calculation – What's wrong with my solution at $ int sec ^ 2 x tan x , dx $?

I tried to find a generic antiderivative for

$$ displaystyle int sec ^ 2x tan x mathop {dx} $$

But I think there is something wrong with my solution because it does not match what I got through an online calculator.

What am I doing wrong?

Below is my solution.

We will use the substitution:

$$ u = sec x qquad du = sec x tan x , dx $$

We substitute and apply the power rule:

$$ int ( sec x) ( sec x tan x , dx) = int u , du = frac {1} {2} u ^ 2 + C = frac { sec ^ 2x} {2} + C $$

The solution I found with the online calculator is:

$$ frac { tan ^ 2 x} {2} + C $$

The steps in the online solution also make sense, so I'm not sure what's going on.

The only thing I have doubts about is whether I derived the $ du $ since $ u = sec x $ correctly. But it seems fine. I used implicit differentiation with $ x $.

Test verification of sec. 24 "Topology" of Munkres, the long line can not be embedded in real

I'm doing this exercise in the topology of Munkres (~ paraphrasing). I have two motivations for reading this book: 1, the topology is really pretty, but the main one, 2, is that I'm not yet an undergraduate student; therefore, I have never taken any kind of "evidence-based" course and would like to learn how to write legible tests. Appreciate:

  1. A verification of my test.
  2. Any advice on how to make my test more readable.

Remember that $ S _ { Omega} $ Denotes the minimum set ordered, uncountable. Let L denote the ordered set $ S _ { Omega} times[01)$[01)$[01)$[01)$ in the order of the dictionary, with its smallest element removed. The set L is a classic example in topology called Long line.

Theorem: the long line is connected to the route and locally homeomorphic to $ mathbb {R} $, but it can not be embedded in $ mathbb {R} $.

The structure of the test is given by Munkres.

(a) Let X be an ordered set; leave $ a <b <c $ Be X points. Show that $[ac)$[ac)$[ac)$[ac)$ has the order type of $[01)$[01)$[01)$[01)$ Yes, both $[ab)$[ab)$[ab)$[ab)$ Y $[bc)$[bc)$[bc)$[bc)$ have the order type of $[01)$[01)$[01)$[01)$.

Assume $[ab)$[ab)$[ab)$[ab)$ Y $[bc)$[bc)$[bc)$[bc)$ have the order type of $[01)$[01)$[01)$[01)$. Then there are order isomorphisms. $ f:[ab)rightarrow[0frac{1}{2})$[ab)rightarrow[0frac{1}{2})$[ab)rightarrow[0frac{1}{2})$[ab)rightarrow[0frac{1}{2})$ Y $ g:[bc)rightarrow[0frac{1}{2})$[bc)rightarrow[0frac{1}{2})$[bc)rightarrow[0frac{1}{2})$[bc)rightarrow[0frac{1}{2})$, why $[01)$[01)$[01)$[01)$ has the same kind of order that $[01/2)$[01/2)$[01/2)$[01/2)$. Define $ h:[ac)rightarrow[01)$[ac)rightarrow[01)$[ac)rightarrow[01)$[ac)rightarrow[01)$ as follows:

$$ h (x) = begin {cases}
f (x) text {if} x in[a, b) \
g (x) text {if} x in[BC)
end {cases} $$

$ h $ it's an order-isomorphism, like $ p <q $ it implies $ h (p) <h (q) $.

Conversely, it assumes $[ac)$[ac)$[ac)$[ac)$ has the order type of $[01)$[01)$[01)$[01)$. Then there is an order-isomorphism. $ f:[ac)rightarrow[01)$[ac)rightarrow[01)$[ac)rightarrow[01)$[ac)rightarrow[01)$. Restricting the domain to $[ab)$[ab)$[ab)$[ab)$ gives a new order-isomorphism $ g:[ab)rightarrow[0d)$[ab)rightarrow[0d)$[ab)rightarrow[0d)$[ab)rightarrow[0d)$, where $ d in (0.1) $. $[0d)$[0d)$[0d)$[0d)$ is of the order type of $[01)$[01)$[01)$[01)$; multiplying by $ 1 / d $ It is an order of isomorphism. Similarly, restricting the domain of $ f $ to $[bc)$[bc)$[bc)$[bc)$ gives order-isomorphism $ h:[bc)rightarrow[d1)$[bc)rightarrow[d1)$[bc)rightarrow[d1)$[bc)rightarrow[d1)$, where $ d in (0.1) $.

