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# Mathematica won't give me a numerical result

Posted 9 years ago
 Hello everyone, I have a problem with Mathematica 9. For my thesis I have to work with a rather big, but not too complicated equation, but Mathematica just won't give me numerical results for it, even if I assign values to every single variable. Even when I use the N-function, it still won't give me a simple numerical result, but a product containing a square root and some sums. Has anyone else encountered such a problem and does anybody know a solution? Thank you in advance!
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Posted 9 years ago
 On a side note, we could let MMa try and simplify the expression, and replace decimals with fractions, i.e. 43.566 with 43566/1000. the result is more compact though not necessarily what you want.we get f[a_, b_] := N[1/(4 Sqrt) Sqrt[((107500 + 21783 a + 18668 b) (224 b + 3 a (70 + 17 b)))/( a b)]] Paul.
Posted 9 years ago
 Ah great. Thank you guys, this actually helps a lot. I'll just remove it then.
Posted 9 years ago
 Hi Daniel,Is there any reason you include the TextCell[""]? removing it then prints a numeric answer. I am suspecting that using TextCell is preserving the format.Paul.
Posted 9 years ago
 TextCell[" "] is the problem. f[a_, b_] := 10 Sqrt \[Sqrt]((215. + 43.566 a + 37.336 b) (0.25 (5/(2 a) + 1500 (1/800 - (-1 + a)/(1000 a)) ) + 0.25 (5/(2 b) + 1250. (1/700 - (-1 + b)/(1000 b)) ))) So f[1,1] yields: 565.986You rule assignment would work for the expression, e.g. f[u, v] /. {u -> 1, v -> 1} 
Posted 9 years ago
 Hi Paul,so this is the equation with all the variables except for a and b. 10 Sqrt \[Sqrt]((215. + 43.566 a + 37.336 b) (0.25 (5/(2 a) + 1500 (1/800 - (-1 + a)/(1000 a)) TextCell[""]) + 0.25 (5/(2 b) + 1250. (1/700 - (-1 + b)/(1000 b)) TextCell[""]))) Now if I set a=1 and b=1 Mathematica gives me this: 384.644 Sqrt[ 0.25 (5/2 + 1.78571 TextCell[""]) + 0.25 (5/2 + (15 TextCell[""])/8)] Using the N-function results in: 384.644 Sqrt[ 0.25 (2.5 + 1.78571 TextCell[""]) + 0.25 (2.5 + 1.875 TextCell[""])] But that doesn't really help me. Do you know where the problem is? Thanks.
Posted 9 years ago
 Hi Daniel, would it be possible for you to post the equation, It is very difficult to give any meaningful answers if we can't see anything.Paul.