Delta-rational chord: Difference between revisions

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=== Partially DR ===

Suppose we wish to approximate a target delta signature of the form <math>+\delta_1 +? +\delta_3</math> with the chord <math>1:~~(1 + E_1)~~:~~(1 + E_2)~~:~~(1 + E_3)~~</math> (where the +? is free to vary). By a derivation similar to the above, the least-squares problem is

Suppose we wish to approximate a target delta signature of the form <math>+\delta_1 +? +\delta_3</math> with the chord <math>1:f_1:f_2:f_3</math> (where the +? is free to vary). By a derivation similar to the above, the least-squares problem is

<math>

\displaystyle {\underset{x,y}{\text{minimize}} \sqrt{\bigg(\frac{x + \delta_1}{x} - ~~E_1~~ \bigg)^2 + \bigg(\frac{x+\delta_1 + y}{x} - ~~E_2~~ \bigg)^2 + \bigg(\frac{x+\delta_1 + y + \delta_3}{x} - ~~E_3~~ \bigg)^2 }}.

\displaystyle {\underset{x,y}{\text{minimize}} \sqrt{\bigg(\frac{x + \delta_1}{x} - f_1 \bigg)^2 + \bigg(\frac{x+\delta_1 + y}{x} - f_2 \bigg)^2 + \bigg(\frac{x+\delta_1 + y + \delta_3}{x} - f_3 \bigg)^2 }}.

</math>

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d2 = sympy.Symbol("\\delta_{2}", real=True)

d3 = sympy.Symbol("\\delta_{3}", real=True)

E1 = sympy.Symbol("~~E_1~~", real=True)

f1 = sympy.Symbol("f_1", real=True)

E2 = sympy.Symbol("~~E_2~~", real=True)

f2 = sympy.Symbol("f_2", real=True)

E3 = sympy.Symbol("~~E_3~~", real=True)

f3 = sympy.Symbol("f_3", real=True)

err_squared = (d1 - ~~E1*x~~) ** 2 + ((d1 + y) - ~~E2*x~~) ** 2 + ((d1 + y + d3) - ~~E3*x~~) ** 2

err_squared = ((x + d1 ) / x - f1) ** 2 + ((x + d1 + y) / x - f2) ** 2 + ((x + d1 + y + d3) / x - f3) ** 2

err_squared.expand()

err_squared_x = sympy.diff(err_squared, x)

@@ Line 79: / Line 79: @@
 === Partially DR ===
-Suppose we wish to approximate a target delta signature of the form <math>+\delta_1 +? +\delta_3</math> with the chord <math>1:(1 + E_1):(1 + E_2):(1 + E_3)</math> (where the +? is free to vary). By a derivation similar to the above, the least-squares problem is
+Suppose we wish to approximate a target delta signature of the form <math>+\delta_1 +? +\delta_3</math> with the chord <math>1:f_1:f_2:f_3</math> (where the +? is free to vary). By a derivation similar to the above, the least-squares problem is
 <math>
-\displaystyle {\underset{x,y}{\text{minimize}} \sqrt{\bigg(\frac{x + \delta_1}{x} - E_1 \bigg)^2 + \bigg(\frac{x+\delta_1 + y}{x} - E_2 \bigg)^2 + \bigg(\frac{x+\delta_1 + y + \delta_3}{x} - E_3 \bigg)^2 }}.
+\displaystyle {\underset{x,y}{\text{minimize}} \sqrt{\bigg(\frac{x + \delta_1}{x} - f_1 \bigg)^2 + \bigg(\frac{x+\delta_1 + y}{x} - f_2 \bigg)^2 + \bigg(\frac{x+\delta_1 + y + \delta_3}{x} - f_3 \bigg)^2 }}.
 </math>
 <!--
@@ Line 94: / Line 94: @@
 d2 = sympy.Symbol("\\delta_{2}", real=True)
 d3 = sympy.Symbol("\\delta_{3}", real=True)
-E1 = sympy.Symbol("E_1", real=True)
+f1 = sympy.Symbol("f_1", real=True)
-E2 = sympy.Symbol("E_2", real=True)
+f2 = sympy.Symbol("f_2", real=True)
-E3 = sympy.Symbol("E_3", real=True)
+f3 = sympy.Symbol("f_3", real=True)
-err_squared = (d1 - E1*x) ** 2 + ((d1 + y) - E2*x) ** 2 + ((d1 + y + d3) - E3*x) ** 2
+err_squared = ((x + d1 ) / x - f1) ** 2 + ((x + d1 + y) / x - f2) ** 2 + ((x + d1 + y + d3) / x - f3) ** 2
 err_squared.expand()
 err_squared_x = sympy.diff(err_squared, x)