I am told that we can't predict whether we shall get a head or tail. We can only say that for an unbiased coin there is 50% probability for either. But coin is not case of Quantum Physics!  I have heard people say that such and such cannot be predicted, but it is obvious that if we have enough information about the event we can use the laws of universe and predict what will happen. So, is it possible to predict whether we get a head or tail in practice? I know we shall have to take into consideration the angular momentum of the coin, law of gravity, friction with air among many other things which I can't think of right now, ah perhaps the friction of the surface and how much bouncy it is.
I'm not sure if this is the right place to ask this question. I realise that this maybe a borderline philosophical question at this point in time, therefore feel free to close this question if you think that this is a duplicate or inappropriate for this forum. Anyway, I'm an electrical engineer and I have some basic knowledge of quantum mechanics. I know that Schrödinger's equation is deterministic. However, quantum mechanics is much deeper than that and I would like to learn more. If this question is not clearly answerable at this point than can anyone point out some recognized sources that try to answer this question. I would appreciate it if the source is scientific and more specifically, is related to quantum theory.
Is this group solvable? Let $G=H\times K$ (a direct product of $H , K$) where $H$ is an abelian 2-group and $K$ is a non-abelian simple group. Is $G$ solvable? why?  These three answers are true, and I thought so. But in a book I found a remark that made me confused: T.M. Gagen, Topics in Finite Groups, London Math. Soc. Lecture Note Ser., vol. 16, Cambridge Univ. Press, Cambridge, 1976, Remark, Theorem A on p. $40 $ and Definition $11.3$ on p. $39$. Please see it and say me your ideas. When using wish to replicate a spell, what school of magic does it count as? Let $A$ and $B$ be $n\times n$ matrices. Suppose $A$ is similar to $B$. Prove $\operatorname{trace}(A) = \operatorname{trace}(B)$.  I'm not sure where to go on this. So far I have this:  If $A$ is similar to $B$, then      $B=P^{-1}AP$ and $A=PBP^{-1}$   This implies that:     $\operatorname{trace}(B) = \operatorname{trace}(P^{-1}AP)$ and $\operatorname{trace}(A) = \operatorname{trace}(PBP^{-1})$   Not sure where to go from here I have QGIS 1.8.0 installed via OSGeo Installer and work on win 7. All I need is to upload the shapefile to the database via this plugin. Unfortunately I am receiving an error:  Something's wrong: ["'shp2pgsql' is not recognized as an internal or external command,\n", 'operable program or batch file.\n']   How to install and configure shp2pgsql to make this plugin work fine? SQL Server default collation vs database with different collation - potential problems? I'm struggling with the design of new database that has to use a different collation (Y Collation) than the SQL Server Instance default collation (X Collation). I'm looking for potential problems.  Temp Tables by default will be created with X Collation so any joins on text data types will fail unless I create temp table with Y collations on columns.  What about SQL Server string functions: REPLACE, SUBSTRING, etc.? Will they always be returned in X collation? What about when I use REPLACE(...) COLLATE Y? Do you see any potential data loss or maybe any different problems? The definition of "Lie group" typically restricts to a smooth manifold.  If we instead define a "Lie group" to be a topological manifold such that multiplication and inversion are continuous, is the manifold necessarily smooth?  Is the smooth structure unique if we want a smooth Lie group?  