EURUSD: risks have risen through the hour-long trend line – Forex News & Analysis

On Friday the 3rd of August, trading on the euro closed down. High volatility was observed in light of the publication of the the US labour market report.

July data on the number of those employed in the non-agricultural sector of the US did not meet market expectations. Although the data was below 189 thousand, the report is not bad, as the average hourly salary has grown and the indicators for May and June have been revised upwards. US 10-year bond yields fell on news of the report, with many major currencies closing in positive territory on Friday as a result.

As a result of last week, major currencies closed in the red zone against USD. The greatest decline was shown by the British pound (-0.84%). Then came the euro (-0.75%), the New Zealand dollar (-0.68%), the Japanese yen (-0.23%), the Australian dollar (-0.03%), and the Swiss franc (-0.02%). The Canadian dollar was the only currency to record growth (+0.55%).

US data:

The number of new jobs was 157 thousand (forecast: 189 thousand). May figures were revised from 244 thousand to 268 thousand, and in June – from 213 thousand to 248 thousand. The overall revision amounted to +59 thousand.

The unemployment level fell to 3.9% (previous: 4.0), which coincided with expectations.

The average hourly earnings index was 0.3% (forecast: 0.3%, previous: revised from 0.2% to 0.1%).

The ISM business activity index for the service sector for July was 55.7 (forecast: 59.0, previous: 59.1).

Day’s news (GMT+3):

  • 9:00 Germany: factory orders s.a. (MoM) (Jun).
  • 11:30 Eurozone: Sentix Investor Confidence (Aug).

06.08.18_EURUSD_H1_5b67f467dadff.png

Fig 1. EURUSD hourly chart. Source: TradingView

Current situation:

Friday’s multidirectional fluctuations once again confirm that it’s pointless to make market forecasts on payrolls day. The 157th degree acted as a support. The price bounced off that area three times and now sellers are trying to test it below 1.1550.

I see the pair is poised to rebound to 45 degrees (1.1558). The Stochastic Oscillator isn’t favouring buyers at the moment, so it will only be safe to enter long positions if the trend line gets broken. The balance line (Lb) will act as an intermediate resistance. Now it is passing through 1.1600. The economic calendar is looking pretty scarce. There’s nothing to stop buyers from inducing a correction.

.(tagsToTranslate)forex(t)eurusd

EUR/USD Technical Analysis: Fresh Attempts for Upward Correction – Forex News & Analysis

After a harsh week in which the EUR/USD was exposed to downward momentum, pushing it towards the 1.1688 support, and with the start of this week’s trading, the pair tried to rebound higher with gains to 1.1793 before settling around the 1.1775 level at the time of writing.

However, TopAsiaFX has stated that attempts to rebound lacked strong momentum to stabilize above the 1.1800 resistance, and the efforts of the Euro are hindered by concerns about the strength of the second wave of the Corona pandemic and the measures of European countries to contain the outbreak of the deadly disease.

These restrictions directly affect the European economy, which is still suffering from the consequences of the first wave.

At the same time, and with these concerns, the European Central Bank monitors the economic performance of the bloc to determine its appropriate policy.

In this regard, European Central Bank Executive Board member Yves Mersch said yesterday that policymakers will carefully examine the economic data received before the next policy session to ensure that the impact of coronavirus containment measures is not repeatedly considered in light of extremely unstable expectations.

“Looking to the future, in the current environment of high uncertainty, the European Central Bank’s Governing Council will carefully evaluate the information received, including developments in the exchange rate, while ensuring that this information received, such as information related to containment measures that were included already at our baseline, were only counted once in our assessment.”

He also said that the economic recovery in the Eurozone remains incomplete and prone to setbacks. The policymaker emphasized that the Board of Directors continues to prepare to adjust all its tools, as appropriate, to bring inflation back to its target level.

The UK is leading European efforts to contain the pandemic. As authorities across the UK impose new restrictions on business and social interactions as COVID-19 infections are soaring in all age groups, where parts of the country’s hospital beds and intensive care wards are filling up.

One of their main goals is to reduce the pressure on the NHS ahead of the winter flu season. In this regard, public health experts say that the lockdown could help reset the epidemic to a lower level, giving doctors time to treat patients and providing breathing space for the government to improve its response.

