0
|
12457 Views
|
6 Replies
|
0 Total Likes
View groups...
Share
GROUPS:

# Two problems - Limit and ContourPlot

Posted 10 years ago
 In Mathematica 8 the code Limit[ Sin[Sqrt[x^2-1]]-Sin[Sqrt[x^2+1]]] gives the correct answer 0 but Mathematica 9 and WolframAlpha both give Interval[{-2,2}], what is wrong here?? Another problem I encountered is in plotting implicit relation with ContourPlot[ x^(2/3) + y^(2/3) == 4, {x, -7, 7}, {y, -7, 7}]. It only gives the quarter of the graph!!
6 Replies
Sort By:
Posted 10 years ago
 Looks like something is missing from the Limit function: Limit[Sin[Sqrt[x^2 - 1]] - Sin[Sqrt[x^2 + 1]], x -> x0] where x0 is set to the desired value. The function in question also has multiple values of x0 that result in a limit/value of zero (although it doesn't appear that it's necessary to use Limit - but I must be missing something). Plot[Sin[Sqrt[x^2 - 1]] - Sin[Sqrt[x^2 + 1]], {x, -20, 20}, PlotRange -> {{-20, 20}, {-1, 0.5}}] results in
Posted 10 years ago
 About the contour plot...I think that the following explains the result: x^(2/3) /. x -> -7 N[x^(2/3) /. x -> -7] (x^2)^(1/3) /. x -> -7 N[(x^2)^(1/3) /. x -> -7] with corresponding results (-7)^(2/3) -1.82965 + 3.16905 I 7^(2/3) 3.65931 
Posted 10 years ago
 Sorry! I have omitted the limit point: Limit[Sin[Sqrt[x^2 - 1]] - Sin[Sqrt[x^2 + 1]], x -> Infinity]. By using the command Plot[Sin[Sqrt[x^2 - 1]] - Sin[Sqrt[x^2 + 1]], {x, 0, 100}, PlotRange -> All] we can see the limit is equal to 0; but Mathematica 9 gives Interval[{-2,2}]!
Posted 10 years ago
 Thank you for your very nice point! Now, the command ContourPlot[ (x^2)^(1/3) + (y^2)^(1/3) == 4, {x, -7, 7}, {y,-7,7} ] gives the correct answer.
Posted 10 years ago
 Uh, this isn't the kind of answer I like to get but for Mathematica 10.0.1 (Windows 7 and a more recent version than 8 and 9 that you mention), I get the correct answer: In[1]:= Limit[Sin[Sqrt[x^2 - 1]] - Sin[Sqrt[x^2 + 1]], x -> Infinity] Out[1]= 0 And for whatever it's worth I get the same response as you for version 9.
Posted 10 years ago
 Likewise for Surd[x, 3]^2 + Surd[y, 3]^2 == 4.