I believe the answer is yes, since a connected Lie group structure is determined by the Lie algebra, but my search attempts failed. When dragging an object, there is a greater start-up force than the force it takes to keep it moving. Why is this? Why are there two different values for friction? How to place a float at the top of a floats-only page? Got French Schengen visa but no "VIS" after giving fingerprints I had my French Schengen visa interview 3 days ago (got the stamped visa and passport within two days, super efficient!). However there's no "VIS" written on the visa, even though I gave my fingerprints during the interview in US. I had presumed VIS would be written on the visa since I gave my fingerprints, or is it for something else? I do remember it took a while to take the fingerprints because the machine looked old and was taking a long time to read my fingerprints.  The next time I want to apply for a visa, do I need to go for an interview again just because the visa has no VIS on it? Clarify an idiom in a political text Please carefully read the text below:     On 24 November, 1993, a meeting of Leftist intellectuals occurred in   London under the auspices of the Institute for Public Policy   Research (IPPR), which is a Labour-leaning think-tank.  A short document   was circulated in advance of the said meeting to clarify its   purpose.        Among other things, the document declared that the task of the   IPPR was: "To do what the Right did in the seventies, namely to break   through the prevailing parameters of debate and offer a new   perspective on contemporary British politics."    The explanatory document also said, "Our concern is not to engage in a   philosophical debate about foundations of socialism."      If this meant that those foundations were not the appropriate thing to   talk about at the 24 November meeting, then that might have been   right; not every thing has to be discussed at every meeting.  But if   what was meant was that discussion of philosophical foundations is not   what the Left now needs, then I disagree, and if that indeed is what   was meant, then it is curious that the breakthrough by the Right should   have been invoked as an achievement for the left to emulate, for if   there is a lesson for the Left in the Right's breakthrough, it is that   the Left must repossess itself of its traditional foundations on   pain of continuing along its present, politically feeble, reactive   course.  If the Left turns its back on its foundations, it will be   unable to make statements that are truly its own.   Since I've heard only one definition of the part identified in bold, I am at a dead end.  As you know, and according to Oxford's dictionary, the verb "repossess" means: To take back property or goods from SB who has arranged to buy them and can not pay.  Here, the object of verb is a thing not a person, but in the text, first,itself comes before of and its traditional foundations. As a result, I thought that this phrase might be an idiom and so has a different meaning with the original verb.  Please explain the meaning of this phrase. How is a pitch supporting fast bowlers or spin bowlers? Computing the cardinality of a combinatorial set Define two sets: $S_1$ and $S_2$ where we have that $a$ is an odd natural number and $b\in \mathbb{N}$, $S_1 = \Big\{ \frac{2}{a},\frac{2}{a-2},\frac{2}{a-4},...,\frac{2}{a-(a-1)},$ $\frac{2+4}{a+2},\frac{2+4}{a},\frac{2+4}{a-2},...,\frac{2+4}{a-(a-1)},$ $\frac{2+8}{a+4},...,\frac{2+8}{a-2},...,\frac{2+8}{a-(a-1)},$ $\vdots$ $\frac{2+4\lfloor\frac{b}{2}\rfloor}{a+\lfloor\frac{b}{2}\rfloor},\frac{2+4\lfloor\frac{b}{2}\rfloor}{a+\lfloor\frac{b}{2}\rfloor-2},...,\frac{2+4\lfloor\frac{b}{2}\rfloor}{a-(a-1)},$ $\frac{2+4(\lfloor\frac{b}{2}\rfloor+1)}{a+\lfloor\frac{b}{2}\rfloor},\frac{2+4(\lfloor\frac{b}{2}\rfloor+1)}{a+\lfloor\frac{b}{2}\rfloor-2},...,\frac{2+4(\lfloor\frac{b}{2}\rfloor+1)}{a-(a-1),}$ $\frac{2+4(\lfloor\frac{b}{2}\rfloor+2)}{a+\lfloor\frac{b}{2}\rfloor-2},\frac{2+4(\lfloor\frac{b}{2}\rfloor+2)}{a+\lfloor\frac{b}{2}\rfloor-4},...,\frac{2+4(\lfloor\frac{b}{2}\rfloor+2)}{a-(a-1)}$ $\vdots$ $\frac{2+4b}{a},\frac{2+4b}{a-2},...,\frac{2+4b}{a-(a-1)} \Big\}$ and $S_2 = \Big\{ \frac{2}{\lceil\frac{a}{2}\rceil}, \frac{2}{\lceil\frac{a}{2}\rceil-1},\frac{2}{\lceil\frac{a}{2}\rceil-2},...,\frac{2}{1}$ $\frac{2+2}{\lceil\frac{a}{2}\rceil+1},\frac{2+2}{\lceil\frac{a}{2}\rceil},\frac{2+2}{\lceil\frac{a}{2}\rceil-1},...,\frac{2+2}{1},$ $\vdots$ $\frac{2+2(\lfloor\frac{b}{2}\rfloor-1)}{\lceil\frac{a}{2}\rceil+(\lfloor\frac{b}{2}\rfloor-1)},\frac{2+2(\lfloor\frac{b}{2}\rfloor-1)}{\lceil\frac{a}{2}\rceil+(\lfloor\frac{b}{2}\rfloor-2)},...,\frac{2+2(\lfloor\frac{b}{2}\rfloor-1)}{1}$ $\frac{2+2\lfloor\frac{b}{2}\rfloor}{\lceil\frac{a}{2}\rceil+\lfloor\frac{b}{2}\rfloor},...,\frac{2+2\lfloor\frac{b}{2}\rfloor}{1},$ $\frac{2+2(\lfloor\frac{b}{2}\rfloor+1)}{\lceil\frac{a}{2}\rceil+\lfloor\frac{b}{2}\rfloor-1},...,\frac{2+2(\lfloor\frac{b}{2}\rfloor+1)}{1},$ $\frac{2+(\lfloor\frac{b}{2}\rfloor+2)}{\lceil\frac{a}{2}\rceil+\lfloor\frac{b}{2}\rfloor-2},...,\frac{2+(\lfloor\frac{b}{2}\rfloor+2)}{1}$ $\vdots$ $\frac{2+2b}{\lceil\frac{a}{2}\rceil-1},...,\frac{2+2b}{1} \Big\}$ The question is, how do I determine the cardinality of the intersection of these two sets i.e. how would I compute the value $|S_1 \cap S_2|$? Note that duplicates may appear within both sets such as $\frac{1}{2}=\frac{2}{4}$ etc. Therefore we must make sure that for each fraction $\frac{n}{m}$ we have $\text{gcd}(n,m)=1$. I am aware of the existence of , but not entirely sure if that identity would be useful here. The question says to solve this equation: $(z+1)^5 = z^5$  I did. Just want to find out if I did it properly and if my run-around logic makes sense.  First I begin my writing the equations as:  $$ (z+1)^5 = z^5$$ $$ \mathbf{e}^{5 \mathbf{Log}(z+1)} = \mathbf{e}^{5 \mathbf{Log}(z)} $$ So $$  \mathbf{Log}(z+1) = \ln|z+1| + \mathbf{Arg}(z+1)i $$   $$  \mathbf{Log}(z) = \ln|z| + \mathbf{Arg}(z)i $$   Now, because the natural logarithm is one-to-one, I write:  $$ \ln |z| = \ln |z+1| \Rightarrow |z| = |z+1|$$   So assign $ z = a +bi$   So that $|z| = \sqrt{a^2 +b^2} = |z+1| = \sqrt{(a+1)^2 +b^2} \Rightarrow a^2 +b^2 = (a+1)^2 +b^2 \Rightarrow a^2 = (a+1)^2 \Rightarrow a = -\frac12 $  So, $z = -\frac12 + bi$ and $z+1 = \frac12 + bi$ for some $b \in \mathbb R$  Now to find $b$  $$ \mathbf{Arg}(z+1) = \mathbf{Arg}(z)$$ $$ \tan^{-1} \frac{b}{\frac12} = \pi - \tan^{-1}\frac{b}{-\frac12}$$  I have a feeling this last part isn't quite right, so I just want to find out if I'm approaching this question properly?  Ultimately, I get $ z = -\frac12$ which upon inspection...is wrong... Integrating Floor Functions $\int_0^{1102} \lfloor{x^{1/2}}\rfloor \mathrm{d}x$ I want to find a general formula to calculate the integral of such function (x&gt;0) Could someone kindly help me? $$\int_0^{1102} \lfloor{x^{1/2}}\rfloor \mathrm{d}x$$ Thank you! How to detect scene strips in the VSE on frame changes? I need to detect scene strips in the VSE on the current frame, meaning on what strip is the play head at this moment? However I can't seem to find the right type(?) to attach the ".strip_elem_from_frame" function to, in order to get the info/name/type on the strip. Here's a test script which must run in a screen with a sequencer with scene and movie strips in it:  import bpy  #update at frame change def currentStrip(*pArgs):     print("Current Strip: "+str(bpy.context.sequences[0].strip_elem_from_frame(bpy.context.scene.frame_current))) #Doesn't detect any strips at all.  #add frame change handle def attachAsHandler():     for f in bpy.app.handlers.frame_change_pre:         bpy.app.handlers.frame_change_pre.remove(f)     bpy.app.handlers.frame_change_pre.append(currentStrip)  currentStrip() attachAsHandler() equivalent norms in Banach spaces of infinite dimension write some AMPscript to skip a record if its empty I have a data extension with a CustomerId(PK) that has multiple Productid's as Productid1, ProductId2, ProductId3...ProductId10.   Some of the Productid's are empty for some Customers and I'm trying to write some AMPscript if the productIdx is empty to show the next one.  I'm trying to display 6 products in an email and at the moment when I preview the email, if from the first six Products one doesn't have a value(is empty) I get the error "The subscriber preview failed to generate. Review the details, correct all issues, and try again. Product1 not found"  I want to skip the product that is NULL and go to the next one available.  %%[ SET @customerId = [customerId]  SET @customer_rows = LookupRows("PRL","customerId",@customerId) IF Rowcount(@customer_rows) &gt; 0 THEN SET @customer_row = ROW(@customer_rows, 1)  SET @product_count = 0  SET @productId1 = FIELD(@customer_row, "productId1")  IF NOT EMPTY(@productId1) THEN SET @product_rows = LookupOrderedRows("PI_UK", 1,"Normal_SalePrice Desc", "ProductCode", @productId1)  IF RowCount(@product_rows) &gt; 0 THEN  SET @product_count = ADD(@product_count,1) ELSE  SET @product_count = ADD(@product_count,0) ENDIF ELSE SET @product_count = ADD(@product_count,0) ENDIF    SET @productId2 = FIELD(@customer_row, "productId2") IF NOT EMPTY(@productId2) THEN SET @product_rows = LookupOrderedRows("PI_UK", 1,"Normal_SalePrice Desc", "ProductCode", @productId2) IF RowCount(@product_rows) &gt; 0 THEN SET @product_count = ADD(@product_count,1) ELSE SET @product_count = ADD(@product_count,0) ENDIF ELSE SET @product_count = ADD(@product_count,0) ENDIF  SET @productId3 = FIELD(@customer_row, "productId3") IF NOT EMPTY(@productId3) THEN SET @product_rows = LookupOrderedRows("PI_UK", 1,"Normal_SalePrice Desc", "ProductCode", @productId3) IF RowCount(@product_rows) &gt; 0 THEN SET @product_count = ADD(@product_count,1) ELSE SET @product_count = ADD(@product_count,0) ENDIF ELSE SET @product_count = ADD(@product_count,0) ENDIF  and so on till ProducId10...   I have uploaded a photo with the PRL Data extension How to make crs transformation of a QgsVectorLayer? I need to transform QgsVectorLayers in python from one crs into another crs. By googling I found the following solution (), which works for points only:  crsSrc = QgsCoordinateReferenceSystem(4326)    # WGS 84 crsDest = QgsCoordinateReferenceSystem(32633)  # WGS 84 / UTM zone 33N xform = QgsCoordinateTransform(crsSrc, crsDest) pt1 = xform.transform(QgsPoint(18,5))   Is there a way to make a crs transformation of an entire layer? Vitali-type set with given outer measure Resources for Blender Since asking for tutorials, videos, resources, links to other content or downloads is considered off topic here, this is a specifically created Community Wiki which gathers resources for Blender and it has been approved by the .  Just write in the appropriate answer/section. If you have concerns, questions, post a  question, so we don't clutter the comments, but you can link your meta question from the comments.  Questions regarding such resources are not allowed anymore, except for very specific and on topic requests (ask on  if you're unsure about your question). Follow the instructions made in the question about how to post, what can be posted, etc.  Learning             Assets    (also see: )   Reference Images/Blueprints (also see: )    Tools    (also see: )      Other   Special Interest ()     Also see the .