Britain has the most serious outbreak of the Coronavirus in Europe, with more than 43,700 confirmed deaths.

According to the technical analysis of the pair: We are still waiting for the EUR/USD to stabilize above the 1.1800 resistance, for an opportunity for a stronger correction upwards, and we expect the pair to remain stable between the 1.1660 support and the 1.2000 psychological resistance for a period of time, as shown on the daily chart.

The psychological resistance at 1.2000 brings condemnation from monetary policy officials at the European Central Bank, as the Euro’s high exchange rate harms the European economy, which depends on exports at a time when it suffers from the effects of the pandemic. At the same time, the 1.1660 support raises buying interest among currency traders.

As for the economic calendar data, today: The German producer price index will be announced, then the Eurozone’s current account will be released. During the American session, building permits and the housing starts in the United States will be announced.

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calculus and analysis – Partial derivative of an integral

I have the following function (it is the incomplete elliptic integral of first kind)
$$ F(b,g) = int_{0}^{b} frac{dx}{sqrt{(1-x^2)(1-gx^2)}} $$
I would like to compute
$$frac{partial F}{partial g} , frac{partial F}{partial b} , frac{partial^2 F}{partial g^2} , frac{partial^2 F}{partial b^2} , frac{partial^2 F}{partial bpartial g}$$
so I defined

F(b_,g_):= Integrate(1/Sqrt((1 - x^2)*(1 - g*x^2)), {x, 0, b})

and tried the command

D(F(b,g),g)

but Mathematica cannot compute it.
Am I doing something wrong or is there a way to do it?

real analysis – Equivalent definitions of Open Set

this might be trivial to you but I am at the beginning of a topology course and I need to verify that $(i) implies (ii)$ of these alternative statement of a Open Set (they are equivalent):

$(i)$ If all convergent sequence $(y_n)$ s.t. $y_n in A^{c}$ for all $n in mathbb{N}$ and $y_i neq y_j forall ineq j$, then we have $lim_{n rightarrow infty} y_n notin A$

$(ii)$ If for all $a in A$ and all $(x_n)$ real sequence s.t. $lim_{n rightarrow infty} x_n=a$, then there exists $n_0 in mathbb{N}$ such that ${x_n colon n geq n_0} subseteq A$

$(i) implies (ii)$

If all convergent sequence $(y_n)$ s.t. $y_n in A^{c}$ for all $n in mathbb{N}$ and $y_i neq y_j forall ineq j$, we have $lim_{n rightarrow infty} y_n notin A$, then for all $a in A$ and all $(x_n)$ real sequence s.t. $lim_{n rightarrow infty} x_n=a$, exists $n_0 in mathbb{N}$ such that ${x_n colon n geq n_0} subseteq A$

Do you have any advice into how to prove that? I tried directly but it was not so easy. Should I try using contrapositive statement? Thank you.

cv.complex variables – Real integrals with complex analysis

I don’t have a clear formal viewpoint on this problem.
Resolving the Euler-Lagrange equations for the string with a point mass perturbation:
$$ frac{partial^2 phi }{partial x^2} = delta (x-a)$$
I encountered the following integral:
$$ I = int_{ – infty}^{+ infty} dk f(x,k) = int_{ – infty}^{+ infty} dk frac{e^{ik(x-a)}}{k^2} $$
Now, during the lessons we computed it as:
$$ I = lim_{ epsilon longrightarrow 0}int_{ gamma_+ cup gamma_-} dk frac{e^{ik(x-a)}}{(k – i epsilon)^2} $$
with $gamma_+$ being the semicircle in the upper half plane, counterclockwise, and $gamma_-$ being the semicircle in the lower half plane, clockwise. We found that
$$ I = 2 pi (x-a) theta(x-a) $$
My first question is:
If we compute it with choosing another pole shift such as $(k+i epsilon)$, we have:
$$int_{ gamma_-} dk frac{e^{ik(x-a)}}{(k-iepsilon)^2} = – 2 pi (x-a) theta(a-x) $$

In general every shift can give a different result.
Is it due to the fact that
$$ lim_{epsilon longrightarrow 0 } int neq int lim_{epsilon longrightarrow 0 }$$
and I can’t commute the operations (because for instance the integrand $f(x,k,epsilon)$ isn’t dominated by a function $g(x,k)$ )???
Moreover, how can I show I can’t commute the two? And in which case I can commute?
My second question is:
Is there another method that uniquely provide an answer to the integral $I$ above?

real analysis – Please check my solution(Uniformly convergence of the functions)

Here is the question in my real analysis textbook.

