The derivative functions D and Dt don't appear to be functioning as advertised.
Following the first example in the "Properties and Relations" section of http://reference.wolfram.com/mathematica/ref/Constants.html I have:
In[1]:= {Dt[ax^2 + b, x, Constants -> {a, b}], D[ax^2 + b, x]}
Out[1]= {2 ax Dt[ax, x, Constants -> {a, b}], 0}
I've duplicated the input, but the output is totally different. How do I get the expected output { 2 a x, 2 a x}?
I am using Mathematica 8.0.1.0 64-bit as installed at Rutgers University.
You need a space between a and x, otherwise it thinks you're talking about a variable named ax:
In[2]:= {Dt[a x^2 + b, x, Constants -> {a, b}], D[a x^2 + b, x]}
Out[2]= {2 a x, 2 a x}
(I realize this isn't really answering the OP's question. But given the level of the question, along with OP's desire to use the Contants option, the following info may prove useful for others in the future.)
My 2 cents on Dt.
IMO, using the Constants option is less than ideal---mainly because it produces messy output. For example:
In[1]:= Dt[x^a y^b, Constants -> {a, b}]
Out[1]= a x^(-1 + a) y^b Dt[x, Constants -> {a, b}] +
b x^a y^(-1 + b) Dt[y, Constants -> {a, b}]
Am I the only one who finds the above behavior annoying/redundant? Is there a practical reason for this design? If so, please educate me... :)
Alternative approaches:
If you don't want to use the Constants option, here are some alternative approaches.
Use UpValues to force constants.
In[2]:= Remove[a, b];
a /: Dt[a] = 0;
b /: Dt[b] = 0;
Dt[x^a y^b]
Out[5]= a x^(-1 + a) y^b Dt[x] + b x^a y^(-1 + b) Dt[y]
Use Attributes. (i.e., give certain symbols the Constant Attribute.
In[6]:= Remove[a, b];
SetAttributes[{a, b}, Constant];
Dt[x^a y^b]
Out[8]= a x^(-1 + a) y^b Dt[x] + b x^a y^(-1 + b) Dt[y]
Use Rules to alter the output of the main Dt[] expression.
In[9]:= Remove[a, b];
Dt[x^a y^b] /. Dt[a] -> 0 /. Dt[b] -> 0
Out[10]= a x^(-1 + a) y^b Dt[x] + b x^a y^(-1 + b) Dt[y]
Related
I am very new to Mathematica, and I am trying to solve the following problem.
I have a cubic equation of the form Z = aZ^3 + bZ^2 + a + b. The first thing I want to do is to get a function that solves this analytically for Z and chooses the minimal positive root for that, as a function of a and b.
I thought that in order to get the root I could use:
Z = Solve[z == az^3 + bz^2 + a + b, z];
It seems like I am not quite getting the roots, as I would expect using the general cubic equation solution formula.
I want to integrate the minimal positive root of Z over a and b (again, preferably analytically) from 0 to 1 for a and for a to 1 for b.
I tried
Y = Integrate[Z, {a, 0, 1}, {b, a, 1}];
and that does not seem to give any formula or numerical value, but just returns an integral. (Notice I am not even sure how to pick the minimal positive root, but I am playing around with Mathematica to try to figure it out.)
Any ideas on how to do this?
Spaces between a or b and z are important. You can get the roots by:
sol = z /. Solve[z == a z^3 + b z^2 + a + b, z]
However, are you sure this expression has a solution as you expect? For a=0.5 and b=0.5, the only real root is negative.
sol /. {a->0.5, b->0.5}
{-2.26953,0.634765-0.691601 I,0.634765+0.691601 I}
sol = z /. Solve[z == a z^3 + b z^2 + a + b, z];
zz[a0_ /; NumericQ[a0], b0_ /; NumericQ[b0]] :=
Min[Select[ sol /. {a -> a0, b -> b0} ,
Element[#, Reals] && # > 0 & ]]
This returns -infinty when there are no solutions. As sirintinga noted your example integration limits are not valid..
RegionPlot[NumericQ[zz[a, b] ] , {a, -1, .5}, {b, -.5, 1}]
but you can numerically integrate if you have a valid region..
