This question is not enlightening nor is it difficult. But some part of my notation or how I have defined my functions is messing up the D[f,x] function.

In[224]:= Clear[x]

In[299]:= Attributes[a] = {Constant}
Attributes[k1] = {Constant}
Attributes[k2] = {Constant}

In[275]:= values = {x_ -> 0, y_ -> 0}

U[x_, y_] := 
 k1/2 (Sqrt[(a - x)^2 + (a/Sqrt[3] - y)^2] - (2*a)/Sqrt[3])^2 + 
  k1/2*(Sqrt[(-a - x)^2 + (-(a/Sqrt[3]) - y)^2] - (2*a)/Sqrt[3])^2 + 
  k2/2*(Sqrt[(-(a/Sqrt[3]) - x)^2 + (a - y)^2] - (2*a)/Sqrt[3])^2 + 
  k2/2*(Sqrt[(a/Sqrt[3] - x)^2 + (-a - y)^2] - (2*a)/Sqrt[3])^2

In[199]:= Simplify[U[0, 0], Reals]

Out[199]= 2 ((k1/2)[(2 (-a + Sqrt[a^2]))/Sqrt[3]]^2 + (k2/2)[(
    2 (-a + Sqrt[a^2]))/Sqrt[3]]^2)

(**The above output is the correct form that I should be seeing**)

D[U[x, y], x] /. values

Out[258]= 0

In[303]:= D[U[x, y], y] /. values

Out[303]= 0

(**The above are correct outputs according to the problem statement**)

In[304]:= Simplify[D[U[x, y], x, y] /. values, Reals]

Out[304]= 0

(**The above derivative should give:(Sqrt[3]/2k1-Sqrt[3]/2k2) \
according to the problem statement 
But for some reason the D[] function is also taking derivates of k1 \
and k2 - which are set as constants

I know this problem has something to do with my notation or the constants k1 and k2. I am new to mathematica, so this may just be a syntax error.

Thanks in advance.

Update: here is the result of my changes, which did output the correct values for my taylor expansion

In[352]:= D[U[x, y], y, x] /. values

Out[352]= (Sqrt[3] k1)/2 - (
 3 (-((2 a)/Sqrt[3]) + (2 Sqrt[a^2])/Sqrt[3]) k1)/(4 Sqrt[a^2]) - (
 Sqrt[3] k2)/2 + (3 (-((2 a)/Sqrt[3]) + (2 Sqrt[a^2])/Sqrt[3]) k2)/(
 4 Sqrt[a^2])

In[354]:= Simplify[D[U[x, y], x, y] /. values, Reals]

Out[354]= (Sqrt[3] a (k1 - k2))/(2 Sqrt[a^2])
  • $\begingroup$ Regardless of the brackets etc... Why do you expect your Out[303] to be different from Out[304]? If you wrap Simplifyaround a zero it will still be zero? (You made a comment below Out[303]saying that this is correct $\endgroup$ – Lukas Apr 20 '16 at 5:20
  • $\begingroup$ that comment should have been placed one line above. A lot about this question was bad. I need to find a better way to go from my notebook file to this site in the future $\endgroup$ – Daniel Schulze Apr 20 '16 at 8:06

Braces [], brackets {}, and parenthesis () all have different meanings in mathematica.

Sorry for the post. I will read the documentation and tutorials more completely.

| improve this answer | |
  • $\begingroup$ Eh, we've all been there. I'm glad you were able to figure it out. $\endgroup$ – MarcoB Apr 20 '16 at 3:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.