How can I define a function f(x) in Mathematica that gives 1 if x is in [-5, -4] or [1, 3] and 0 otherwise? It's probably something simple but I just can't figure it out!
The basic construction you want is Piecewise, in particular the function you were asking for can be written as
f[x_] := Piecewise[{{1, -5 <= x <= -3}, {1, 1 <= x <= 3}}, 0]
or
f[x_] := Piecewise[{{1, -5 <= x <= -3 || 1 <= x <= 3}}, 0]
Note that the final argument, 0 defines the default (or "else") value is not needed because the default default is 0.
Also note that although Piecewise and Which are very similar in form, Piecewise is for constructing functions, while Which is for programming. Piecewise will play nicer with integration, simplification etc..., it also has the proper left-brace mathematical notation, see the examples in the documentation.
Since the piecewise function you want is quite simple, it could also be constructed from step functions like Boole, UnitStep and UnitBox, e.g.
UnitBox[(x + 4)/2] + UnitBox[(x - 2)/2]
These are just special cases of Piecewise, as shown by PiecewiseExpand
In[19]:= f[x] == UnitBox[(x+4)/2] + UnitBox[(x-2)/2]//PiecewiseExpand//Simplify
Out[19]= True
Alternatively, you can use switching functions like HeavisideTheta or HeavisidePi, e.g.
HeavisidePi[(x + 4)/2] + HeavisidePi[(x - 2)/2]
which are nice, because if treating the function as a distribution, then its derivative will return the correct combination of Dirac delta functions.
For more discussion see the tutorial Piecewise Functions.
Although Simon's answer is the canonical and correct one, here are another two options:
f[x_] := 1 /; IntervalMemberQ[Interval[{-5, -3}, {1, 3}], x]
f[x_?NumericQ] := 0
or
f[x_] := If[-5 <= x <= -3 || 1 <= x <= 3, 1, 0]
Edit:
Note that the first option depends on the order that the definitions were entered (thanks Sjoerd for pointing this out). A similar solution that does not have this problem and will also work correctly when supplied an Interval as input is
f[x_] := 0 /; !IntervalMemberQ[Interval[{-5, -3}, {1, 3}], x]
f[x_] := 1 /; IntervalMemberQ[Interval[{-5, -3}, {1, 3}], x]
All is good and well but as a general rule of the thumb one should try always the simplest approach and keep away as possible from the sophisticated high level programming. In this particular situation I mean the following:
f[x_ /; -5 <= x <= -3] = 0 etc ... etc
Related
I aim to calculate and preserve the results from the maximization of a function with two arguments and one exogenous parameter, when the maximum can not be derived (in closed form) by maximize. For instance, let
f[x_,y_,a_]=Max[0,Min[a-y,1-x-y]
be the objective function where a is positive. The maximization shall take place over [0,1]^2, therefore I set
m[a_]=Maximize[{f[x, y, a], 0 <= x <= 1 && 0 <= y <= 1 && 0 <= a}, {x,y}]
Obviously m can be evaluated at any point a and it is therefore possible to plot the maximizing x by employing
Plot[x /. m[a][[2]], {a, 0.01, 1}]
As I need to do several plots and further derivations containing the optimal solutions x and y (which of course are functions of a), i would like to preserve/save the results from the optimization for further use. Is there an elegant way to do this, or do I have to write some kind of loop to extract the values myself?
Now that I've seen the full text of your comment on my original comment, I suspect that you do understand the differences between Set and SetDelayed well enough. I think what you may be looking for is memoisation, sometimes implemented a bit like this;
f[x_,y_] := f[x,y] = Max[0,Min[a-y,1-x-y]]
When you evaluate, for example f[3,4] for the first time it will evaluate to the entire expression to the right of the :=. The rhs is the assignment f[3,4] = Max[0,Min[a-y,1-x-y]]. Next time you evaluate f[3,4] Mathematica already has a value for it so doesn't need to recompute it, it just recalls it. In this example the stored value would be Max[0,Min[a-4,-6]] of course.
I remain a little uncertain of what you are trying to do so this answer may not be any use to you at all.