(b) Let X be an ordered set. Leave $ x_0 <x_1 <… $ Be a growing sequence of points of X; suppose $ b = sup {x_i } $. Show that $[x_0b)$[x_0b)$[x_0b)$[x_0b)$ has the order type of $[01)$[01)$[01)$[01)$ if each interval $[x_ix_i+1)$[x_ix_i+1)$[x_ix_i+1)$[x_ix_i+1)$ has the order type of $[01)$[01)$[01)$[01)$.

This is simple, having already tried. $ (a) $. Begin with $ i = 1 $: $[x_0x_1)[x_1b)$[x_0x_1)[x_1b)$[x_0x_1)[x_1b)$[x_0x_1)[x_1b)$ It is a partition of a set in the same way that we saw in $ (a) $; therefore, each must have the same type of order that $[01)$[01)$[01)$[01)$. We proceed by induction, considering now the whole. $[x_1b)$[x_1b)$[x_1b)$[x_1b)$ and the sequence $ x_1 <x_2 <… $ The test is the same.

(c) Leave $ a_0 $ denotes the smallest element of $ S _ { Omega} $. For each element $ a $ of $ S _ { Omega} $ different from $ a_0 $, shows that the interval $[a_0times0atimes0)$[a_0times0atimes0)$[a_0times0atimes0)$[a_0times0atimes0)$ of $ L $ has the order type of $[01)$[01)$[01)$[01)$.

The[[Innuendo: Proceed by transfinite induction. Either has an immediate predecessor in $ S _ { Omega} $, or there is a growing sequence $ a_i $ in $ S _ { Omega} $ with $ a = sup {a_i } $]

How can I order this ?: $[a_0a)$[a_0a)$[a_0a)$[a_0a)$ is accountable by the definition of $ S _ { Omega} $. Following the trail: yes $ a $ it has an immediate predecessor, $ b $, so $[btimes0atimes0)$[btimes0atimes0)$[btimes0atimes0)$[btimes0atimes0)$ has the order type of $[01)$[01)$[01)$[01)$ (All points in this set are of the form $ b times x $, where $ x en[01)$[01)$[01)$[01)$).

By induction backwards. $ a_0 $ any finite section of $ S _ { Omega} $of which $ a $ is the largest element, is the isomorphic order for $[01)$[01)$[01)$[01)$.

Yes $ a $ is not the largest element of a finite section of $ S _ { Omega} $, then there is a growing sequence $ a_i $ in $ S _ { Omega} $ with $ a = sup {a_i } $, this sequence is exactly every element of $ S _ { Omega} $ strictly less than $ a $, in order (which must be countable). Every $ a_ {i + 1} $ In this sequence has an immediate predecessor, namely: $ a_1 $, then each $[a_{i}times0a_{i+1}times0)$[a_{i}times0a_{i+1}times0)$[a_{i}times0a_{i+1}times0)$[a_{i}times0a_{i+1}times0)$ must be isomorphic order to $[01)$[01)$[01)$[01)$ for the previous argument. Then for $ (b) $, then you must $[a_0times0atimes0)$[a_0times0atimes0)$[a_0times0atimes0)$[a_0times0atimes0)$.

(d) Show that $ L $ The road is connected.

Extensible $ (c) $ to include sets of the form $[ab)abenL$[ab)abinL$[ab)abenL$[ab)abinL$ It's trivial – just add / subtract the sets of the ends – I'll assume this. This implies that $[a,b]$ has the same kind of order that $[0,1]$. So between two points $ a, b in L $, there is an order-isomorphism $ f:[0,1] correct arrow [a,b]$; which is necessarily a homeomorphism and therefore a continuous map.

Is that last step valid? How can I make that clearer?

(e) Show that each point of $ L $ It has a homeomorphic neighborhood with an open interval in. $ mathbb {R} $.

Leave $ x $ be our point, assume $ x neq a_0 $. Then there are some points, $ a, b in L $, such that $ a <x <b $. By $ (d) $, $ L $ The path is connected, so there is some homeomorphism. $ f:[c,d] correct arrow [a,b]$ such that $ f (c) = a $, $ f (d) = (b) $. Then, restricting the range of $ f $ to $ (c, d) $ we get a homeomorphism $ f & # 39; 🙁 c, d) rightarrow (a, b) $.

And finally…

(f) Show that L can not be embedded in $ mathbb {R} $or indeed in $ mathbb {R} ^ n $ For any $ n $.