NIntegrate[zz[a, b], {a, -.5, -.2}, {b, .8, .9}] ->> 0.0370076
Edit ---
there is a bug above Select in Reals is throwin away real solutions with an infinitesimal complex part.. fix as:..
zz[a0_ /; NumericQ[a0], b0_ /; NumericQ[b0]] :=
Min[Select[ Chop[ sol /. {a -> a0, b -> b0} ],
Element[#, Reals] && # > 0 & ]]
Edit2, a cleaner approach if you dont find Chop satisfyting..
zz[a0_ /; NumericQ[a0], b0_ /; NumericQ[b0]] :=
Module[{z, a, b},
Min[z /. Solve[
Reduce[(z > 0 && z == a z^3 + b z^2 + a + b /.
{ a -> a0, b -> b0}), {z}, Reals]]]]
RegionPlot[NumericQ[zz[a, b] ] , {a, -2, 2}, {b, -2, 2}]
NIntegrate[zz[a, b], {a, 0, .5}, {b, 0, .5 - a}] -> 0.0491321
I had a math problem I solved like this:
In[1]:= Solve[2x(a-x)==0, x]
Out[1]= {{x->0}, {x->a}}
In[2]:= Integrate[2x(a-x), {x,0,a}]
Out[2]= (a^3)/3
In[3]:= Solve[(a^3)/3==a, a]
Out[3]= {{a->0}, {a->-Sqrt[3]}, {a->Sqrt[3]}}
My question is if I could rewrite this to compute it in one step, rather than having to manually input the result from the previous line. I could easily replace the integral used in step three with the Integrate command from step two. But what I can't figure out is how I would use the result from step 1 as the limits of integration in the integral.
You could combine step 1 and 2 by doing something like
Integrate[2 x (a - x), {x, ##}] & ## (x /. Solve[2 x (a - x) == 0, x]);
If you agree to delegate the choice of the (positive oriented) domain to Integrate, by means of using Clip or Boole:
In[77]:= Solve[
Integrate[
Clip[2 x (a - x), {0, Infinity}], {x, -Infinity, Infinity}] == a, a]
Out[77]= {{a -> 0}, {a -> Sqrt[3]}}
or
In[81]:= Solve[
Integrate[
2 x (a - x) Boole[2 x (a - x) > 0], {x, -Infinity, Infinity}] ==
a, a]
Out[81]= {{a -> 0}, {a -> Sqrt[3]}}
The reason only non-negative roots are found, is that Integrate will integrate from the smallest root to the largest root, i.e. from {x,0,a} for positive a and {x,a,0} for negative a.
I have a long expression that I would like to split into a collection of terms. For example say I have:
a + b - c + d + 4*e - 3*f
I want to split the expression by addition/subtraction into:
{a, b, -c, d, 4*e, -3*f}
My motivation for this is that I want to deal with the original expression term by term. Is this possible?
Edit: The examples given are VERY simplistic compared to what I'm actually dealing with in Mathematica, it's just that I'm not sure how to write Math around here.
To split the expression, you need to use Level. Level gives you a list of subexpressions and you can specify the level at which you want the subexpressions returned. In this case, you need levelspec 1.
In[1]:= expr = a + b - c + d + 4 e - 3 f;
In[2]:= Level[expr, 1]
Out[2]= {a, b, -c, d, 4 e, -3 f}
An example with a slightly more complicated expression:
In[3]:= expr2 = a^2 + 5 bc/ef - Sqrt[g - h] - Cos[i]/Sin[j + k];
In[4]:= Level[expr2, 1]
Out[4]= {a^2, (5 bc)/ef, -Sqrt[g - h], -Cos[i] Csc[j + k]}
Since no one else has mentioned it, equivalent to Yoda's Level[expr, 1] construction is to use Apply to replace the head of an expression with List:
In[1]:= expr = a + b - c + d + 4 e - 3 f;
In[2]:= List ## expr
Level[expr, 1] == %
Out[2]= {a, b, -c, d, 4 e, -3 f}
Out[3]= True
In[4]:= expr2 = a^2 + 5 bc/ef - Sqrt[g - h] - Cos[i]/Sin[j + k];
In[5]:= List ## expr2
Level[expr2, 1] == %
Out[5]= {a^2, (5 bc)/ef, -Sqrt[g - h], -Cos[i] Csc[j + k]}
Out[6]= True
The two methods do basically the same thing and have identical timings (using my version of a average timing function)
In[1]:= SetOptions[TimeAv, Method -> {"MinNum", 80000}, "BlockSize" -> 20000];
In[7]:= List ## expr // TimeAv
Total wall time is 0.244517, total cpu time is 0.13
and total time spent evaluating the expression is 0.13
The expression was evaluated 80000 times, in blocks of 20000 runs. This yields
a mean timing of 1.625*10^-6 with a blocked standard deviation of 2.16506*10^-7.
