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Replacement using pattern matching?
Highlighting text with StringReplacePart but also using Style, Subscriptreplacement rules from a pattern and a matching expressionString pattern matching & replacementHow to Make a change of variablesPattern matching & simple replacementTransformation rules to distinguish common variable with Subscript and OverBar $x,x^2,barx,barx^2, x_p,x_p^2, barx_p, barx^2_p $Pattern matching an expression involving TimesReplace custom functions, leave built in functions untouched?Symbol Level pattern matching and rule replacementHaving problems with using TableForm to format a table
$begingroup$
Starting with the expression
eq1 = 5 Exp[1 + 2 x + 3 x^2 + 4 x^3];
I want to implement the following transformation:
eq1 /. Exp[a_ x^2 + b_ x^3] -> f[a/b]
However, I am not getting the desired output
$5 f[3/4]exp(1+2 x)$.
Further, I also need to perform the following transformation.
eq2 = c Exp[a SuperStar[[Alpha]] + b [Alpha]];
eq2 /. d1_ Exp[d2 SuperStar[[Alpha]_]] -> f[d2]
I am providing the above expressions again in latex font for clarification.
$$
texteq2=c exp left(a alpha ^*+alpha bright);
$$
$$
texteq2text/., textd1$_$ exp left(alpha _^* textd2right)to f(textd2)
$$
Any help will be appreciated.
pattern-matching replacement
asked May 9 at 9:43
Mark RobinsonMark Robinson
1847
$endgroup$
add a comment |
$begingroup$
Starting with the expression
eq1 = 5 Exp[1 + 2 x + 3 x^2 + 4 x^3];
I want to implement the following transformation:
eq1 /. Exp[a_ x^2 + b_ x^3] -> f[a/b]
However, I am not getting the desired output
$5 f[3/4]exp(1+2 x)$.
Further, I also need to perform the following transformation.
eq2 = c Exp[a SuperStar[[Alpha]] + b [Alpha]];
eq2 /. d1_ Exp[d2 SuperStar[[Alpha]_]] -> f[d2]
I am providing the above expressions again in latex font for clarification.
$$
texteq2=c exp left(a alpha ^*+alpha bright);
$$
$$
texteq2text/., textd1$_$ exp left(alpha _^* textd2right)to f(textd2)
$$
Any help will be appreciated.
pattern-matching replacement
asked May 9 at 9:43
Mark RobinsonMark Robinson
1847
$endgroup$
add a comment |
$begingroup$
Starting with the expression
eq1 = 5 Exp[1 + 2 x + 3 x^2 + 4 x^3];
I want to implement the following transformation:
eq1 /. Exp[a_ x^2 + b_ x^3] -> f[a/b]
However, I am not getting the desired output
$5 f[3/4]exp(1+2 x)$.
Further, I also need to perform the following transformation.
eq2 = c Exp[a SuperStar[[Alpha]] + b [Alpha]];
eq2 /. d1_ Exp[d2 SuperStar[[Alpha]_]] -> f[d2]
I am providing the above expressions again in latex font for clarification.
$$
texteq2=c exp left(a alpha ^*+alpha bright);
$$
$$
texteq2text/., textd1$_$ exp left(alpha _^* textd2right)to f(textd2)
$$
Any help will be appreciated.
pattern-matching replacement
asked May 9 at 9:43
Mark RobinsonMark Robinson
1847
$endgroup$
Starting with the expression
eq1 = 5 Exp[1 + 2 x + 3 x^2 + 4 x^3];
I want to implement the following transformation:
eq1 /. Exp[a_ x^2 + b_ x^3] -> f[a/b]
However, I am not getting the desired output
$5 f[3/4]exp(1+2 x)$.