Simple approach
results = Table[{x, y, a} /. m[a][[2]], {a, 0.01, 1, .01}]
ListPlot[{#[[3]], #[[1]]} & /# results, Joined -> True]
(The Set = is ok here so long as 'a' is not previosly defined )
If you want to utilise Plot[]s automatic evaluation take a look at Reap[]/Sow[]
{p, data} = Reap[Plot[x /. Sow[m[a]][[2]], {a, 0.01, 1}]];
Show[p]
(this takes a few minutes as the function output is a mess..).
hmm try this again: assuming you want x,y,a and the minimum value:
{p, data} = Reap[Plot[x /. Sow[{a, m[a]}][[2, 2]], {a, 0.01, .1}]];
Show[p]
results = {#[[1]], x /. #[[2, 2]], y /. #[[2, 2]], #[[2, 1]]} & /# data[[1]]
BTW Your function appears to be independent of x over some ranges which is why the plot is a mess..
I am trying to get Mathematica to approximate an integral that is a function of various parameters. I don't need it to be extremely precise -- the answer will be a fraction, and 5 digits would be nice, but I'd settle for as few as 2.
The problem is that there is a symbolic integral buried in the main integral, and I can't use NIntegrate on it since its symbolic.
F[x_, c_] := (1 - (1 - x)^c)^c;
a[n_, c_, x_] := F[a[n - 1, c, x], c];
a[0, c_, x_] = x;
MyIntegral[n_,c_] :=
NIntegrate[Integrate[(D[a[n,c,y],y]*y)/(1-a[n,c,x]),{y,x,1}],{x,0,1}]
Mathematica starts hanging when n is greater than 2 and c is greater than 3 or so (generally as both n and c get a little higher).
Are there any tricks for rewriting this expression so that it can be evaluated more easily? I've played with different WorkingPrecision and AccuracyGoal and PrecisionGoal options on the outer NIntegrate, but none of that helps the inner integral, which is where the problem is. In fact, for the higher values of n and c, I can't even get Mathematica to expand the inner derivative, i.e.
Expand[D[a[4,6,y],y]]
hangs.
I am using Mathematica 8 for Students.
If anyone has any tips for how I can get M. to approximate this, I would appreciate it.
Since you only want a numerical output (or that's what you'll get anyway), you can convert the symbolic integration into a numerical one using just NIntegrate as follows:
Clear[a,myIntegral]
a[n_Integer?Positive, c_Integer?Positive, x_] :=
a[n, c, x] = (1 - (1 - a[n - 1, c, x])^c)^c;
a[0, c_Integer, x_] = x;
myIntegral[n_, c_] :=
NIntegrate[D[a[n, c, y], y]*y/(1 - a[n, c, x]), {x, 0, 1}, {y, x, 1},
WorkingPrecision -> 200, PrecisionGoal -> 5]
This is much faster than performing the integration symbolically. Here's a comparison:
yoda:
myIntegral[2,2]//Timing
Out[1]= {0.088441, 0.647376595...}
myIntegral[5,2]//Timing
Out[2]= {1.10486, 0.587502888...}
rcollyer:
MyIntegral[2,2]//Timing
Out[3]= {1.0029, 0.647376}
MyIntegral[5,2]//Timing
Out[4]= {27.1697, 0.587503006...}
(* Obtained with WorkingPrecision->500, PrecisionGoal->5, MaxRecursion->20 *)
Jand's function has timings similar to rcollyer's. Of course, as you increase n, you will have to increase your WorkingPrecision way higher than this, as you've experienced in your previous question. Since you said you only need about 5 digits of precision, I've explicitly set PrecisionGoal to 5. You can change this as per your needs.
To codify the comments, I'd try the following. First, to eliminate infinite recursion with regards to the variable, n, I'd rewrite your functions as
F[x_, c_] := (1 - (1-x)^c)^c;
(* see note below *)
a[n_Integer?Positive, c_, x_] := F[a[n - 1, c, x], c];
a[0, c_, x_] = x;
that way n==0 will actually be a stopping point. The ?Positive form is a PatternTest, and useful for applying additional conditions to the parameters. I suspect the issue is that NIntegrate is re-evaluating the inner Integrate for every value of x, so I'd pull that evaluation out, like
MyIntegral[n_,c_] :=
With[{ int = Integrate[(D[a[n,c,y],y]*y)/(1-a[n,c,x]),{y,x,1}] },
NIntegrate[int,{x,0,1}]
]
where With is one of several scoping constructs specifically for creating local constants.
Your comments indicate that the inner integral takes a long time, have you tried simplifying the integrand as it is a derivative of a times a function of a? It seems like the result of a chain rule expansion to me.
Note: as per Yoda's suggestion in the comments, you can add a cacheing, or memoization, mechanism to a. Change its definition to
d:a[n_Integer?Positive, c_, x_] := d = F[a[n - 1, c, x], c];
The trick here is that in d:a[ ... ], d is a named pattern that is used again in d = F[...] cacheing the value of a for those particular parameter values.
