What's wrong with this Latex equation? Missing $? and undefinded control sequence
3
documentclass[10pt]{article}
usepackage[usenames]{color} % Farbunterstützung
usepackage{amssymb} % Mathe
usepackage{amsmath} % Mathe
usepackage[utf8]{inputenc} % Direkte Eingabe von Umlauten und anderen Diakritika
usepackage{braket}
(-1)^p=(-1)^{$P^{prime}$+$P^{primeprime}$}
math-mode equations
edited Dec 6 '18 at 18:56
Peter Mortensen
54736
asked Dec 6 '18 at 17:15
tqinktqink
161
add a comment |
3
documentclass[10pt]{article}
usepackage[usenames]{color} % Farbunterstützung
usepackage{amssymb} % Mathe
usepackage{amsmath} % Mathe
usepackage[utf8]{inputenc} % Direkte Eingabe von Umlauten und anderen Diakritika
usepackage{braket}
(-1)^p=(-1)^{$P^{prime}$+$P^{primeprime}$}
math-mode equations
edited Dec 6 '18 at 18:56
Peter Mortensen
54736
asked Dec 6 '18 at 17:15
tqinktqink
161
4
Welcome! You cannot have(-1)^p=(-1)^outsidemathmode. Trydocumentclass{article} begin{document} $(-1)^p=(-1)^{P^{prime}+P^{primeprime}}$ or $(-1)^p=(-1)^{{P^{prime}+P^{primeprime}}}$ end{document}.
– marmot
Dec 6 '18 at 17:16
4
You're also missing both abegin{document}and anend{document}directive.
– Mico
Dec 6 '18 at 17:17
add a comment |
3
3
3
documentclass[10pt]{article}
usepackage[usenames]{color} % Farbunterstützung
usepackage{amssymb} % Mathe
usepackage{amsmath} % Mathe
usepackage[utf8]{inputenc} % Direkte Eingabe von Umlauten und anderen Diakritika
usepackage{braket}
(-1)^p=(-1)^{$P^{prime}$+$P^{primeprime}$}
math-mode equations
edited Dec 6 '18 at 18:56
Peter Mortensen
54736
asked Dec 6 '18 at 17:15
tqinktqink
161
documentclass[10pt]{article}
usepackage[usenames]{color} % Farbunterstützung
usepackage{amssymb} % Mathe
usepackage{amsmath} % Mathe
usepackage[utf8]{inputenc} % Direkte Eingabe von Umlauten und anderen Diakritika
usepackage{braket}
(-1)^p=(-1)^{$P^{prime}$+$P^{primeprime}$}
math-mode equations
math-mode equations
edited Dec 6 '18 at 18:56
Peter Mortensen
54736
asked Dec 6 '18 at 17:15
tqinktqink
161
edited Dec 6 '18 at 18:56
Peter Mortensen
54736
asked Dec 6 '18 at 17:15
tqinktqink
161
edited Dec 6 '18 at 18:56
Peter Mortensen
54736
edited Dec 6 '18 at 18:56
Peter Mortensen
54736
edited Dec 6 '18 at 18:56
Peter Mortensen
54736
54736
asked Dec 6 '18 at 17:15
tqinktqink
161
asked Dec 6 '18 at 17:15
tqinktqink
161
asked Dec 6 '18 at 17:15
tqinktqink
161
161
4
Welcome! You cannot have(-1)^p=(-1)^outsidemathmode. Trydocumentclass{article} begin{document} $(-1)^p=(-1)^{P^{prime}+P^{primeprime}}$ or $(-1)^p=(-1)^{{P^{prime}+P^{primeprime}}}$ end{document}.
– marmot
Dec 6 '18 at 17:16
4
You're also missing both abegin{document}and anend{document}directive.
– Mico
Dec 6 '18 at 17:17
add a comment |
4
Welcome! You cannot have(-1)^p=(-1)^outsidemathmode. Trydocumentclass{article} begin{document} $(-1)^p=(-1)^{P^{prime}+P^{primeprime}}$ or $(-1)^p=(-1)^{{P^{prime}+P^{primeprime}}}$ end{document}.
– marmot
Dec 6 '18 at 17:16
4
You're also missing both abegin{document}and anend{document}directive.
– Mico
Dec 6 '18 at 17:17
4
4
Welcome! You cannot have
(-1)^p=(-1)^ outside math mode. Try documentclass{article} begin{document} $(-1)^p=(-1)^{P^{prime}+P^{primeprime}}$ or $(-1)^p=(-1)^{{P^{prime}+P^{primeprime}}}$ end{document}.– marmot
Dec 6 '18 at 17:16
Welcome! You cannot have
(-1)^p=(-1)^ outside math mode. Try documentclass{article} begin{document} $(-1)^p=(-1)^{P^{prime}+P^{primeprime}}$ or $(-1)^p=(-1)^{{P^{prime}+P^{primeprime}}}$ end{document}.– marmot
Dec 6 '18 at 17:16
4
4
You're also missing both a
begin{document} and an end{document} directive.– Mico
Dec 6 '18 at 17:17
You're also missing both a
begin{document} and an end{document} directive.– Mico
Dec 6 '18 at 17:17
add a comment |
1 Answer
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6
I believe this is what you're looking for. (I would also argue that writing ' is easier than ^{prime} -- and that writing '' is a lot easier than ^{primeprime}.)
documentclass{article}
begin{document}
$(-1)^p=(-1)^{(P'+P'')}$
end{document}
answered Dec 6 '18 at 17:22
MicoMico
275k30373761
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
I believe this is what you're looking for. (I would also argue that writing ' is easier than ^{prime} -- and that writing '' is a lot easier than ^{primeprime}.)
documentclass{article}
begin{document}
$(-1)^p=(-1)^{(P'+P'')}$
end{document}
answered Dec 6 '18 at 17:22
MicoMico
275k30373761
add a comment |
6
I believe this is what you're looking for. (I would also argue that writing ' is easier than ^{prime} -- and that writing '' is a lot easier than ^{primeprime}.)
documentclass{article}
begin{document}
$(-1)^p=(-1)^{(P'+P'')}$
end{document}
answered Dec 6 '18 at 17:22
MicoMico
275k30373761
add a comment |
6
6
6
I believe this is what you're looking for. (I would also argue that writing ' is easier than ^{prime} -- and that writing '' is a lot easier than ^{primeprime}.)
documentclass{article}
begin{document}
$(-1)^p=(-1)^{(P'+P'')}$
end{document}
answered Dec 6 '18 at 17:22
MicoMico
275k30373761
I believe this is what you're looking for. (I would also argue that writing ' is easier than ^{prime} -- and that writing '' is a lot easier than ^{primeprime}.)
documentclass{article}
begin{document}
$(-1)^p=(-1)^{(P'+P'')}$
end{document}
answered Dec 6 '18 at 17:22
MicoMico
275k30373761
answered Dec 6 '18 at 17:22
MicoMico
275k30373761
answered Dec 6 '18 at 17:22
MicoMico
275k30373761
answered Dec 6 '18 at 17:22
MicoMico
275k30373761
275k30373761
add a comment |
add a comment |
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4
Welcome! You cannot have
(-1)^p=(-1)^outsidemathmode. Trydocumentclass{article} begin{document} $(-1)^p=(-1)^{P^{prime}+P^{primeprime}}$ or $(-1)^p=(-1)^{{P^{prime}+P^{primeprime}}}$ end{document}.– marmot
Dec 6 '18 at 17:16
4
You're also missing both a
begin{document}and anend{document}directive.– Mico
Dec 6 '18 at 17:17