Further, I also need to perform the following transformation.
eq2 = c Exp[a SuperStar[[Alpha]] + b [Alpha]];
eq2 /. d1_ Exp[d2 SuperStar[[Alpha]_]] -> f[d2]
I am providing the above expressions again in latex font for clarification.
$$
texteq2=c exp left(a alpha ^*+alpha bright);
$$
$$
texteq2text/., textd1$_$ exp left(alpha _^* textd2right)to f(textd2)
$$
Any help will be appreciated.
pattern-matching replacement
pattern-matching replacement
asked May 9 at 9:43
Mark RobinsonMark Robinson
1847
asked May 9 at 9:43
Mark RobinsonMark Robinson
1847
asked May 9 at 9:43
Mark RobinsonMark Robinson
1847
asked May 9 at 9:43
Mark RobinsonMark Robinson
1847
asked May 9 at 9:43
Mark RobinsonMark Robinson
1847
1847
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
eq1 /. Exp[a_. x^2 + b_. x^3 + c_.] -> Exp[c] f[a/b]
5 E^(1 + 2 x) f[3/4]
eq2 /. d1_. Exp[d2_. SuperStar[α_] + d3_.] -> Exp[d3] f[d2]
E^(b α) f[a]
answered May 9 at 9:56
kglrkglr
193k10214435
$endgroup$
2
$begingroup$
If you use defaults in the replacement pattern (dots after the patterns), they become more versatile:Exp[a_. x^2 + b_. x^3 + c_.] -> Exp[c] f[a/b]. This will also work witheq1a = 5 Exp[x^2 + 4 x^3], for example. Also,d1_. Exp[d2_. SuperStar[α_] + d3_.] -> Exp[d3] f[d2]for the second one.
$endgroup$
– Roman
May 9 at 12:29
$begingroup$
@Roman, great point. Updated with default patterns.
$endgroup$
– kglr
May 9 at 12:49
$begingroup$
@Shadowray, thank you; I added the needed_.s.
$endgroup$
– kglr
May 9 at 13:52
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
eq1 /. Exp[a_. x^2 + b_. x^3 + c_.] -> Exp[c] f[a/b]
5 E^(1 + 2 x) f[3/4]
eq2 /. d1_. Exp[d2_. SuperStar[α_] + d3_.] -> Exp[d3] f[d2]
E^(b α) f[a]
answered May 9 at 9:56
kglrkglr
193k10214435
$endgroup$
2
$begingroup$
If you use defaults in the replacement pattern (dots after the patterns), they become more versatile:Exp[a_. x^2 + b_. x^3 + c_.] -> Exp[c] f[a/b]. This will also work witheq1a = 5 Exp[x^2 + 4 x^3], for example. Also,d1_. Exp[d2_. SuperStar[α_] + d3_.] -> Exp[d3] f[d2]for the second one.
$endgroup$
– Roman
May 9 at 12:29
$begingroup$
@Roman, great point. Updated with default patterns.
$endgroup$
– kglr
May 9 at 12:49
$begingroup$
@Shadowray, thank you; I added the needed_.s.
$endgroup$
– kglr
May 9 at 13:52
add a comment |
$begingroup$
eq1 /. Exp[a_. x^2 + b_. x^3 + c_.] -> Exp[c] f[a/b]
5 E^(1 + 2 x) f[3/4]
eq2 /. d1_. Exp[d2_. SuperStar[α_] + d3_.] -> Exp[d3] f[d2]
E^(b α) f[a]
answered May 9 at 9:56
kglrkglr
193k10214435
$endgroup$
2
$begingroup$
If you use defaults in the replacement pattern (dots after the patterns), they become more versatile:Exp[a_. x^2 + b_. x^3 + c_.] -> Exp[c] f[a/b]. This will also work witheq1a = 5 Exp[x^2 + 4 x^3], for example. Also,d1_. Exp[d2_. SuperStar[α_] + d3_.] -> Exp[d3] f[d2]for the second one.
$endgroup$
– Roman
May 9 at 12:29
$begingroup$
@Roman, great point. Updated with default patterns.
$endgroup$
– kglr
May 9 at 12:49
$begingroup$
@Shadowray, thank you; I added the needed_.s.