Say I have a crazy function, f, defined like so:
util[x_, y_, c_] := 0.5*Log[c-x] + 0.5*Log[c-y]
cost[x_, y_, l_] := c /. First[NSolve[util[x, y, c+l] == Log[10+l], c]]
prof[x_, y_] := 0.01*Norm[{x,y}, 2]
liquid[x_, y_] := 0.01*Norm[{x,y}, 2]
f[x_, y_, a_, b_] := cost[a, b, liquid[x,y] + liquid[a-x, b-y]] - Max[a,b]
- cost[0,0,0] + prof[x,y] + liquid[x,y] + prof[a-x, b-y] + liquid[a-x, b-y]
Now I call NMinimize like this:
NMinimize[{f[50, 50, k, j], k >= 49, k <= 51, j >= 49, j <= 51}, {j, k}]
Which tells me this:
{-21.0465, {j -> 51., k -> 49.}}
But then if I actually check what f[50,50,49,51] is, it's this:
0.489033
Which is pretty different from the -21.0465 that NMinimize said.
Is this par for the course with NMinimize?
Floating point errors compounding or whatnot?
Any ideas for beating NMinimize (or some such function) into submission?
It certainly seems to be related to your function f not being restricted to numerical arguments, plus the symbolic preprocessing performed by NMinimize. Once you change the signature to
f[x_?NumericQ, y_?NumericQ, a_?NumericQ, b_?NumericQ]:=...
The result is as expected, although it takes considerably longer to get it.
EDIT
We can dig deeper to reveal the true reason. First, note that your f (the original one, args unrestricted) is quite a function:
In[1423]:= f[50,50,49.,51.]
Out[1423]= 0.489033
In[1392]:= f[50,50,k,j]/.{j->51.`,k->49.`}
Out[1392]= -21.0465
The real culprit is NSolve, which gives two ordered solutions:
In[1398]:= NSolve[util[x,y,c+l]==Log[10+l],c]
Out[1398]= {{c->0.5 (-2. l+1. x+1. y-2. Sqrt[100.+20. l+1. l^2+0.25 x^2-0.5 x y+0.25 y^2])},
{c->0.5 (-2. l+1. x+1. y+2. Sqrt[100.+20. l+1. l^2+0.25 x^2-0.5 x y+0.25 y^2])}}
The problem is, what is the ordering. It turns out to be different for symbolic and numeric arguments to NSolve, because in the latter case we don't have any symbols around. This can be seen as:
In[1399]:=
Block[{cost},
cost[x_,y_,l_]:=c/.Last[NSolve[util[x,y,c+l]==Log[10+l],c]];
f[50,50,k,j]/.{j->51.,k->49.}]
Out[1399]= 0.489033
So you really have to settle on what is the right ordering for you, and which solution you really want to pick.
I have a basic problem in Mathematica which has puzzled me for a while. I want to take the m'th derivative of x*Exp[t*x], then evaluate this at x=0. But the following does not work correct. Please share your thoughts.
D[x*Exp[t*x], {x, m}] /. x -> 0
Also what does the error mean
General::ivar: 0 is not a valid variable.
Edit: my previous example (D[Exp[t*x], {x, m}] /. x -> 0) was trivial. So I made it harder. :)
My question is: how to force it to do the derivative evaluation first, then do substitution.
As pointed out by others, (in general) Mathematica does not know how to take the derivative an arbitrary number of times, even if you specify that number is a positive integer.
This means that the D[expr,{x,m}] command remains unevaluated and then when you set x->0, it's now trying to take the derivative with respect to a constant, which yields the error message.
In general, what you want is the m'th derivative of the function evaluated at zero.
This can be written as
Derivative[m][Function[x,x Exp[t x]]][0]
or
Derivative[m][# Exp[t #]&][0]
You then get the table of coefficients
In[2]:= Table[%, {m, 1, 10}]
Out[2]= {1, 2 t, 3 t^2, 4 t^3, 5 t^4, 6 t^5, 7 t^6, 8 t^7, 9 t^8, 10 t^9}
But a little more thought shows that you really just want the m'th term in the series, so SeriesCoefficient does what you want:
In[3]:= SeriesCoefficient[x*Exp[t*x], {x, 0, m}]
Out[3]= Piecewise[{{t^(-1 + m)/(-1 + m)!, m >= 1}}, 0]
The final output is the general form of the m'th derivative. The PieceWise is not really necessary, since the expression actually holds for all non-negative integers.
