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12

Go to a new cell in a notebook and write "=", this will create an orange equality sign in the beginning of the cell. This means that you have entered the free form input mode. Write the name of a food item or a type of food and evaluate the cell, Mathematica will then return the entity corresponding to that food item or food type. You can hover the entity to ...


11

The date is different because Wolfram Alpha is wrong about when the Battle of Agincourt took place. Alpha is returning 25 October 1415 in the Gregorian calendar as the date of the battle, but the battle took place on that date in the Julian calendar (the calendar in use at the time). That date corresponds 3 November 1415 in the proleptic Gregorian ...


10

For large enough bounds, NSum is used. Compare timings: NSum[(-1)^(n + 1)/n, {n, 1, 100000}] // AbsoluteTiming {0.004744, 0.693142 - 1.3494*10^-16 I} N[Sum[(-1)^(n + 1)/n, {n, 1, 100000}]] // AbsoluteTiming {1.84727, 0.693142}


9

I remember from last years conference (Wolfram Technology Conference 2015) that there was a presentation on food entities. It doesn't seem to have been recorded but you can get the presentation (Food in the Wolfram Language). It shows a great deal on how to access foods and gives details on some of their properties. The food data is somewhat deep and very ...


9

Here's a start.... x = "healed"; y = "pay"; Intersection[ WolframAlpha["synonyms of " <> y, {{"Synonyms:WordData", 1}, "ComputableData"}, InputAssumptions -> {"*MC.%7E-_*WordData-"}], WolframAlpha["rhymes with " <> x, {{"Rhyme:WordData", 1}, "ComputableData"}, PodStates -> {"Rhyme:WordData__More"}, InputAssumptions -> ...


7

Ctrl+= gasoline heat of combustion * density results in: Quantity[32.7, "Kilojoules"/"Centimeters"^3] To get energy per gallon: UnitConvert[%, "Joules"/"Gallons"] (* Quantity[1.23782965*^8, "Joules"/"Gallons"] *)


6

As pointed out by Michael Seifert in a comment GeoGraphics[Polygon[Ctrl+=Italy]] should work (it worked for me), but if you want do it without using Wolfram's server, try GeoGraphics[Polygon[Entity["Country", "Italy"]], GeoBackground -> None]


6

As Guesswhoitis already suggested, this is a machine precision issue. So let us do your computation with arbitrary precision numbers. For doing so, all machine numbers have to be replaced with arbitrary precision numbers, otherwise the computation falls back to machine numbers. In the following command I have done this by placing `30 after each machine ...


5

Facebook has deprecated the API that Mathematica relied on to gather this information. Official source.


5

Something like this perhaps? bibOriginal = Import["http://www.gutenberg.org/cache/epub/10/pg10.txt"]; (* load the bible *) bib = StringDrop[bibOriginal, 658]; (* drop the introductory text *) bib = StringDrop[bib, -18733];(* drop the legalistic closeout at the end*) bibC = Map[{#[[1]], Length[#]} &, Gather[Sort[ToLowerCase[Characters[bib]]]]]; bibC = ...


5

Your input should be: Series[ (I x + y Exp[I p]) Tanh[Pi (I x + y Exp[I p])] Log[(I x + y Exp[I*p])^2 + a^2] I y Exp[I p], {y, 0, 2}, Assumptions -> {x > 0, y > 0, a > 0, p ∈ Reals} ] Asterisks can be used for multiplication, but a space is more usual. I removed unnecessary parentheses. The more important points are all functions (Exp, ...


5

Case closed but the topic should be left here for future visitors who use V9. So: Bug introduced in 9.0 and fixed in 10.0.0


5

Note This answer was meant for a bit of fun but instead does a good job of managing expectations for Entity in the WL at present. Image Identify and Google Image Search Wolfram have heavily promoted their ImageIdentify project (http://blog.wolfram.com/2015/05/13/wolfram-language-artificial-intelligence-the-image-identification-project/) which might be a ...


5

airports = {"ABE", "ABI", "ACT"}; AirportData["Properties"] data = AirportData[#, {"IATACode", "Name", "Cities", "Latitude", "Longitude"}] & /@ airports; data // Grid airports = {"ABE", "ABI", "ACT"}; EDIT: A quicker method would be to download a .csv file and filter it. data = Import["http://ourairports.com/data/airports.csv"]; The ...


4

You're pretty close. The following code should do it: Entity["Airport", #]& /@ airports


4

This is a documentation oversight. The function FlightData is not available in Mathematica as of this time.