Out[7]= {1.625*10^-6, {a, b, -c, d, 4 e, -3 f}}
In[8]:= Level[expr, 1] // TimeAv
Total wall time is 0.336927, total cpu time is 0.16
and total time spent evaluating the expression is 0.16
The expression was evaluated 80000 times, in blocks of 20000 runs. This yields
a mean timing of 2.*10^-6 with a blocked standard deviation of 3.53553*10^-7.
Out[8]= {2.*10^-6, {a, b, -c, d, 4 e, -3 f}}
You might also be able to use MonomialList, if you expression is a polynomial:
In[56]:= MonomialList[a + b - c + d + 4*e - 3*f]
Out[56]= {a, b, -c, d, 4 e, -3 f}
(Doesn't work on non-polynomials, such as Yoda's expr2.)
You could also use Replace:
In[65]:= Replace[a + b - c + d + 4*e - 3*f, HoldPattern[Plus[a___]] :> {a}]
Out[65]= {a, b, -c, d, 4 e, -3 f}
You need to use HoldPattern (or some equivalent trick) to prevent Plus[a__] from evaluating to a__, which has the result of just wrapping the first argument in a list instead of creating a list of the arguments to Plus.
Say I have a list of Rules
rules = {a -> b, c -> d};
which I use throughout a notebook. Then, at one point, it makes sense to want the rules to apply before any other evaluations take place in an expression. Normally if you want something like this you would use
In[2]:= With[{a=b,c=d}, expr[a,b,c,d]]
Out[2]= expr[b, b, d, d]
How can I take rules and insert it into the first argument of With?
Edit
BothSome solutions fail do all that I was looking for - but I should have emphasised this point a little more. See the bold part above.
For example, let's look at
rules = {a -> {1, 2}, c -> 1};
If I use these vaules in With, I get
In[10]:= With[{a={1,2},c=1}, Head/#{a,c}]
Out[10]= {List,Integer}
Some versions of WithRules yield
In[11]:= WithRules[rules, Head/#{a,c}]
Out[11]= {Symbol, Symbol}
(Actually, I didn't notice that Andrew's answer had the Attribute HoldRest - so it works just like I wanted.)
You want to use Hold to build up your With statement. Here is one way; there may be a simpler:
In[1]:= SetAttributes[WithRules, HoldRest]
In[2]:= WithRules[rules_, expr_] :=
With ## Append[Apply[Set, Hold#rules, {2}], Unevaluated[expr]]
Test it out:
In[3]:= f[args___] := Print[{args}]
In[4]:= rules = {a -> b, c -> d};
In[5]:= WithRules[rules, f[a, c]]
During evaluation of In[5]:= {b,d}
(I used Print so that any bug involving me accidentally evaluating expr too early would be made obvious.)
I have been using the following form of WithRules for a long time. Compared to the one posted by Andrew Moylan, it binds sequentially so that you can say e.g. WithRules[{a->b+1, b->2},expr] and get a expanded to 3:
SetAttributes[WithRules, HoldRest]
WithRules[rules_, expr_] := ReleaseHold#Module[{notSet}, Quiet[
With[{args = Reverse[rules /. Rule[a_, b_] -> notSet[a, b]]},
Fold[With[{#2}, #1] &, Hold#expr, args]] /. notSet -> Set,
With::lvw]]
This was also posted as an answer to an unrelated question, and as noted there, it has been discussed (at least) a couple of times on usenet:
A version of With that binds variables sequentially
Add syntax highlighting to own command
HTH
EDIT: Added a ReleaseHold, Hold pair to keep expr unevaluated until the rules have been applied.
