the notion of norm that was given to me and I feel that it has a key role to play. Feedback on my approach?