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Mathematics Algebra Wolfram Language

Using ReplaceAll (/.) with complex variables?

Posted 8 years ago

I have hopefully a simple question. (I'm using home version of Mathematica 7) If define a simple complex matrix test = {{a + Ib, b + Ic}, {d + Ie, e + If}} and do a simple replacement test /. {e -> 1, f -> 2} How come the complex parts using the symbols e and f are not replaced? The above yields {{a + Ib, b + Ic}, {d + Ie, 1 + If}} If I do something simple test2 = a + I b test2 /. {a -> 4, b -> 2} Yields: 4 + 2 I as expected, but if I call up the [[2,2]] element of the array above and try the ReplaceAll it doesn't work test6 = test[[2, 2]] Yields: e + If but test6 /. {e -> 3, f -> 5} Yields: 3 + If So how come the complex part doesn't get replaced? How can I do this replacement?

POSTED BY: Paul Hillman

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Posted 8 years ago

Without any disrespect, whats the fuzz? The OP's replacements does seem to work (provided there is a space left between I and the variable symbol): In[1]:= test = {{a + I b, b + I c}, {d + I e, e + I f}} Out[1]= {{a + I b, b + I c}, {d + I e, e + I f}} In[2]:= test /. {e -> 1, f -> 2} Out[2]= {{a + I b, b + I c}, {I + d, 1 + 2 I}}

POSTED BY: Hans Milton

Posted 8 years ago

Hint: Ie is a single symbol. Ie or I e is the product of two things. 2I or 2I or 2 I also happen to be the product of two things. That should be enough.

POSTED BY: Bill Simpson

Posted 8 years ago

This is not the whole story. `ReplaceAll` (and related functions) use the FullForm for pattern matching, and to replace a part of the complex number one needs to use this form. It is true that `3 + 4 I` is atomic but one may use the head of this expression to get into its parts FullForm[3 + 4 I] returns Complex[3,4] therefore 3+ 4 I /. Complex[x_,4]:> Complex[x,12] returns 3 + 12 I as expected

POSTED BY: Yehuda Ben-Shimol

Posted 8 years ago

Sorry about the coloring, don't really know how that happened. a,b,c,d,e,f were suppose to be unknown symbols.

POSTED BY: Paul Hillman

Posted 8 years ago

You could've noticed it from the coloring already; unknown symbols are blue (or green inside e.g. Module) (a,b,c,d,e,f), known symbols are black (I).... Note that you can't use replace(all) on complex numbers: 3 + 4 I /. 4 -> 5 as they are treated as an 'atom', so it will just return 3 + 4I.

POSTED BY: Sander Huisman

Posted 8 years ago

Thanks Bill, for some reason I thought that I was reserved, so Ie and If would be a product to two things.

POSTED BY: Paul Hillman

Posted 8 years ago

Your thinking that I is reserved is correct, but that isn't sufficient to override the principle that anything beginning with a letter followed by any number of letters and digits will be considered to be one symbol. This kind of problem shows up again and again and again and again...