The[[Innuendo: any subspace of $ mathbb {R} ^ n $ It has an accounting basis for its topology.

Suppose there was an imbedding $ f: L rightarrow mathbb {R} ^ n $, therefore a homeomorphism $ f & # 39 ;: L rightarrow Y $, $ Y subset mathbb {R} ^ n $ obtained by restricting the range of $ f $. $ Y $ It must have an accounting basis because $ mathbb {R} ^ n $ is metrizable, so $ L $ It must also have an accounting base for its topology, being homeomorphic with $ Y $.

Assume $ L $ I had an accounting basis, call this base $ U_ {i} $ where $ i in mathbb {Z} ^ {+} $. Each open set of $ L $ It must contain, in its entirety, at least one. $ U_i $. Leave $ x _ { alpha} $ Be the collection of all the points of the form. $ y times 0 $ where $ y in S $ {Omega} $. $ x _ { alpha} $ It is uncountable, being indexed by $ S _ { Omega} $. Then for (e), there is a countless collection $ V _ { alpha} $ of neighborhoods around $ x _ { alpha} $. Keep in mind that we can choose, with more force, $ V _ { alpha} $ in such a way that it is paired-disunion.

That last step feels neglected; Any suggestions?

Then each contains some $ U_ {i} $; So, there is some injective function. $ f: S _ { Omega} rightarrow mathbb {Z} ^ {+} $ defined by $ f ( alpha) = i $ as long as $ U_i subset V _ { alpha} $. This contradicts that $ S _ { Omega} $ it is countless

$ therefore $

mysql – error in the update of the index entry sec in

My environment:

# cat / etc / redhat-release
Red Hat Enterprise Linux Server version 6.10 (Santiago)
# rpm -q mysql-server
mysql-server-5.1.73-8.el6_8.x86_64
# 

/var/log/mysqld.log:

InnoDB: error in the update of the index entry sec.
InnoDB: index `totalcount` of the table` pressflow``node_counter`
InnoDB: tuple DATA TUPLE: 2 fields;
0: len 8; hex 000000000001c3b4; asc ;; 1: len 4; hex 80001a2f; asc / ;;

InnoDB: record PHYSICAL RECORD: n_fields 2; compact format; information bits 0
0: len 8; hex 000000000001c3b3; asc ;; 1: len 4; hex 80001a2f; asc / ;;

TRANSACTION 0 2174487909, ACTIVE 0 sec, process no 2253, update of the 140295045822208 operating system thread update or deletion, the thread was declared within InnoDB 499
mysql tables in use 1, locked 1
2 locking structures, stack size 368, 1 row lock (s), undo registry entries 1
MySQL thread ID 2770, query ID 1362014 X.X.X X.X.X.X X Updating
UPDATE node_counter SET daycount = daycount + 1, totalcount = totalcount + 1, timestamp = 1559172381 WHERE nid = 6703

InnoDB: send a detailed error report to http://bugs.mysql.com

Please advise

Show the inequality $ sec x ge 1+ (x ^ 2/2) $

Test inequality

$ sec (x) ge 1+ (x ^ 2/2) $

How do I distinguish the meaning of the SQL Performance Counters as Transactions / sec. And Errors / sec? [duplicate]

This question already has an answer here:

  • Batch requests / sec. Reported by the DMV millions of times larger than the Activity Monitor

    3 answers

I have been storing a lot of SQL performance counters in the last 10 days or so. I am looking to make a dashboard in the health of the server with this data. My question arises when I start looking at my meters that involve a "metric" / sec (Transactions / sec., Errors / sec., Etc.).

According to my understanding, the name implies that these counters should show me an aggregated statistic according to the time consulted on how many errors or transactions occur per second. Everything I have read online about these performance counters seems to affirm this description. However, you will notice in the image below that the number is high (423,683 transactions / sec), and it grows significantly every day. It seems to me that you are adding the transaction counter per second to the previous sums, if that makes sense. Is this the expected behavior? If so, to get the value I'm looking for, do I need to subtract the days to get the Transactions / sec for that day?

Transactions / sec For a single database registered for many days.

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Please let me know which is the most flexible, profitable and safe. How do I select the best option for my requirement?

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Trigonometric equation including sec

What are the values ​​of x in degrees when:
$$
sec ^ 2 (x) -2 sqrt {2} sec (x) + 2 = 0
$$

I think you should use derivatives of $ sec (x) $ but I'm fighting.
Any help would be appreciated.