One problem with Andrew's solution is that it maps the problem back to With, and that does not accept subscripted variables. So the following generates messages.
WithRules[{Subscript[x, 1] -> 2, Subscript[x, 2] -> 3},
Power[Subscript[x, 1], Subscript[x, 2]]]
Given that With performs syntactic replacement on its body, we can set WithRules alternatively as follows:
ClearAll[WithRules]; SetAttributes[WithRules, HoldRest];
WithRules[r : {(_Rule | _RuleDelayed) ..}, body_] :=
ReleaseHold[Hold[body] /. r]
Then
In[113]:= WithRules[{Subscript[x, 1] -> 2,
Subscript[x, 2] -> 3}, Subscript[x, 1]^Subscript[x, 2]]
Out[113]= 8
Edit: Addressing valid concerns raised by Leonid, the following version would be safe:
ClearAll[WithRules3]; SetAttributes[WithRules3, HoldRest];
WithRules3[r : {(_Rule | _RuleDelayed) ..}, body_] :=
Developer`ReplaceAllUnheld[Unevaluated[body], r]
Then
In[194]:= WithRules3[{Subscript[x, 1] -> 2, Subscript[x, 2] -> 3},
Subscript[x, 1]^Subscript[x, 2]]
Out[194]= 8
In[195]:= WithRules3[{x -> y}, f[y_] :> Function[x, x + y]]
Out[195]= f[y_] :> Function[x, x + y]
Edit 2: Even WithRules3 is not completely equivalent to Andrew's version:
In[206]:= WithRules3[{z -> 2}, f[y_] :> Function[x, x + y + z]]
Out[206]= f[y_] :> Function[x, x + y + z]
In[207]:= WithRules[{z -> 2}, f[y_] :> Function[x, x + y + z]]
Out[207]= f[y$_] :> Function[x$, x$ + y$ + 2]
I'm currently doing some normalization along the lines of:
J = Integrate[Psi[x, 0]^2, {x, 0, a}]
sol = Solve[J == 1, A]
A /. sol
For this type of normalization, the negative square root is extraneous. The result of this calculation is:
In[49]:= J = Integrate[Psi[x, 0]^2, {x, 0, a}]
Out[49]= 2 A^2
In[68]:= sol = Solve[J == 1, A]
Out[68]= {{A -> -(1/Sqrt[2])}, {A -> 1/Sqrt[2]}}
Even if I try giving it an Assuming[...] or Simplify[...], it still gives me the same results:
In[69]:= sol = Assuming[A > 0, Solve[J == 1, A]]
Out[69]= {{A -> -(1/Sqrt[2])}, {A -> 1/Sqrt[2]}}
In[70]:= sol = FullSimplify[Solve[J == 1, A], A > 0]
Out[70]= {{A -> -(1/Sqrt[2])}, {A -> 1/Sqrt[2]}}
Can anyone tell me what I'm doing wrong here?
I'm running Mathematica 7 on Windows 7 64-bit.
ToRules does what the box says: converts equations (as in Reduce output) to rules. In your case:
In[1]:= ToRules[Reduce[{x^2==1,x>0},x]]
Out[1]= {x->1}
In[2]:= {ToRules[Reduce[{x^2==1},x]]}
Out[2]= {{x->-1},{x->1}}
For more complex cases, I have often found it useful to just check the value of the symbolic solutions after pluging in typical parameter values. This is not foolproof, of course, but if you know there is one and only one solution then it is a simple and efficient method:
Solve[x^2==someparameter,x]
Select[%,((x/.#)/.{someparameter-> 0.1})>0&]
Out[3]= {{x->-Sqrt[someparameter]},{x->Sqrt[someparameter]}}
Out[4]= {{x->Sqrt[someparameter]}}
Solve doesn't work like this. You might try Reduce, instead, e.g.
In[1]:= Reduce[{x^2 == 1, x > 0}, x]
Out[1]= x == 1
It's then a little tricky to transform this output to replacement rules, at least in the general case, because Reduce might use arbitrary many logical connectives. In this case, we could just hack:
In[2]:= Solve[Reduce[{x^2 == 1, x > 0}, x], x]
Out[2]= {{x->1}}