Frequency and period are inversely related; thus
T = 1/f.
Multiple both sides by f to get
f T=1
then divide by T to get
f = 1/T.
statement with expressions separated from text and with bindings between math and text made explicit
Frequency and period are inversely related; thus
expression 1: T = 1/f.
Multiple both sides of expression 1 by f to get expression 2
expression 2: f T=1
then divide both sides of expression 2 by T to get expression 3
expression 3: f = 1/T.
statement with inference rules made explicit
claim: Frequency and period are inversely related; thus
inference rule: declare initial expression
expression 1: T = 1/f.
inference rule: Multiple both sides of expression 1 by f to get expression 2
expression 2: f T=1
then
inference rule: divide both sides of expression 2 by T to get expression 3
expression 3: f = 1/T.
inference rule: declare final expression
use of a Computer algebra system to implement inference rules
The following expansion requires
- conversion of Latex to SymPy
- correctly implemented inference rules
>>> import sympy
>>> from sympy import *
>>> from sympy.parsing.latex import parse_latex
claim: Frequency and period are inversely related; thus
inference rule: declare initial expression
expression 1: T = 1/f.
To confirm consistency of representations, the input Latex expression can be converted to SymPy and then back to Latex using
>>> latex(eval(sympy.srepr(parse_latex('T = 1/f'))))
'T = \\frac{1}{f}'
We'll work with the SymPy representation of expression 1,
>>> sympy.srepr(parse_latex('T = 1/f'))
"Equality(Symbol('T'), Pow(Symbol('f'), Integer(-1)))"
Rather than using the SymPy, use the raw format of expression 1
>>> expr1 = parse_latex('T = 1/f')
inference rule: Multiple both sides of expression 1 by f to get expression 2
expression 2: f T=1
Although we can multiply a variable and an expression,
>>> expr1*Symbol('f')
f*(Eq(T, 1/f))
what actually needs to happen is first split the expression, then apply the multiplication to both sides
>>> Equality(expr1.lhs*Symbol('f'), expr1.rhs*Symbol('f'))
Eq(T*f, 1)
Application of an inference rule (above) results in the desired result, so save that result as the second expression (below).
>>> expr2 = Equality(expr1.lhs*Symbol('f'), expr1.rhs*Symbol('f'))
inference rule: divide both sides of expression 2 by T to get expression 3
expression 3: f = 1/T.
>>> Equality(expr2.lhs/Symbol('T'), expr2.rhs/Symbol('T'))
Eq(f, 1/T)
Again, save that to a variable
>>> expr3 = Equality(expr2.lhs/Symbol('T'), expr2.rhs/Symbol('T'))
>>> latex(expr3)
'f = \\frac{1}{T}'
inference rule: declare final expression
statement with inference rules and numeric IDs for symbols
To relate the above derivation to any other content in the Physics Derivation Graph, replace T and f with numeric IDs unique to "period" and "frequency"
>>> import sympy
>>> from sympy import *
>>> from sympy.parsing.latex import parse_latex
claim: Frequency and period are inversely related; thus
inference rule: declare initial expression
expression 1: T = 1/f.
>>> expr1 = parse_latex('T = 1/f')
>>> eval(srepr(expr1).replace('T','pdg9491').replace('f','pdg4201'))
Eq(pdg9491, 1/pdg4201)
Save the result as expression 1
>>> expr1 = eval(srepr(expr1).replace('T','pdg9491').replace('f','pdg4201'))
inference rule: Multiple both sides of expression 1 by f to get expression 2
expression 2: f T=1
>>> feed = Symbol('f')
>>> feed = eval(srepr(feed).replace('f','pdg4201'))
>>> Equality(expr1.lhs*feed, expr1.rhs*feed)
>>> Equality(expr1.lhs*feed, expr1.rhs*feed)
Eq(pdg4201*pdg9491, 1)
>>> expr2 = Equality(expr1.lhs*feed, expr1.rhs*feed)
inference rule: divide both sides of expression 2 by T to get expression 3
expression 3: f = 1/T.
>>> feed = Symbol('T')
>>> feed = eval(srepr(feed).replace('T','pdg9491'))
>>> Equality(expr2.lhs/feed, expr2.rhs/feed)
Eq(pdg4201, 1/pdg9491)
>>> expr3 = Equality(expr2.lhs/feed, expr2.rhs/feed)
Convert from numeric ID back to Latex symbols in Latex expression
>>> latex(eval(srepr(expr3).replace('pdg9491','T').replace('pdg4201','f')))
'f = \\frac{1}{T}'
inference rule: declare final expression
removal of text, pure Python
The above steps can be expressed as a Python script with two functions (one for each inference rule)
from sympy import *
from sympy.parsing.latex import parse_latex
# assumptions: the inference rules are correct, the conversion of symbols-to-IDs is correct, the Latex-to-SymPy parsing is correct
def mult_both_sides_by(expr, feed):
return Equality(expr.lhs*feed, expr.rhs*feed)
def divide_both_sides_by(expr, feed):
return Equality(expr.lhs/feed, expr.rhs/feed)
# inference rule: declare initial expression
expr1 = parse_latex('T = 1/f')
expr1 = eval(srepr(expr1).replace('T','pdg9491').replace('f','pdg4201'))
feed = Symbol('f')
feed = eval(srepr(feed).replace('f','pdg4201'))
expr2 = mult_both_sides_by(expr1, feed)
feed = Symbol('T')
feed = eval(srepr(feed).replace('T','pdg9491'))
expr3 = divide_both_sides_by(expr2, feed)
latex(eval(
srepr(expr3).replace('pdg9491','T').replace('pdg4201','f')))
# inference rule: declare final expression
How would the rigor of the above be increased?
To get beyond what a CAS can verify, a "proof" would relate each of the two functions to a set of axioms. Given the two arguments (an expression, a "feed" value), is the returned value always consistent with some set of axioms?
The set of axioms chosen matters. For example, we could start with Zermelo–Fraenkel set theory
That would leave a significant gap between building up addition and subtraction and getting to calculus and differential equations. "Theorems of calculus derive from the axioms of the real, rational, integer, and natural number systems, as well as set theory." (source)
