| name |
latex |
input count |
feed count |
output count |
used in derivation |
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declare assumption
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Eq.~\ref{eq:#1} is an assumption. |
0 |
0 |
1 |
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indefinite integrate RHS over
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Indefinite integral of RHS of Eq.~\ref{eq:#2} over $#1$; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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indefinite integrate LHS over
|
Indefinite integral of LHS of Eq.~\ref{eq:#2} over $#1$; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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X dot both sides
|
Take inner product of $#1$ with Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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expand magnitude to conjugate
|
Expand $#1$ in Eq.~\ref{eq:#2} with conjugate; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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multiply LHS by unity
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Multiply LHS of Eq.~\ref{eq:#2} by 1, which in this case is $#1$; yields Eq.~\ref{eq:#3} |
1 |
1 |
1 |
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multiply both sides by
|
Multiply both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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select real parts
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Select real parts of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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replace scalar with vector
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Replace scalar variables in Eq.~\ref{eq:#1} with equivalent vector variables; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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subtract expr 1 from expr 2
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Subtract Eq.~\ref{eq:#1} from Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#2}. |
2 |
0 |
1 |
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change three variables in expression
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Change of variable $#1$ to $#2$ and $#3$ to $#4$ and $#5$ to $#6$ in Eq.~\ref{eq:#7}; yields Eq.~\ref{eq:#8}. |
1 |
6 |
1 |
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declare guess solution
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Judicious choice as a guessed solution to Eq.~\ref{eq:#1} is Eq.~\ref{eq:#2}, |
1 |
0 |
1 |
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add zero to LHS
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Add zero to LHS of Eq.~\ref{eq:#2}, where $0=#1$; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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substitute LHS of three expressions into expression
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Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} into Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}. |
4 |
0 |
1 |
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multiply expr 1 by expr 2
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Multiply Eq.~\ref{eq:#1} by Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
2 |
0 |
1 |
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factor out X from RHS
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Factor $#1$ from the RHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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swap LHS with RHS
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Swap LHS of Eq.~\ref{eq:#1} with RHS; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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separate two vector components
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Separate two vector components in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2} and Eq.~\ref{eq:#3} |
1 |
0 |
2 |
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boundary condition
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Boundary condition: Eq.~\ref{eq:#2} when Eq.~\ref{eq#1}. |
1 |
0 |
1 |
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subtract X from both sides
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Subtract $#1$ from both sides of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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separate vector into two trigonometric ratios
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Separate vector in Eq.~\ref{eq:#2} into components related by angle $#1$; yields Eq.~\ref{eq:#3} and Eq.~\ref{eq:#4}. |
1 |
1 |
2 |
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declare identity
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Eq.~\ref{eq:#1} is an identity. |
0 |
0 |
1 |
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function is even
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$#1$ is even with respect to $#2$, so replace $#1$ with $#3$ in Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}. |
1 |
3 |
1 |
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declare final expression
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Eq.~\ref{eq:#1} is one of the final equations. |
1 |
0 |
0 |
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indefinite integral over
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Indefinite integral of both sides of Eq.~\ref{eq:#2} over $#1$; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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claim LHS equals RHS
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Thus we see that LHS of Eq.~\ref{eq:#1} is equal to RHS. |
1 |
0 |
0 |
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substitute LHS of five expressions into expression
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Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} and LHS of Eq.~\ref{eq:#4} and LHS of Eq.~\ref{eq:#5} into Eq.~\ref{eq:#6}; yields Eq.~\ref{eq:#7}. |
6 |
0 |
1 |
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LHS of expr 1 equals LHS of expr 2
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LHS of Eq.~\ref{eq:#1} is equal to LHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
2 |
0 |
1 |
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both sides cross X
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Take cross product of Eq.~\ref{eq:#2} and $#1$; yields Eq.~\ref{eq:#3} |
1 |
1 |
1 |
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sum exponents
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Sum exponents on LHS and RHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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substitute LHS of six expressions into expression
