3
$\begingroup$

I have a complicated expression which I want to simplify following a rule. For example, if I want to make the following replacement:

A[t] - B*A[t - d1] = Y[t]

for any values of t and d1, for example,

A[t - 4] - B*A[t - 4 - d1] = Y[t-4]

Using the wildcard underscore notation, I've been able to get a simple version to work,

A[t-4] - B*A[-d1 + t-6] /. A[t_] - B*A[d_] ->Y[t]

which correctly returns Y[t-4]. However, if everything is premultiplied, then I cannot get the right result:

(x1*A[t - 4] - x1*B*A[-d1 + t - 4]) /. {A[t_] - B*A[d_] -> Y[t]}

which should give x1*Y[t-4]. I understand that if I look at the FullForm that Mathematica recognizes it as something different, but I need a way around that. I don't think I can easily collect terms as my expressions are something like:

(x1*A[t - 4] - x1*B*A[-d1 + t - 4]) + (x1*x2*A[t] - x1*x2*B*A[-d1 + t])

and I need a rule which changes that to x1*Y[t-4] + x1*x2*Y[t]. I was thinking about using another wildcard before the terms, but this doesn't appear to be working:

(x1*A[t - 4] - x1*B*A[-d1 + t - 4]) /. {x_*A[t_] - x*B*A[d_] -> x*Y[t]}

Any help or a simple way to make these replacements would be much appreciated!

$\endgroup$

1 Answer 1

4
$\begingroup$
rule = {a_. A[t_] - a_. B*A[d_] :>  a Y[t]};

Use it with ReplaceAll or ReplaceRepeated:

(A[t - 4] - B*A[-d1 + t - 4]) /. rule

Y[-4 + t]

(x1*A[t - 4] - x1*B*A[-d1 + t - 4]) /. rule

x1 Y[-4 + t]

(x1*A[t - 4] - x1*B*A[-d1 + t - 4]) + (x1*x2*A[t] - x1*x2*B*A[-d1 + t]) //. rule

x1 Y[-4 + t] + x1 x2 Y[t]

$\endgroup$
2
  • $\begingroup$ This is great, thank you! very clean solution $\endgroup$
    – mike_s
    Commented Oct 30, 2018 at 18:23
  • $\begingroup$ @mike_s, my pleasure. Thank you for the accept and welcome to mma.se. $\endgroup$
    – kglr
    Commented Oct 30, 2018 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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