सिद्धान्तशिरोमणि: गणिताध्याय (भास्कराचार्य - ग्रहगणित, मध्यमाधिकार व स्पष्टाधिकार सटीक)

351 Differentiating 2 K δ K = 2 r δ r + 2 R cos m δ r ie. δ K = [δ r (r + R cos m)] / K . But from fig. 65 (which is a portion of the epicyclic figure) M̂₁ = 180—m Ô₁ = θ = Mandaphala M̂₂ = m—θ Fig. 65 r = K cos m̅—̅θ̅ — R cos m so that r + R cos m = K cos (m — θ). Substituting in the above, δ K = [δ r × K cos m̅—̅θ̅] / K = δ r cos m̅—̅θ̅ But δ r from (1) is [r (K—R)] / R ∴ δ K = [r (K—R)] / R cos m̅—̅θ̅. But from the tri- angle of fig. 65. K = R cos θ + r cos m̅—̅θ̅ so that r cos m̅—̅θ̅ = K—R cos θ. But θ being small cos θ may be taken to be unity so that r cos m̅—̅θ̅ = K—R. Again substituting in the above δ K = [(K—R)²] / R (2) Now as per the formulation of Bhāskara K¹ = R² / (2 R—K) = R² / (R + R — K) = R / [1 + (R—K)/R] = R / [1 — (K—R)/R] = R (1 — (K—R)/R)⁻¹ Since |(K—R)/R| < 1, expanding binomially,