Pharma Industry Players Leverage AI For Drug Discovery By Using Precise Molecular Target Identification As Per The Business Research Company's Artificial Intelligence (AI) In Drug Discovery Global Market Report 2022...
finance.yahoo.com > newsTBRC’s market research report covers artificial intelligence in drug discovery market size, artificial intelligence in drug discovery market forecasts, major artificial intelligence in drug discovery companies and their market share, key strategies to undertake, and more.LONDON, Jan. 18, 2022 (GLOBE NEWSWIRE) -- According to The Business Research Company’s research report on the artificial intelligence in drug discovery market, Artificial Intelligence (AI) tools are now widely used for molecular... ...
STRATA Skin Sciences Expands into Israel with Leading Medical Device Distributor
finance.yahoo.com > newsEnables STRATA to Commercialize and Market its XTRAC Excimer Lasers to an Estimated 300,000 Psoriasis and 160,000 Vitiligo Patients in IsraelHORSHAM, Pa., Jan. 18, 2022 (GLOBE NEWSWIRE) -- STRATA Skin Sciences, Inc. (NASDAQ: SSKN), a medical technology company dedicated to developing, commercializing and marketing innovative products for the treatment of dermatologic conditions, announced that it has entered into an agreement with JuvenIL, the dermatology and aesthetics portion of Trimaco, a lea... ...
How to be a god: We might one day create virtual worlds with characters as intelligent as ourselves
techxplore.com > newsMost research into the ethics of Artificial intelligence (AI) concerns its use for weaponry, transport or profiling. Although the dangers presented by an autonomous, racist tank cannot be understated, there is another aspect to all this. What about our responsibilities to the AIs we create?...
Bill von Hagen, computer scientist and Pittsburgh punk pioneer, dies at 66
post-gazette.com > news > obituaries > storiesBill von Hagen, computer scientist and Pittsburgh punk pioneer, dies at 66 Pittsburgh Post-Gazette...
Is this isomorphism true for all square-free integer k?
math.stackexchange.com > questionsIs this true for all square-free integer k? $$ \mathbb{Z}/(a^2+b^2)\mathbb{Z} \space \cong \space \mathbb{Z}[\sqrt{k}]/(a+b\omega) $$ where $\omega=\sqrt{k}$ for $k \equiv 2,3\space (\text{mod}\space 4)$ and $\omega=(\sqrt{k}+1)/2$ otherwise. I have tested this idea for $k=-1$ and $5$, but don't know whether it is true for all square-free integers. If it is true, how to prove it? I only know the proof of $k=-1$ case, which is easy.... ...
Show that $Pleft ( left | sum_{i=1}^{n}X_i right |leq 2sqrt{n} right )to 1-2Phi(-2)$ - solution explanation
math.stackexchange.com > questionsShow that $$P\left ( \left | \sum_{i=1}^{n}X_i \right |\leq 2\sqrt{n} \right )\to 1-2\Phi(-2)$$ I have been given that $EX_i=0$ and $Var(X_i)=1$. Now use CLT and thus $$P\left ( \left| \sum_{i=1}^{n}X_i \right| \leq x\sqrt{n} \right )=P\left ( \left|\frac{1}{n} \sum_{i=1}^{n}X_i-0 \right| \leq x\frac{1}{\sqrt{n}} \right )\to 1-2\Phi(-x)$$ Insert $x=2$ and we are done Questions: Why does the convergence follow as that? From my understanding we can rewrite the expression inside $P(*)$ and notice that it is the ... ...
Watson's Lemma by altering an integral to the right form
math.stackexchange.com > questionsThe integral from 0 to pi/2 of (cost)^(1/2). Exp(-lambda.cost)dt as lambda tends to infinity. I am asked to find a suitable substitution to get the integral in the form where the Watsons Lemma can be used to find an asymptotic approximation of the integral, I know this means the integral contains an exp(-lambda.s) and the lower limit of the integral is 0 but I can not work out what the substitution is. I tried s=cost but it becomes extremely tedious. Once I have the right u substitution I believe I will be able to proceed with ... ...
If $D_{O_K/mathbb{Z}}(x_1, . . . , x_n)$ is square-free, why is $ lbrace x_1, . . . , x_nrbrace$ is a basis of $O_K$ over Z?
math.stackexchange.com > questionsI am stuck on this problem: Let K be a number field of degree n and let $ \lbrace x_1, . . . , x_n\rbrace$ be a basis of K over $\mathbb{Q}$ contained in the ring of integers $O_K$ of K. Show that if $D_{O_K/\mathbb{Z}}(x_1, . . . , x_n)$ is square-free, then $ \lbrace x_1, . . . , x_n\rbrace$ is a basis of $O_K$ over Z. Can somebody help me?...
An Introduction to Adversarial Machine Learning - DZone AI
dzone.com > linkAdversarial machine learning is concerned with the design of ML algorithms that can resist security challenges, the study of the capabilities of attackers, and the understanding of attack consequences. Adversarial Machine Learning states that there are four types of attacks that ML models can suffer....
Identity and nation state - Opinion - Al-Ahram Weekly
english.ahram.org.eg > newscontent > opinionIdentity and nation state - Opinion - Al-Ahram Weekly Ahram Online...