Thanks to your update, it's clear what's happening here. Mathematica doesn't actually calculate the derivative; you then replace x with 0, and it ends up looking at this:
D[Exp[t*0],{0,m}]
which obviously is going to run into problems, since 0 isn't a variable.
I'll assume that you want the mth partial derivative of that function w.r.t. x. The t variable suggests that it might be a second independent variable.
It's easy enough to do without Mathematica: D[Exp[t*x], {x, m}] = t^m Exp[t*x]
And if you evaluate the limit as x approaches zero, you get t^m, since lim(Exp[t*x]) = 1. Right?
Update: Let's try it for x*exp(t*x)
the mth partial derivative w.r.t. x is easily had from Wolfram Alpha:
t^(m-1)*exp(t*x)(t*x + m)
So if x = 0 you get m*t^(m-1).
Q.E.D.
Let's see what is happening with a little more detail:
When you write:
D[Sin[x], {x, 1}]
you get an expression in with x in it
Cos[x]
That is because the x in the {x,1} part matches the x in the Sin[x] part, and so Mma understands that you want to make the derivative for that symbol.
But this x, does NOT act as a Block variable for that statement, isolating its meaning from any other x you have in your program, so it enables the chain rule. For example:
In[85]:= z=x^2;
D[Sin[z],{x,1}]
Out[86]= 2 x Cos[x^2]
See? That's perfect! But there is a price.
The price is that the symbols inside the derivative get evaluated as the derivative is taken, and that is spoiling your code.
Of course there are a lot of tricks to get around this. Some have already been mentioned. From my point of view, one clear way to undertand what is happening is:
f[x_] := x*Exp[t*x];
g[y_, m_] := D[f[x], {x, m}] /. x -> y;
{g[p, 2], g[0, 1]}
Out:
{2 E^(p t) t + E^(p t) p t^2, 1}
HTH!
I want an arbitrary function p[x] that integrates to 1 and for all x, 0 <= p[x] <= 1. Some kind of transformation rule?
You could use ProbabilityDistribution for this together with an undefined function of x:
dist = ProbabilityDistribution[p[x], {x, -Infinity, Infinity}];
It now knows a few rules to apply:
continuous probability density: probability of a single value is zero
In[26]:= Probability[x == 0, x \[Distributed] dist]
Out[26]= 0
the probability of having a value at all
In[28]:= Probability[x > 0 || x <= 0, x \[Distributed] dist]
Out[28]= 1
The CDF at - infinity
In[29]:= CDF[dist][-\[Infinity]]
Out[29]= 0
The CDF at + infinity
In[30]:= CDF[dist][\[Infinity]]
Out[30]= 1
The PDF
In[32]:= PDF[dist][x]
Out[32]= p[x]
However, it doesn't assume the PDF of the distribution is normalized:
In[33]:= Integrate[PDF[dist][x], {x, -Infinity, Infinity}]
Out[33]= Integrate[p[x], {x, -Infinity, Infinity}]
The latter can be taught, defining an UpValue for p:
p /: Integrate[p[x], {x, -Infinity, Infinity}] = 1;
Now it can integrate the PDF:
In[4]:= Integrate[PDF[dist][x], {x, -Infinity, Infinity}]
Out[4]= 1
You know that your second requirement, i.e. 0 <= p[x] <= 1, is not generally true for probability density functions, do you?
In case you're just asking for examples of density functions (PDFs) that match your criteria, here are two (out of uncountably many):
p(x) = 1 if 0 < x < 1
0 otherwise
p(x) = x/2 if 0 < x < 2
0 otherwise
We could even generalize those slightly:
p(x) = 1/k if 0 < x < k
0 otherwise
p(x) = 2x/k^2 if 0 < x < k
0 otherwise
The latter works for k >= 2.
We can even generalize that with another parameter to get a class of such functions with arbitrary exponent
p(x) = (a+1)/k^(a+1)*x^a if 0 < x < k
0 otherwise
which works for all a > 1 and k > a+1.
For more interesting examples I think you'll need to give more criteria.
You mention a transformation rule so perhaps you'd like to take an arbitrary bounded function on R1 and translate/scale it so that it's always between 0 and 1 and integrates to 1.
That will have a straightforward answer as long as you can get the min, max, and integral of the given function.
Go ahead and edit the question to ask that if that's indeed what you're looking for.