real analysis – Show that a solution remains in $ ( Bbb R ^ +) ^ 2 $

We consider the following differential system:

$$
begin {align}
dfrac {dx} {dt} & = (1-x) – dfrac {xy} {1 + x} \
dfrac {dy} {dt} & = ( dfrac {2xy} {1 + x} – 1) times y
end {align}
$$

For $ t geq 0 $ and $ (x (0), y (0)) in ( Bbb R +) ^ 2 $

CA watch $ (x (t), y (t)) in ( Bbb R +) ^ 2 quad forall t> 0 $.

Let's assume the existence of $ t> $ 0 such as $ (x (t), y (t)) notin ( Bbb R +) ^ 2 $
Then there are two cases:

  • (1): the path cuts the axis $ x = $ 0.
    Then there is $ t_0 $ such as $ x (t_0) = $ 0 and $ x (t) <0 $ for $ t in) t_0, t_0 + epsilon ($ for a certain $ epsilon $. L & # 39; writing $ x (t) = (t-t_0) dfrac {dx} {dt} (t_0) + o (t-t_0) = (t-t_0) + o (t-t_0) $ we get a contradiction because $ t-t_0> $ 0 and $ x (t) <0 $.
  • (2): the path cuts the axis $ y = 0 $. That's where I do not know how to conclude. We also have $ t_0 $ such as $ y (t_0) = 0 $. Because $ dfrac {dy} {dt} = ( dfrac {2xy} {1 + x} – 1) times y $, it is to see, by induction, that $ dfrac {d ^ ky} {dt ^ k} (t_0) = 0 quad forall k in Bbb N ^ * $. But that does not mean that the function must be zero on a neighborhood of $ t_0 $, does he?

On the second point above. I do not know where the contradiction is.

Singapore lowers its leverage ceiling – News & Analysis

Singapore has lowered its leverage ceiling by more than half. That means the country's forex traders will now have access to leverage of 1:20, instead of 1:50 as it was before the changes.

The decision was made by the Monetary Authority of Singapore. However, given the regulatory changes made by the European Securities and Markets Authority, certain shortcomings allow for a more generous leverage under strict conditions.

Investors accredited in Singapore will have access to the initial leverage. However, as in Europe, this would involve meeting strict capital requirements.

Among the requirements, one needs a personal net worth of more than 2 million Singaporean dollars (1.5 million dollars) or proof of an annual income of more than 300 000 Singapore dollars. Traders with more than a million Singapore dollars in cash can also benefit from greater leverage.

calculation and analysis – The resolution command does not solve this equation!

I've tried to solve the following equation with Mathematica:

$ left (2-x ^ 2 right) left (n left (x ^ 4-x ^ 2 + 1 right) – pi left (x ^ 2-2 right) right) sinh ( pi x) cosh (nx) + sinh (nx) left ( left (2-x ^ 2 right) left ( pi left (x ^ 4-x ^ 2 + 1 right) -n left (x ^ 2-2 right) right) cosh ( pi x) -x left (x ^ 4 + 2 x ^ 2 right) sinh ( pi x) right) = $ 0

but the answer is:
"This system can not be solved with the methods available to solve."

I've also tried Maple, the result has been a long relationship in terms of RootOf.
How can I get an explicit solution for $ x $ in terms of $ n $?

complex analysis – Bilinear transformation $ f (z) = 9 e ^ {i theta} frac {(z – 1)} {(z – 9)}, theta in [ 0,2 pi ] $

begin {array} {l} { text {Consider the bilinear transformation} f (z) = 9 e ^ {i theta} frac {(z – 1)} {(z – 9)}, theta in (0,2 pi)} \ { text {then which of the following statements is / are correct}} \ { text {(a)} | f left (3 e ^ {i theta} right) | = 3, forall theta in (0,2 pi)} \ { text {(b)} | f left (3 e ^ {i theta} right) | = 1, forall theta in (0,2 pi)} \ { text {(c)} | f left (3 e ^ {i theta} right) | = 2, forall theta in (0,2 pi)} \ { text {(d)} f (z) text {is consistent on} mathbb {C} – {9 }} end {array}
begin {array} {l} { textbf {My attempt: -}} \ { text {(d) Correct, since f (z) is a mobious transformation}}} \ { text {(c) and (b) Can be eliminated by taking $ theta = 2 pi $}} \ { text {It is clear that f (z) is a rotation of} theta text {degrees}} \ { text {and dialing by} 9 text {, So how} | f left (3 e ^ {i theta} right) | = 3? } end {array}

actual analysis – almost complete proof that $ int_A f_n to int_A f $

This problem comes from Bass. I have a problem with my proof that I could not solve it. It may not even be the right way to solve it, but that's what I proposed.