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Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} and LHS of Eq.~\ref{eq:#4} and LHS of Eq.~\ref{eq:#5} and LHS of Eq.~\ref{eq:#6} into Eq.~\ref{eq:#7}; yields Eq.~\ref{eq:#8}. |
7 |
0 |
1 |
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apply operator to bra
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Apply operator in Eq.~\ref{eq:#1} to bra; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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integrate
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Integrate Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}. |
1 |
0 |
1 |
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sum exponents RHS
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Sum exponents on RHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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change five variables in expression
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Change of variable $#1$ to $#2$ and $#3$ to $#4$ and $#5$ to $#6$ and $#7$ to $#8$ and $#9$ to $#10$ in Eq.~\ref{eq:#11}; yields Eq.~\ref{eq:#12}. |
1 |
10 |
1 |
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divide expr 1 by expr 2
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Divide Eq.~\ref{eq:#1} by Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
2 |
0 |
1 |
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factor out X
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Factor $#1$ from Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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simplify
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Simplify Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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apply divergence
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Apply divergence to both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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X cross both sides by
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Take cross product of $#1$ and Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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change six variables in expression
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Change of variable $#1$ to $#2$ and $#3$ to $#4$ and $#5$ to $#6$ and $#7$ to $#8$ and $#9$ to $#10$ and $#11$ to $#12$ in Eq.~\ref{eq:#13}; yields Eq.~\ref{eq:#14}. |
1 |
12 |
1 |
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distribute conjugate to factors
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Distribute conjugate to factors in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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raise both sides to power
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Raise both sides of Eq.~\ref{eq:#2} to $#1$; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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conjugate both sides
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Conjugate both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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apply function to both sides of expression
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Apply function $#1$ with argument $#2$ to Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4} |
1 |
2 |
1 |
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normalization condition
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Normalization condition is Eq.~\ref{eq:#1}. |
0 |
0 |
1 |
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function is odd
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$#1$ is odd with respect to $#2$, so replace $#1$ with $#3$ in Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}. |
1 |
3 |
1 |
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square root both sides
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Take the square root of both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2} and Eq.~\ref{eq:#3}. |
1 |
0 |
2 |
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add X to both sides
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Add $#1$ to both sides of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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apply gradient to scalar function
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Apply gradient to both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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solve for X
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Solve Eq.~\ref{eq:#2} for $#1$; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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expand RHS
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Expand the RHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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claim expr 1 equals expr 2
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Thus we see that Eq.~\ref{eq:#1} is equivalent to Eq.~\ref{eq:#2}. |
2 |
0 |
0 |
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separate three vector components
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Separate three vector components in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2} and Eq.~\ref{eq:#3} and Eq.~\ref{eq:#4}. |
1 |
0 |
3 |
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substitute LHS of expr 1 into expr 2
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Substitute LHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
2 |
0 |
1 |
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expand LHS
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Expand the LHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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sum exponents LHS
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Sum exponents on LHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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both sides dot X
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Take inner product of Eq.~\ref{eq:#2} with $#1$; yields Eq.~\ref{eq:#3} |
1 |
1 |
1 |
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expand integrand
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Expand integrand of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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indefinite integration
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Indefinite integral of both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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factor out X from LHS