$endgroup$
– kglr
May 9 at 13:52
add a comment |
$begingroup$
eq1 /. Exp[a_. x^2 + b_. x^3 + c_.] -> Exp[c] f[a/b]
5 E^(1 + 2 x) f[3/4]
eq2 /. d1_. Exp[d2_. SuperStar[α_] + d3_.] -> Exp[d3] f[d2]
E^(b α) f[a]
answered May 9 at 9:56
kglrkglr
193k10214435
$endgroup$
eq1 /. Exp[a_. x^2 + b_. x^3 + c_.] -> Exp[c] f[a/b]
5 E^(1 + 2 x) f[3/4]
eq2 /. d1_. Exp[d2_. SuperStar[α_] + d3_.] -> Exp[d3] f[d2]
E^(b α) f[a]
answered May 9 at 9:56
kglrkglr
193k10214435
edited May 9 at 13:51
answered May 9 at 9:56
kglrkglr
193k10214435
answered May 9 at 9:56
kglrkglr
193k10214435
answered May 9 at 9:56
kglrkglr
193k10214435
193k10214435
2
$begingroup$
If you use defaults in the replacement pattern (dots after the patterns), they become more versatile:Exp[a_. x^2 + b_. x^3 + c_.] -> Exp[c] f[a/b]. This will also work witheq1a = 5 Exp[x^2 + 4 x^3], for example. Also,d1_. Exp[d2_. SuperStar[α_] + d3_.] -> Exp[d3] f[d2]for the second one.
$endgroup$
– Roman
May 9 at 12:29
$begingroup$
@Roman, great point. Updated with default patterns.
$endgroup$
– kglr
May 9 at 12:49
$begingroup$
@Shadowray, thank you; I added the needed_.s.
$endgroup$
– kglr
May 9 at 13:52
add a comment |
2
$begingroup$
If you use defaults in the replacement pattern (dots after the patterns), they become more versatile:Exp[a_. x^2 + b_. x^3 + c_.] -> Exp[c] f[a/b]. This will also work witheq1a = 5 Exp[x^2 + 4 x^3], for example. Also,d1_. Exp[d2_. SuperStar[α_] + d3_.] -> Exp[d3] f[d2]for the second one.
$endgroup$
– Roman
May 9 at 12:29
$begingroup$
@Roman, great point. Updated with default patterns.
$endgroup$
– kglr
May 9 at 12:49
$begingroup$
@Shadowray, thank you; I added the needed_.s.
$endgroup$
– kglr
May 9 at 13:52
2
2
$begingroup$
If you use defaults in the replacement pattern (dots after the patterns), they become more versatile:
Exp[a_. x^2 + b_. x^3 + c_.] -> Exp[c] f[a/b]. This will also work with eq1a = 5 Exp[x^2 + 4 x^3], for example. Also, d1_. Exp[d2_. SuperStar[α_] + d3_.] -> Exp[d3] f[d2] for the second one.$endgroup$
– Roman
May 9 at 12:29
$begingroup$
If you use defaults in the replacement pattern (dots after the patterns), they become more versatile:
Exp[a_. x^2 + b_. x^3 + c_.] -> Exp[c] f[a/b]. This will also work with eq1a = 5 Exp[x^2 + 4 x^3], for example. Also, d1_. Exp[d2_. SuperStar[α_] + d3_.] -> Exp[d3] f[d2] for the second one.$endgroup$
– Roman
May 9 at 12:29
$begingroup$
@Roman, great point. Updated with default patterns.
$endgroup$
– kglr
May 9 at 12:49
$begingroup$
@Roman, great point. Updated with default patterns.
$endgroup$
– kglr
May 9 at 12:49
$begingroup$
@Shadowray, thank you; I added the needed
_.s.$endgroup$
– kglr
May 9 at 13:52
$begingroup$
@Shadowray, thank you; I added the needed
_.s.$endgroup$
– kglr
May 9 at 13:52
add a comment |
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