Very very basic question about Mathematica expressions
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From: Sam Takoy on 20 Jul 2010 03:42 Hi, I find that I don't understand a very basic thing about the building blocks of Mathematica. When the first command is s = x + h I figured that (string?) "s" represents (string?) "x+h". This theory pans out when one tries s^2 or D[s, x] or Integrate[s, {x, 0, h}] and s /. h -> 1 But then by doesn't this work: Manipulate[ Plot[s, {x, 0, h}], {h, 0.1, 1}] Many thanks in advance!
From: Themis Matsoukas on 21 Jul 2010 07:12 The control variable (h) is internal to Manipulate but you can link it to an external variable. The following works: s := x + h; Manipulate[Plot[s /. h -> H, {x, 0, H}], {H, 0.1, 1}] Notice the := in the definition of s. If you replace it with =, the above will still work but only as long as h has not been assigned a numerical value. However, if you do assign a numerical value, say h=3, the graph will plot x+3, regardless of the value you pick for H. For this reason it is a good idea to use the delayed assignment (:=) rather than the regular =. Themis
From: Helen Read on 21 Jul 2010 07:12 On 7/20/2010 3:42 AM, Sam Takoy wrote: > Hi, > > I find that I don't understand a very basic thing about the building > blocks of Mathematica. > > When the first command is > > s = x + h > > I figured that (string?) "s" represents (string?) "x+h". To make it a string, i.e., a sequence of text characters, you would enter the following. s="x+h" Strings are useful for lots of things, but I don't think a string is what you intend. What you have instead is an expression. > But then by doesn't this work: > > Manipulate[ Plot[s, {x, 0, h}], {h, 0.1, 1}] > > Many thanks in advance! Manipulate needs to know that s is a function of both x and h. Clear[s] s[x_,h_]:=x+h Also you should put an explicit PlotRange in your Plot, so that Manipulate doesn't go changing the scale on you. Try it without a PlotRange: Manipulate[ Plot[s[x, h], {x, 0, h}] And see how much better it is with one. Manipulate[ Plot[s[x, h], {x, 0, h}, PlotRange -> {{0, 1}, {0, 1}}], {h, 0.1, 1}] Defining functions is a very basic idea in Mathematica, and worth getting used to. For example, suppose you want to find the extrema of the function f(x)= -84 x + 9 x^2 + 4 x^3 + 25 So define it as a function. f[x_] := -84 x + 9 x^2 + 4 x^3 + 25 Plot[f[x], {x, -10, 10}] Might want to see the 1st derivative along with it. Plot[{f[x], f'[x]}, {x, -10, 10}] Solve[f'[x] == 0, x] f[-7/2] f[2] -- Helen Read University of Vermont
From: Yanping on 21 Jul 2010 07:12 Hi, I am a newbie. For your question, if you check the PLOT function, Plot[f,{X,Subscript[X, min],Subscript[X, max]}] the f has to be a function of X, while in your case, s=x+h, s is NOT a function of x. If you use s[x_,h_]:=x+h; Manipulate[ Plot[s[x,h], {x, 0, h}], {h, 0.1, 1}] It would work the way you want. best, yp On Jul 20, 3:42 am, Sam Takoy <sam.ta...(a)yahoo.com> wrote: > Hi, > > I find that I don't understand a very basic thing about the building > blocks of Mathematica. > > When the first command is > > s = x + h > > I figured that (string?) "s" represents (string?) "x+h". > > This theory pans out when one tries > > s^2 > > or > > D[s, x] > > or > > Integrate[s, {x, 0, h}] > > and > > s /. h -> 1 > > But then by doesn't this work: > > Manipulate[ Plot[s, {x, 0, h}], {h, 0.1, 1}] > > Many thanks in advance!
From: dr DanW on 21 Jul 2010 07:14
I ran into this problem yesterday. I don't know exactly why it happens, I think it has something to do with the way Manipulate localizes variables. To solve it, I use a trick I found that lets me take an expression built up of global symbols and localize the symbols. Your trivial example: s = x + h Make a function out of it. The Evaluate[] is necessary to evaluate s, which replaces it with x+h sfnc = Function[{x, h}, Evaluate[s]] Now the Manipulate[] works fine Manipulate[Plot[sfnc[x, h], {x, 0, h}], {h, 0.1, 1}] I find myself using this trick a lot. Regards, Daniel |