Explanation Given: The formula is, F ( 1 , 1 ; 2 ; x ) = − x − 1 ln ( 1 − x ) . Approach: The standard result of Gaussian hypergeometric function is, F ( α , β ; γ ; x ) = 1 + ∑ n = 1 ∞ ( α ) n ( β ) n n ! ( γ ) n x n ... ( 1 ) The factorial function ( t ) n is defined as, ( t ) n = t ( t + 1 ) ( t + 2 ) ... ( t + n − 1 ) , n ≥ 1 ( t ) 0 = 1 , t ≠ 0 . The power series expansion of ln ( 1 + x ) around x = 0 is, ln ( 1 + x ) = x − x 2 2 + x 3 3 − x 4 4 + ⋅ ⋅ ⋅ , − 1 < x < 1 . Calculation: Consider the given formula, F ( 1 , 1 ; 2 ; x ) = − x − 1 ln ( 1 − x ) . Compare the above formula to the standard result, then, α = 1 β = 1 γ = 2 Thus, F ( 1 , 1 ; 2 ; x ) = 1 + ∑ n = 1 ∞ ( 1 ) n ( 1 ) n n ! ( 2 ) n x n ... ( 2 ) Evaluate the factorial function ( t ) n by the given formula. Thus, ( 1 ) n = 1 ( 1 + 1 ) ( 1 + 2 ) ... ( 1 + n − 1 ) = ( 1 ) . ( 2 ) . ( 3 ) ... ( n ) = n ! ( 2 ) n = 2 ( 2 + 1 ) ( 2 + 2 ) ... ( 2 + n − 1 ) = ( 2 ) . ( 3 ) . ( 4 ) ... ( n + 1 ) = ( n + 1 ) ! Substitute n ! for ( 1 ) n and ( n + 1 ) ! for ( 2 ) n in the equation ( 2 ) . F ( 1 , 1 ; 2 ; x ) = 1 + ∑ n = 1 ∞ n ! n ! n ! ( n + 1 ) ! x n = 1 + ∑ n = 1 ∞ n ! ( n + 1 ) ! x n = 1 + ∑ n = 1 ∞ n ! ( n + 1 ) n ! x n = 1 + ∑ n = 1 ∞ 1 ( n + 1 ) x n Continue to simplify the above equation as,

Fundamentals of Differential Equations and Boundary Value Problems

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ISBN: 9780321977106

Author: Nagle, R. Kent

Publisher: Pearson Education, Limited

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Solution Summary: The author explains how the formula F(1,1;2;x)=-x -1mathrmln (1-x

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To verify:

The given formula.