Explanation Given: The function is f ( t ) = e − t 2 . Approach: 1) Taylor series of a function g ( t ) about the point t = a is given by, g ( t ) = g ( a ) + ( t − a ) g ′ ( a ) + ( t − a ) 2 2 ! g ″ ( a ) + ... 2) The Laplace transforms satisfies the property: L { t n } ( s ) = n ! s n + 1 . Calculation: Consider, f ( t ) = e − t 2 so, f ′ ( t ) = − 2 t e − t 2 f ′ ( 0 ) = 0 f ″ ( t ) = 4 t 2 e − t 2 − 2 e − t 2 f ″ ( 0 ) = − 2 f ‴ ( t ) = 12 t e − t 2 − 8 t 3 e − t 2 f ‴ ( 0 ) = 0 f 4 ( t ) = 12 e − t 2 − 48 t 2 e − t 2 + 16 t 4 e − t 2 f 4 ( 0 ) = 12 Thus the Taylor series for f ( t ) = e − t 2 about t = 0 is, f ( t ) = f ( 0 ) + t f ′ ( 0 ) + t 2 2 ! f ″ ( 0 ) + t 3 3 ! f ‴ ( 0 ) + t 4 4 ! f 4 ( 0 ) +

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Chapter 7.RP, Problem 18RP

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The Taylor series for f(t)=e−t2 about t=0 and expansion for L{f}(s) in powers of 1s.

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Chapter 7.RP, Problem 17RP

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The Taylor series for f ( t ) = e − t 2 about t = 0 and expansion for L { f } ( s ) in powers of 1 s .

Solution Summary: The author explains the Taylor series for f(t)=e-t2 and the Laplace transforms satisfies the property.

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The Taylor series for f(t)=e−t2 about t=0 and expansion for L{f}(s) in powers of 1s.

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