Right here, $ int $ is equivalent to $ int_ mathbb {R} $

Problem:

Let $ (X, mathcal {A}, mu) $ to be a measurement space. Suppose you have the functions $ f_n $ and $ f $ which are integrable and not negative (for all $ n $ in the case of $ f_n $). In addition, suppose that $ f_n to f $ almost everywhere, and $ int f_n to int f $. Prove that for every $ A in mathcal {A} $.

$$ int_Af_n mathrm {d} mu to int_A f mathrm {d} mu $$

My attempt:

We can rewrite the integral of $ f_n $ as

$$ int_A f_n mathrm {d} mu = int f_n chi_A mathrm {d} mu $$
or $ chi_A $ is the indicator function for the whole $ A $.

For everyone $ n $, define the functions

$$ g_n (x): = f_n (x) chi_A (x) =
begin {cases}
f_n (x), & x in A \
0, & x notin A
end {cases}
$$

For each $ n $, $ g_n (x) geq 0 $, so $ | g_n (x) | = g_n (x) leq f_n (x) $

Now we get the limit

$$ lim_ {n to infty} g_n = lim_ {n to infty} (f_n (x) chi_A (x)) = chi_A (x) lim_ {n to infty} f_n ( x) = f (x) chi_A (x) $$ for everyone $ x $ such as $ f_n (x) to f (x) $.

Since $ f_n to f $ almost everywhere, it follows that $ g_n to g = f chi_A $ almost everywhere.

Moreover, by hypothesis, since each $ f_n $ is integrable, we get

$$ g_n leq f_n text {a.e.} Rightarrow | g_n | leq | f_n | text {a.e.} Rightarrow int | g_n | mathrm {d} mu leq int | f_n | mathrm {d} mu < infty $$

So for all $ n $ we have that $ g_n $ is integrable.

Functions $ g_n $ are measurable since the function of the indicator and $ f_n $ are measurable. Therefore, the multiplication of the two functions is also measurable.

Therefore, we found that $ g_n $ are measurable, and that $ g_n to g $ almost everywhere. It means that I have almost all conditions (without wanting to play pun) of the conditions required to use the dominated convergence theorem. I miss finding an embeddable feature $ h: X to (0, infty) $ such as, $ | g_n (x) | leq h (x) $ A.E.

The closest I got to this was the previous statement in this answer that $ | g_n (x) | = g_n (x) leq f_n (x) $ A.E.

I do not know how to find a function that absolutely links $ g_n $ and does not depend on $ n $and, since we have not assumed $ f_1 leq f_2 leq … $ I can not say that $ f_n leq f $ which would solve this problem. What do I miss?

(Finally, I will finish the proof assuming that I found such a function $ h $ this allows me to use the DCT)

If we invoke the DCT, we have that

$$ lim_ {n to infty} int_A f_n mathrm {d} mu = lim_ {n to infty} int f_n chi_A mathrm {d} mu = lim_ {n to infty} int gnn mathrm {d} mu = int lim_ {n to infty} g_n mathrm {d} mu = int g mathrm {d} mu = int f chi_A mathrm {d} mu = int_A f mathrm {d} mu $$

calculation and analysis – The defined integral of a fractional polynomial contains sin (x) and x ^ n

I meet an integral defined as $ int _ {- infty} ^ {+ infty} frac { sin ( text {x0} omega) – sin (x omega)} {( sin (x omega) – sin ( text {x0} omega)) ^ 2+ (x- text {x0}) ^ 2} , dx $so i'm trying

Integrate((- Sin(x (Omega)) + 
    Sin(x0 (Omega)))/((x - x0)^2 + (Sin(x (Omega)) - 
     Sin(x0 (Omega)))^2), {x, -(Infinity), +(Infinity)}) // 
Simplify

However, mathematica can find the result.
So could you help me?
Thank you.

calculation and analysis – Real and imaginary parts of a complex logarithm

I need to get the real and imaginary parts of the expression below with Mathematica:

$ (- i a – m ^ 2) ln ( frac {i m ^ 2} {2 a}) $,

or $ a $ and $ m $ are real.

