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

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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,