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Factor $#1$ from the LHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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substitute RHS of expr 1 into expr 2
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Substitute RHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
2 |
0 |
1 |
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multiply RHS by unity
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Multiply RHS of Eq.~\ref{eq:#2} by 1, which in this case is $#1$; yields Eq.~\ref{eq:#3} |
1 |
1 |
1 |
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differentiate with respect to
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Differentiate Eq.~\ref{eq:#2} with respect to $#1$; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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evaluate definite integral
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Evaluate definite integral Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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partially differentiate with respect to
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Partially differentiate Eq.~\ref{eq:#2} with respect to $#1$; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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conjugate transpose both sides
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Conjugate transpose of both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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expr 1 is true under condition expr 2
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Eq.~\ref{eq:#1} is valid when Eq.~\ref{eq:#2} occurs; yields Eq.~\ref{eq:#3}. |
2 |
0 |
1 |
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replace constant with value
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Replace constant $#1$ with value $#2$ and units $#3$ in Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5} |
1 |
3 |
1 |
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add zero to RHS
|
Add zero to RHS of Eq.~\ref{eq:#2}, where $0=#1$; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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make expr power
|
Make Eq.~\ref{eq:#2} the power of $#1$; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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combine like terms
|
Combine like terms in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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substitute LHS of two expressions into expression
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Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} into Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4}. |
3 |
0 |
1 |
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expr 1 is equivalent to expr 2 under the condition
|
Eq.~\ref{eq:#1} is equivalent to Eq.~\ref{eq:#2} under the condition in Eq.~\ref{eq:#3}. |
2 |
0 |
1 |
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maximum of expression
|
The maximum of Eq.~\ref{eq:#2} with respect to $#1$ is Eq.~\ref{eq:#3} |
1 |
1 |
1 |
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take curl of both sides
|
Apply curl to both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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change four variables in expression
|
Change of variable $#1$ to $#2$ and $#3$ to $#4$ and $#5$ to $#6$ and $#7$ to $#8$ in Eq.~\ref{eq:#9}; yields Eq.~\ref{eq:#10}. |
1 |
8 |
1 |
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drop non-dominant term
|
Based on the assumption $#1$, drop non-dominant term in Eq.~\ref{#2}; yeilds Eq.~\ref{#3} |
1 |
1 |
1 |
|
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assume N dimensions
|
Assume $#1$ dimensions; decompose vector to be Eq.~\ref{eq:#2}. |
0 |
1 |
1 |
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substitute LHS of four expressions into expression
|
Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} and LHS of Eq.~\ref{eq:#4} into Eq.~\ref{eq:#5}; yields Eq.~\ref{eq:#6}. |
5 |
0 |
1 |
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boundary condition for expression
|
A boundary condition for Eq.~\ref{eq:#1} is Eq.~\ref{eq:#2} |
1 |
0 |
1 |
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RHS of expr 1 equals RHS of expr 2
|
RHS of Eq.~\ref{eq:#1} is equal to RHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
2 |
0 |
1 |
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change variable X to Y
|
Change variable $#1$ to $#2$ in Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4}. |
1 |
2 |
1 |
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distribute conjugate transpose to factors
|
Distribute conjugate transpose to factors in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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select imaginary parts
|
Select imaginary parts of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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replace summation notation with vector notation
|
Replace summation notation in Eq.~\ref{eq:#1} with vector notation; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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replace curl with LeviCevita summation contravariant
|
Replace curl in Eq.~\ref{eq:#1} with Levi-Cevita contravariant; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
|
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integrate over from to
|
Integrate Eq.~\ref{eq:#4} over $#1$ from lower limit $#2$ to upper limit $#3$; yields Eq.~\ref{eq:#5}. |
1 |
3 |
1 |
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apply operator to ket
|
Apply operator in Eq.~\ref{eq:#1} to ket; yields Eq.~\ref{eq:#2}. |
1 |
0 |
1 |
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divide both sides by
|
Divide both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
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add expr 1 to expr 2
|
Add Eq.~\ref{eq:#1} to Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#2}. |
2 |
0 |
1 |
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declare initial expression
|
Eq.~\ref{eq:#1} is an initial equation. |
0 |
0 |
1 |
|
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change two variables in expression
|
Change variable $#1$ to $#2$ and $#3$ to $#4$ in Eq.~\ref{eq:#5}; yields Eq.~\ref{eq:#6}. |
1 |
4 |
1 |
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conjugate function X
|
Conjugate $#1$ in Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}. |
1 |
1 |
1 |
|
|
magnitude of vectors
|
Absolute value of both sides |
1 |
0 |
1 |
|