That between any two consecutive zeros of J 0 ( x ) there is precisely one zero of J 1 ( x ) by using Rolle’s Theorem. Theorem used: Suppose that a function f ( x ) is continuous on the closed interval [ a , b ] and differentiable on the open interval ( a , b ) . Then if f ( a ) = f ( b ) , then there exists at least one point c in the open interval ( a , b ) for which f ' ( c ) = 0 . The derivative of J v ( x ) with respect to x can be expressed by J v − 1 ( x ) or J v + 1 ( x ) by the formula [ x v J v − 1 ( x ) ] ′ = x v J v − 1 ( x ) and [ x − v J v ( x ) ] ′ = − x − v J v + 1 ( x ) , the recurrence relations are J v − 1 ( x ) + J v + 1 ( x ) = 2 v x J v ( x ) and J v − 1 ( x ) − J v + 1 ( x ) = 2 J v ' ( x ) . Calculation: From the theorem [ x − v J v ( x ) ] ′ = − x − v J v + 1 ( x ) it can be written as follows: d d x ( x − v J v ( x ) ) = − x − v J v + 1 ( x ) x − v J v ' ( x ) − v x − v − 1 J v ( x ) = − x − v J v + 1 ( x ) Taking x − v common on the both sides in above equation. x − v ( J v ' ( x ) − v x − 1 J v ( x ) ) = x − v ( − J v + 1 ( x ) ) J v ' ( x ) − v x − 1 J v ( x ) = − J v + 1 ( x ) For v = 0 and substitute it into above equation. J 0 ' ( x ) − ( 0 ) x − 1 J 0 ( x ) = − J 0 + 1 ( x ) J 0 ' ( x ) = − J 1 ( x ) In order to calculate between any two consecutive zeros of J 0 ( x ) , therefore the two consecutive be x 1 and x 2 such that J 0 ( x 1 ) = 0 = − J 1 ( x 2 ) . Hence, it is clear that between any two consecutives zeros of J 0 ( x ) there is precisely one zero of J 1 ( x ) .

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