ipsec – Creating / Configuring a sec IP tunnel

Hope this isn't too much of a problem to get help, but here we go.

I am trying to create a sec IP tunnel connecting two devices (VM – Linux Centos). Now, by mistake, I thought that IP sec Tunnel meant VPN … so I used openVPN (Application and Protocol) and got it working fine.

The only problem is that they told me it is completely wrong, since they are two different things. VPNs are one thing and IP Sec tunnels are another.

My first question would be … what's the difference? I guess encryption is different, but I'm not sure what else.

The second (and more important at least to me) … is there a step-by-step guide on how to make one? – tunnel vs transport I would take the tunnel, but at this point any step-by-step guide would help. Also, since this must be done on virtual machines (one host is static IP and the others can change) … having a Cisco router guide or something like that would not help me.

My biggest confusion was that when I wrote create ip sec tunnels on youtube / google I get guides for VPN and it's wrong.

By step by step guide I mean:

sudo yum install Xyz or sudo apt-get install Xyz
– insert more Linux commands here, etc …–

Thank you for all of your help!

What is the integral of arctan (x) ^ 3 * sec (x) ^ 4 dx? [closed]

I have trouble calculating this integral, is there any chance that someone can explain how to do it or at least start it?

ssd – SanDisk USB Speed ​​Claims 150 mb / sec

I get very different speeds when I write instead of writing from USB to my desktop.

My desktop is an SSD, class 40 and when I write data from the desktop to USB, the speeds are around 20 mb / s, but when I write from the USB to the desktop, they are at 150 mb / s.

Any idea why that is the case? See both screenshots in the links below. I used the same file type and approximately the same file size.

desktop to USB transfer speed

USB to desktop transfer speed

public key: what are the DER signature and the SEC format?

There are different formats used to encode public keys and signatures in binary (octet sequences). They are defined in the Standards for Efficient Cryptography 1 (SEC).

A public key is a point on an elliptical curve, which consists of a x Y y to coordinate. There must be a standard way to serialize these parts and deserialize them later. The standard defines a compressed and uncompressed format for this serialization. Approximately, the uncompressed format uses the prefix byte 0x04 followed by serialization x coordinated, followed by serialized y to coordinate. The lengths of x Y y They are determined by the curve. For bitcoin, the curve secp256k1 It is always used, so the coordinates are 32 bytes. The resulting uncompressed public key is 65 bytes in total.

The compressed version prefixes the octet flow with 0x02 if the y coordinate is even and 0x03 if the coordinate and is odd. the x The coordinate is then serialized, which results in a 33-byte public key. This information is sufficient because the y coordinate can be retrieved if x It is known and the sign is known (Technically, we do not need the sign either).

The signatures consist of a r value (the x coordinate of a point on the curve) and a s value, which is an integer (32 bytes for secp256k1). The signature can be encoded in only 64 bytes in this compact form.

The previously linked document also provides descriptions of these formats in ASN.1 (abstract syntax notation 1). This is a textual interface description language that has been used to describe interoperability protocols for a long time. Another standard, DER, or Distinguished Coding Rules, describes a process for serializing data in octet streams in accordance with the ASN.1 specification for the data type.

The ASN.1 for signatures is (simplified):

ECDSA-Sig-Value ::= SEQUENCE {
    r INTEGER,
    s INTEGER
}

According to DER rules: a SEQUENCE is of type 0x30 and is followed by a length (1 byte for length <128), then the data for the sequence itself. INTEGER has type type 0x02 followed by its length, followed by the integer in binary. These integers are also signed, which means that if their value is> = 2 ^ 31, a byte with an additional zero value must be added. The resulting DER encoding for a Secp256k1 signature resembles something like this (assuming 32 and r bytes of s):

0x30 0x44 0x02 0x20  0x02 0x20 

As you can see, this is quite wasteful compared to the concise 64-byte serialization of (r, s). The uncompressed version of public keys is also a waste because every byte in a transaction is valuable. The Segwit update to bitcoin introduced some restrictions on what formats can be used, and in addition, the next Taproot update will require 32-byte public keys (only x), which are not part of the SEC standard.

Closed form for the integral $ int _ {- pi / 2} ^ { pi / 2} W ( sec ( varphi)) d varphi $?

I am interested in the integral $$ I = int _ {- pi / 2} ^ { pi / 2} W ( sec ( varphi)) d varphi $$where $ W $ is the product registration function, defined so that $ W (x) e ^ W (x)} = x $. The numerical value for $ I $, as wolfram alpha produced it, is $$ 2.824240767667472 ldots $$

I would love a closed representation of this value, or proof that there is no such representation. I know that the indefinite integral has no solution in terms of standard functions, but I hoped that the defined integral named could simply.

An equivalent integral that can be a little easier (?) Is $$ I = 2 int_ {W (1)} ^ { infty} frac {1 + x} { sqrt {(xe ^ x) ^ 2-1}} $$
that was reached by the replacement of $ xe ^ x = sec ( varphi) $, and converges extremely fast in the area in question.

And that's all I really have. I am not sure how to attack this problem, and I would love to know. At least, I hope you find it interesting to think for a few minutes!

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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) $