4

This =Solve[64((x-Sqrt[x^2-1])/(x+1))^2-25(1-Sqrt[x-1]/(x+1))^2+9==0, x] and this WolframAlpha["Solve[64((x-Sqrt[x^2-1])/(x+1))^2-25(1-Sqrt[x-1]/(x+1))^2+9==0,x]"‌​] solve immediately. Note: One reason for using WolframAlpha to handle some calculations is that you can sometimes see the steps involved, see additional information, make use of curated ...


4

I will start by just copying how Jason B imported the bible: bible = Import["http://www.gutenberg.org/cache/epub/10/pg10.txt"]; bible = StringSplit[bible, { "*** START OF THIS PROJECT GUTENBERG EBOOK THE KING JAMES BIBLE \ ***", "*** END OF THIS PROJECT GUTENBERG EBOOK THE KING JAMES BIBLE ***" }][[2]]; After that I will use WordCounts and ...


4

We can import the bible directly from the web and convert it to a list of words fairly easily. Here I'm going to remove the chapter/verse headings as well as extraneous punctuation marks. biblewords = Import["http://www.gutenberg.org/cache/epub/10/pg10.txt"] // StringSplit[#, {"*** START OF THIS PROJECT GUTENBERG EBOOK THE KING JAMES BIBLE ...


4

You can import the prices from the Energy Information Administration. RCLC1d = Import["http://www.eia.gov/dnav/pet/hist_xls/RCLC1d.xls"]; c1 = RCLC1d[[2, All, {1, 2}]]; c1prices = Cases[Drop[c1, 3], {_, _?NumberQ}]; DateListPlot[c1prices, Joined -> True, PlotLabel -> "WTI Oil Price since " <> DateString[ c1prices[[1, 1]], {"Day", " ", ...


3

Based on the documentation, I would say that this is a bug. Note, however, that you can perhaps get at the data you want via the call EntityValue[Entity["Country", "UnitedStates"], EntityProperty["Country", "ElectricityTotal", "Date" -> 2007]] though the relationship between "ElectricityTotal" and "ElectricityProduction" is less than clear. I would ...


2

I just tried it, it works on Alpha as is. It just did not like the Exclusions -> None but it still took it. Most of the times, Wolfram Mathematica language will also work in Wolfram Alpha as well. I like to think of it as the Wolfram Alpha language being an almost superset of the Wolfram Mathematica language. Plot[Piecewise[{{10*x - 0.5*5*x^2, 0 < x ...


1

Edit, still the lazy way: Go the lazy way and try the folling


1

This answer suggested by @Hubble07 works. subsetQ[list1_, list2_] := WolframAlpha[ "is " <> ToString[list1] <> "a subset of " <> ToString[list2], {{"Result", 1}, "ComputableData"}]


1

I found that wolfram|Alpha wasn't good enough, by RhymeZone is a great site for this, and they even support near rhymes. Here's an example: Rhymes[r_] := Rhymes[r] = Module[{url, links, rhy}, url=StringTemplate["http://www.rhymezone.com/r/rhyme.cgi?Word=`1`&typeofrhyme=perfect&org1=syl&org2=l&org3=y"][r]; ...


1

It looks like you're crossing types of input For Me, '=' and ctl + '=' return no pod, just an entity- and '==' always returns a pod.


1

Perhaps this unsophisticated attempt is a starting point. whales = {"have hair", "live birth", "breathe air", "live in water", "have fins", "can swim"}; fish = {"lay eggs", "have scales", "breathe water", "live in water", "have fins", "can swim"}; tint = Intersection[whales, fish]; w = Complement[whales, tint]; f = Complement[fish, tint]; r1 = ...


1

You can use Rationalize to convert numbers to exact numbers gammaex = 0.2506 // Rationalize[#, 0] &; omega[t_] = 2.43163218375*10^7*Exp[1700*(1/298.15 - 1/(273.15 + t))] // Rationalize[#, 0] &; w[t_] = (3.414105049212413*10^12)/(omega[t]) // Rationalize[#, 0] &; v[t_] = Sqrt[661.6469313477045*(t + 273.15)] // Rationalize[#, 0] &; ...


1

Okay, I'm slowly going through the questions involving elliptic integral evaluations. Yet again, none of the software mentioned in this thread have managed to produce a "clean" expression. For the benefit of future readers, here's a tidier closed form for your perusal: N[2 (7 Sqrt[5] + 2 (10 EllipticE[2 π/3, 11/12] + EllipticF[2 π/3, 11/12])/Sqrt[3])/9, 20] ...



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