$f$ is a continuous on $mathbb{R}$

Say $f_n(x) = {n over 2}int_{x-{1 over n }}^{x+{1 over n }} f(t) dt$

Show the $f_n$ is a uniformly converge to $f(x)$ on $(0,1)$


In fact, the textbook itself solve this by M.V.T for integral.

But I tried a different way like the below.

$(sol)$ For $forall x in (0,1)$, consider the $I_n = (x- {1 over n }, x+{1 over n })$

Since $f$ is a continuous on $I_n$, f is uniformly continuous on $I_n $

Therefore, by the definition of the uniform continuity

$(1)$ $exists {2 over n} leq delta $ s.t. $Vert x-y Vert < delta$ $Rightarrow$ $Vert f(x) – f(y) Vert <{2epsilon over n } $

$(2)$plus, By Archimedes $exists k in mathbb N s.t. n geq k Rightarrow {2epsilon over n } < epsilon$

By $(1)$ and $(2)$
$Vert f_n(x) – f(x) Vert = Vert {n over 2}int_{x-{1 over n }}^{x+{1 over n }} (f(t) – f(x)) dt Vert leq {n over 2}int_{x-{1 over n }}^{x+{1 over n }} Vert f(t) – f(x) Vert dt leq {n over 2 } {2 over n } epsilon =epsilon $

Hence, $exists k s.t. n geq k Rightarrow Vert f_n(x) – f(x) Vert <
epsilon$
(Uniformly convergence.)

I don’t have a confidence my solution is right or not. Please check my idea and solution.

Thank you.

real analysis – Is a locally invertible weak limit of injective maps injective almost everywhere?

Let $Omega_1,Omega_2 subseteq mathbb R^2$ be open, connected, bounded, with non-empty $C^1$ boundaries.

Let $f_n:barOmega_1 to barOmega_2$ be Lipschitz injective maps with $det(df_n)>0$, and suppose that $f_n$ converges to a $C^1$ function $f: barOmega_1 to barOmega_2$ weakly in $W^{1,2}$, and that $det(df)>0$ everywhere on $barOmega_1$.

Is it true that $|f^{-1}(y)| le 1$ a.e. on $Omega_2$?

Does the answer change if we assume in addition that $f_n|_{K} to f|_{K}$ strongly in $W^{1,2}$ for every $K subset subset Omega_1$?

Seeking recommendations/warnings for creating a server for malware sample upload and analysis

I am trying to conceptualize a server where users can upload potentially dangerous malware samples to a server. They would not need to be stored in the file system per se but kept in memory long enough to be analyzed by other programs/libraries. (Similar to: VirusTotal)

I’m not opposed to incorporating other open source solutions/libraries, but my main concern is reducing risk to the overall system between when a user uploads the file to a web service and getting it to other programs for analysis.

  1. If the file is never saved, and remains only an input stream (java), would that be a strong mitigating factor?
  2. Do you know of any resources that are relevant/talk about sandboxing while still exposing some ports, in constant use by a program, for exchanging analysis information with another local server?
  3. Do you have any recommendations or warnings for accepting malware sample submissions? (If "don’t try", please elaborate on the pitfalls)

calculus and analysis – Approximation of a function

I am just wandering if it is possible to demonstrate that

enter image description here

      Integrate(
 1/(E^((((Xi) - (Eta)^2*(Tau))^2 + (Psi)^2)/
     (2*(Eta)^2*(1 + (Tau))))*Sqrt((Tau))*
   (1 + (Tau))), {(Tau), 0, Infinity})

tends to

enter image description here

   Integrate(Exp((-(1/(2*(Eta)^2)))*
     ((((Xi) - (Eta)^2*(Tau))^2 + (Psi)^2)/
        (Tau)))/(Sqrt((Tau))*(Tau)), 
    {(Tau), 0, Infinity})

in the limit of both ξ and ψ going to infinity.
In this case, it is needed to introduce a measure of the error of the approximation, as the integral of the difference between the two functions? Can this be viewed as a norm in some space?
Thanks in advance.