So we have:

Refine(Re((-I a - m^2) Log((I m^2)/(2 a))), {Element(a, Reals), Element(m, Reals)})

However, Mathematica returns the command again. How can I proceed?

electrum – Esplora on local elementsregtest – analysis failed: data is not fully consumed during explicit deserialization

this is related to esplora (the block explorer) and its back-end electrs API. Is it possible to run esplora for a local element test?

When I run electrs it is the error I get back:

DEBUG - Server listening on 127.0.0.1:44224
DEBUG - Running accept thread
INFO - NetworkInfo { version: 180101, subversion: "/Elements Core:0.18.1.1/" }
INFO - BlockchainInfo { chain: "liquidregtest", blocks: 1, headers: 1, bestblockhash: "9cc7c8fb1c8e2e1e8ed184f5e31548eb5859b74e7552ba8841c41aeeb24d0ae3", pruned: false, verificationprogress: 0.334, initialblockdownload: Some(false) }
DEBUG - opening DB at "./db/liquidregtest/newindex/txstore"
DEBUG - 0 blocks were added
DEBUG - opening DB at "./db/liquidregtest/newindex/history"
DEBUG - 0 blocks were indexed
DEBUG - opening DB at "./db/liquidregtest/newindex/cache"
DEBUG - downloading all block headers up to 9cc7c8fb1c8e2e1e8ed184f5e31548eb5859b74e7552ba8841c41aeeb24d0ae3
TRACE - downloading 2 block headers
ERROR - server failed: Error: failed to parse header 000000a021cab1e5da4718ea140d9716931702422f0e6ad915c8d9b583cac2706b2a9000ac20a615d9b0d4df3e3ac2cb7018a07bd314d6bb715a57adead7c03e208b3658890e9d5d01000000012200204ae81572f06e1b88fd5ced7a1a000945432e83e1551e6f721ee9c00b8cc332604b00000000010151
Caused by: parse failed: data not consumed entirely when explicitly deserializing

The command used to launch electrs is:

cargo run --features liquid --release --bin electrs -- -vvvv --daemon-dir ~/.elements/elements-0.18.1.1/elementsdir/ --daemon-rpc-addr 127.0.0.1:18886 --cookie user:password --network liquidregtest -v

And there is an elementd running on this 127.0.0.1.118886 whose name is liquidregtest (in the configuration file is the line: chain = liquidregtest), in fact with the elements-cli j & I successfully requested an address and executed a generated command, which returned the "9cc7c8fb1c8e2e1e8ed184f5e31548eb5859b74e7552b8841c41aeeb24d0ae3" that you see in the debug (which means that the block was created).

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ClamAV, Exim and antivirus analysis

Which antivirus system do most administrators use for email on their servers these days? Or do you even use an anti-virus system on your mail system?

Most of our servers are cPanel. And since a very long time, we use clamAV integrated with Exim for the search of viruses on these servers.

However, ClamAV has become a real memory machine, occupying almost 1 GB of memory by itself. Any definitions it has to load can slow down the restart of clamd. And I do not know how effective it is. While browsing the logs of two of our servers, clamd blocked last week 55 messages, due to viruses / malware, including 40 phishing messages, which I wonder if SpamAssassin would have been intercepted anyway.

This makes me wonder … what is the current prevalence of viruses in e-mail attachments? Most of the problems in email today are probably phishing-related, blurring the line between viruses / malware and spam.

Is there a better and more effective antivirus solution for Exim? Maybe too many things creak, and that's why the number of my journals is so low. Although I do not receive any complaints from our clients.

1GB of memory is really not enough to hurt things on a dedicated server. But for smaller VPS where memory is more important, 1 GB can represent a considerable part of the total memory allocated. And does Clamd really do anything?

Has anyone ever thought about this?